good day viewers welcome back to our

Channel today we have the following

definite integral problem integral from

0 to 4 16 all 16 + x² DX touch let's see

how we're going to solve this problem

take a note integral of 1 over X2 + a

2 the x is = 1 all a * the tan inverse

of x a + C

so a here is a real number and which is

greater than zero okay so we're going to

use this here so I can write this limit

as follows 16 * integral from 0 to

4 1 all

over X2 plus of 16 here so 4 2 DX so we

can see that this is applicable here now

so we have here 16 all over you know

4 the tan inverse

of X All 4 close bracket with our limit

0 to 4 so now putting our upper limit so

we get 16 / 4 is 4 the tan inverse of 4

all over 4 is

1us 4 the tan inverse of 0 All 4 is 0

and this is equal to 4 tan inverse of 1

so this is equal to 4 * the tan inverse

of 1 is pi 4

P four canel four here so we finally get

pie wow thank you for viewing and please

make sure you subscribe to my YouTube

channel for more exciting videos

Definite integral. The result is so so good.


good if you was welcome back to my

channel if you have not subscribed to

this Channel please

subscribe share our videos and comment

thank you today we have the following

limits we have a solution here we're not

going to use the so-called L hobby to

scroll that is differentiating up and

down so we'll be approaching this

through algebra okay

so how do we solve this problem remember

if we make direct optician here we have

four minus four

so if you make that resolution we get

four minus one zero all over you know

square root of 4 here is two then we

have 2 minus two this is equal to zero

all over zero which is not permitted in

mathematics this is an indeterminate

form okay now since we have we can e we

can solve this question in different

ways okay so we can rationalize or make

substitution but I will go for making

substitution because that will be easier

okay so solution let's get started

now I'm going to approach this problem

by first making substitution let's

square root of x equals y okay so you

know X is going to be y resp Square two

so we have here x is equal to Y raised

by 2 okay so we have our limit now as

limit

as Y is approaching so let's see what

why we approach as X is approaching 4

okay so here we have square root of x if

you substitute into x 4 we have square

root of 4 which is 2. so Y is

approaching 2. so up here we have

instead of X we have y squared minus one

and down here we have e sub square root

of x we get y minus two okay so we'll be

using the difference of two

squares formula we have a raised to

power 2 minus b e raised part 2 equal is

equal to a minus B

times a plus b okay so we have a squared

minus B squared equal to a minus B times

a plus b okay so y squared minus 4 we

can write as follows y squared minus you

know

to sweat okay so using this property

above so we get we get y minus two

then times y plus two okay so here we

have our limit

as Y is approaching two so up there we

have here y minus 2

we have here y plus two

we have here y minus 2.

then close brackets so we can do

cancellation here so y minus 2 and Y

minus 2 cancel out so we have this equal

to limit

so we have limits as Y is approaching

two

of Y Plus 2. so here we have as Y is

Operation so we'll get to that

substitute into y two so we get two plus

two and this is equal to four so this is

the right answer to the problem thank

you for viewing and please make sure you

subscribe to our YouTube channel for

more interesting videos see you in our

next video

Limit. Avoid L'Hospital rule. Here we showed the algebraic approach.


hello welcome back today we have to find

this limit here so this is read as limit

as X is approaching one of this function

here x² -1 all over x

-1 so before we proceed let us try to

make a direct substitution here so if we

make direct substitution here so we're

going to

have X to be equal to 1 so we have 1 s

right then - 1 all over so down here we

have 1 - 1 and 1 s is 1 1 - 1 we have

here 0 over 1 - 1 here this is 0 so 0/ 0

is an indeterminate form and this is not

allowed in mathematics right so but the

most interesting part of this solution

is that I'm not going to make use of the

L hope to so let us see a very nice way

to solve this problem so we're going to

recall the difference of two squares

formula so recall we have this a

2us b 2 is =

to a + b * a - B right so now let us

start to factorize this numerator here

so here we're going to have x^2 minus

you know 1 is the same as 1 2 so going

to rewrite this in this form so this is

going to be equal to x + 1 all over X -

1 right so our limit now is going to be

limit

as X is approaching

one of up here we have x + 1 * by x

-1 all over x -1 right so we can see

that x -1 and x -1 will cancel out so

this becomes

limit as X is approaching one of x + 1

right so this is going to be equal to

you know put x = 1 so here we have 1 + 1

and this is equal to two and that's the

right answer to this problem thank you

for watching please do wa to like

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Interesting Limit

Interesting limit.


in this place we are given the following

equation 2 raised by x minus 2 is equal

to 100 then we're asked to find the

value of 2 raised to power x

if you want to solve this equation to

solve for the value of x here we're

going to use the natural logarithm but

we just use some algebraic manipulation

approach here we have 2 raised for S

minus 2 equals 100 2 raised by S

according to indices law over two raised

by 2 is equal to 100

is 4 times 100 therefore 2 raised to S

is equal to 400 please subscribe to this

Channel and turn on the navigation Bell

if you find our videos interesting bye

Algebra is beautiful


hello welcome back once again in our

today's video we're going to solve the

following given Olympia mathematics

problem so from the left hand side we

have 27 to the^ X so this is equal to

from the right hand side we have 1 / X

now for us to get rid of this power here

we're going to raise both sides of this

equation to the power of the reciprocal

of X right so here raise this to the

power of the recip of X which is 1 / X

and here 2 1 / X so we can see x * 1 / X

is 1 so from there we have 27 to^ 1 this

is = to 1 /x ra^ 1/ x right okay so from

here let's make the left hand side the

right hand side and the right hand side

the left hand side so get 1 /x to^ 1 /x

this is equal to 27

okay so from here we get 1 /x to^ 1 /x

this isal 7 same as 3 to the^ 3 right so

we have this power 1 /x = 3 from the

base 1 / x = 3 so this implies from here

that 1 /x is = to 3 right so if the

reciprocal of X is equal to 3 then this

implies

that X is = 1 / 3 right or if you want

to find out you can put this over one

and cross multiply x * 3 is

3x this is = to 1 * 1 is 1 divide by the

coefficient of

x this will cancel this and here we get

X is = 1 / 3 and this is the correct

answer thank you for watching please

kindly subscribe to my channel alsoo

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Olympiad Math | Can you solve ?


hello good viewers welcome back once

again today we're going to solve for the

value of x in the following given

equation we have 2^ of 3 to ^ of 4 to ^

of X this is equal to

512 if you notice this you will know

that 512 can be expressed in powers of

two right and from here we have the base

here to so let's go ahead and express

512 as power of

2 recall 512 is exactly equal to 2 the

Power of 9 right so here we have 2^ of

3^ 4 x this will be equal to 2^ of

9 now carefully looking at this equation

we can see that we have equal bases so

we can equate the powers so equating the

powers this implies that 3^ 4^ X this is

equal to 9 again we can express 9 in

index powers of three right so here we

have 3 ^ 4^ x this is equal to 3 2 from

here once again we can see that we have

equal bases so we can equate the powers

from here this implies that 4 to^ of X

is equal to 2 do not forget that 4 is

equal 2 2 right then with this power

here remember we have 2 2 then to the^

of X this is equal to 2 you know 2 * X

is 2 X so we have 2 ^ 2x this is equal

to 2 ^ of 1 again we have equal bases so

we can equate the power so we have here

2x is = 1 dividing both side by 2 we get

the solution that X is = 1 over two

amazing and this is a right answer thank

you for watching please kindly subscribe

to my my channel also like comment and

share bye-bye

China | Interesting exponential equation | #maths #mathematics


hello welcome back once again today we

need to discuss something very

important a lot of students fails to

provide all solutions to this

problem many would think like

this we

have x - 9 to the^ 2 which is equal to

81 many we go ahead and expand this

then transforming everything to a

quadratic

equation but that is really a long way

to go and many would think that the best

way to solve is just to take root of

both sides in this form here take root

of the left hand side that would be x -

9 which will be equal to < TK 81 then we

get x - 9 is equal to > of 81 is 9 add 9

to both sides so we get X is = to9 9 + 9

which is equal to

18 yes of course this is a

solution but this only provide one

solution so this is not the best way to

solve this

problem there's something the person

really

forget when we're taking roots of both

sides the right hand side should carry

plus or minus so the best to the best

solution to this problem should be we

have x - 9 2 which is equal to 81 now

taking root of both sides so we get x -

9 this should be plus or minus square

root of

81 yes this is 100%

true

okay so from here we get x - 9 is equal

to plus or minus squ of 81 is

9 so now we have two linear equations to

solve the first one is x - 9 is = POS 9

and the second one is x - 9 is

9 so from here add 9 to both sides so we

get X is = 9 + 9 which is

18 yes this is indeed a solution now add

n to both Sid from here we get X is = -9

+ 9 here we get another solution x is =

0 we can see now with this approach we

found all solutions x equal to 0 and x

equal to

18 thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

It's easy but we need to be careful | Interesting algebraic problem


hello good day everyone welcome back

once again to my YouTube Channel please

if you have not subscribed to this

Channel please kindly subscribe to this

channel

subscribe to this YouTube channel and

also share this video here we giv the

following problem that A + B + C is = 1

a ^ of 2 + B to the ^ of 2 + c^ of 2 is

= 3 and right over here we asked to find

the value of a * b + B * C + a *

C okay now let's see how to solve this

problem let us consider expanding the

following

binomial A +

B + C ra^ of 2 if we expand this

binomial we're going to

have

a² Plus

b 2 + C 2 + 2 a b +

2 b c then + 2 a c right so here we have

a + b + C ra^ of 2 is = to a 2 + b 2 + C

2 now the whole three terms here we can

see that two is Comm on here so we're

going to factor out two so here we have

+ 2 Open Bracket a * b + B * C + a * C

right so we can see that we can make

substitutions from here we given that A

+ B + C is equal 1 and right over here

we have a + b + C ra to the power of 2

so this going to become 1^ of 2 is equal

to a 2 + b 2 + c² the value is given in

the problem as three so here we're going

to have this to be replaced with three

then plus 2 Open brackets a * b + B *

C+ a c close bracket right so now we can

make the right hand side to be the left

hand side and the left hand side to be

the right hand side since we are solving

for a * b + B * C + a * C so we're going

to have here two Open brackets a * b + B

* C + a * C then + 3 is = 1 power of 2

so we can see that this does not affect

the question right so here we're going

to have two Open Bracket a * b + B * C +

a * C then + 3 is = to 1 2 is 1 so we're

going to subtract three from both sides

right so here we have two Open brackets

a * b + B * C + a * C is equal to you

know 1 - 3 that

is 2 right so from here since we are

looking for the value of a * b + B * C +

a * C so we can divide both sides of the

equation by the coefficient of a * b + B

* C + a * C so here we divide by 2 and

here we divide by two so this cancel out

so we get the value for a * B plus b * C

+ a * C to be equal to -2 2 and that is

-1 and this is the right answer thank

you for watching please do where to like

please comment and also share this video

bye-bye

Interesting Olympiad Math. Involving three variables with only two equations.


