hello welcome back once again today we

have this Olympiad mathematics problem

we're given the following Quantic

equation that x ^ 4 - 8 is equal to0 so

here we're going to evaluate the C power

of X that is x ^ 6 right okay from the

given equation we have x ^

4 - 8 which is equal to 0 right so here

we have X to the^ 2^ 2 right X to the^ 4

can be re written in this form because

we know 2 * 2 here is equal 4 then minus

so let's look for a number that when we

Square it it will give us 8 so I'm

thinking of 2 < tk2 but how did I get

this we can say that 8 is equal to < TK

8 squ right so which is equal

to < TK 4 * 2 all 2 which is equal to of

4 is 2 so we get here 2 < tk2 2 so you

can see so 2 < tk2 2 will give us 8 so

here we write as 2un of 2^ 2 which is

equal

to0 okay so from here we're going to

make use of the difference of two

squares formula a s - b 2 is = to a - b

* A + B so make use of this difference

of two squares so we get x^2 - 2 < tk2 *

X2 + 2 < tk2 and this is equal to Z so

these two products are equal to Z so

this implies from here that either X2 -

2 < tk2 is = 0 or X2 +

2 < tk2 is = 0 so from here we get x2 is

equal to add 2k2 to both side so this

will become pos2 <

tk2 so we leave this here so that is

going to help us in this video so X2

here will be 0 - 2 < tk2 will give us -2

tk2 remember we are solving for x to

the^ of 6 right okay so from the

original equation we have x ^ 4 - 8

which is = 0 so we have x ^ 4 is = 0 + 8

which is 8 now let's multiply both side

by x^ 2 since we are looking for x^ 6 so

we can see from here x to the of record

this

property oh that is an initial law right

a to the^ M MTI a the^ n since they have

the same base so we need to add the

powers right so this will be n+ + n so

here 2 + 4 will give us 6 so we can see

the x^ 6 we are looking for is already

right over here so it will be X it will

be 8 * X2 so this will be equal to 8 *

X2 is pos2 < tk2 and -2 < tk2 so when

you combine them so they will be written

in this form plus or minus 2 < tk2 right

so from here we get our final answer x^

6 will be equal to if we multiply this

will give us plus or minus 8 * 2 is 16

then < tk2 so this here is the right

answer thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Olympiad Math problem | Can you solve ?


hello everyone welcome back once again

please if today is your first day of

visiting this Channel please subscribe

to this Channel Please Subscribe and

share our videos thank you today we're

going to evaluate this expression of

number here 2^ 10 - 1 all/ 2^ 5 + 1

before we proceed I'm going to recall

the difference of two squares formula

here we are going to make use of this

idea identity a 2 - b 2 = A + B * Aus B

right okay now let us try to rewrite

this numerator here so this is equal to

you know this identity that a of B * C

is the same as a of B in bracket C right

so this is going to be

2 of 5 * 2 - 1 all over

2^ of 5 + 1 right so this is equal to

make use make use of this identity here

so we have 2 power of 5 in bracket then

squ minus you know 1 is the same as 1

squ then all/ 2 5 + one right so now

we're going to make use of this identity

right over here right so here we have 2

of 5 the base then plus the second base

+ 1 * by 2^ 5 - 1 all over 2^ 5 + 1

right so we're going to cancel out 2^ 5

+ 1 and 2 5 + 1 so this is equal to 2 5

- 1 and this is equal to you know 2^ of

5 is 32 - 1 and this is equal to 31 and

this is the right answer please

subscribe to this Channel and share our

videos thank you

Simplification of numbers. How faster can we do this?


good day viewers welcome back once again

today we have an interesting exponent of

irrational numbers

right this is amazing right we have here

cube root of 6 to the^ of Cub root of 6

rais power of three and we are going to

evaluate this without calculators okay

now let's get into the video here we

have to rewrite the cube roots of

something as follows if we

recall this property when we have the

end root of a this is equal to a ^ 1 / n

right so from here we have -6 ^ 1 / 3

right ra the^ of -6 to^ 1 / 3 then ra

power of 3 now from here let us straet

this power first we have a negative

let's put it -1 * 6^ 1 / 3 of course

then of 3 this is equal to let us recall

another property a * b^ C this is equal

to a^ C * b^ C so from here we have -1 3

* 6^ 1 / 3^ 3 from here we all know

three is odd -1 to the power of an odd

number is 1 so we have1 times here we

need to multiply these Powers together

right we know this property a b c this

is equal to a b * C right so if we

multiply these Powers together this

three will cancel this three and here we

have six from here we have -6 amazing

right now remember that this this base

here is -6 to the^ of 1 / 3

1 3 3 and here we evaluated this power

right over here to be 6 so here we have

ra^ -6 from here we write as -1 * 6^ 1 /

3^ of -6 now we distribute -1 to^ of -6

* by 6 1 / 3 to^ of

-6 from here we need to recall another

property once again when we have a^ of n

this is equal to 1 a^ of n so from here

we

have -1 to^ of -6 * 6^ 1 /

3-6 this is the same as 1 -1 ^ 6 MTI 1

6^ 1 / 3 to^ of 6

and from here 6 is even so -1 to the

power of an even number is 1 1 over 1

times here we multiply these Powers

together so we have 1/

6 power of 1/ 3 * 6 this is equal to 1/

1 * 3 here 1 and here 2 1 all over 6 2

and this is equal to 1 over 36 and this

is the right answer thank you for

watching please kindly subscribe to my

YouTube channel also like comment and

share bye-bye

Interesting exponential problem involving irrational numbers | #maths #olympiad


hello welcome back once again my lovely

people today we're going to find the

value of x + y + z given that x² + y s +

z s = 1 x * y + y * Z + x * Z =

1 the solution to this that is the value

for x + y + z must be two right I not

mean that it is equal to two I mean we

should have two values for x + y + z you

know why take a look at this let's

assume that what what was given is X2 +

Y2 + Z2 = 3 and XY + y z + x z = 3 as

well so we can see we have TR solution X

= Y this = Y and this = Z meaning x = 1

y = 1 Z = 1 take a look 1 s + 1 + 1 is 3

1 * 1 + 1 * 1 + 1 * 1 is three right and

you can also see that X can

be1 y can be 1 and Z can be1 so with

this we can say that x + y +

z is going to be3 and positive 3 in this

case right so here we're expecting two

values for x + y + z so let's call x + y

+ z m right so squaring both sides M2

this is going to be x + y + z Z 2 so

from here M2 is going to be for expand

this we get x2 + y^ 2+ z^ 2 plus I can

factor out

to then y z + x z so here make

substitutions so M2 equals here the

value is given us 1 + 2 MTI the value

for this is also 1 so from here we get

M2 = 1 + 2 which is

take root of both side we get M = plus

orus theun 3 and here we come up with

the solution and this is a very quick

solution plus or minus root3 thanks to

God who invited mathematics right okay

mathematics is awesome and that's my

best subject thank you for watching

please kindly subscribe subscribe like

comment and share bye-bye

Interesting Olympiad Math


hello welcome back once again today

we're going to solve this exponential

equation so from here we have the cube

root of

x of the cube root of x of the cube root

of x this will be equal to now 16 is the

same as 2 ra^

4 so from the left we have the cube root

of XIs by the cube root of XIs by the

cube root of x this will be equal to

here two now take note that this four

right over here is same as 2 to the^ of

two right so this will be 2 power of 2

okay now from here we're going to

compare this as follow here we have

something of the same thing of the same

thing equals another thing of the same

thing of the same thing right so either

we say that the cube of x = 2 or we say

the cube root of x = 2 or from here as

well the cube root of x = 2 right so

this implies that the cube root of x = 2

now from here Cube both

sides we know the cube and the cube root

will cancel and here we're left with X

which is equal to 2 ^ of 3 which is 8

and here is the right answer thank you

for watching please kindly Subs subbe to

my channel also like comment and share

bye-bye

Interesting exponential equation with cube root truncated powers.


hello good day viewers today we have

another Olympia ma problem here we given

that Aus B is equal to 2 a * B is equal

to 1 you we ask to find the sum of A and

B please if you have not subscribed to

this YouTube Channel please kindly

support this channel by subscribing and

also do not forget to share our

videos here I will be presenting a very

nice algebraic identity which I'm going

to use to solving this problem here we

have Aus b s is equal to a + b

s minus 4 a b right so this identity is

going to help us so much here here

recalling that a minus B is equal 2 so

here we're going to replace a minus B

with 2 so here we have 2 squ right CU

Aus B is s this is equal to remember we

are looking for a + b right so here we

have a +

b 2 so here we have minus 4 * a remember

a is equal to 1 so here we

have 4 * a is going to be just four

right so now the idea here is to solve

for a plus B right so now we can

exchange the sides of this equation so

this will become a + b 2 - 4 is = 2 of 2

right so here we have a + b 2 - 4 is = 2

2 is equal to 4 so you know this4

crossing over will BEC positive right

so here we

have a + b s is equal to here we have 4

+ 4 and that's equal 8 so here we take

root of both sides so we get a + b is

equal to plus or minus < TK 8 and this

is exactly equal to plus orus < TK 4 * 2

you know of 4 is 2 so this is exactly

equal to plus or minus

S 4 that is 2 then < tk2 and this is the

final answer thank you for watching

please do well to subscribe to our

YouTube channel and share our videos

Olympiad Math. Algebra Equations. #maths #olympics


hello good day viewers in this video we

have a very nice problem here we're

given that 8 of M is equal to 27 here we

asked to evaluate 4 to the^ of M without

first finding the value of M right

please if you have not subscribed to my

Channel please kindly support this

channel by subscribing please do way to

like comment and share here we're given

that 8 R of M is equal to 27 right so

from here 8 is the same as 2^ of 3 then

here we have a power of M here and this

is equal to 27 right okay now let us

recall this law that a of B into bracket

of C is the same as a of C then of B

right so from here we're going to have 2

of M then R of 3

and this is equal to 27 right from here

we're going to take the cube root of

both sides of this equation so from here

we get 2 of M because from here cube

root of this here we have cube root of

27 you know the cube root and Cube

cancel out so we left 2 of M is equal to

the cube root of 27 that is equal to 3

so from here we find out that 2 of M is

equal to 3 okay now let us consider the

expression we are about to evaluate from

here so here we have 4^ of M you know

this is same as 2^ of 2 then power of M

right so this is equal to using this law

here so this going to be 2^ of M then of

2 and from here we have 2 of m = 3 so

this is going to be 3^ of 2 and that is

equal to 9 and this is the final answer

thank you for watching please kindly

subscribe to my channel like comment and

share bye-bye

Interesting exponential problem.


