hello welcome back once again today

we're going to solve this beautiful

equation with fractional

powers and here we'll be solving for

both real and complex values for X now

let's get

started okay now from the left hand side

we have X to the^ 3 / 2 this is equal to

5 the^ 3 /

2 when we say that x equal to 5 here

obviously this is a solution

but the question here is is that the

only solution the answer is no but this

happens to be the only real solution and

we have two more complex

Solutions okay now let's get into this

video now let's go ahead to let's go

ahead and square both sides so that we

can eliminate this two right over here

so we have here x^ 3 2 2

this will be equal to 5 3 2

2 so we can see from here 3 2 * 2 2

cancels 2 here 2 cancels 2 so we have X

the^ of 3 this is = to 5^

3 so subtract 5^ 3 from both side we get

x^ 3us 5^ 3 this is equal to Z from here

we're going to make use of the

difference of two cubes formula a cub -

B Cub this is equal to Aus B multip by a

2 + a b + b

2 so from here this will become x - 5

into braet X2 + 5 * X then + 5 2 which

is

25 and this will be equal to

zero so from here if this time this

equal to 0 so we have either this is

equal to 0 or this is equal to 0 so here

x - 5 is = 0 here we found the solution

that x = 5 which is the only real

solution and here we have X2 + 5x + 25

this is equal 0 now using the general

quadratic formula we get x = b will be5

plus or minus otk of b 2 is 5 S which is

25 - 4 * a * C 4 * 1 * 25 will give us

100 then /

2 so from here we get X is = -5 plus or

minus sare < TK of -75 right then / by 2

so from here x = to -5 plus or minus

already square root of1 is iot right and

do not forget that 75 is the same thing

as 25 *

3 then / 2 so here we get X = -5 + or

minus of 25 is 5 so we get 5 I root of 3

then / 2 and here we have the other two

compx solutions for X thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye I

Interesting Olympiad Math | Can you find all solutions ?


hello welcome back once

again today we're going to solve the

following given exponential

equation and here we be solving for the

value of x that will satisfy this

equation from the left hand side we have

5 to the power of x * 5^ x * 5^ X we

know this is equal to 10 now let us

recall this law of algebra we know a to

the of B * a power of C * a power of D

this will be equal to a the power of b +

C + D since we have the same basis right

so we can see from the left hand side we

can apply this law and that will become

5^ x + x + x and then this is equal to

10 from here we have 5 to the ^ 3x X

this is equal to 10 now we're going to

apply logarithm on both sides of this

equation so let's log both sides so here

we have log of 5^ 3x this will be equal

to log of 10 we know this property of

logarithm log of a to the^ B is equal to

B MTI the log of a so this power of 5

here which is 3x will come down here so

we get 3x * the log of 5 this will be

equal to log of we know 10 is same as 5

* 2 we know this Lo this logarithmic law

that log of a * B is equal to log of a +

log of B so we can see that this law

here we can apply it at the right hand

side so remember the left hand side is

still 3x * log of 5 then this will be

equal to log of 5 plus log of 2 since

you're soling for X let's divide both

side by the log of

5 by the log of

5 so this will cancel this and we're

left 3x isal to splitting this

denominator we get log of 5 / log of 5

+ log of 2

/ log of

5 we know this here we cancel to one

right because log of 5 log of 5 is 1 so

we have 3x is = 1 + log of 2 / log of 5

so we know this another

important logarithmic law we have that

the log of a

/ the log of B is equal to the log of a

to the base B so this law we're going to

apply it right over here so this will

become 3x is =

1+ so that will be log of 2 to the base

5 so let's divide both by the

coefficient of

x so this will cancel this so from here

we have our solution that X is equal to

split this denominator that is 1 / 3

plus log of 2 to the base 5 then / 3 and

this is our solution thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Germany | Olympiad Math | Can you solve ?