hello welcome back once again today we

have this interesting equation here with

a negative base we have the negative pi^

x = a and we have two interesting

constants here we have pi and the

constant e how can pk^ X be equal to a

and carefully looking at this equation

one can easily say there's no real

solution right yes I strongly believe

that the only solution for X must be

complex Solutions right now let's get

into this video in this video we're

going to solve for just a complex

solution we're not here to solve for

general form or the general solution

right okay now this go into this video

and we're going to start by natural

loging both sides of this equation so

here we have the natural log of piun to

the^ of X this is equal to the natural

log of a

now with the property of the natural

logarithm remember this power here is

crossing over here we have X MTI by the

natural log

ofk is same as -1 * pi right and this is

equal to the natural log of e is one

right because the base of the natural

logm is

e okay now from here let us take note of

this property of the natural logarithm

when we have the natural log of a * B

this is same as the natural log of a

plus the natural log of

B so from here this will be X into

bracket so this will be natural log of

-1 plus the natural log of

Pi and this is equal to one so from here

let's take note of something really very

important and amazing

from -1 is equal to e to power of I pi

right now let's natural log both sides

so we have natural log of1 this will be

equal to natural log of e to the^ of I

pi and with the property of the natural

logarithm so this power here is coming

down here so here we have the natural

log of1 to be equal to I Pi MTI the

natural of e and we all know that the

natural log of e is equal to 1 so we get

this is equal to I Pi * 1 which is I pi

and therefore we have X into bracket so

instead of natural logarithm of1 so we

replace that with I Pi then plus the

natural log of Pi then this is equal to

1 since we are solving for x let's go

ahead and divide both sides by this

coefficient of x which is I pi plus the

natural log of

[Music]

Pi and from here this will cancel this

right and here we get the solution X is

= 1 all over I Pi + the natural log of

Pi and this is the complex solution

thank you for watching please kindly

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comment and share bye-bye

Interesting equation involving the two most interesting constants.


hello welcome back once again today we

have this nice question here we have

this cubic equation x Cub + X2 = 11 and

here we're going to evaluate this hexic

expression x ^ 6 + 2 x ^ 5 + x ^ 4 - 2 X

Cub - 2 x^ 2 +

15 how are we going to do

this okay now let us see from the given

equation we have X cubed +

X2 which is equal to

11 right okay now from here let's go

ahead and subtract one from both sides

so here we have X cubed + x^2 - 1 this

will be equal to 11 -

1 so from here we have X Cub + + x^2 - 1

this will be = 11 - 1 which is 10 let's

Square both sides of this equation from

here so from the left that is same as X

Cub + X2 - 1 * X Cub + X2 - 1 this is =

to 100 so from here x Cub * X Cub will

give us X to the^ 6

+ x * X2 will give us x ^ 5 - x * -1

that ISX cubed so continue with the

second term here we get x² * X Cub which

is x^ 5 plus x * X2 X ^ 4 - x 2 * 1 here

will give us X2 then -1 we multiply

everything right so we getx Cub - x 2

then + a 1 then this equals

100 so from here let us simplify we have

here x to the^

6 Plus from here x ^ 5 + x ^ 5 will give

us 2 x ^ 5 right and from here we have +

x ^ 4 here x Cub - x Cub will give us

-2X Cub here x^2 - X2 will give us -2X 2

and here we have + 1

then this equals 100 okay now let's go

back to the expression we asked to

evaluate so if you look at this this is

almost the same with the Hees expression

we asked to evaluate the only difference

here is that we have plus one here and

here we have + 15 so what happens when

we add 14 to both sides of this equation

you will find out that the left hand

side will automatically turn to the heic

expression right so let's go ahead and

do that so let us add 14 to both sides

so here we get X cubed + 2 x ^ 5 + x ^ 4

- 2 X

Cub - 2 x^ 2 plus here we're adding 14

so is that 14 yes so 1 + 14 will give us

15 right and here this equals 100 + 14

which is

114 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 back once again

today we're going to solve these systems

of

equation so we have from equation 1 we

have that 8 + B is = 1 we have from

equation 2 that b + C is = 2 and we have

from equation 3 that a + C is =

3 okay now let's get

started let us start by subtracting

equation 3 from equation 1 so here we

have equation 1

A + B is = to 1 and we have the equation

3 A+ C is equal 3 so let's subtract

these

equations

okay so now here we have a - A that will

give us 0 then we have B minus C from

the left hand side so we have B minus C

this is equal to 1 - 3 that will give us

-2 so now we have our equation 4 as b -

c is = to -2 so this is our equation 4

now let us go back to equation 2 we can

see from equation 2 we have B and C at

the left hand side and this equation 4

also contain B and C so let's make use

of equation 2 and equation 4 so this is

equation 4 equation 2 I mean b + C I

mean is equal to 2 right so now what are

we going to do are we going to subtract

or are we going to add obviously we're

going to

add so now b + B will give us 2 b

then C + C will give us 0 and this is

equal to -2 + 2 will give us 0 from here

let's solve for b divide by the

coefficient of

B this will cancel this from here this

implies that b is equal to

Z okay now from this equation two we can

easily solve for C right we have that b

+ C

is = 2 now substitute the value for B

then we have 0 + C is equal to 2

subtract 0 from both sides so we get C

is = to 2 - 0 which is 2 so we have

found a solution for c c is equal to 2

then we can use any of the three

equations to solve for a so let's make

use of equation one we know that

equation one we have A +

B which is equal to 1 remember that b is

equal to Z so we substitute the value of

B in terms

of B oh is that in terms of no exactly

equal to Zer right so B is equal to Z so

we have a + 0 is = 1 so when you

subtract 0 from both sides so we get a

is = 1 - 0 which is

one so we have our

solution a

is = to

1 B is equal to Z and C is equal to 2

right thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting systems of equations | Can you solve ?


hello good viewers welcome back once

again please kindly subscribe to my

channel also like comment and share in

this video we're going to solve for x in

the following exponential equation right

now let's get started here we have 2 ^

of

X+ here we're going to write this second

term at the left hand side using the

property a to the power of b + C this is

equal to a power B * a^ of C

so this will be 2^ x * 2^ 3 this is

equal to

right so from here we have 2 ^ of x + 2

^ x * 2 cubed which is 8 this is equal

to

144 so let us factor out 2 the^ of X and

here we are left with 1 + 8 right this

is equal to 144

so from here this is equal to 2 x * 9

this is equal to 144 so from here let's

divide both sides by

9 so this cancel this from here we have

2 x this is equal to you know 144 / 9 we

are going to

have 16 right so from here we have 2^ of

x is equal to 16 is the same as 2 ^ 4 so

here we have equal bases we can equate

the powers and from here we get X isal

to 4 and that's the right answer thank

you for watching please don't foret to

subscribe to my channel also like

comment and share bye-bye

Very interesting exponential equation #maths #education


hello good day viewers welcome back once

again to another video please kindly

support my channel by subscribing also

like comment and share today we're going

to find the following limits limit as X

is approaching -2 of 3x + 6 all X Cub +

right here we're going to make use of

algebraic manipulations we are not going

to use the lhop to scroll right okay now

we're going to manipulate this function

here and I think we're going to make

some nice cancellations because from

here if you if we do direct substitution

here we are going to have 0/ 0 and this

is in indeterminate form right so let's

see what we're going to do so now let's

talk about this function here and see

the behavior we have here 3x + 6 all

over X Cub + 8 you know is that from the

numerator we can factor out three and we

have x + 2 right and this denominator

here we can write as follows X Cub plus

you know 8 is the same as 2 cubed okay

now from the denominator here we're

going to make use of the sum of two

cubes right and we have a formula for

that when we have a cubed + B Cubed you

know this is equal to a +

b multip by a 2 sorry

sorry this is minus from here

right oh so this is equal to a + b

* a 2 - A B then + b^ 2 right so from

the denominator here we have X Cub + 2

Cub you know this is going to be equal

to x + 2 * X2

- 2 * X which is 2 X then + 2^ 2 and

that's 4 right so from here we have 3x +

6 all X Cub + 8 you know this is equal

to here 3 into braet x + 2 then all

over x + 2 * X2 - 2x + 4 right so from

here we cancel out x + 2

and here we have this is = to 3 all over

x^2 - 2x + 4 now we can take the limit

so we have the limit as X is approaching

-2 of 3 x + 6 all X Cub + 8 you know

this is equal to limit as X is

approaching

-2 of 3 all over x^2 - - 2x + 4 and here

we make direct sub from here we have 3

all over here -2 2 - 2 * -2 + 4 right so

let us simplify this properly so here we

have this equals to 3 all over you -2

here is RA to the power of an even

number so we're going to have a positive

result so here we have -2 s which will

be four and here minus * minus is plus

here 2 * 2 which is 4 and here we have

four as well so this simplifies to 3 all

over 12 which is equal to 1/ 4 and this

is the right answer thank you for

watching please do wa to subscribe to my

channel also like comment and share

bye-bye

Interesting limit. Can you do this without L'hospital's rule ?  #maths #education