hello my wonderful viewers here we're

going to solve this cubic equation we

have X Cub =

16x now let's get started we have a x

cubed then we're going to subtract 16 x

right and this is equal to Z so from

here we have a common factor right and

that is X so we factor out X and we're

left with X2 - 16 this is equal to zero

now from here this implies that here x x

is = 0 right or from the bracket we have

x^2 - 16 = 0 now let's go ahead and

solve for x in this equation here we

already found a solution for X right and

that is xal 0 so from here we're going

to write this equation as x^2 - 16 is

the same as 4 2 and this is equal to0

now we're going to make use of the

difference of two squares formula here

right so we have a 2 - b 2 this is equal

to a + b * a - b right so make use of

that here we have x + 4 right then MTI x

- 4 this is equal to Z so from here we

have that x + 4 is = 0 the other bracket

here x - 4 = 0 so we have two linear

equations here and from here x = to4 4

from the other side here x = 4 and here

we found the solutions that x is = 0 we

also have that X is = to -4 and we also

have that X is = to 4 and this are the

solutions thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Nice cubic equation | #maths #mathematics


hello welcome back once again to my

YouTube Channel please kindly support

this YouTube

channel by subscribing to this YouTube

Channel Please Subscribe and share this

video here we have this limit problem

limit as x ts to Infinity of this

function here 1 / x^ of 1 all X right

okay now before we proceed let us take

note of this follow

algebraic

identity when we have a

m of n so this is same as a of n over M

of n right okay right over here we

have 1

x root of 1 x so this is same as 1^ 1 /

X all over x^ of 1 all X you know this

is equal to 1 1 / X is 1 then all X of

1x right okay let us take note of this

property again a power of 1 / n is the

same as the nend root of a right so this

actually mean that 1 over x^ 1 / X is

going to be 1 all over the X root of X

right so now our limit that is limit as

X is approaching Infinity of 1 x 1

= limit as X is approaching Infinity of

1 xot x you know we can say rewrite this

as follows 1 all over the

limit as X is approaching Infinity of

the X root of x so if you check our

recent

videos you will see where we find this

limit limit as X is approaching Infinity

of xot of X check our previous videos

this limit is equal to one right so this

is going to be 1 over 1 and this is

equal to 1 and this is the right answer

thank you for watching do wa to

subscribe to this YouTube channel and

share our videos

An interesting limit. How to deal?


hello good day viewers welcome back once

again please kindly support my channel

by subscribing also like comment and

share today we are going to solve for x

in the following exponential equation

here we have x to the power of x to the^

of 4 this is equal to 64 here we're

going to find the value of x right so

here we have X x^ 4 this is equal to 64

now let us raise both sides of this

equation to the power of 4 and here we

raise this to the power of 4 right so

from here we're going to have you know

this property a to the^ of

b^ of C this is equal to a of B * C

right so from here the power of X here

is x ^ of 4 so we're going to multiply

the powers together that is x^ 4 * 4

right so here we

have X of x^ of 4 multip by 4 right so

this is equal to you know in the bracket

here 64 is the same as 8 to the^ of 2

and here ra power of 4 right okay now

let us recall this property again when

we have a to the power of B * C this is

same as

a power of C ra power of B right so from

here we're going to have X power of this

four then power of x to the^ 4 right so

this is equal to you know from here

we're going to multiply these Powers

together so we have 8 power of 2 * 4

right so from here we have

x^

4^ x^ 4 this is equal to 8 power of 8

right so you know we can put this in

bracket now we can see what is going on

here we have here x^ of 4 ra of x^ 4

this is equal to 8 to^ of 8 so from here

this implies that x^ 4 is equal to 8

right so from here we have x^ 4 this is

equal to 8 you know from here we're

going to take the fourth root of both

sides so here we get X is equal to the 4

root of 8 and this is the exact answer

but when we use calculator to

approximate it we actually have

63 sorry 681 I mean 6 81 7 99 3 and this

is the right answer thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Olympiad Math | Another interesting exponential equation. #education #math #mathematics


good viewers welcome back to my Channel

today

we have the following

functional equation

so we have F of seven all over 2 raised

to power x minus one

equals 10 over x squared plus one then

we're going to evaluate F of one okay

now

remember that

we are going to look for a real value of

x such that when you substitute into X

here

this inside this bracket here becomes

one so then the responding value here at

the right hand side should be the value

of f of one okay so in this case we'll

be equating this to be equal to one okay

so we're going to say let's 7 all over

2 raised to power x minus 1 equals one

okay

now put this over one then from here we

cross multiply so here we have two

raised power x minus one times one

so this gives us 2 raised to x minus one

equals seven times one so that's seven

so you know this negative one crossing

over so we have two raised power x

equals

eight okay now we have here two raised

by x equals eight is the same as two

over two two respiratory so from here we

get x equals three okay

so we're going to substitute into X here

as three so now we have F of

minus one is equal to 10

over three respirator plus one

okay

so from here we have F of

7 over 8 minus one close bracket is

equal to

10 all over you know here

nine

plus one so here we have F of

you know here we have H minus one as the

denominator here so that is seven and

seven seven

by seven is one so we have F of one is

equal to ten all over ten and this is

equal to one and this is the right

answer to the question please if you

enjoyed this video please do well to

subscribe please

subscribe to our channel for more

interesting videos thank you

A nice functional equation. f(1) = ?  Here is a unique way to solve this problem


hello welcome back once again today we

have this interesting exponential

equation we have 3 ^ x + 3^ x + 3^ x

this is equal to 1 and here we'll be

solving for the value of x now let's get

started from the left hand side we have

3 to the power of x + 3 to the power of

x + 3 to the^ of X this is equal to 1

okay now 3 to^ x is common here so let's

factor that out so we have 3 ^ of X into

bracket 3 ^ x / 3^ X is 1 + 3^ x / 3^ X

is 1 + 3^ x / 3^ X is 1 right and here

we have this is equal to 1 okay so from

here we have 3 to^ of X multiply by here

the sum of this one one and 1 is

right and here we have this is equal to

one now let's go ahead and divide both

sides by three and here by 3 so this

cancel this and here we have 3^ X is =

to now from the right we have 1 all over

3 remember the power of this three here

is one right now let's take note of this

1 / a to the power of n is = to a of n

so here this will be 3 of1 now we can

see that we have equal bases right so we

can equate the power and from here this

imply that X is = to -1 and this is the

correct answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Interesting exponential equation


hello good day welcome back to another

video here we're given that x + 6 * 9 -

x is = to 5 and right over here we're

going to evaluate x + 6 ^ 2 + 9 - x ^ 2

please kindly support my channel by

subscribing to this Channel please also

like comment and share here we're going

to use this identity that a

2 + b 2 is equal to

a + b 2 - 2 a b right so right

here x + 6 the^ of 2 + 9 - x to the^ of

2 is going to be equal

to the sum of x + 6 and 9 - x x + 6 then

plus 9 - x right

squar then - 2

then the product of x + 6 and 9 - x that

is x + 6 and 9 - x right that is making

use of this identity here so this going

to be equal to you know here x and this

negative X cancel out here we have 6 + 9

and that is

s - 2 * the produ of x + 69 - x is 5 as

given so here we have five so this is

going to be equal to 15 squ from here so

that is going to give us

225 so here we have 2

25 then - 2 * 5 that is 10 and from here

we get

215 and this is the right answer thank

you for watching please do wor to

subscribe to my Channel please like

comment and share bye-bye

If and only if you know the trick.


hello good day viewers welcome back once

again to another video please subscribe

to this channel also like comment and

here here we're given the following

Olympia M problem here we're given that

x + 1 /x is = 2 right over here we ask

to evaluate x ^ of

2024 - 1 / x ^ of 2 and 24 right okay

carefully looking at this equation one

can easily say that the value of x is 1

right because here 1 + 1 / 1 is = 2

right so another way is to manipulate

this equation so by multiplying both

sides by X we

get x2 + 1 is = 2X and from here we get

x^2 - 2x + 1 this is equal to Z and here

this is a perfect square that is the

square of x -1 which is equal to Z from

here we we we can see again that X is

equal to 1 right so we can easily get

the value of this which is x^ of

2024 - 1 / x^ 2024 which is equal to 1 -

1 / 1 and this is equal to zero and this

is the right answer thank you for

watching please do wa to subscribe to

this channel also like comment and share

bye-bye

Interesting and easy problem.


hello welcome back once again today

we're going to solve this interesting

exponential equation so from the left we

have 10 to the ^ x + 10 to^ x + 10 to

the^ of X this is equal to

600 so we can see from the left hand

side 10^ x is common so we're going to

factor it out so we have 10^ X into

bracket this / this is 1

plus this ID by this is 1 plus this ID

by this is 1 then this is equal to

600 from here we get 10^ x * 1 + 1 + 1

is 3 then this is equal to

600 so in our next step let's divide

both side by

three yes so this cancels this so from

the left hand side we left with 10^ X so

this will be equal to 600 ID 3 will give

us

200 so from here we're going to log both

sides of this equation so from the left

hand side we have log 10 to the^ of X so

this will be equal to log of 200 so with

the property of log reading we know that

the log of a the^ b is equal to B MTI

log of a right so this will become X MTI

log log of 10 this will be equal to log

of we know 200 is the same as 100 *

2 let us apply this law at the right

hand side we know that log of a MTI B is

equal to log of a + log of B so here

this will be x * the log of 10 this will

be equal to log of 100+

log of

2 so here we have x * the log of 10 this

is equal to log of 100 is same as 10

2+ log of

2 so from here x m the log of 10 this is

equal to you know this property here so

this will be 2 MTI the log of 10 plus

log of 2 so divide both by the

coefficient of x which is log of 10

here by the log of 10 so this will

cancel this and from here we get X isal

to splitting this denominator we get 2

log of 10 / log of 10

plus log of 2 / log of 10 so let us

record this property again we have log

of a / log of B if you want to write it

as a ling as a single logarithm I mean

so this will be log of a to the base B

so we're going to use it here so here

log 10 we cancel log 10 and here we have

our solution X is = to 2 plus use this

here so this will become log 2 to the

base 10 so here is our

solution thank you for watching please

do not forget to subscribe

please

like

comment and please share our videos

thank you

Olympiad Math problem | Can you solve ?