hello welcome back once again welcome

back to shid online math class please if

today is the first day of visiting this

Channel please kindly support us by

subscribing to our channel so we can see

it over here subscribe please support

this our Channel thank you now we're

going to evaluate this the fourth root

of two all to the power of8 right so let

us see how to do this let us recall that

the M root of n is equal to n of 1 all n

right so this expression will

become in bracket 2 of 1 over 4 then

close bracket R of 8 right so let us

recall a r of M in bracket R power of n

is equal to a power of M * n right so

here this is equal to

2 of 1 / 4 * 8 right okay this is equal

to

2 of you know this is 8 all over one so

here 4 1 and here 2 2 of 2 and this is

equal to 2 * 2 and this is equal to 4

and this is the final answer 10 you

4th root radical Expression


hello good viers welcome back once again

today we're going to solve for the

values of X in the following given

equation x^ 4 is equal to x^ of X yes of

course we one can easily say that X is

equal to 1 right but there's a way to

solve for the values of X course I found

out that X has two real values okay

let's go ahead and figure them out we

have to take the natural log of both

sides so we have here the L L of x^ of 4

this is equal to the Ln of X power of

X now recall the property of logarithm

which STS that the Ln of a to the power

of B this is equal to B MTI the Ln of a

right so here this is 4 * the Ln of X

this isal to x * the Ln of X now let's

subtract this Ln

X Ln of x from both sides so here we

have 4 * the Ln of x - x * the Ln of X

this is equal to Z Now from here we can

factor out the Ln of X and here we have

4 - x then this is equal to Z so from

here we have that Ln of X is = to Z that

is this implies okay or here we have 4 -

X8 is equal to0 so you know from here x

is = to e to the^ of 0 or from here x =

4 - 0 so from here we get the value x =

1 or here we have X is equal to 4 and

these are the real solutions thank you

for watching please kindly subscribe to

my YouTube channel also like comment and

share bye-bye

An interesting exponential equation | #maths #olympiad #mathcompetition


hello welcome back here we have an

interesting summation

problem here we're given the sum as n

goes from 0 to Infinity of1 of n all n +

one so here I'm going to make use

of integral calculus here so so how do

we find this sum how can we find the

exact value of this

sum okay now let us recall that the

integral of x of

n DX is equal to x n + 1 all over n + 1

+ C right but in the definite

form so we have integral from 0 to 1 of

x

n d x so this is going to be x n + 1 / n

+ 1 from 0 to 1 so if inside the limit

we're going to have 1 over n + 1 right

so here this sum is the same as sum as n

goes from 0 to Infinity of NE 1 n * 1 n

+ 1

so right over here this is going to be

sum as n goes from 0 to

Infinity 1 n so 1 / n + 1 is equal to

this definite integral here right so

here we're going to

have integral from 0 to 1 of x to the^

of n

DX okay so from here this is going to be

you know we can write out this

integral then here we have some as n

goes from 0 to Infinity of -1 n * x n

then

DX so we all know that this is just an

infinite geometric series here right so

and right over here this is exactly

equal to 1/ x +

1 so this is going to be equal to

integral from 0 to 1 1 /x + 1 DX so you

know the integral of this is just the

natural log of x + 1 and here we have

the limit 0 to 1 so here if you insert

the upper limit first we're going to

have the natural log of

two right then minus if inside the lower

limit we're going to have the natural

log of one we all know that the natural

log of one is zero so this is going to

be the natural log of 2 - 0 and that is

the natural log of 2 therefore the Su is

exactly equal to the natural log of two

thank you for watching please do well to

subscribe to this channel like comment

and share

Interesting Sum


hello welcome back once again today we

have this nice Olympia mathematics

problem we're given that x + y is equal

3 X2 + y^2 is = 5 and right over here

we'll be evaluating x - y okay now let's

get started from equation one we have x

+ y this is equal to 3 now let's go

ahead and square both sides of this

equation one

now you know how to expand this binomial

so we have

X2 +

y² + 2 * x * y then this will be equal

to 3 2 which is 9 now take note of this

X2 + y^2 has a value of 5 as you can see

in equation 2 so here we replace this

with five so we have here x² + y s which

is 5

then plus 2X y this will be equal to 9

okay so from here this five is positive

so if it's crossing over it will become

negative so we have here 2x Y is = to 9

- 5 so from here we have 2X Y is = to 9

- 5 which is 4 from here divide both

side by the coefficient of x y so here

we get theu of X and Y which is equal to

amazing okay now let's go back to the

original problem remember we solving for

x - y now let us recall this nice

algebraic

identity that

a squar + b s is equal to a - b s

then plus 2 * a * B okay so this is a

very powerful identity that will be

making use of here so if you expand this

from here okay if you try to prove this

so when you expand a minus b s it will

give you a 2 + b 2 - 2 a b and when you

add this to it that will be + 2 a so we

can see that -2 a + 2 a is zero so

therefore we have a + b² which is equal

to a² + b² so this identity holds okay

so now from equation 2 we have X2 + y^ 2

which is equal to 5 now making use of

this identity from here we get here x -

y^ 2 then + 2 x y this will Beal to 5 so

here we have the value for 2x y right

which is 2 so so here we replace this

with 2 so we get here x - y^ 2 + 2 * 2

which is 4 this is equal to 5 so

subtract four from both side we get x -

y^ 2 this is = to 5 - 4 which is 1 now

taking root of both sides so we get x -

Y is = plus orus < TK 1 which is = to+

or - 1 so therefore we have our solution

that x - Y is = to plus orus one thank

you for watching please kindly subscribe

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Interesting Olympiad Math problem


hello welcome back once again today

we're going to solve this exponential

equation with fractional basis right

here we have 1 /

2^x which is equal to 1 all 4 now let's

get started let us take note of this

property we know 1 / a to^ n is equal to

a to the power of n right so let us

rewrite our equation as 1 over 2 here

has a power of 1 then raise the power

ofx then this will be equal to 1 all

over 4 is same as 2 to the^ 2 okay so

from here in the bracket right over here

make use of this property so we have 2

to^

of-1 then

ofx this will be equal to here this will

be 2 to the^ of -2 there's another

interesting algebraic property remember

a b ra C is equal to a^ B * C so from

here this will be 2^ of-1 * - x will

give us POS X then this is equal to 2 to

the^ of -2 we can see we have the same

base from both sides of this equation so

we can kindly equate the powers right so

this implies from here that X is equal

to -2 and this is the right answer thank

to my channel also like comment and

share bye-bye

Simple exponential equation | tricky one with fractional bases


hello welcome back once again today we

have an interesting problem here we're

going to evaluate I to the^ of 5,000 -

one for those of us that are not

familiar with this I it is also known as

the Iota right which is the imaginary

unit so this is the imaginary unit so I

here is equal to according to Leonard a

1 right so if you square both sides we

found out that

i² will be equal

to1 okay so from here

5,000 is a multiple of two right so here

I to the^

5,000 -

1 we can write as I to the^ 2 multip by

5,000 / 2 give us

2,500 then we have here minus1

[Music]

right okay so from here let us take note

of this property you know a to^ b * C

this is same as a to the^ B in bracket

then ra power C so from here this will

be equal to I squar in bracket then

power of 2,500

then - 1 so from here this will be equal

to i² remember that is -1 so have1 to

the^ of

2,500 - 1 so we know that -1 to the

power of an even number is one right so

this power here is even so therefore

this will be one so we have 1 - 1 which

is equal to0 so therefore our right

answer is0

thank you for watching please kindly

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comment and share bye-bye

Can you evaluate ?  i^5000 - 1 = ?


hello welcome back once again today let

us solve for the integral value of x in

the following given nice exponential

equation from the left we have X to the^

X then this equals here let us take note

of this a to^ m + n is = a^ M * a^ n so

here this will be 5^ x * 5 to the^ 25

from here let us divide both sides by 5

to^ of X and here by 5 to the^ X we can

see this we cancel this from here and

from the left here remember we have the

same Powers right so let us take note of

this once again a^ m / b^ M this will be

equal to a / B all to the power M so

here this will be same as x / 5 all to

the^ X X then this equals 5^

25 now here let us manipulate the right

hand side of this equation as

follow here 5 is the same as 25 / 5 then

r^ of 25 now from here let us compare

here we have five and we have five here

x = 25 from the power x is also equal to

25 right so whatever position X is 25 5

is right over there right so from here

this implies that X is equal to 25 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 | Nice exponential equation


hello good viewers welcome back once

again today we have the following simple

exponential equation and here we'll be

solving for the value of x for the left

hand side we have 6^ x + 6^ x + 6^ x and

this is equal to 36 from the left hand

side we can see that 6^ X is common so

we Factor it out and that become 6^ X

into bracket this ID by this is 1 plus

this/ this is 1 plus this/ this is 1

then this is equal to

36 from here this is 6^ X mtip by this

plus this plus this is 3 then this is

equal to 36 so from here let's go ahead

and divide both side by three this will

cancel this and here we have 6^ X is =

36 / 3 is 12 right okay now let us make

use of the log rthm so here we're going

to simplify the answer in terms of log

RM but it will be well simplified so

we're going to take the log on both

sides so we have the log of 6^ x this is

equal to the log of 12 now with the

following properties of logarithm we

have the log of a the^ b isal to B MTI

the log of

a and we also have that the log of a * B

is equal to the log of a plus the log of

B so we're going to make use of these

amazing properties of logarithm so here

this will come over here and that will

become X multi log of 6 then this will

be equal to the log of instead of 12 we

write it as 2 *

6 okay so from here we have x * the log

of 6 this is equal to the log of 2 + the

log of

6 okay now from here we're solving for x

let's go ahead and divide both side by

the coefficient of x which is log of 6

and here we divide by the log of 6 this

will cancel this from here so we have X

is = to we split that denominator that

we become log of 2 / log of

6+ log of 6 / log of

6 now from here this will cancel this

and here we have one again let us recall

another interesting property of the

logarithm the log of a / the log of B is

equal to this logm we can write it in in

a single logarithm so that will become

the log of a to the base B right so from

here we can see we have our solution for

X that X is equal to let's start from

here remember this divid by this is one

so we have one plus so here we have that

to rewrite it in a single logarithm as

the log of 2 to the base 6 and here we

have the solution to the problem thank

you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Watch this special method | Interesting exponential equation.


hello you're welcome today we're going

to solve for the R value of x in the

following given radical equation so we

can see from the left hand side we have

the square root of x MTI theare root of

x and this is equal to 64 right okay so

let us go ahead and square both sides we

can see we have root sign at the left

hand side and we know when we Square

both Sid this root sign will get cancel

out right so Square both

sides so now we know the square always

cancel the square root right so we have

here x * the root of x and this is equal

to 64 to the^ 2 now let's go ahead and

square both sides one more time so

Square

here and square the right hand

side so at the left hand side we're

going to make use of this property we

know that 8 * n m to the power of B for

R values A and M this is equal to a to

the power B MTI M to the^ of B right and

again we know that A to the^ B to the^ C

is equal to A to the^ B * C so the left

hand side this is going to be this

x² multip by < TK of X2 Square will

cancel the square root so we have just X

so this is equal to 64 to the^ of 2 * 2

which is equal to 4 so from the left

hand side you know the power here is one

since we are multiplying we're going to

add the powers so we get x to^ 3 this

will be equal to 64 as the base here is

4 to^ 3 then ra the power of 4 right yes

okay now we can interchange these Powers

here so this is going to be x^ 3 isal to

so that can be written as 4 to the^ 4

then to the^ of 3 so you can see this

one goes out and this one comes in right

okay so now from here we have something

cubed and we also have something cubed

so this implies from here that the base

X is = 4 to ^ 4 so from here we get our

real solution that X is equal to

256 and this is the correct answer right

okay now we equated the bases here

because we are looking for the real

solution if the equation ask to solve in

total both the real and the complex

solution no need for only equating the

bases we still going to do some algebric

manipulations to get the two remaining

complex Solutions right so anytime you

arrive at this point unless the question

asks for only R solution then you can

equate the bases but if the question

asks you to solve in total you are going

to solve the cubic equation thank you

for watching this is the correct answer

x equal to 256

Tricky Olympiad Math | Can you solve ?