hello good viewers welcome back once

again in this video we'll be solving for

the value of a * B given that A + B is =

4 and a minus B is = 1 right in the

video I'm going to present two different

methods let's start with the first

method but you know what the first

method is awesome why the second one is

amazing right so here we have a + b this

is equal to 4 actually from here if we

try solving for B in terms of a we get B

is = 4 - A let's keep this in mind then

we have here A minus B this is equal to

1 let's go ahead and solve this equation

simultaneously if we add both equations

B will be eliminated right we have here

a + a which is 2 a this is equal to 4 +

1 that is 5 from here this is telling us

or implying that a is = 5 2 for B we

already have that 4 - A will give us B

which is 4 - 5 over 2 and this will give

us 3 all 2 we can come up with the

solution a * B will be 5 2 * 3 2 and

this will give us 15 all over

4 now let's see how the second method

would

be for the second method I'll be making

use of algebraic identity you guys know

me very well I'm a fan of that we have

to make use of this

property a * B is a + b 2 - a - b 2 all

/ 4 right so from here we have a * B

this is equal to a + b from the original

problem you know that is 4 then we have

4 2 minus a minus B is 1 we have 1 2

then all over 4 we know 4 2 is 16 - 1 2

which is 1 / 4 and here we come up with

the solution again 15 all over 4 and

this is the answer thank you for

watching please kindly subscribe to my

YouTube channel also like comment and

share bye-bye

A junior Olympiad Math | Two different methods | #maths #olympiad


hello welcome back once again today we

have a form problem on modular

arithmetic here 2 to the^ of 16 is

Congress to what in mod

257 and what does this mean the better

way to interpret this is what is the

remainder when 2 to the^ of 16 is

divided by

257 Okay so

let's take note that

2 to the power

of yes that is

8 is equal to

256 right and 256 is congr

to -1 mod

257 now from this equation here we found

out that 2 to the^ of 8 is Congress to

mod

257 now let's raise both side to the

second power that will be 2 ^ of 8 SAR

this will be Congress to -1 SAR mod

257 Now using this power rule a to the

power of B of C this this = a ^ B * C

here we become 2 ^ 8 * 2 which is 16

this will be Congress to remember this

power here is given so this will result

to positive one which is in mode

257 therefore the remainder when 2 to

the^ of 16 is ID

257 is one and this is the right answer

thank you for watching please kindly

subscribe to my channel Channel also

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Number Theory | Modular Arithmetic |


hello good viewers welcome back once

again let's solve this exponential

equation very quickly so here we have

square root of 3 < TK X this is equal to

3 of 1 /un x now let us recall that the

square root of M is same as m^ of 1 /

2 right so here the root of 3^ ofun x

will be same

as 3 of root

x 1 / 2 then this is equal to 3 of 1 /un

x again we have this

property that when we have something

like this we rewrite it as a b * C so

here this will be same as 3un x * 1 2

will beunk x 2 this is equal to 3 of 1

/un x right so now we have the same

basis right so we can equate the powers

so this implies that root of x / 2 is =

1 allun of x from here let's go ahead

and cross multiply root x * root X will

beunk x^2 right then this is equal

to so this is equal to oh sorry

shouldn't that should not be written

like this right but that would be

absolute value so root x * root X from

here

that will give

us root X like this squared right then

this is equal to 2 * 1 that will give us

2 so from

here the square and the root sign cancel

out right so here we're left with X isal

to 2 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

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Interesting equation with radical powers | #maths #mathematics


hello good viewers welcome back once

again today we're going to solve the

given equation x^ 1X = < tk2 here we are

going to solve for positive integers X

right okay here we have x^ 1

/x this is equal to < TK 2 we have here

x^ of 1 /x this is equal to < tk2 is the

same as 2 to^ of 1 / 2 now we can see

from here that obviously X is = 2 is a

solution

right okay now let's try manipulating

this equation to find out if we have

more positive integers of X that will

satisfy this equation so from here I'm

going to raise both sides to the power

of the LCM of X and 2 and that will be

2x right and from the right we have a 2

power of 1 / 2 to the power of 2

x now from here with this property a to

the^ of B to the power C this is equal

to a^ B * C now from here 1 x * 2X x

will cancel X and we have X the^ of 2

and this will be equal to 1 2 * 2 X2 we

cancel 2 and we have 2 to the^ of X okay

now from here we can actually make use

of the lambat W function but we have

some obvious Solutions here we all know

that 2^ 2 = 2^ 2 which we have found

earlier that x = 2 and again we know

that 4^ 2 = 2^ 4 which is an obvious

solution from here here that x =

4 from here we have the two positive

solutions that is positive integers for

X so we have x = 2 and we have x = 4 and

these are the two positive integers of X

that will satisfy the given

equation I know most of you guys may ask

if there are more positive integers for

X but when I cross

this equation I found out that the

equation has only two intersection

points which is at two and four and

these are the only two positive integers

that satisfy the equation thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Interesting exponential equation


hello welcome back once again today we

have this interesting optic equation x^

8 + x^ 4 is equal to 2 and here we're

going to solve for the real value of x

right so let's get started from the left

hand side we have x to^ 8 + x^ 4 from

the right hand side this is equal to

2 now we have X to the^ 8 + x ^ 4 this

is equal to 2 is same as 1 +

1 so here we write x^ 8 - 1 the first

one here then plus x^ 4us the other one

this is equal to Z Now from here let us

rewrite x^ 8 will be x^ 4 2ar we all

know 4 * 2 is 8 minus here 1 2

then plus x^ 4 - 1 this is equal 0 From

Here let us apply the difference of two

squares formula here remember that a 2 -

b 2 is equal to a + b * Aus

B so from here that will become x ^ 4 +

1 * x^ 4 - 1 then Plus this let's put it

in bracket x^ 4 - 1 this is equal to 0

we can see x ^ 4 - 1 is common so Factor

it out we get x^ 4 - 1 then into bracket

this will be x ^ 4 + 1 left and here +

one left this is equal to Z so we have

these two products which is equal to Z

so this implies from here that either x^

4 - 1 is = 0 or X x^ 4 + 1 + 1 is 2 is

equal 0 so obviously there is no real

solution for X here right yes so no real

solution for X right over here so this

equation here we provide the real

solutions for X so let's solve this

equation so we have x^ 4 - 1 which is

equal 0 so we have this is same as x^ 2

s - 1 same as 1 2 this is equal to Z so

applying the difference of two squares

formula this will be X2 + 1 * x^2 - 1

this is equal to0 so we have these two

products which is equal to 0 so this

also implies from here that either X2 +

1 is = 0 or X2 - 1 is equal to Z and

again no real solution for this equation

here this equation only has imaginary

Solutions so here take one to the other

side we get x² is = to 1 take the square

root of both side we get X is equal to

plus or

minus < TK 1 so here we have the only

solution is that only solution only

solutions so plus or minus one right so

therefore the only two real solutions

are X is = to Plus minus

one thank you for watching please kindly

subscribe to my channel also like

comment and share

Harvard University 🎓 Entrance Exam | Can you solve ?


hello good viewers welcome back once

again to this YouTube Channel please

kindly subscribe to this channel here we

have another interesting exponential

equation 1 / 8 ^ x + 2 is = 32 ^ of X

here we are asked to solve for the value

of x right okay now just like we did in

the previous videos all we need to do is

to make sure we have equal bases from

both sides of this equation so that can

equate the powers right okay having it

known that 1 / 8 is the same as 8 to

the^ of-1 right that is 1 / 8 this is

equal to 8 to the^ of -1 and again 8 is

the same as 2 to the^ of 3 then ra of1

so you know this law we are going to

multiply this three And1 together so we

have 1 / 8 is the same as 2^ of -3 right

and from the right

32 is the same as 2^ 5 okay now the

equation is totally transformed

to 2 ^

-3 power x + 2 this is equal 2^ 5 x so

we can multiply the powers together from

here so we have 2^ -3 into braet x + 2

this is equal to 2 5 * X right so we

have equal bases so we can equate the

powers right from here so we have -3

into braet x + 2 this is equal to 5x so

let us open this bracket we have a -3x -

6 this is equal to 5x right so 5x is

crossing over and -6 is crossing over to

the right hand side so here we get -3x -

5x this is equal to 6 right so from here

we get

-8x this is equal to 6 so divide both

side by 8 and here by 8 so here we get X

is equal to you know this is same as -6

all 8 and here simplifying we

get3 / 4 and this is the right answer

thank you for watching please kindly

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comment and share bye-bye

Interesting exponential equation.#maths #education


hello good day viewers welcome back once

again in this video we're going to solve

for the values of X in the following

quatic equation we have x ^ 4 + X2 this

is equal to 20 here let us solve for the

values of X for X to ^ of 4 we are going

to rewrite it

as x^ of 2 power of 2 right because 2 *

2 here is 4 then plus

X2 then - 20 this is equal Z Now from

here let us make the following

substitution we're going to let x^ 2 = y

from here we have y^ 2 + y - 20 this is

equal to 0 we are very lucky that from

here this quadratic equation is

factorizable we have y^2 plus y here we

can write it as 5 y - 4 y here - 20 this

is equal to Zer but my question here is

where is 5 Y and 4 y coming from now the

product of y^2 and -20 is -20 y^ 2 right

now we think about two terms that when

we multiply

them the will give us -2 y^2 and when we

add add them they will give us positive

y these two terms are 5 Y and -4

y so that is where 5 Y and this 4 Y come

from so from here we have a common

factor here Y into braet y + 5 from here

we have a common factor of -4 then we

have y + 5 this is equal to0 so from

here this implies that y - 4 * y + 5

this is equal to 0 from here we have

either y - 4 is = 0 or y + 5 is = 0 so

from here we have y is = 4 or Y is =

to5 now recall again X2 is = y

now when

Y is equal to 4 we get x² is = 4 if you

take root of both sides we get X is = to

plus or minus the RO TK of 4 and this we

have plus or minus

2 now when

Y is equal

to5 we get x2 is = -5 if we take root of

both side this implies that X = to Plus

or - <

tk-5 which is equal to plus orus IUN 5

from here we come up with the

solutions we have X is = to plus or

minus 2 so we have two right we have X

is = to

-2 we have X is equal

to positive IUN 5 and we have X isal to

IUN 5 and these are the solutions thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

A Homemade Quartic Equation | Interesting problem | #maths #olympiad


hello good viewers welcome back once

again today we're going to solve this

equation this equation looks very simple

actually it is very simple but do not be

surprised that this equation has two

solutions but not just one because of

how simple it looks we have here x 4

Plus 8 all X and this is equal to 3 we

can put all over 1 now we're to multiply

each of the terms from both sides by

something and that something will be the

LM of the whole denominators for x and y

and fortunately the

L of 4 x and 1 is = to 4 x right so here

we have 4X multi by x 4 plus here 4X * 8

X this is = to 4X * 3 / 1 so this 4 and

this 4 x x will give us

X2 then plus this X and this x 4 * 4

here will give us 32 and this is equal

to for x * 3 will give us a 12

x so from here we have x² we subtract

this

12x then + 32 this is equal to zero okay

now this is a quadratic equation of

course this quadratic equation is

factorizable we look for two terms that

when we multiply that they will give it

2x2 and when we add them they give -2X

such terms should be both negative right

so we have -8x and -4x so the -2X will

write it in terms

of 8 s and oh in terms of 8 X and -4x

right okay so here this is equal to

Z now we're going to factor out the

common factor here which is X and we

have this and here minus 4 should be

factored out we have x - 8 and this is

equal to

Z so from here we have x - 4 into braet

x - 8 this is equal to 0 so from here

this implies that x - 4 is = 0 or from

here we have x - 8 is = 0 so solving for

x we have x = 4 and we also have this

solution that X isal to 8 and these are

the solutions thank you for watching

please kindly subscribe to my YouTube

channel also like comment and share

bye-bye

Interesting fractional equation that later turned into a polynomial of order 2 |  #mathematics


hello welcome back once again today we

have this interesting exponential

equation with irrational bases otk 2^ X

is = root 3^

X let's divide both sides of this

equation by root of 3 to the power of

X this will cancel out here and we left

it one right now remember we have the

same Powers here let us apply this law

we know a^ m / b^ m is equal to a / B

all to the^ n so therefore this will

become otk of 2 / Ro of 3 all to the

power of X and this is equal to 1 from

here we know this one right over here we

can write it in index power of root of 2

/ 3 remember we need to get the same

base from both sides so here we have

root of 2 / root 3 to the power of X so

this will be equal to we know rooot of 2

/un 3 to the^ of 0 will be equal to 1 so

from here we have the same base so we're

going to equate the power so from here

this implies that X is equal to zero and

this is the right

answer thank you for watching please

kindly please subscribe to my channel

also like comment and share bye-bye

Interesting exponential equation with irrational bases | Can you solve  ?


hello good viewers welcome back once

again today we're going to evaluate this

expression here x^ of

999 given that x^2 - x + 1 is =

0 now let's get into the video before we

proceed let us recall the sum of two

cubes formula we have a cubed + B Cubed

this is equal to a

that is A + B into braet a 2 - A B + b

2 okay now if we let a = x and we let b

= 1 then we obtain a cubed will be X

cubed plus B Cubed will be 1 cubed which

is 1 this is equal

to a + b which is x + 1 into bracket A 2

will be X2 - A which is x * 1 which is x

+ b 2 which is 1 2 which is

1 so we can see from the right hand side

of this equation from here we have X2 -

x + 1 and from the original equation

that is exactly the left hand

side so what is this trying to let us

know this is telling us that when we

multiply x^2 - x + 1 by x +1 this will

automatically turn to X Cub + 1 now

let's go ahead and do that here we have

from the equation

X2 - x + 1 this is equal to0 now let's

multiply x + 1 to both sides of this

equation so here we have x^2 - x + 1 * x

+ 1 this will be equal to 0 MTI x + 1

from the left hand side we all know that

will be X Cub + 1 and this is equal 0

from here this implies that X Cub is = 0

- 1 which is1

what about thetion we are trying to

evaluate luckily

999 is equal to 3 * 3 3 3 so from this

equation here if we raise both sides to

the power of 3 3 3 we will obtain X to

the^ of 9 999 from the left hand side

now let's go ahead and do that we have a

x cubed rais to the power of 3 3 3 this

will corres corly be equal

to1 3 3 3 now let's record this property

a to the^ of B to the^ of C this is

equal to a of B *

C so from here 3 * 3 3 3 will give us x^

of 9 9 9 this is equal to you know the

power here is odd -1 ra the power of odd

will give us -1 and and this is the

right answer thank you for watching

please kindly subscribe to my YouTube

channel also like comment and share

please support my channel by joining my

membership or you can send me a super

tank if you find this video really

interesting and helpful bye-bye see you

in the next video

Olympiad Math | Interesting problem | #maths #olympiad #mathcompetition


hello you are welcome here we're going

to find the value of root of X2 + Y2 + 1

given the following two systems of

equations x^2 = 19 x + y y^2 = 19 y + x

let's call this our equation 1 and let's

call this our equation

2 so let's start like this we're going

to subtract equation two from equation

one right so so we have to subtract oh

sorry that is equation 1 minus equation

2 so equation one the left hand side is

X2 minus equation 2 the left hand side

is y^

2 then this will be equal to for the

right hand side so we have 19 x +

Yus the equation 2 the right answer

which is 19 y + x so from here we have

- y^ 2 this equals here 19 x - x will

give us 18

x then here 9 that is y - 19 y that will

give us -8 y so here remember this is

the difference of two squares so we

write it as x - y * x + y and from here

18 is common so we factor that out and

here we have x - y so remember that X is

not equal to Y right so here we cancel

this from both sides so here we get x +

y = 18

amazing okay now let us add the two

equations so we can derive something

very beautiful one more time so here we

have equation 1 plus equation 2 this

will imply from the left that would be

X2 + Y 2 that will be equal to 19 X + y

+ 19 y +

x okay so from here we have X2 + y^2 =

here 19 x + x that will give us 20 x

then plus y + 19 y will give us 20 y so

from here we have X2 + y^2 = here factor

out 20 and here we have x + y so from

here x + y is 18 so we get 20 * 18 right

and that will give us

360 so from here we have something

similar to the expression we asked to

evaluate which is < TK of

X2 + y^2 + 1 and this will be equal to

the square root of x + y s is 360 right

so we have 360 then we have + 1 1 this

equals to

361 right which square root is 19 and

here is the right answer thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Interesting Olympiad Math


hello you're welcome today we're going

to solve this exponential equation from

here we have

256 root x = 64 to^ of

X now there's something really nice we

need to take note of here we have 256

and remember this is exactly equals 2 to

the^ of

8 and then from the other side the base

there which is 64 is the same as 2 to

the power of 6

so we're going to write the bases in

terms of the powers of two right so from

the left instead of 256 we write it as 2

to the^ of 8 then ra the power of root X

this equals 64 we write it as 2 to the

power of 6 then to the power of X now

let's take note of this a to the power

of B to the power of C this is same as a

to the power B * C so from here we have

2^ 8 * otk X will give us 8un X then

this is equal to here 2^ 6 * X will give

us 6 x so we can see that we have

obtained equal bases from both sides of

this equation so from here this implies

that 8un

x = 6 x that is equating the powers

right okay so from here let us Square

both sides to get rid of this root sign

so here we have 8un x s this will be

equal to 6 x^ 2 now take note of this a

* B to the^ of M this is same as a to

the power of M * B to the power of

M we can see from here 8 2 will give us

64 and the root x s Square will cancel

the root sign we have just X then this

be equal to 62 will give us 36 then

X2 now let's change the sides of this

equation so we can write as 36

x^2 = 64x but this does not affect the

original equation

right okay so from

here we have to subtract 64x from both

sides so we get 36 x^ 2 - 64x

then this equals z we have XX is common

so we factor that out so we have 36 x -

64 then this equal 0 so from here this

implies that x = 0 or 36 x - 64 = 0 add

64 to both side we get 6 36 x = 64 then

from here we divide both sides by 36

6 here by 36 36 cancel 36 and here we

left with X = here let us divide by 4 64

/ 4 will give us 16 then all over 36 / 4

will give us 9 so here we come up with

the solutions that x = 0 and also x = 16

/ 9 and these are the only two real

solutions

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Interesting exponential equation


good evening viewers welcome back to my

channel

today we have the following interesting

problem

if two raised for X all over three plus

two raised by S is equal to one over two

then eight raised per X all over three

plus eight raised by x c was Dutch

okay

wow please if you have not subscribed to

this my Channel please subscribe turn on

the notification Bell so that you'll be

notified anytime we upload new video

okay let's see how we're going to solve

this problem

this expression I'm going to rewrite it

as follows

so I will be writing this expression as

we have here H raised by X

all over three loss eight raised by X

okay

so

we all know that this age can be

expressed as two raised by three then

raised by X

all over

3 plus you know two raised battery

raised by X now we're going to use the

following in this is law

that's when we have a to the power B

in bracket raised by m

we can rewrite as follows a raise for M

and bracketing risk YB okay

so here we have

here to raise power x respiratory

all over

3 plus

2 raised by X

raised by three okay

wow

now I'm going to make a substitution

here so let me say

let's to raise power x equals k

now we have change

the Outlook of this question so

what we're having here now our new

equation here

now is going to be

okay

all over 3 plus k equals 1 over 2 of

which will be evaluating

of which we'll be evaluating

K Cube

all over 3 plus K Cube and this is

exactly equals to the value of the

expression we're about to evaluate a to

raise power x all over 3 plus 8 raised

by X okay

wow that's nice okay now cross multiply

here so if you cross apply we have two

times K so we have 2K is equal to you

know 3 plus K plus times this one here

so we have three plus okay you know

subtracting this K from both sides so we

get K is equal to three so finally we

get K is equal to three so we can easily

evaluate this expression which will also

give us the right value of this

expression here so since K is equal to 3

so this expression here is going to be

all over you know here 3 plus 27 and

here we have what 27 all over 30 and

simplifying this so simplifying the

using these three to divide both sides

to divide up and down I mean so we get 9

all over ten and this is the right

answer to the question thank you for

viewing and please make sure you

subscribe to our YouTube Channel please

subscribe

Please Subscribe share our video and

comment thank you

An Olympiad Math Problem


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Interesting surdic equation.                    #olympiad #maths #education


and we're going to find the last digit

of this

of this number here so we're going to

find the last digit using the

exponential method so we have two raised

for 2008. now 2008

is the same as four times five zero two

so now when we are looking for the last

digit of any number is equally the same

as divided the number by 10 so we have

two raised to the powerful this is just

application of common indices law now 2

raise power 4 is equal to 16 so we have

in Brackets 16 raise power 502 in the

bracket here the last digit here is six

so take it out we have six raised to

power five zero two remember six raised

for Any number greater than zero we

always have the last digit to be six and

this is division by 10 that is a model

ten so the last digit of two principle

2008 is equal to what six thank you for

viewing and make sure you subscribe to

my YouTube channel for more interesting

videos

The fastest method


hello welcome back my lovely people

today we're going to solve this

exponential equation we're given 6^ X2 =

15 so we're going to solve this problem

by using the natural logarithm so we're

going to Ln both sides of this equation

so Ln of 6^ of X2 this equals the Ln of

15 now with the property of the natural

log so this power here is going to cross

over to this place and that will become

x² MTI the Ln of 6 this equals the Ln of

15 right so from here remember we're

suming for X right so we're going to

divide both side from here by the Ln of

6 here by the Ln of 6 so this will

cancel this and here we get x2 equal the

Ln of 15 / the Ln of 6 from here we're

going to take root of both sides and

here we come up with the solution X =

plus or minus the square root of Ln of

/ the Ln of six and this is the right

answer thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting exponential equation