hello good viewers welcome back once

again let's find a solution to the

following given exponential equation we

have x^ x^ 5 this is equal to 100 now

let's get

started we're going to raise both sides

of this equation to the power of five so

from the left hand side we have x to the

power of x^ 5 then rais power of 5 from

the right hand side this is equal to 100

100 is the same as 10 power of two right

then we have a power of

five

okay now from here we'll be making use

of this

property when we have a r of B and

bracket of C we can interchange the

powers as a a to the power of C then ra

power of

B and similarly we can still rewrite it

as a to the power of B *

C now from the left hand side the power

of X here is X to the^ of 5 and we have

this power outside this bracket here now

make use of this property we can

interchange that as x to the power of 5

then ra power of x to the^ of 5 right

right then this is equal to from here we

make use of this property here so this

will become 10 to the^ of 2 * 5 which is

10 right so from here we

have x^ of 5 then of x^ of 5 this is

equal to 10 the power of 10 from here we

can see we have the same pattern of the

basis with the X exponents x^

5^ x^ 5 this is equal to 10^ 10 so this

implies that X to the^ of 5 is equal to

10 right so from here we take the fifth

root of both sides since we're

interested in the real solution we get X

is equal to the fifth root of 10 and

this is the right answer thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

China| Nice Olympiad Math | Interesting exponential equation | #maths #mathematics


hello good viewers welcome back once

again today we're going to solve for the

value of a s + b s given that a cub - B

Cub is = 56 Aus B is =

2 now we're going to make use of this

property a cubed - B Cubed this is equal

to Aus B

Cubed + 3 a b into braet a minus

B this tells us that we're going to

solve for a * B from here we're going to

make

substitutions a cubus b Cub is 56 this

is equal to A- B that would be 2 then

cubed plus 3 a b * Aus b

2 from the right hand side 2 cubed is 8

we have 8 plus 3 * 2 * a b that will

give us 6 a b this is equal to

56 let us subtract eight from both sides

we get 6 a b is = to 56 - 8 that will

give us 48 from here we divide both

sides by

6 and from here we get a * B is = to 48

/ 6 that will give us 8

right okay now a 2 + b

2 can be written in terms of a minus b s

then + 2 A B from here A minus B is 2 we

have 2 2 + 2 * a b which is 2 * 8 this

is equal to 4 + 2 * 8 is 16 right and

from here we get this is equal to 20 and

this is right answer thank you for

watching please kindly subscribe to my

YouTube channel also like comment and

share bye-bye

Junior Olympiad Math | Interesting algebraic problem | #maths #olympiad


hello welcome back once again today we

have this interesting Olympiad

mathematics problem and here we're going

to be solving for the value of x right

now let's get

started now let us manipulate the right

hand s side of this equation we know

that 3 <

tk3 is equal to 3 here isun

9 then multiply by this

root3 so since we have the products of

two numbers under root sign so we can

write as root of 9 * 3 so which gives

root of 27 so root of 3 < tk3 I mean is

equal to root of

27 right okay so

again root of

27 is equal to otk of 3 to the^ 3

because 3 to the^ 3 is 27 right now let

us make use of this identity we know

that the sare root of a to the power of

B is same

as < TK of a in bracket ra the power of

B so from here this is going to be equal

to squ < TK 3 in bracket ra^

3 and do not forget that this we can as

well rewrite as this three as Square <

TK 3 in braet squared because Square

always cancel the square root so now

let's go back to the original equation

our original equation is x^ x^ 2 this is

equal to 3 <

tk3 so using this idea here we can write

the right hand side we have X x^ 2 this

will be equal to < tk3

R of < tk3 then squar now let's compare

we have X here we have root3 we have X

we have root3 we have square here we

have square so this implies from here

that X is equal to square root of 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 Olympiad Math | Can you solve ?


hello welcome back once again today we

have interesting cubic equation for

Olympiad and here we'll be solving for

the values of X in the given equation so

here we have x^ 3 + x this is equal to

10 is same as 8 + 2 okay so here we have

X to the^ 3 then - 8 + x here - 2 this

is equal 0 so from here we have x ^ 3 -

8 is as 2^ 3 then here we have + x - 2

this is equal to Z so from here we're

going to make use of the difference of

two cubes formula remember when we have

a cubus B Cubed then this can be

factored as Aus B multip by a 2 + a b +

b 2

so make use of this property from here

we

have x - 2 into bracket x 2 + 2 * X

which is 2 x + 2 2 which is 4 then plus

here we have x - 2 this is equal to0 so

from the left hand side x - 2 is common

so let's Factor it out and we are left

with X2 +

2x + 4

then plus here we left with one then

this is equal to Z so from here we have

x - 2 * X2 + 2x + 5 then this is equal

to 0 so from here we have these two

products which is equal to Z so we this

implies from here that either x - 2 is =

or X2 + 2 x + 5 is = 0 so from here add

two to both sides so we get our real

solution that X is = 2 so from here

let's make use of the general quadratic

formula so we have X = to B will be

2+ or minus sare otk of b 2 is 4 - 4 * a

* C is 20 then we are dividing by 2 so

here x will be equal = to -2 plus or

minus this right over here is

-16 so the root of -6 first of all

square of this the netive one there is

Iota and the square root of 16 is 4 and

we are dividing by 2 so if you split

this denominator here we get our complex

solution x = -2 / 2 will give us -1 plus

or minus 4 I / 2 will give us 2

I and here we have the complex Solutions

thank you for watching please kindly

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comment and share bye-bye

Oxford University 🎓 Entrance Exam | Can you solve ?


hello good day viewers welcome back once

again to another video please subscribe

to this channel also like comment and

share here we are asked to evaluate this

interesting limit limit as n is

approaching Infinity of n ^ 4 - 2 n 2 +

1 all 3 n ^ 4 - n 2 + 5 so mind you that

this is an easy problem to

solve so what do we do what we're going

to do here is to divide both upstairs

and downstairs by the first term here

which has the highest power right so

this limit is the same as

limit as n is oppressing infinity so

here we have to

divide n ^ 4 - 2 n 2 + 1 then over n^ 4

then everything over 3 n^ 4 - n 2 + 5

then Over N the^ of 4 right so from here

we need to split so here we have limit

as n is approaching Infinity so here we

have n ^ 4/ n ^ 4 - 2 n 2 / n ^ 4 + 1 /

n ^ 4 then we have all over 3 n^ 4 / n ^

4 - n 2 / n ^ 4 + 5 / n ^ of 4 so from

here we're going to simplify where

necessary so here this is equal to

limit as n is approaching Infinity so

you know this will cancel out so we have

1 minus you know this reduces to 2 / n^

of 2 then + 1 / n ^ 4 then all over this

will cancel out with left to 3 minus you

know this reduces to 1 n 2 and here we

have + 5/ n^ of 4 right so this is equal

to 1us you know 2/ n² as n is

approaching Infinity is zero right

because the button will keep getting

larger larger larger larger and larger

right so and that will T to zero right

so here we have plus here the bottom is

larger right so we're going to see have

zero as n is approaching Infinity right

that will turn to zero and here we have

3 minus here we have 0 then + 0 and here

we have 1 / 3 and this is the right

answer thank you for watching please do

well to subscribe to this channel also

like comment and share bye-bye

Advanced limit. Interesting limit.


hello welcome back once again to my

YouTube Channel please if today is your

first day of visiting my Channel please

kindly support us by subscribing to our

Channel please and please subscribe to

this Channel and share our video today

we are asked to evaluate the double

factorial of

five we are going to evaluate this the

double factorial of five by definition

let us know what is double factorial is

all about so here we have by definition

if n is even the double factorial of n

is = to n * N - 2 * N - 4 all the way

till 4 *

2 then if n is OD the double factoral of

n is equal to n N - 2 * N - 4 * all the

way till * 3 * 1 and we have seen this

here that the double factor of n is not

equal to the N factor in bracket then

another factorial right okay now here we

have to evaluate the

five double factorial right here we can

see that five is an odd number right so

we're going to make use of this prop to

here so this is going to

be 5 * 5 -

2 right which is 3 * 5 -

4 and that is one right so this is going

to be 5 * 5 - 2 that is 3 * 5 - 4 that

is 1 so we can see this property holds

here this is odd number and this is

equal to 15 and this is the right answer

please can you support us by subscribing

to our Channel

Double Factorial.


hello welcome back once again today we

have this nice Olympiad mathematics

problem we're given a to the^ x + 1 ^ x

= 9 and here at this ugly expression

here we asked to evaluate it 2^ 9x + 1 /

9x now let's get started now the base

here is 8 which we can express in index

power of two right so 8 is as 2^ of 3

right then with this power of X here 8

power x will be equal to 2^ 3x right so

here we have 2 to^

3x + 1 / 2^ 3x this is = to 9 right and

this expression here we can actually

rewrite it as 2 3x then to the power of

something xse right so we have to take

note of this that 9x is = to

3x *

3 right so that expression will

become 2 ^ 3x in bracket 3 + 1 / 2 ^ 3x

in bracket 3 and this is what what we're

going to evaluate so to clear the

complicated stuff here we need to make

the following substitution we're going

to let 2^ 3x = M therefore the original

equation will become m + 1 / m is = 9

and therefore what are we going to

evaluate we're going to evaluate M Cub +

1 / M Cub hope you get it right so here

let us take note of this famous identity

in

algebra

that X cubed + 1 / X Cub is = x + 1 / X

Cub - 3 into x + 1 /x with the help of

this identity I think we are almost done

right so therefore MB + 1 / M Cub will

be = n + 1 / M Cub - 3 into n + 1 / M

and from here this is equal to 9 Cub - 3

* 9 so this is equal to 9 cubed which is

729 - 3 * 9 which is 27 from here we get

70 2 right so in our conclusion now

we're going to say that

2^ 9 x + 1 / 2^ 9x is =

72 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Interesting Olympiad Math | Here is the trick !


hello everyone welcome back once

again please if you have not subscribed

to our Channel please do not forget to

subscribe and share our videos today we

have this beautiful exponential

problem e r of 2x is equal to 16 we are

asked to find the value of

x

solution now let us recall the natural

logarithm of X is equal to the logarithm

of x to the base e right and again the

natural logarithm of e is just equal to

one right because the base is what e we

can see it here the natural L of e is

equal to the log of e to the base of e

which is equal to 1 right okay now we

have e of

2x =

16 we're going to to take the natural

log of both sides right so here we have

the natural log of e of

2x is equal to the natural log of 16 so

now let us use

this

logarithmic

identity so we have here that the

natural log of K power of let's say B is

equal to what this B crossing over to

multiply the entire log so we have B *

the natural log of K

right so here this is coming over here

so we have 2X * the natural log of e

equal to the natural log of 16 so right

over here we know this is equal to 1 and

2x * 1 is = 2x then this is equal to the

natural log of

16 so we can divide both side by the

coefficient of x to get the value of x

so here over two and here over two so

this cancel this well with X is just

equal to the natural log of 16 / 2 but

we can simplify

further knowing that the natural log of

16 is the natural log of 4 of 2 then

over two so you know this property here

so this coming over so we have 2 * the

natural log of 4 / 2 so this two cancel

this two and we are left

with the natural log of four and this is

the right answer thank you for watching

do to subscribe to this Channel and

share our videos

An exponential equation. e^(2x) = 16, x = ?