hello good day viewers welcome back once

again today we're giv the following

equation x + 1 / x = 1 and here we're

going to evaluate x^2 - 1 /

x^2 before we proceed let us recall the

difference of two squares

formula we have a s - b 2 this is equal

to a + b MTI Aus B now this expression

here we are asked to evaluate we can

Factor it

as x + 1 /

X MTI by X - 1 /x right okay now from

the given equation we already have the

value for x + 1 /x so our major goal

here is to find the value of of x - 1 /

X from the given equation we have x + 1

/x this is equal to one now let's Square

both sides of this

equation from the left hand side we'll

be making use of the following identity

a + b 2 this is equal to a - b

s + 4 a b now make use of this this

identity the left hand side of this

equation will be X - 1 /x

2+ 4 a here will be 4 * x * 1 x which is

4 this is equal to 1 so from here we

have x -1 / x^ 2 this is = to 1 - 4

which is -3 from here we have X - 1 /x

this is = to plus or minus root of -3

from here we have x -1 / X this is equal

to plus or minus you know root of1 is I

then root of

3 therefore the expression we asked to

evaluate x^2 - 1 /x2 which we factored

as x + 1 /x * X - 1 /x this will be

equal to x + 1 / X is given us 1 then

multip X - 1 / X which is equal to plus

or minus IUN of 3 and from here we get

the solution as plus or minus I < tk3

and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Very tricky homemade math problem | #maths #mathematics


again today we're going to solve for the

value of x in the following exponential

equation involving root signs right okay

let's get to the

video here we're going to square both

sides of this equation to get rid of the

first root

sign squar this will be 9 squ so from

here this one and the first one here we

cancel out and we're left with square <

TK of 1/ 9 to^ x this is equal to um

that is 9 squ right which is 8

to1 and let's go ahead and square both

sides

again this one and this one we cancel

out and we left with 1/ 9 to the^ of X

this is equal to um that is

81 2 so 81 is the same as 3 power of 4

then of two so when we multiply this

four and two together you know this

property m

n z this isal m n * Z so this will give

us 3 4 * 2 which is

8 now 1/ 9 let us recall again 1 m n = m

n so 1 / 9 is 1 / 3 2 x this is = to 3 3

8 now make use of this property here we

have in the bracket 3 -2 x this = to 3 8

so from here we multiply these Powers

together we have 3 -2 x this is equal to

3 of 8 we have equal bases so actually

we can equate the powers right so we

have -2X this is equal to 8 now let's go

ahead and divide both side by

-2 so here we have S is = to -8 / 2 that

will give us -4 and here is the solution

thank you for watching please kindly

subscribe to my YouTube channel also

like comment and share bye-bye

Interesting exponential equation | #maths #olympiad


hello good viewers welcome back once

again today we're going to solve the

Bangladesh Olympia math we're given the

equation I think that's quadratic right

X2 - 3x + 1 = 0 here we're going to

evaluate this expression x^ 6 + 1 x^ 6

from this given equation we can divide

each time from both sides by X and see

what happens x² / x - 3x / x + 1 / X

this is equal to 0 /

X from here x² / X that will reduce to x

- 3x / X that is 3 + 1 /x this = 0 now

let's move this -3 to the right so we

have here x + 1 /x =

3 now from here we're going to CU both

sides of this equation

from the left let us make use of this

property A + B Cubed this equals a cubed

+ B Cubed + 3 * a * B into bracket A + B

and mind you when we Cube this second

expression here it will look like this

1X Cub this will be equal to 1/ X cubed

because I may not explain this in this

video okay now from the left we have the

first term cubed which is X cubed plus

the second term cubed which is 1 / X

cubed then plus 3 * the first term * the

second term first term which is x * 1 /

X will give us 1 so we left with three

into bracket the sum of X and the

reciprocal of X then this will be equal

to 3 Cub which is 27

from here remember that x + 1 x = 3 so

we have here x Cub + 1 / X Cub plus here

will be 3 * 3 right then this is equal

to

27 from here we have X Cub + 1 / X Cub +

3 * 3 which is 9 this =

27 subtract 9 from both side we arrived

at X Cub + 1 / X cub

= 18

right from here let us Square both

sides from the left we're going to make

use of this a + b 2 this equal a 2 + b 2

+ 2 * a * B so from here the first term

squar will be X Cub 2 plus this one will

be same as 1 / X cub Cub s right because

1 2 is 1 then plus 2 * the first time x

Cub * the second term 1 / X cub and this

equals 18 s which is

324 from here 3 * 2 here is 6 so we have

x ^ 6

+ 1 / x^ 6 and we can see the expression

we asked to evaluate has already popped

up right right so we have here plus this

we cancel this and we have two and this

is equal to

324 subtract two from both sides and we

have the solution x 6 + 1 / x 6 =

322 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Bangladesh Olympiad Math | Interesting Algebra problem |


hello good viewers welcome back once

again please kindly subscribe to this

channel also like comment and share here

we're given the following interesting

limit limit as x ts to 0 of < TK 2 - <

TK 2 - x all all over 2 here we asked to

evaluate this limit right if we try to

make direct substition here that is

we're going to have theare < TK of 2

minus the < TK of 2 - 0 right then all

over 0 and you know that division by 0

is undefined even though we simplify the

upstairs here we're going to have > of 2

- < TK of 2 / 0 and this equal 0/0 so

this is in in determinant form right so

let us go ahead and manipulate this

which we have in this bracket right over

here okay so what we're going to do here

is we are going to multiply the upstairs

and downstairs by the conjugate form of

this upstairs here so that is

limit as X is approaching zero so here

we have the < TK of 2 minus the < TK of

2 -

x then we multiply by the conjugate form

of this which is < TK of 2 plus the < TK

of 2 - x the same thing happens

downstairs so here we have X multip by

the < TK 2 + the < TK 2 - x right so

here this is equal to

limit as X is approaching zero so you

know here we're going to make use of the

difference of 2 squ formula and what is

this formula when we have a minus B * A

+ B this is going to give us a 2 - b 2

right so up here we have the < TK of 2

2us the < TK 2 - x squared right then

all / X into braet < TK 2 + the < TK of

2 -

x okay so here you know this square and

this square root will cancel out square

and square root will cancel out so you

know this is equal to

limit as X is approaching zero so upset

now we have 2 minus then into bracket

what we have there 2- X right then all

over X Open Bracket < TK 2 plus the sare

otk of 2 - x right so you know this is

equal to

limit as X is approaching zero so here

you know 2 - 2 here then plus X that is

what we have because of this negative

sign here so 2 - 2 is 0 and 0 + x is X

so here we have X then all over X into

braet < TK 2 + < TK 2 - x right you know

this will cancel out and from here we're

going to take our limit now so this is

going to be 1 all over you know here we

have < TK 2 + S < TK of 2 -

0 so this is equal to 1 all over the <

TK 2 + the < TK of 2 and this is equal

to 1 all over 2 < tk2 and this is the

right answer thank you for watching

please do well to subscribe to my

channel also like comment and share

bye-bye

Interesting limit. How to evaluate without L'hospital's rule ?


hello welcome back once again today

we're going to solve this simple

exponential equation from the left hand

side eight can be Express an index power

of 2 as 2 the^ 3 so here we have a

bracket 2^ 3 then ra to the power x this

will be equal to we know square root

means the^ of 1 / 2 so theare root of 2

is as 2^ of 1 / 2 from here with this

property a to the^ b^ C this is equal to

a^ B * C so this will become 2^ 3x which

is equal to here 2^ 1 / 2 we can see

from both sides we have the same base so

we're going to equate the power so this

implies from here that 3x is equal to 1

/ 2 right so here cross multiply so this

here we multiply this and this here we

multiply this that is the meaning of

cross

multiplication so here 3x * 2 here will

give us 6 x this will be equal to 1 * 1

here will give us 1 from here let's

divide both side by the coefficient of x

6 cancel 6 from here so here we have our

solution X is = 1 / 6 and this is the

right answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Simple exponential equation |