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Implicit differentiation. #olympiad #education #math


hello welcome back once again today

we're going to evaluate this infinite

series we have this Su as n goes from 0

to Infinity of n / 5^ n and if you check

my

thumbnail I said there that I did this

within seconds and that is

true now here is how we're going to

evaluate

this let us remember that 1 iD 1 - x

which is an infinite geometric series

equals sum as n goes from 0 to Infinity

of x to the^ of n right okay now we're

going to differentiate both sides so

here we have D

DX of 1 / 1 - x this will be equal to

the partial derivative with respect to X

of some as n goes from 0 to Infinity of

X the^ of

n now here we're going to make use of

the quent rule right remember that the D

DX of these two functions of x g of x /

the H of X this equals here G Prime of x

m the H of xus G of x * the H Prime of x

then divided by h of

X2 right okay now from the left when you

differentiate this or when you when you

find the derivative of 1 / 1 - x we're

going to have 1 / 1 -

x

s and this equals here the partial

derivative with respect to X that's

going to be Sol from 0 to Infinity the

power here which is n m x 12 the^ minus1

right okay now let's just take note of

something like this when we have a nus n

this is equal to a m / a of n so here we

have 1 / 1 - x^ 2 this equals Su as n

goes from 0 to Infinity of n MTI x^ n

and here / x^ 1 which is X

multiply both side by

X from the left here we're going to have

x / 1 - x^ 2 and this going to be here

this we cancel this Su from 0 to

Infinity of n x^ of n now let us recall

that the sum the given sum equals

this right so here we're going to let X

exactly equal 1 5 right so when we let X

= 1 5 here we get from the right sum

from 0 to Infinity of n if s = 1 5 that

is going to be / 5^ n this will be equal

to 1 5 / 1 - 1 5 2 which is equal to 1/

5 / 1 - 1 5 is 4 5 and when you square

that that is 16 / 2 25 thisal 1 5 like

this ID 16 all

25 and from there this equals 1/ 5 *

then the reciprocal 25 all over

16 right and from here this equal 5/ 16

and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Interesting advanced infinite series


hello good viewers in this equation

we're going to solve for the R value of

values of X we're given a 2^ 2 x + 2^ x

this is equal to

20 now let's get

started but before we proceed let us

recall that a to the power of b c this

can be written as a to the power of B

then ra power of C we can actually write

it as a to the power of C ra power of B

so for 2^ 2 x we can write it as 2^

x^ 2 then + 2 power of X right then this

is equal to 20 so from here we're going

to make a substitution we're going to

let 2 X = Y now for a real solution for

X here 2 xal to Y we're expecting y to

be greater than zero right

so in the equation we derive now in

terms of Y y^ 2 + y is = 20 we

interested in solving the positive value

of y so let's proceed we have y^2 + y -

20 this equal to0 so y here I can write

it as y^ 2 + 5 y - 4 y - 20 this is

equal to Zer so we have a common factor

of Y we Factor it out we are left to y +

5 in the bracket and here -4 should be

factored out we left with y + 5 and from

here we get y - 4 * y + 5 this is equal

0 so from here y - 4 is = 0 or we have

here y + 5 is = 0 so from here Y is = 4

from here we have y isal to -5 for this

we don't have a real solution for X

right so we're interested in the

positive value of y so let us

recall that 2 X isal Y since Y is = 4 we

have that 2 x = 4 so here we have 2 x =

4 is 2^ s so since we have obtain equal

bases from both sides so this implies

that we can equate the powers and that's

we have the solution X is equal to 2

thank you for watching please kindly

subscribe to my YouTube channel also

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Exponential equation | #maths #olympiad #mathcompetition


hello welcome back once again today we

have this interesting equation x^ X is =

12 and here we're going to solve for the

value of x that will satisfy this

equation so let's get started from the

left hand side we have X to the^ X then

this is equal to 12 so here we're going

to log both sides of this equation so

let's take the natural log we have the

natural log of x x to^ x this will be

equal to the natural log of 12 right we

know this property of logarithm this

power here we come down here right so

this will become X MTI the natural log

of x this is equal to the natural log of

12 right now this x here we can actually

rewrite it using this property that a

can be written as e to the power of Ln

of a we know that the the base of the

natural logarithm is e so e to the^ of L

of anything will be that right so here

this x here let's rewrite it so that

will become e^ of Ln of X then multiply

by that Ln of X at the side this will be

equal to the Ln of 12 now we're going to

make use of the Lamb function but you

need to recall that lamb function of x *

e^ X where the power of E and the

coefficients are exactly equal that is

coefficient right so this simplifies to

X so now we can see that the power of a

here natural log of x and coefficient

natural log of x they exactly equal so

this allows us to use the lambar W

function so we have the W of the left

hand side e^ Ln of X time the Ln of X

this will be equal to W of the right

hand side the Ln of 12 so the left hand

side simplifies to the Ln of X this is

equal to the W of the Ln of 12 so from

here so from here recall that if the Ln

of X is equal to a then this implies

that X isal to e of a right so from here

L of X is equal to this so that implies

that X is equal to e to the power of the

Lamb function of Ln of 12 and we find a

very nice approximation for x x is

approximately equal to

2955 and this is the correct answer

thank you for watching please kindly

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hello good viewers welcome back once

again please do wa to subscribe to my

channel also like comment and share

today we have this math Olympiad problem

x + y is = 3 X2 + Y2 is = 7 right over

here we are going to evaluate x^ 4 + y ^

4 right okay let's get started from

equation one we have x + y and this is

equal to three now we're going to square

both sides of this equation and from the

left we're going to make use of this a +

b 2 this is = to a 2 + b 2 + 2 * a * B

from the left we have

X2 + Y 2 + 2 x y this is equal to 3²

which is 9

right so from here you know X2 + y^2 is

= 7 right so here we have 7 + 2x y this

is equal to 9 so we're going to subtract

7 from both sides and we get 2x Y is = 9

- 7 which is 2 and this implies from

here that x * Y is = 2 / 2 which is one

right okay

now we're going to find the value of x

the power of 4 + B to the power of 4 +

y^ 4 I mean right so from equation 2 we

have X2 + Y 2 and that is equal to 7 so

let's go ahead and square both

sides so from the left we make use of

this property as well so we have here X2

^ 2 which is x ^ 4

+ y^ 2^ 2 which is y ^ 4 then here + 2

X2 y² and this is equal to 49 right so

here we write as follows x ^ 4 + y ^ 4

plus here 2 into braet x y^ 2 and this

is equal to 49 so remember that x * Y is

= 1 right so from here we get x^ 4 + y^

4 + 2 * 1 2 this is = to 49 so and from

there x^ 4 + y^ 4 then here + 2 is = to

49 so let's go ahead and subtract two

from both side and here we find the

value for x^ 4 + y^ 4 which is equal to

49 - 2 and that is equal to 47 and 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. I think algebraic manipulations are going to play roles here. #maths


[Music]

hello good viewers welcome back once

again please kindly support my channel

by subscribing also like comment and

share in this video we are going to find

a solution to this problem here we have

the root of three mtip by the root of 3

multip by the root of 3 multi by the

root of 3 and so on till Infinity we

asked to find the exact value

right okay now we're going to

let we're going to let x

equals

this root of three multip by root of 3

multip by root of

3 multi by and so on right so from here

we're going to square both sides so here

we have

X2 this will be exactly the whole of

this squared right so from here we have

x² is equal to you know this square and

this root here we get cancelled out

right and here we have 3 multi by the

root of 3 MTI by the root of 3 multi by

the root of 3 multi by and so on right

now carefully looking at this equation

from the right you will observe that

here we have three multiply by X right

because this is exactly what we started

with so here we have X2 = 3 * X so from

here we subtract 3 X from both sides we

get x^2 -

3x this is equal to Zer so from the left

we factorize X into 2 bracket x - 3 this

is equal to0 so from here we get X is =

to 0 or x - 3 is = 0 that is we have x =

0 or X is = to 3 but we all know that

the value of this is going to be greater

than zero right which actually mean that

the whole of that

expression cannot be equal to zero right

so this value is an extraneous value so

from here we go with this value here so

from here we can see

that the root of 3 MTI the root of 3 m

the root of 3 and so on is equal to 3

and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

An infinite nested radical | Interesting problem #education #mathematics #math


hello good day viewers welcome back once

again today let's find the solution to

the following given nice functional

equation F of 2x -1 + f of 3 this is =

to 4x - 1 from here we're going to find

F of 2 from the given equation we're

going to let x =

2 so from here 2 * 2 here is 4 4 - 1 is

3 right so we have F of

3+ F of 3 this is equal to remember X is

= 2 so we have 4 * to 8 8 - 1 that is 7

so from here F of 3 + f of 3 this will

give us 2 F of 3 right this is equal to

7 so from here divide both sides by

two 2 cancel two so this implies that F

of 3 is = to 7 / 2 so from the original

equation we can then rewrite it as F of

2x - 1 plus now instead of f of 3 we use

7 over 2 then this is equal to 4X -1 now

let's think about that value of x which

will make 2x - 1 to be 2 right so for us

to get this value of x we should make 2x

- 1 = 2 so we need to let 2x - 1 = 2 now

when you sol for X here x is = to X is =

2 so from here we're going to let x = 3

/ 2 so we get F of 2 * 3 / 2 - 1 + 7 2

this is = to 4 * 3 2 - 1

so from here 2 / 2 here is 1 so we left

with f of 3 * 1 is 3 then 3 - 1 will

give us 2 then plus 7 2 this is equal to

from here 4 / 2 here that will give us 2

then 2 * 3 is 6 6 - 1 will give us 5 so

from here let's solve for f of 2 so we

have F of 2 is = 5 / 1 - 7

/ 2 so 5 1 is same as 10 / 2 - 7 / 2 so

from the numerators 10 - 7 is 3 then

over 2 and this is the right answer

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Interesting functional equation | #maths #mathematics


hello good viewers welcome back once

again today we're going to find the

square root of 641 squ + 641 +

642 and we're going to do this without

calculators now I will start like this

I'm going to

let

641 = X right so if we add one to both

sides from the left we have 641 + 1

which is 642 and from the right we have

x + 1 therefore this expression in terms

of X is going to be 641 squ is the same

as X squ right plus 641 which is

X+

642 which is x + 1 right so from here

this is equal to the square root

of

X2 + x + x that will give us 2X + 1 and

I'm very sure we're very familiar with

this quadratic expression here that

quadratic expression is exactly the

square of x + 1 right so this is equal

to the square root of the square of x +

1 and do not forget when we have

something like this this is the absolute

value of x +1 remember that X is a ray

value right so the absolute value of x

+1 is = x + 1 right so therefore square

root

of

x² +

x + x + 1 which is equal to the square

root of X2 + 2x + 1 is = x + 1 now let

us recall that let us recall that X is

equal to 6 41 so when we substitute here

we get that the square root of

641 SAR

plus here

642 right that this is equal to 641 + 1

which is equal to

642 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

China| Can you evaluate this ? | No calculators please | #maths #mathematics


hello welcome back once again today

we're going to solve the following given

exponential equation we have here 2^ 4

to^ 3^ x this is equal to here 32 is the

same as 2^ 5 right now from the left the

base here is 2 y 4^ 3^ x is the Power of

Two And from the right here the base is

two why just five is the power right so

we're going to to the power since we

have obtained equal bases right so from

here this implies that 4 to^ 3 to^ x = 5

now from here we're going to make use of

log reading let us log both sides so

here we have the log of 4^ 3^ x this

equals the logm of 5 now let us take

note of this the log of a to the power

of P same as the power here crossing

over becoming B the log of a right so

here the Bas is 4 the power is 3 to^ x

so this power will cross over so that

will become 3 to the^ X multi by the log

of four this is equal to the log of five

now from here let's go ahead and divide

both side by the log of four so divide

here by the log of four also divide here

by the log of four

now from here this will cancel this we

have 3 to the power of X this will be

equal to the

log of 5 / the log of four okay now from

here let us log both sides again let us

log both sides again so here we have the

log of 3 to the^ X this is equal to the

log of log of 5 / log of four

now with the same property of logi this

power of three here which is X will

cross over that will become X log of

three this is equal to the

log of log of 5 / the log of 4 now from

here we're solving for x so divide both

side by log of 3 that is the coefficient

of x right so here by log of three this

will cancel this so from here we get the

exact solution which is the

log of log of 5/ the log of four then

the all of this ided by the log of 3

right so which is approximately equal to

87 and here is the right answer thank

to my channel also like comment and

share bye-bye

Interesting exponential equation |

#maths #physics #exam #chemistry


hello welcome back once again here we

have another question here we're given

that 2 ^ of X is equal to 3 we're asked

to evaluate 16 to the power

ofx please if you have not subscribed to

this Channel please kindly subscribe to

this Channel and also share this video

many may think that the best way to

evaluate this is to find the value of x

in this equation but mind you that

before you find the value of x here we

need to make use of logarithm but to

avoid this I'm going to show you an

easier way to get this problem solved so

here we're given

16 ra power

ofx so you know this is equal to 16 is

the same as 2 power of 4 then power

ofx so these powers are going to

multiply together so here we're going to

have 2 of -4x

which we can also rewrite as follows 2

power of X then ra power of -4 so mind

you that 2 power of X is already equal

to 3 right so right over here we're

going to replace this with 3 so we have

16 ofx is going to be equal to 3^

of4 so you know with this negative sign

up here we're going to have 1/ 3^ 4 so

this is = to 1 / 3 ^ 4 is 3 * 3 * 3 * 3

and that is 81 this is the final answer

thank you for watching please do well to

like comments and share

Interesting exponential problem. #maths #olympics


hello good viewers today we have this

nice interesting exponential

equation okay now the equation looks a

little bit

but believe me is a very simple question

23 to the^ 27 x in bracket 27 x again

this is equal to 23 ^ 9^ 3 before we

proceed let us make a substitution let

us let this 27 of x equals something

it's a y and here we have 23 to^

y y this is equal 23 9 3 let us record

this property m n z this isal to m n * Z

so here we multiply Y and Y we have 23

y^ 2 this is equal to 23 9 3 we have

equal bases right you know what is going

to happen we're going to to equate the

powers so here we have y^ 2 this is

equal to here we have 9^ of 3 right so

but 9 here that is 9 3 so we have y^ 2 =

9 is the same as 3 S then of

3 so here we have Y 2 is equal

to 3 S of 3 here making use of this

property this becomes 3

of 2

* 3 which is 6 so from here we take the

root of both side we have y is equal to

the square otk of 3^

6 so recall that the N root of M is = m

1 / n so here Y is

= 3 I mean 1 / two and from here Y is

equal to you know 6 * 1/ 2 is 3 so we

get 3 power of

3 okay now let us recall

that 27 x was = to Y 27 here is 3 3 so

we have 3 3 right then R of X this is

equal to Y which is 3 to the^ of

3 so from here we multiply these Powers

together here we have 3 3x this is equal

to 3 3 we have equal bases right so this

implies that the powers we can equate

them 3x is = 3 divide both sides by

3 and here we come up with the solution

X is = 3 / 3 which is 1 and this is the

solution thank you for watching please

kindly subscribe to my YouTube channel

also like comment and share bye-bye

Can you solve this complicated looking exponential equation? | #maths #olympiad #mathcompetition


[Music]

hello good day viewers welcome back once

again today we have an interesting

problem and this is an Olympiad mat we

are given the following equation 36 x ^

4 + 1 that this is equal to 0 right over

here we're going to evaluate 3x + 1 / 2x

from the given equation we can re write

it

as 9 x^ 2 * 4 x 2 right CU 9 * 4 is 36

x² * X will give us x^ 4 from here then

+ 1 this is equal to zero now from both

sides we're going to divide each term by

X2 so here 9 X2 * 4 X2 all over 4 x 2 +

1 over 4 X2 this is equal to 0 all over

4x2 so from here this and this will

cancel out and here we have 9

x2+ here 1 /

4x2 this is equal to 0 so from here this

is same as 3x^ 2 then plus here 1 / 2x^

2 this is equal to zero now let us

recall this property of algebra a a 2 +

b 2 this is same as a + b 2 - 2 * a * B

right so using this property from here

we have 3x right which is the first time

here squ plus the second term squ here

which is 1 / 2x then s - 2 * by the

first term 3x multi by the second term 1

/ 2x 2 1/ 2x I mean sorry for that so

that is 1/ 2x that is the second term

here and this is equal to

0 so from here we have 3x + 1 / 2x^ 2

then minus here 2 and two we cancel out

x and x will cancel out and we're left

with -3 this is =

Z from here we have 3x + 1 / 2X x 2 this

is equal 0 + 3 that will give us 3 so

here we arrive at the solution that 3x +

1 / 2x is = to + orus otk of 3 and this

is the right answer thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Japanese | Can you solve ? | Nice Olympiad Math | #maths #mathematics


hello welcome back once again today

we're going to solve this exponential

equation we have 5^ X2 = 150 and here

we'll be solving for the values of X

right okay now let's begin by Ling both

sides of this equation okay so here we

have the Ln of 5 to^

X2 this will be equal to the Ln of 150

right so now with the property of

logarithm this power will come over here

so we have x² * the Ln of 5 this will be

equal to from here Ln of 150 is same as

Ln of 5 * 30 right and with the property

of logarithm this is same as Ln of 5

plus the Ln of

30 okay so from here we have the Ln of 5

plus the Ln of 30 so in our next step

let's divide both side by the Ln of 5 so

we have x² MTI the Ln of 5 this is equal

to Ln of

5+ Ln of 30 so the by the Ln of

5 so here this will cancel this we have

here x² is equal to here we split this

denominator so we have Ln of 5 / Ln 5

plus Ln 30 ID Ln of 5 so from here Ln of

5 Ln of 5 is 1 then plus Ln of 30 / the

Ln of 5 right okay so from here once

again Ln of 30 is some as Ln of 5 * 6

which is same as Ln of 5 + Ln of

6 so from here we have X2 is = to 1+

that will be Ln of 5 plus Ln of 6 then /

by the Ln of 5 so we have here X2 is = 1

+ Ln 5/ Ln 5 is

1+ Ln of 6 / the Ln of

5 let us take note of this the Ln of a

divid by the Ln of B is equal to the

logarithm of a to the base B right so

we're going to apply that law here so we

have here X2 is = 1 + 1 which is 2 then

plus so this will be the log of 6 to the

base of 5 so remember we're solving for

x so let's take the root of both sides

we have X is equal to plus or minus the

square root of

2+ the log of 6 to the base 5 and here

we have have the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Nice exponential equation | Can you solve ?


hello good day everyone welcome back

once again please if today is your first

day of visiting this Channel please

kindly support us by subscribing to our

YouTube channel and sharing our videos

thank you today we have this interesting

limits right over here we have limit as

n is approaching Infinity of this

function right over here n² + 2 all/ 4

n² - 3 if we try to make a direct

substitution right over here we're going

to get Infinity S which is infinity + 2

is infinity IDE 4 * Infinity 2 is

infinity - 3 which is infinity and this

is not allowed in mathematics

right okay now let us try to see what we

can do here let us try to divide both

the numerator and denominator by n^ of

2 so this is equal to

limit as n is approaching

Infinity n 2 + 2 all/

n² / 4 n 2 - 3 all/ n 2 right so this is

equal to

limit as n is approaching Infinity

so let us try to split the denominator

we have N2 / n 2 + 2 / n 2 all/ 4 n 2 /

n 2 - 3 / n² right now let us make a

little bit cancellation

here so we have here n 2 canel n 2 and

here we have n² canel n 2 so we are left

with

limit as n is approaching

Infinity here we are left it 1 + 2 n 2

over here we leftt with 4 - 3 all n of 2

so right over here we have now if n is

approaching Infinity 2 all n 2 as n is

getting larger larger

larger and larger is approaching zero

right so here we have 1 + 0 all over

here we have 4 minus you know 3 n 2 as n

is approaching Infinity meaning the

bottom is getting larger larger larger

and larger so that way approach zero

right so we are left with just one all

over four and this is the right answer

thank you for watching please do well to

to subscribe to this Channel and share

our videos bye-bye

Limit of a function. interesting limit


hello welcome back once again today we

have this amazing exponential equation

and here we'll be solving for the value

of x we have here 2 to^ of x^ of X this

is equal to 3 we need to take note of

something really very important here the

base here is 2 and X the^ of X is the

power of

two now we're going to natural log both

sides of this equation so here we have

the natural log of 2^ x^ X then this

will be equal to the natural log of 3

now since we know that x the^ x is the

power of two then this power right over

here we can move it to the front right

and that will become X to^ of X MTI the

natural log of 2 and this is equal to

the natural log of 3 at this point we

need to divide both side by the natural

log of two and here

log of two this will cancel this and

from the left hand side we have x to the

power of X this is equal to the natural

log of 3 / the natural log of

2 okay now again let us natural log both

sides of this equation one more time so

here we have the natural log of x^ X

this is equal to the natural log of the

natural log of 3 ID by the natural log

of

two now with power rule so this we cross

over and here we have X MTI the natural

log of x this is equal to the natural

log of the natural log of 3/ natural log

of

2 okay now from here this x here we can

rewrite it as e to the^ of Ln of X we

all know that this e to the^ of Ln

cancel out to be one and here we're left

with just X right and here we're

multiplying by the natural of x then

this is equal to the natural log of the

natural log of 3 / the natural log of

2 okay now from here we're going to make

use of the lumber W

function okay but we need to understand

this function first the lamb function

with the following argument e to the

power of x m x is equal to

X so we can see that the power of E here

Isa equal to the coefficient so here

let's take the lamb function on both

sides so here we have the

lambat function of e to power of Ln X

MTI the Ln X this is equal to the W of

Ln of 3 / the Ln of 2 from here remember

the power of E here is equal to the

coefficient so this simplifies to just

the of X and this is equal to the W oh I

think I missing something here yes so

this is the natural log of the all of

this right so sorry for that and here we

have the W of the natural log of the

natural log of 3 / the natural log of 2

so from here x is equal to e to the

power of the W of the natural log of the

natural log of 3 / the natural log of

two and using my calculator this is

approximately equal to

1392 1

26 612 3 1 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 | The Lambert W Function is a good way, trust me.