hello good day viewers welcome back once

again today we have an improper integral

here and we're going to evaluate this we

have the integral from 0 to Infinity of

e^ of -5x * X to ^ of two in this video

I'll be making use of this special

function is a very beautiful one gamma

function and it has the integral

representation as the integral from 0 to

Infinity of

e^x * x^ n

-1 DX so our goal here is to transform

this integral to

look exactly like this in this form I

mean and then we can make use of the

gamma function remember that the gamma

of n = n -1

factorial we can just say that this

gamma function is chose the extension of

the factorial it's an amazing function

now let's get into this video from here

the power of E here is just --x and from

here the power of E here is -5 * X so

we're going to make a substitution by

letting 5x = y from here divide B by 5

we get X = Y / 5 then differentiate both

side we get DX = the Y / 5 then our

integral

becomes the integral from 0 to Infinity

yes 5 * 0 is 0 5 * infinity is infinity

then e to the^ of Y from here then * x 2

will be y / 5 2 which is y^ 2 all 25 *

the X which is the y/ 5 then from here

25 * 5 is 125 so we put it out that is 1

125 the integral from 0 to Infinity e y

multip by here y^ 2 we can write as y^ 3

- 1 then d

y now if you compare this with this here

we can see that here we have the gamma

of three which is equal to 3 - 1

factorial

right and that we give us 2 factorial

which is equal to 2 * 1 and that's equal

2 so from here our right answer becomes

1 all 125 * 2 / 1 which is equal to 2

all over

125 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

The Gamma Function


good day viewers welcome back once again

to my YouTube Channel please if you have

not subscribed to our Channel please

kindly support us by subscribing to our

Channel thank you here we have this

infinite

next radical right we have x equal to

theare < TK of 5+ the < TK of 5+ the <

TK of 5+ and so on and this goes on

forever right so we call this an

infinite nested radical right okay now

we asked to find the value of this x

right so let us get started here we have

X is = to the square < TK of 5 + the

square < TK of 5 + the < TK of 5 plus

and so on right okay so this is equal

to x = square root of you know here we

have 5+ so here root of 5 + and so on is

just X right because X is equal to this

so here we have squ root of 5 plus and

here we have from here we have exactly

what is here which is X so we replace

that with X so right over

here this is what we have now < TK 5 + x

is = X right so we're going to square

both sides to just eliminate this Square

sign so let us Square Square both

sides so this will cancel this right so

here we have X2 is just equal to 5 + x

so subtracting 5 + x from both sides so

we just get x^2 - 5 - x = 0 so let us

put this equation well x^2 - x - 5 is

equal to Z so right over here I'm going

to make use of the

braga's quadratic formula right so we

have that X is just equal to B plus or

minus < TK b ^ 2 - 4 a c all/ 2 a so

right here the coefficient of X2 is a

right and that is what 1 here B is equal

to coent of X here -1 and C is equal to

-5 which is the constant right so this

is equal

B

is1 which is + 1 then plus or minus

square otk of b 2 is1 2 which is 1 - 4 a

c means 4 * 1 * -5 which is

-20 all 2 a 2 * 1 which is 2 so this is

equal to 1+ or minus Square Ro TK of you

know this opening this bracket we have

positive sign and 1 + 20 is just 21 or

by two right so now take note

that

this this infinite next

radical has a result which is positive

and not negative right so which means

that the negative value for x will be

ignored right so here we have X = to 1 +

the < TK of 21 / 2 or 1us the RO of 21 /

2 and this value is negative because

sare root of 21 is bigger than one right

so to subtract this from one will give

us a negative result so this is

cancelled and this is the right answer

thank you for watching do wa to

subscribe to our YouTube channel and

share our videos

An infinite nested radical


hello good viewers welcome back once

again today we have this interesting

problem here and here we're going to

solve this

problem we have here that 98% people

fail this

problem obviously this is true most

people may think that the only solution

here is X isal to

3 but is something very important we

need to consider here let us recall

about the absolute value forun function

of X remember that the absolute value I

mean the sare root of x² I mean is equal

to the absolute value of

x and

again we have another interesting

property of this absolute value function

the absolute value of

x is equal

tox if and only if x is less than zero

right and again the absolute value of x

is equal to X if an only if x is greater

than or equal to

zero now let's take a look at this

problem here we have square root of

x²

remember square

root of x² is not equal

to < TK X

squared these are not equal because this

is where many students get

confused right now here we can see we

have a different tax and that is

something like this so the left hand

side here is just the absolute value of

x right and this is equal to three now

let's take note of this if the absolute

value of

x is equal to a then therefore

X is = plus or minus

a so from here we can see the solutions

that X is equal to+ or - 3 and here we

have the solution to the problem x = 3

and x = -3 thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting problem | Can you find all solutions ?


hello good viewers welcome back once

again let's find the solution to the

following given radical

equation there are different ways of

solving this problem but in this video I

will start by making a

substitution first of all I will let

root of x = Y when you square both sides

we get X is = Y

2 now from here we have root X and that

is y so we write

y+ X here will be y^

2 and this is equal to 2 * root X will

be 2 *

y now from here let's try

simplifying from the left hand side we

have y^ 2 then + y from the right hand

side will be Min - 2 y right and this is

equal to zer

now from here we get y^ 2 now y - 2 y

will give us - Y and this is equal to Z

so let's go ahead and factor out y we

have Y into braet y - 1 this is equal to

Z and from here this implies that Y is

equal to0 or in the bracket y -1 is = 0

so from here we get solutions for y y is

= 0 or Y is =

1 now let's recall that X is actually

equal to Y2 right so

when Y is = 0 we get x is = 0 2 and that

is equal to

Z and

again when Y is = 1 X is = 1 2 and that

is one so from here we have Solutions we

have X is equal to Z and we have that

also X is = to 1 and these are the

solutions thank you for watching please

comment and share bye-bye

Indians | Interesting surdic equation | #maths #mathematics


hello good viewers in this video I'll be

providing a solution to this nice

Olympia mat problem here here we given

that the < TK of 7 to the^ of < TK of 24

- < TK X is = 49 here we asked to find

the value of values of X please if you

have not subscribed to my Channel please

kindly support this channel by

subscribing please like comment and

share here we're giving < TK of 7 to

the^ ofun of 24 - < TK of 8 X is = to 49

here we can put this in bracket otk of 7

is the same as 7^ of 1 all/ 2 right then

ra the power of 24 - < TK X and this is

equal to 49 right so according to the

law of indices you know we can multiply

these Powers here together so we're

going to have

7 of square root of 24 - < TK of X all

all over 2 this is equal to 49 is the

same as 7^ of 2 so we can see from both

sides we have equal bases so we can

easily equate the powers right so from

here we have square root of 24 - < TK x

all/ 2 this is equal to 2 so cross

multiply so here we get sare < TK of 24

- < TK X is = 2 * two here that is four

right so from here we're going to square

both sides of this equation here we

Square here and here we Square here

right so we can see that this square and

square root cancel out so we're left

with 24 minus theare root of x this is

equal to 4 S that is 16 so from here

we're going to subtract 16 sorry from

here we're going to collect L ter so

here we're going to have < TK of X is

also equal to 24 -

16 right so from here we getun X is = 24

- 16 that is equal 8 okay so from here

now we have the square root of x is

equal to 8 so we're going to square both

sides here we have of X2 this is equal

to 8 2 so square and square root cancel

out so here we get x = 82 and that is 64

and this is the right answer thank you

for watching please kindly support to my

channel subscribe like comment and share

bye-bye

Interesting problem for Olympiad.


hello good day everyone welcome back

once again today we have this nice and

interesting exponential equation we have

2^ root X is = 16 ^ x / 4 and here we'll

be solving for the values of X that we

satisfy this awesome looking

equation now let's get into this video

from the left hand side we have 2 to the

power of root of x then this will be

equal to remember 16 can be expressed in

index power of 2 so therefore 16 is 2 to

the^ 4 then remember we have 4 or I mean

that is X right yes so x / 4 as the

power here now let's take note of this

property a the^ B to the^ C this is

equal to a^ B * C so from here we have

from the left hand side 2 the^ < TK X

then this will be equal to 2 to the^ of

4 multi by x /

4 okay so from here we have this nice

cancellation remember this is 4 all over

one so here this four we cancel this

four here so we have 2 to the^ of otk X

this will be equal to 2 to the^ of

X so from here we have equal basis so

from here we're going to equate the

powers so this implies from here that

that the square root of x is equal to

X from here to get rid of this root sign

let's go ahead and square both sides of

this equation so we know that the square

and the root signs are en right so the

square and the root sign they are en so

they will cancel out from here so we

have here x is =

x² now let's switch the sides of this

equation so from the right hand side let

us make it the left hand side so x² is

equal to X subtracting X from both side

we get x^2 - x this is equal 0 so we can

see from here x is common so we Factor

it out and we have x -1 which is equal

to0 okay so from

here we have x * x - y = 0 so we have

either x is = 0 or X - 1 is = 0 and from

here x is = 1 so therefore we have only

two real solutions that satisfy the

given equation so we have xal 0 or X is

equal to 1 thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting exponential equation | can you solve ?


hello good day viewers welcome back once

again today let us solve this simple

exponential equation here we be solving

for the value of x so from here we have

8^ of x + 8 to ^ of X which is equal to

1 right so from the left hand side h8

to^ of X is common so let's factor out

8^ X so that will become 8 ^ of X into

bracket 8 a^ x / a^ X we left with 1

plus a^ x / 8^ X that would be 1 right

then this is equal to 1 so from here we

have 8 to the^ X MTI by 1 + 1 from here

is 2 then this is equal to 1 so from

here let's divide both sides by two so

we have 8 to^ x x * 2 / 2 this will be

equal to 1 / 2 so from here 2 we cancel

out 2 so we have here 8 to^ x which is

equal to 1 / 2 so let's take note of

this property we know 1 all a to the^ n

this is equal to a to the^

n right so here 1 / 2 is = to 1 / 2^ 1

which is equal to using this algebraic

identity here so this will be 2

of1 so from here we have 8 to the^ of X

which is equal to 2 ra power

of1 from here we know this base here can

be expressed in index power of 2 so that

is 2 to the^ 3 which is equal to this 8

then we have a power of X this is equal

to

2^1 from here let us consider another

algebraic identity We Know a^ M in

bracket n is = to a^ M *

n so from here this is equal to 2^ 3 * X

which is 3x then this is equal to 2

of1 we can see from both sides of this

equation we have the same base right so

we're going to equate the power so this

implies from here the power of two here

3x is equal to the power of two from the

right hand side -1 since we're solving

for x let's divide both side by the

coefficient of x soide by 3/ 3 so this

will cancel this and here we have our

solution X is = -1 / 3 and this is the

right answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Tricky exponential equation


hello good viewers welcome back once

again today we have an indefinite

integral here we're going to evaluate

this using integration by parts

formula I know we're very familiar with

this property here the integral of U DV

this is equal to u v minus the integral

of V duu the right over here we have

this indefinite integral here integral

of x * e^ X DX X now let us make the

following substitutions in terms of u

and v here we're going to let u

= x right so if you differentiate this

we have du this is equal to the

derivative of x with respect to X is 1

and here we have

DX okay the next substitutions in terms

of V we have V is equal to now the

integral of e to the power of

X the integral of e to the power of X DX

so if we differentiate this we actually

have DV is equal to the integral of e

that is e sorry that is d v if you

differentiate the integral of e to^ x

actually here we have V is = e to the

power of

X so here if if you differentiate both

side we get DV this is equal to e to^ x

DX Now using this property here we have

u v and here this is U and this is V so

here we have x * e to the power of X

then minus the integral of v and what is

v v is e to the power of X and we have

here

DX okay because du here is equal to DX

as you can see here so this from here we

have x * e^ of x minus the integral of

e^ x remains e^ of X right and here we

have plus C and this is the right answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye but

Integration by parts | #maths #mathematics


hello welcome back once again in this

video let us solve for the value of x

that will satisfy this equation from the

left hand side we have the < TK of X

plus the < TK of X and this is equal to

the < TK of

8 from the left hand side we know < TK

of x + < TK X will give us 2 * the < TK

X then this is equal to the square root

of

8 now we know that squ < TK of 8 we can

actually rewrite it right from the left

hand side here we have 2 * root of x so

this will be equal to the square Ro TK

of 8 here is as 4 *

2 then we have here 2un x is = to so

we're going to write this as < TK 4 MTI

the < TK of 2 so from here we have 2un X

this is equal of 4 is 2 then multiply by

< tk2 right so from here two we cancel

two from both sides we have your < TK X

is = to < TK 2 to solve for x simply

Square both

sides Square cancel square root square

cancel square root and here we get our

right answer that X is = to 2 and this

is the correct answer thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Simple radical equation | Can you solve ?