hello dear friends welcome back once

again today we're going to solve this

cubic equation and we will solve for

both real and complex Solutions right

okay let's start we have here

solution so we have here x to the^ 3

minus 1 here is as 1 to^ 3 and this is

equal to 0 so from the left hand side

we're going to make use of the

difference of two cubes formula remember

a Cub - B Cub is = to a - b * a 2 + a b

b² okay so make use of this property the

left hand side will become X - 1 * x 2 +

x * 1 which is x + 1 2 which is 1 and

this is equal

to0 so if these two products is equal to

0 so we have either x

-1 is = 0 or from this we have X2 + x +

1 is = 0 so from here we have the r

solution that X is = 1 right okay so

from here let us solve this quadratic

equation we're going to make use of the

general quadratic formula so X will be

equal to B plus or minus Square < TK of

b 2 - 4 a c then / by 2 * a right so if

you if you carefully look at this

expression on the left hand side of this

equation so we realize that a is equal

to the leading

coefficient the leading time the

coefficient which is one and B here is

the coefficient of x which is one and C

the constant is also one okay so here we

have X is = to B will be-1

plus or minus otk of b 2 will be 1 2

which is 1 - 4 * a * C which is 4 then /

2 * 1 which is 2 so here we have the

complex solution -1 plus or minus < TK

of -3 / 2 okay so from here we have X is

= to1 plus or minus first of all of1 is

Iota right then we have root of three

right over there then / by two so these

are the two complex Solutions thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Interesting cubic equation | can you find all solutions ?


good day everyone welcome back once

again today we're going to solve for the

value of x in the

following um fractional

exponential equation right oh if my

English is too bad sorry for that so

here we have 6^ x - 2 x / 2 x this is =

to 8 right okay 6 of X the base we can

write as 2 * 3 then of x - 2 x / 2^ X

this is equal to 8 now let's recall this

property M * n power of Z the same as M

of Z * n power of

Z so here we have 2^ of x * 3^ x - 2^ x

/ 2^ X this is is equal to 8 so for the

numerator here we have a common factor

of 2 x so we have 2 x into bracket 3 X -

1 / 2 x this is equal to 8 you know from

here this two get cancel out and we left

with 3 X - 1 this is equal to 8 add one

to both side we get 3 x is equal to

9 from here we have three of X is equal

to 9 is the same as 3 S since we have

obtained equal bases so we can equate

the powers and X becomes two thank you

for watching please kindly subscribe to

my YouTube channel also like comment and

share bye-bye

A very tricky exponential equation | #maths #olympiad


hello welcome back once again today

we're going to solve for the value of x

in the following given equation so here

we have root of

2x + x this is equal to 1 now from here

let's subtract X from both sides so we

get < TK of 2x this is equal to 1 - x

right okay now to get rid of this root

sign here we're going to square both

sides of this equation so we Square here

and from here Square we cancel this root

sign we have 2 x left then this is equal

to Here let us take note of this a - b 2

this is equal to a 2 + b 2 - 2 * a * B

right okay so here 1 2 is 1 + x

2 - 2 * a * B in this case will be 2 * 1

* X which is

2x okay now let's make the right hand

side the left hand side so we have 1 +

X2 - 2x then this is equal to from the

left 2x so subtract 2 X from both Sid

let's start with this we have x² - 2x -

right then we have + 1 this is = z so

from here we have x^2 - 4x + 1 1 this is

equal to Z at this point this is a

quadratic equation which we're going to

solve using the general quadratic

formula so we have xal to B will be 4

because B is negative so it's negative

will be positive then plus orus otk of b

s b s is -4 which is 16 - 4 * a * B 4 *

1 * 1 which is 4 then / 2 so this is

equal to 4 plus or minus square root of

12 right then ID by two so here we have

X is = to 4 plus or minus remember this

12 right over here is 4 * 3 so theun of

12 will beun of 4 * 3un of 4 is 2 and we

have root3 then we are dividing this by

two so from here let's split the

denominator 4 / 2+ or minus 2 < tk3 / 2

of which we can see that 4 3 is 2 plus

or minus 2K3 / 2 2 cancel two we left

with

root3 okay so now when I checked the

values which one satisfy this equation I

found out that X is = 2 + < tk3 is an

extrus

root it is is called an estrous root

because it does not satisfy the given

equation and therefore the only

satisfying root here is 2 minus < tk3

and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Interesting radical equation.


hello welcome back once again today we

have this simple exponential equation

and here we're going to solve for the

value of x from the left we're given 4

to the ^ of x + 4 to the^ X then this

equals

16 okay now take note of this when we

have

A+ a you know this is equal to 2 okay

right so from the left remember 4^ X is

the same as 4^ X so to add this together

it will be same as 2 * 4 to the power of

X right then this equals

16 from here let us divide both sides by

two divide this by two and divide here

by two this will cancel this then we are

left with 4 to the^ X this equals 2 now

16 / 2 that will give us

8 now in our next step we're going to

rewrite four as 2^ two right because 4 =

2 to the^ 2 and from the right 8 which

is equal to 2 to the^ 3 so here this

will be 2 to the^

2 x then this equals 2 to the power of 3

now take note of this when we have a to

the power of B then to the power of C

this is same as a to the power of B *

C so here we're going to apply this Rule

and here we get 2 to the^ of 2 * X which

is 2x this equal 2 to the^ 3 from here

we have the same basis right so let's go

ahead and equate the powers so from here

this implies that 2X

= 3 remember we're solving for x so

let's divide both sides by two so divide

this by two and divide here by two this

cancel this and here we get the solution

that x = 3 all

2 thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Exponential equation |


hello good day viewers welcome back once

again today we're going to provide

solution to the following given

exponential equation we have 8^ x + 8^ x

+ 8^ x / 4 x + 4 x this is equal to

12 now let us recall something very

important and very nice you know when we

suming three variables that is same

variables this will be same as three

multip by that Vari right or when we

have b + B this will be same as 2 * B so

from here we have 8 x + 8 x + 8 x that

would be same as 3 * 8^ of X then

divided by here that will be 2 * 4 to

the^ X then this is equal to 12 from

here we have 3 multi by 8 is same as 2

cubed right

that is 2 cubed then of x / 2 * 4 is as

s x then this is equal to

12 you know from here with this property

a^ b c this is same as a b * C so this

will become 3 * 2^ 3

x / 2 * 2^ 2x this is equal to 12

now take a note the left hand side we

can write it as 3 all 2 then multiply by

now let us recall another interesting

property when we have a^ B / a^ C this

is same as a the^ of B minus C since we

have the same bases right so here this

will be same as 2 to the^ of 3x -

2x then this is equal to 12

from here we have 3 2 then multi 2^ of X

this is equal to 12 right so from here

let's go ahead and multiply both sides

by two so here multiply by two and here

MTI two this two and this two so here we

have 3 * 2 to the^ X this will be equal

to that will give us 24 right so from

here divide both side by three

so from here we get 2 to^ x this is

equal to 24 3 will give us 8 right so

from here we have 2 to the^ X this is

equal to 8 is the same as 2^ 3 since we

have equal bases here this implies that

we can equate the power here we have X

is equal to 3 and this is the right

answer thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Exponential equation.


hello welcome back once again to my

YouTube Channel please kindly support us

by subscribing to our

Channel please and please subscribe to

this Channel and share our videos today

we have this exponential equation

problem here we have 6 ^ of x = 13 then

we are asked to find the value of x

right so here we have 6 of x

is equal to

13 so now we're going to make use of the

natural logarithm right so I'm going to

Ln both sides of this equation so right

over here I have the Ln of 6 of X is

equal to the Ln of 13 right so we know

this logarithmic property the power of

this C is going to come over to this

place and multiply the entire

right so here we have x

* the Ln of 6 is equal to the Ln of 13

so here we're solving for x so we're

going to divide both side by the

coefficient of x and the coefficient of

x here is Ln of 6 right so here we have

to divide by Ln of

6 and right over here by Ln of six so

this cancel this so we can see now that

X is equal to the Ln of 13 all/ by the

Ln of 6 so this is approximately equal

to so this is approximately equal

to

1.4 3

1 5 3 right okay now let us cross check

and see if this can satisfy the above

equation

so the equation here is so let us CR

check so the equation here is 6 of x

equal to 13 so now let us insert the

value of x here the approximate value of

x so here we have 6 rais power of 1.