hello good viewers welcome back once

again today we're going to solve for the

all real values of X that will satisfy

this given exponential equation we have

e^ x² this is equal to e^ of X now the

first thing to do is to Ln both sides of

this equation if we have the idea that

the Ln of e to any power is equal to

that power so from here we have the Ln

of e to the of x² this is equal to the

Ln of e to power of X so making use of

this property from here we have the

power of E which is

x² then this is equal to the power of E

here which is X from here do not make

mistake of dividing both side by X if

you do that you will lose a solution so

the best thing to do here is to subtract

X from both sides and here equals zero

from here we factor out X and we're left

with x² / x with X then x / X

is1 and this is equal

to0 so from here we have X is equal to 0

or the one in bracket X - 1 = 0 from

here we have X is = 1 so from here we

have the following solution that x is =

0 and x is = 1 and this are the real

solutions thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Germany | Nice exponential equation | #maths #mathematics


hello welcome back once again today

we're going to evaluate this with

calculator and here is how we're going

to do this we have 6^ 12 / 12 power 6

which we can write as 6^ 12 ided by here

12 is the same as 2 * 6 then power of 6

now with this basic property in algebra

remember when we have a * B all of n we

can express this as a to the^ of n * B

to the power of n right so from here

this is going to be equal to 6^ 12 /

here we get 2 ^ 6 * 6^

6 from here we can express this as 1 all

over here 2^ 6

multiply by here 6^ 12 / 6^

6 now let us take note of this again

when we have a to the power of n / B to

the power of

n i mean a to the^ of n so sorry for

that so we have a the power n ided by

the same base here a to the power of

another power we can actually subtract

the power that will be N - n right so

from here we have this equal 1 all 2^ 6

then multip by here this is going to be

6^ 12 - 6 so from here we get 1 2^ 6

multip by here this will become 6 ^ 6

then from here put this over one 6^ 6 *

1 will give us 6^ 6 / 2^ 6 * 1 will give

us 2^

6 there's another nice property we're

going to take note of

again when we have a^ of m / B the^ of M

since we have the same Powers we can

rewrite it as a / B all to the power of

M right so from here that will

be 6 / 2 all power of 6 and from here 6

2 is 3 we get 3 to the^ of 6 which is

equal to

729 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Can you simplify without calculators ?


hello good day everyone welcome back

once again today we're going to solve

for the value of x in the following

Olympiad M problem yes this is an

interesting equation we are going to

find that number which when four is

raised to the power of that number then

that will give us the reciprocal of that

number and what is this number let's go

ahead and figure it out okay this

question is very simple right so let's

get into the video from the right hand

side I want to re write the equation as

1 /x this is = to 4 of X now the next

thing to do here is

to is to get rid of this power of four

here which is which is X so for us to

get rid of this x we need to raise both

sides of this equation to the reciprocal

of X which is 1 /x

right and here by 1 /x so from here we

get um that is 1

/x of 1 /x this will give us from here

we know we can multiply This Together

recall that M of b n isal m b * n so we

multiply This Together X time the

reciprocal is 1 so here we have

four so from here we have 1 /x of 1 /x

this is equal to 4 is the SU

2 2 so from here this implies that

either we equate the bases or equate the

powers so because we have um a pattern

of exponents here the base 1 / x isal 2

and also the power 1/ x is = 2 so we

have 1 /x this is equal to 2 okay now

let's um cross multiply x * 2 we give us

2x this is equal to 1 * 1 that is 1

divide both by 2 and here we get x = 1 /

2 and this is the solution thank you for

watching please kindly subscribe to my

YouTube channel also like comment and

share bye-bye

Olympiad Math | Interesting exponential equation | #maths #olympiad


good everyone viewers here welcome back

to my Channel today we have the

following equation 3.8 times 2 raised by

x squared is equal to one where X to

find the value of values of X okay

now I'm going to start by logging both

sides of this equation remember that the

logarithm of a times B

is the same as log a plus log B okay now

I have log of 3 raised by S times 2

raised to power x squared is equal to

log of 1.

now I have

log of 3 raised for X here plus log of 2

raised power x squared is equal to log

of 1 is 0. now using the property of

logarithm so this x we multiply the

entire logarithm here and X we multiply

data logarithm here

so I have a x Times log of 3 plus x

squared times log of 2 is equal to 0.

now I'm going to factor out X here so I

have X to be factored out X log 3

divided by S we are left with log 3 plus

x square log 2 divided by X we are left

with X log 2 close bracket is equal to

zero so from here we have either x equal

to 0 or you know log 3 plus X log 2

equals zero and here we have S equals 0

or

you know X log

2 is equal to you know if this log 3

should cross over this side so it will

change to a negative

log three and here we have X is equal to

zero or divided by the question of s

this is over log 2 here

and here over log 2 log 2 cancel log 2

so half s is equal to negative log three

all over log two so we have come to the

end of today's video that

in the above equation

the values of s will have S is equal to

zero or

negative log three all over

log two thank you for viewing and please

make sure you subscribe to my YouTube

channel for more interesting videos

An interesting exponential equation. Feel free to solve !


hello good viewers welcome back once

again today we have another nice

algebraic problem here we have x² + y s

= 58 we have the square of the sum = 10

right over here we're going to evaluate

x -1 all

squar from the equation one we have X2 +

y^2 = 58 now with the one of the

properties of algebra the left hand side

we can write it as

X + Y 2 then - 2x Y and this is equal to

58 right so from equation 2 we already

have that x + y^ 2 is 10 so here this

will be replaced with 10 and here we

have - 2x y this is equal to

58 so from here we have 2X y = 10 - 58

that will give us

-48 now let us multiply both by two I

have reason for this and we're going to

know very soon so here we have 4X y this

will be equal to 96 now the reason for

doing this will come up soon so just let

us pay attention and listening so here

we have to evaluate x - Y 2 which we can

use this identity x + y^ 2 - 4 x y so

the reason why I did this here by

multiply both by two that is finding the

value for 4X y because I know that 4X y

must come up somewhere in the evaluation

of this and we can see it right over

here right okay so from here we can

easily evaluate this remember that x +

Y2 is 10 then minus 4X y

is96 then this will turn out to be 96

positive so here we have to add this up

and from here we get

1006 and this is the right answer thank

you for watching please kindly subscribe

to my Channel please turn on the

notification Bell so that so that you'll

be notified anytime we upload new videos

byebye I'm going to cut it short

Interesting Algebra problem | #maths #mathematics


good day viewers welcome back to my

Channel today we have the following

interesting s mathematical problem

okay now today we're having a first

class on the S problem and anyone who

want to apply or register for this set

you can follow up my channel by

subscribing please subscribe so that the

whole

classes will reach you okay now today we

have the following problem this happens

to be their problem number number 10

okay here we have the function G of X is

equal to ax² + 24 now for the function G

defined above a is a constant and G of 4

= 8 what is the value of G of -4 many PE

many people may initially think that the

answer to this question

is8 because here we have the 4 and

because we have pos4 maybe the

corresponding values will change signs

but that is very wrong okay now let's

get started now the first thing we need

to do we can just substitute into x-4 in

order to find the value of this

expression here G of4 okay we need to

First find this unknown number here

which is a constant here okay how do we

do that now we have our function G of x

is = a

X2 + 24 okay now we are given that g of

4 is equal to 8 okay now let us

substitute into X here for and see what

we're going to have as our corresponding

right hand side value so here we have G

of

four to be equal to you know a 4 2 then

+ 24 and this happens to be equals 8

okay so we need to solve this simple

equation here and find the value of this

constant a here okay here we have here 4

S is equal to 16 so we have here 16 a +

24 here is equal to 8 okay now 6 24 here

is crossing over so we have 16 a is

equal to

16 now dividable side by the coefficient

of a and that is over 16 and here over

16 16 cancel 16 so here we have a = to

-1 okay now our original function here

is G of x = to here a is = to1 so we

have - x 2 + 24 okay now from here we

have gotten what we first need finding

that constant a and that happens to be1

now we can easily find the value of G

of-4 and here we have G of -4 isal

to -4 2 +

24 here we have negative remember that

in bracket -4 s is

16 +

24 here we can switch it as 24- - 16 and

this is equal to 8 and G of -4 is equal

to 8 so option A here is correct thank

you for viewing and please make sure you

subscribe to my YouTube channel for more

interesting videos

SAT math problem. Number (10). In this video, l showed how to prepare for SAT exam.


hello welcome here we have an

interesting question here so we have

here the W of the natural logarithm of

four but the question here is what is

this

W so like is this a

function or this is just a written

expression so but take note that this W

Here is known as the lamber W function

so we can still call it the product

logarithm and here is how the function

works so when we

have the lambar W function of let's say

we have X time a here a is a number then

power of X so this simplifies to X right

so mind you that whatever that is here

must be equal to the power of this e so

now how does this relate to this problem

okay let's see how to solve this problem

so now take

note the natural logarithm of

four is equal to the Natural logi of you

know four is the same as

2 of

2 right so you know the one of the

properties of logarithm right

right so here the power is going to come

over to this place so like this right so

here we're going to have this to be

equal to 2 * the natural logarithm of 2

right okay now this expression here is

going to

be

lambat W function of 2 * the natural

of

two okay but

here we don't have

something looking similar to this right

so here we have e to the power of

something but here we don't have e even

e at all in this bracket so now let us

recall this

property

that e of natural log of x is equal to X

so we're going to use this property here

so we're going to write this as the

natural the lambar W function of you

know two we write two in this form so

we're going to have e of the natural log

of two mind you that the E the natural

log of two will give all this two here

so here we Times by the natural log of

two so we can see that we have

similar expression like what we have at

the the left hand side of this equation

so you know we can still rewrite this as

the lamb W function of the natural log

of 2 right time eal log of 2 so here we

can see that we what we have here is

exactly equal to the power of E so we

can now see that this simplifies to the

natural log of two and this is the right

answer thank you for watching please do

well to like comment and share

The Lambert W Function. ProductLog. how to deal without calculators. #maths #olympics


hello good viewers in this video we're

going to solve for the value of a * b +

B * C + a * C given that A + B + C is =

1 also a s + B sare + c² is = 1 please

if you have not subscribed to my Channel

please kindly subscribe and also like

comment and

share now let's get started from

equation one we have a + b + C this is =

to 1 I would like to square both sides

of this

equation now from the left when you

expand this we have a

2 + b

2 + C 2 + 2 into braet a b + B C + a c

and this is equal to 1 2 that will give

us

one from here we're going to make a

substitution

we already have that aare + B sare + c

sare = 1 so from here we can replace

this

as 1 then we have here + 2 into braet a

b + B C + a c then this is equal to one

so from here this one is crossing over

and that will be negative so we're going

to subract one from both sides so by so

doing we have 2 into bracket a b + B C +

a c then this is equal to 1 - 1 that

will give us Z so from here let's go

ahead and divide both sides by two since

we're interested in finding the value of

a plus b c plus a c so here this will

cancel out

right okay let's finish it up

here so from here two cancel two and

here we have a + B

C+ a c this is equal to 0 / 2 and that

is 0 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Japanese | A nice tricky Olympiad Math | #maths #mathematics


hello good day viewers welcome back once

again to another video today we're going

to find the square root of a sodic

expression the > of 10 + 2un of 21

right

okay we have theare root of 10 + 2 of

root of

21 we can write this as square root of

10 is the same as 7 + 3 right then + 2

root of 21 is same as 7 * 3 so from here

this is equal to the square root of 7

here is same as sare < TK of 7 s then

plus 3 here is same as the < TK of 3

S then

plus 2 multi by you know otk 7 * 3 let

us recall that the root of M * n same as

root of M * root of n so here we have

root of 7 * the root of 3 now from here

let us recall the expansion of this

binomial a + b 2 is = a 2 + b 2 + 2 a b

right and if you compare this with what

we have under this root sign we will

find out that a is equal to root of 7

and B is equal to root of 3 if you

follow the pattern a squ will have

something squ plus b squ you have

another something which is different

from the first something squ plus 2 *

something * the different something and

here we have 2 * a * B right so this

implies that we have the square root of

< TK 7 + < tk3 S and from here this is

same as the absolute value of Ro < TK 7

+ < tk3 which is = to < TK 7 + < tk3 and

this is the right answer thank you for

watching please kindly subscribe to my

YouTube channel also like comment and

share bye-bye

Junior Olympiad Math | Can you find the square root of a surdic expression? | #maths


hello welcome back once again today

we're going to solve this awesome

looking equation and we're going to

solve for all real values of X that will

satisfy this equation right so let's get

started from the left hand side we have

x to^ x to^ 4 this is equal to 64 right

now in our next step let us raise both

sides of this equation to the power of

four so here raise it to the^ 4 and from

the right hand side raise it to the^

4 okay now we're going to make use of

this exponential rule we know when we

have a to the^ B in bracket to the^ C

this is same as a to the^ B * C we're

going to apply this rule at the left

hand side so that will become we know

the power of X here is x^ 4 so that will

become x x x^ 4 * 4 this will be equal

to we know 64 is = 8 2 right so we have

here

² then ra the^ 4 now this exponential

rule here let us apply it here okay so

we have here x^ x^ 4 * 4 this will be

equal to 8^ 2 * 4 which is equal to 8

this exponential rule when we have have

this product we can take either a either

B or C inside bracket and a and the

other one outside don't know what is

wrong with me today okay so here we're

going to write this here you know we

have this product between this power

here we have x^ 4 * 4 so we can actually

write as x ^ 4 then ra x^ 4 then this is

equal to 8 to the^ 8 okay now you can

see we have the same pattern with bases

and exponents we have x ^ 4 is = to 8

from the exponents we have x^ 4 isal 8

so we're going to equate either the

bases or the powers so this implies that

x to^ 4 is equal

to 8 right okay now from here we're

going to solve for x but we

interested in solving for the real value

for x so we write the left hand side as

X2 to the^ 2 then this is equal

to 8 now take the square root of both

sides we get x2 is equal to plus or

minus the square root of

8 right okay so here we have the square

root of 8 now remember we're solving for

for the ray value so we're going to

discard the negative value so for r

x we get x2 will be equal to the square

root of 8 right okay now we're going to

take the square root of both side one

more time so we get X will be equal to

plus or minus this square root of square

otk of 8 we know square root of square

root of something is that to the of 1 4

which is equal to plus or minus the 4

root of 8 so therefore we have our two

real solutions that satisfy this

equation x is equal to plus or minus the

four

root of 8 and this is the correct answer

thank you for watching please kindly

subscribe to my channel also like

comment and share

Oxford University 🎓 Entrance Exam | Can you solve ?


hello welcome back once again today we

have this nice and interesting

exponential equation and here we'll be

solving for the values of X okay now

let's get started first of all let us

take note of this the M

root of a is equal to a to the^ 1 / M

right so from here remember we have the

X root of 5 and everything is RA to the^

1 /x this is equal to

25 so from here make use of this

property so in the bracket we have 5^ 1

/x then 1 / X then this is equal to

25 okay in our next step let us record

this property a to the^ of B in bracket

C this is equal to a^ B * C so from here

that will be 5^ 1X * 1

/x you know that would be 1 / x^ 2 so we

have 5 ^ 1 / x^ 2 this will be equal to

25 is the same as 5 to the^ 2 so we can

see we have the same basis from both

sides so let us equate the power so this

implies from here the 1 /x x

2 is = to 2 so put this over one and

cross multiply so x * 2 will give us 2

X2 this is equal to 1 * 1 which is 1 so

divide both sides by

2 so 2 we cancel two so here we have

X2 is equal to 1 / 2 from here let us

take root of both sides since we're

solving for x so X will be equal to plus

or minus the sare < TK of 1 / 2 which is

equal to plus or minus we have the

square root of the numerator / the

square root of the denominator so from

here this is equal to plus or minus otk

of 1 is 1 / root of 2 and this is the

right answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Interesting exponential equation | Can you solve ?


hello good day viewers welcome back once

again please kindly support this channel

by subscribing also like comment and

share today we have an interesting

exponential equation problem here we're

given that 2^ of 2^ X is equal to 3^ 3^

X and right over here we are solving for

the value of x right okay so the first

thing we're going to do here is to elen

both sides of this equation right so

here we have the Ln of 2^ of 2 to the^ X

you know this is equal to the Ln of 3

to^ of 3 to^ of X right okay now recall

this

property the Ln of a to the power of B

is equal to B * the Ln of a right here

the power of 2 is 2 the^ of X

so we we have from here 2^ x * the Ln of

2 this is equal to 3^ of x * the Ln of 3

right okay so from here we divide both

side by the Ln of two so here we have

over Ln of two and from here over the Ln

of two so this will cancel this here we

have 2^ X is equal to 3^ x * by the Ln

of 3 over the Ln of

2 okay so what what do we do from here

from here we're going to divide both

sides Again by 3 ^ of X so here 3 ^ x

and right over here 3 ^ of X so this

will cancel this and from here we're

going to make use of this property when

we have a to the power of B over another

base with the same power C to the power

B this is same as a/ C all to the power

of B right so from here we're going to

have 2 / 3 to the power of X this is

equal to the Ln of 3 over the Ln of 2

okay so what do we do from here from

here I'm going to Len both sides again

so here we have the Ln of 2/ 3 to the^

of X this is equal to the Ln of the Ln

of 3 over the Ln of 2 so from here this

we cross over here so we have x * the Ln

of 2/ 3

this is equal to the Ln of the Ln of 3

over the Ln of 2 so we're solving for x

so here we divide both by the Ln of 2/ 3

and here Ln of 2/ 3 so this cancel this

and here we get the value of x which is

the Ln of Ln of 3 over the Ln of

2ide by the Ln of 2 over 3 and this is

the right answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Interesting exponential equation. How to solve for real value of  x


from the left hand side cube root of x

is the same as x^ 1 / 3 this is equal to

from the right hand side sare root of x

is same as x^ 1 / 2 we know the LCM of 2

and 3 is equal to 6 raise both side to

the power of 6 so here raise this to

the^ of 6 and here raise this to the

power 6 from the left hand side we have

X to the^ 1 / 3 * 6 is 2 then this is

equal to x^ 1 2 * 3 * 6 I mean is 3 from

here we have X Cub - X2 is = 0 so we

have X2 which is common factor it out we

have X - 1 which is equal to 0 so we

have either x² is = to 0 or X - 1 is = 0

from here take root of both side we get

x = 0 and this is indeed a solution and

here we have x = 1 which is a solution

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Interesting radical equation | Can you solve ?  #maths #mathematics #exam


hello welcome back once again today we

have this interesting

problem we're going to simplify the

following given division of irrational

numbers

right so let's see how we're going to

solve this

problem now let us split the denominator

so we're going to have the square < TK

of 5 / 8 all over theare < TK of 10 then

plus theare < TK of 8 all over

5id by the sare root of

10 okay so from here we're going to make

use of the following rule when we have

the square root of

m / the square root of

n this isal to theun of m /

n so we're going to make use of this

rule here so this will

become the square root of so we have

here 5 all 8 then ID 10 then

plus the square root of we have here 8

all 5 / 10 right okay so now let's talk

about this fraction

we know

that 5 8 / 10 put it like this over one

this is same as 5 all over 8 * then we

take the reciprocal of this fraction

here which is 1 all over 10 so from here

we can see that 5 here is 1 5 here is 2

so we get 1 all over 16

and for the other part here we have 8 /

5 then / 10 / 1 so this is = to 8 / 5 *

then the reciprocal 1 / 10 when you use

two here we get four we used to here we

get five we can see we have 4 all over

25 so this will become the square root

of so the first one we have 1 / 16

then plus the square root of here we get

4 all over

25 so from here remember this rule here

this particular rule here so we're going

to take the square root of the numerator

ID by the square root of the denominator

plus the square root of the numerator ID

by the square root of the denominator so

this will be

of 1 is 1 /un of 16 is 4 plus 4 is

2id of 25 is 5 so here we're going to

simplify this fraction so we know the

LCM of 4 and 5 is 20 so we have 20 20 /

4 is 5 then this one * 5 will give us

5+ 20 / 5 is 4 then this 2 * 4 we give

us

8 right so here we get our solution 13

over 20 and this is the correct answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

[Music]

Germany | Nice Olympiad Math | Simplification of radicals


hello welcome back once again let's find

the solution to this problem we have X2

+ x + 1 this is equal to 0 so from here

we're going to evaluate x^ of

2025 before we proceed let us take note

of the following difference of two cubes

formula we have a cubed minus B Cubed

this is equal to a - b then * by a² + a

b + b

2 so from this from this given

identity let a = x and let b = 1 so we

have X Cub - y Cub which is 1 this is

equal to x

-1 then MTI X2 + a will be x * 1 which

is x + b² will be 1 2 which is 1 and

here from the given equation X2 + x + 1

is equal 0 so here this is equal to0 so

we have X Cub -1 this is equal to x -1 *

0 so from here we have X Cub - 1 this is

equal to0 so from here x X Cub is = 0 +

1 which is 1 now remember we're asked to

evaluate x^ of

2025 right but luckily 2025 is a

multiple of three right remember 2025 is

equal to 3

MTI

675 now what happens if we raise both

sides of this equation to the power of

you will see that from the left hand

side we will have exactly x^ of

2025 so let's go ahead and do that so

here we have X cubed then we have to

raise to the power of

675 this is equal to 1 to the power of

675 so from the left here we have x to

the power of

225 and here comes the answer one and

that's the right answer thank you for

watching please kindly subscribe to my

Channel also like comment and share

bye-bye

Japanese | Nice Olympiad Math | #maths #mathematics