4315 so we can actually approximate it

to four decimal places so this is

approximately equal to 12

9 9 9 right so we can see that this is

approximately equal to what 13 thank you

for watching please do well to subscribe

to my YouTube channel like share and

comment

An exponential equation problem


hello welcome back once again today we

have this interesting equation to solve

and we're going to solve for the real

values of X that will satisfy this

equation so let's get

started we're going to start with the

substitution we're going to let the

square < TK of x = y and when you square

both side we get X is = y^ 2 right so

here we have X which is equal to y^ 2

multi by we have root of x which isal to

Y this will be equal to here we have X

which is equal to Y plus here we have

this I mean X here is equal to y^2 so

sorry for that right yes okay so here we

have plusun X =

Y so from here Y 2 * y will give us y

Cub then - y^

2 then - Y and the whole of this is

equal 0 so we can see from the left hand

side Y is common so Factor it out we get

y^ 2 - y - 1 this is equal Z so from

here this implies that either Y is = 0

or y^ 2 - y - 1 is = 0 so this is a

quadratic equation we're going to solve

it using the general quadratic formula

and when we use that b here will be 1

plus or minus the root of B 2 is 1 which

is 1 - 4 a 4 A is4 1 - - 4 is 1 + 5 1 +

5 I mean and that is 5 and we're very

familiar with the solution to quadratic

equations right so when you solve this

you're going to get 1 + orus otk 5 / 2

now

recall our x value is equal to y^2 right

so here we have when Y is = 0 x will be

= to 0 2 which is equal Z so when Y is =

to the positive 1 +un 5 oh the coding

ratio X will be equal to 1 +un 5 / 2 2

and when you square this you get 3 + <

TK 5 / 2 again we have when y = 1 - < TK

5 / 2 x will be = to 1 - root5 / 2 2 and

when you square this you get 3 - < TK 5

/ 2 but unfortunately this root here is

an exous root because it does not

satisfy the equation so therefore we

have only two real solutions for X we

have xal to0 and we also have X is = to

3 + < TK 5 / 2 thank you for watching

please kindly subscribe to my channel

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Interesting radical equation | Can you solve ?


hello welcome back once again today

we're going to

evaluate the natural log of a negative

number

right if I could recall vividly when I

was in secondary school that I think in

grade 8 and nine right so we're thinking

that this is impossible right well this

is impossible in the real world but very

possible in the complex word right

okay we're going to evaluate the natural

log of

-1 and how are we going to do

this first of all we need to recall this

wonderful Yow identity we

know that E rais power of I

Pi +

1 is equal to Z right if you subtract

one from both

sides we'll find out

that e of I

Pi is equal to

-1 right

okay so the natural log

of1 which we have here the natural log

of1 is going to be equal to the natural

log of e of I

Pi since we know that E power of I Pi is

equal to1 from here right so instead of

writing this netive 1 so we write e to

the power of I Pi now let us recall this

property of

logarithm so we know

that the natural log of a to the power

of B is equal to

B multip by the natural log of a right

so with this property

here so we're going to have from

here I

Pi multiply by the natural

log of e and we know the natural log of

e is equal to 1 right so this will be

equal to I

Pi multi by 1

so which is equal to I pi and this is

the right answer

Is that possible 🙆  ???


good day viewers welcome back to my

Channel today we have the following

addressing problem A + B = 4 a 2 + b² is

= 10 then a 4 + B 4 = dash wow okay now

solution let us recall that a 2 + b 2 is

equal to a + b 2 - 2 a Okay so

here I'm going to rewrite this as

follows a 2 + b 2 so to be written as a

+ b^ 2 - 2 a b = 10 my major priority

here is to find the value of a * B first

okay so I'm going to replace a + b with

four here so I have 4 S - 2 a b = 10 so

4 squ is 16 - 2 a = 10 you know 16 - 10

and 16 is crossing over so we get 2 a is

equal to 6 so from here divide both side

by

two so a * b equal 3 okay now we can

easily get the value of a power

4 then plus b 4 so we writing as a 2 2 +

b 2 squ okay so this is equal to using

this so a 2 + b 2 2 - 2 a 2 b 2 so this

is equal to you know a 2 + b 2 is equal

to 10 so we have here 10^ 2 - 2 remember

a 2 + b 2 is the same as a a * B bracket

squ so this is equal to 10 2 - 2 Open

Bracket you know 8 * B is 3 S and is

equal

to 100 - 2 * 3 S is 9 so this is equal

to this is equal to 100 - 2 * 9

18 and this is equal

to

thank you for viewing and please make

sure you subscribe to my YouTube channel

for more interesting videos

Can you do this without finding the values of a and b ? Watch out. Algebraic Manipulation is best.


hello welcome back once again today

we're going to find the remainder when

29 race report of 999

999999 is divided by 30 then in modular

form that is a modular arithmetic form

this is how it should be

written now let's get started from the

page here

29 is Congress to -1 mode 30 right you

know why when you add the modul we get

-1 + 30 then mod 30 from here1 + 30 will

give us 29 mod 30 right so which

actually mean that 29 is similar to -1

mod 30 therefore the power of this 29

which is 999

999 9 999 will be similar or congr to -1

999 999

mode 30 right okay so from here this

power here must be odd right you know

why the last digit here is odd therefore

this power must be odd And1 to the power

of an odd number will give US1 then we

have here mode 30 from here you know

this simplifies so let's put this

similar so this simplifies to what 29

right so

therefore when 29 to the^ of 999 999

999 is

divided by

30 = 29 amazing and that's the right

answer thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Application of Modular Arithmetic.

#maths #mathematics #olympics


hello welcome back once again today

we're going to solve this radical

equation here and we'll be solving for

the integral values of X now let's get

started so here we have square root of

rootx this is equal to X remember

anytime root sign is involved in

equation what usually come to the mind

is to square the equation right so here

let's go ahead and square both sides so

here we have the square root ofun X

squ so this will be equal to x 2 now

here this Square here we cancel the

first root sign here so here we have

root X is = to X2 right okay in our next

step let's just go ahead and square both

sides one more time so we have root of x

s so this is = to X2 2 so from here

Square cancels the square root sign so

from the right hand side so here we have

just X here then this is equal to x^ 2 *

2 here is 4 now let us make the right

hand side left hand side and left hand

side the right hand side and that does

not affect the equation right so here we

have x^ 4 is = to X now here we get x^ 4

- x this is equal to 0

now X is common here so factor out X so

we get X Cub

-1 this is equal to 0 so if you have

this product to be equal to zero either

this is equal to Z or this is equal to 0

so here we have our first Solution

that's X is = to

0 and here x cubed - 1 is = to 0 we have

X Cub minus let's write this as 1 cub

this is equal to Z making use of the

difference of two cubes

formula a cubus b Cub is same as a minus

B into bracket a 2 + a b + b 2 so here

we have X - 1 into braet x 2 + 1 * X is

x + 1 2 which is 1 this is equal to 0

here we have another product so either

this is equal to0 or this is equal to0

so we have X - 1 = 0 add one to both

side we arrive at another solution that

X is = 1 from here we have this

quadratic expression and we're equating

it to zero right so here checking the

discriminant we find out that

discriminant is less than zero and

therefore no real solution exist here

check it out X here would be B which is

1 plus or minus Ro of b² which is 1 - 4

a c which is3 that is Iota root of 3 / 2

so we can see that this is a complex

solution therefore we don't need the

complex value for x so here we have only

two integral solution X = to 1 and X is

equal to Z thank you for watching please

comment and share bye-bye


hello good viewers welcome back once

again today we have this exponential

equation problem here we have 64 m + 64

M this is equal to 1 here we'll be

solving for the value of M now let's get

started we have 64 m + 64 ^ M this is

equal to 1 from the left hand side 64

the^ m is common so let's Factor it out

so here we have 64 to of M into bracket

this this is 1 plus this ID this is 1

then this is equal to 1 so from here we

have 64 to the^ M * 1 + 1 is 2 then this

isal to 1 now let's canly divide both

side by

two this will cancel this and here we

have 64 the^ M this is equal to 1 / 2

remember the power of this two here is 1

right now let's take note of this 1 / a^

n is same as a^

n so the the right hand side we're going

to write it as 2 to the^

of-1 and again do not forget that 64 can

be expressed in index form with base 2

so here 64 is 2 to^ 6 right so 2^ 6^ M

this is equal to

2^1 so so here we get 2^ 6 * m here is 6

M this is =

2^1 remember we have equal Bas right so

let us equate the powers so from here

this implies that 6 m is =

to1 divide by 6 6 cancel 6 and here we

get m 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

How to solve simple exponential equation


hello welcome back once again today we

have this exponential equation and we're

going to solve for the value of x that

will satisfy this equation so from the

left side we have 64 to the power of X

we know 64 can be expressed in index

power of 8 as 8 to the^ of 2 right so

left hand side will become 8 to the^ 2

then ra to the power of X then this is

equal to we know square root

of 8 is equal to 8 to the^ of 1 / 2

right then the^ of 12 okay now from here

let's take note of this property we have

a the^ b a braet the^ c this is equal to

a^ B * C so from here this will become

8^ 2 * X and this is equal to 8^ of we

know 1 2 *

12 put this over one this will be 12 / 2

which is equal to 6 so here 1 2 * 12 is

6 so we can see from here we have the

same base from both sides so we're going

to equate the powers so this implies

from here that

2x is equal to 6 from here we're solving

for x let's divide by the coefficient of

x so divide this by two and divide here

by two so this two we cancel this two we

can see from the left hand side we are

left with X which is equal to 6 / 2

which is 3 and this is the right answer

to the problem thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Tricky exponential equation | can you solve ?


good day viewers today we having the

following Nigerian oad mathematical

problem X Cub + y Cub =

8,100 x + y = 30 then x² + Y2 = that

okay now to solve out the values of X

and Y before Computing this expression

the value of this expression should not

be the right decision to be made here

okay so let us try to make some

manipulations or introduce some albic

manipulations approaches okay okay now

let us see how we're going to solve this

problem so I'm using the following ideas

that a cub + B Cub is equal to a + b Cub

- 3 A B A + B and again a 2 + b 2 is = a

+ b 2 - 2 a okay so you may ask the

reason why I introduced the following

properties I did this in order to

compute the value of a * B because I've

seen that what I'm about to evaluate the

value of a is needed okay so we only

have the value of x + y so we don't have

x * y so we need to compute the value of

x * y so let us try to write the left

hand side of this first equation here so

writing it we write as follows x + y

Cube - 3 * X Y Open Bracket x + y is =

to 8, 100 okay so now the value of x + y

is 30 right so we have 30 Cub - 3 you

know 3x y * you know x + y here is 30

okay so this is equal to

8,100 so here 30 cu cube is

27,000 so minus 3 * 90 3 * 30 here is 90

so we get X Y = to

8,100 so remember we have 90 x y is = to

27,000 -

8,100 okay so now subtracting this so we

get 90 x y is equal to you know

27,000 -

8,100 so this is giving us

18,900 okay so we'll be dividing both

side by 90 to get the value of x * y so

we get x * y = 1890 0 all over 90 so 0 0

so this is giving us 2 1 0 okay so wow

so we can get the value of X2 + y^ 2

which is the same as X

Y 2 - 2 * x * Y and this = to X + Y 2 is

30 2 - 2 * you know XY is 2 1 0 okay so

this is given us so we have here 30 s so

that is what so we get

900 minus here we have

420 okay okay so and this is exactly

equal to

480 and this is the right answer thank

you for viewing and please make sure you

subscribe to my YouTube channel for more

interesting videos

Nice Olympiad math problem. Do not solve for the values of x and y.