hello good viewers welcome back once

again today we have this interesting

Olympiad mathematics problem and here

we'll be

evaluating 5x + xun x given that x +un x

is =

4 now let's get started from the given

equation we have X Plus root of x this

is equal to

4 now let's go ahead and multiply both

sides by root of x so here we have root

of x into braet

X+ root of

x then this is equal to root of x *

4 okay so from here root of x * X will

give us X root X then we have plus root

of x * root of

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x we give

root of

X2 of which this Square will cancel the

square root and we have just X so right

over here we have X and this is equal to

4ot of X so let's call this our equation

one okay now let's go back to the

original equation we have here x + otk x

this is equal to 4 now let's go ah head

and multiply both sides by 4 so we have

4 into braet x +un x this is = to 4 * 4

so from here we have 4 * X is 4 x then

we have plus 4 * root of x will give us

4un X then this will be equal to

16 now let's call this our equation 2

okay now from equation one we have 4

root X to be equal to xun x + x so right

over here we're going to replace this

with xun x + x so here we get 4X plus so

instead of this we we write as xun x + x

right then this is equal 16 so from here

4x + x will give us

5x then we have plus xun

X okay so this is the left hand side now

let's go back to the original problem

let us know what we are asked to

evaluate so we can see that we have this

expression here to evaluate 5un x + xun

x and right from here we already have

the expression here and this is equal to

16 which is the correct answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Tricky Olympiad Math problem | Can you solve ?


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Junior Olympiad Math. |Simplification of powers of numbers #education #maths


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Junior Olympiad Math. | Interesting problem. #education


hello welcome back once again today we

have this awesome looking problem here

this is an Olympian mathematics problem

yes indeed this problem is very

tricky imagine we have just an equation

here involving two different variables

and we asked to evaluate a / B 2 plus b

/ a

s now let's go ahead and solve this

problem from the given equation let us

expand the left hand side we have here a

- b 2 which is equal to 6 a b expanding

this this will give us a 2 + b^ 2 - 2 a

b and this is equal to 6 a add 2 a to

both sides we get a 2 + b 2 is equal to

6 a + 2 a we give us 8 a

this point let us divide both sides by a

* B so divide this by a b and divide

here by a split this denominator from

here we get a 2 / a plus b 2 / a this

will be equal to a will cancel a we left

with

8 a s / a will give us just a remember

we have over B plus b Sid B we are left

with b and we have over a and this is

equal to 8 wonderful right now let us

recall the expression we asked to

evaluate which is a / B 2 plus b/ a s so

let us manipulate this expression but

before that take note of this property

in algebra a 2 + b 2 is = a + b 2 - 2 *

a *

B so with this identity here remember

something is squared here and something

is squared here so it's going to be into

bracket that's first something a / B

plus the second something B / a then

squared according to this property here

minus 2 * the first something here which

is a / B multip the second something

which is B / a

so from

here this is a / B + B /

a² minus just two because these two

cancel out and from here we have the

solution remember a/ b + B / a b is 8 so

we have 8 2 - 2 8 2 is 64 then - 2 which

is equal to 62 so therefore

a / B 2 + B / a 2 is = 62 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 problem | Can you solve ?


welcome back my

fans here we have this mind

calling problem so we are asked to at

least find all the values of X

right okay now let's see how to solve

this problem but before we proceed

please kindly support this channel by

subscribing to this YouTube

Channel please subscribe to the channel

and share this video here we have x^ 4

is = to 1 we have x^ 4 - 1 is equal 0

you know this x 4 we can write it as

follows X of 2 squar minus you know 1 is

the same as 1 2 and this is equal to Z

right so right from here we can make use

of the difference of two squares mainly

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

right okay here we're going to have

X2 +

1 *

X2 - 1 is = 0 so from here now we can

say X2 + 1 is equal to 0

or x^2 - 1 is = 0

right okay now let us solve this

first we have X2 - 1 is =

0 so we

have X2 is = 1 that's when this1 crosses

over it will have opposite sign right so

we can go ahead and take the root of

both sides so we have xal to plus or

minus < TK 1 so we get X = to+ or

minus 1 right so you can see we have at

least found

two values of X and these are the real

values right here we have + one

or1 now let us go ahead and solve this

problem

here we have x^2 + 1 is =

0 so from here we have X2 is = to -1

since pos1 is crossing over so it will

have opposite sign so we have here x =

to plus or minus root

of1 and we all know that theun of1 is

equal to I that is the imaginary unit

right so this this is going to give us

plus or minus I right so we can go

ahead

and write down the possible values of X

in that equation so we have

1 we have -

1 we have positive

I and then we have I and this are the

right answers thank you for watching

please kindly support this channel by

subscribing to the channel subscribe

please and share this video bye-bye

A very nice equation. x⁴ = 1.


hello wonderful viewers here we have

another interesting question please

subscribe to this channel also like

comment and share here we're given that

x^ of 1 1999 - 1 / x ^ of 1 1999 is

equal to 11 and here we asked to

evaluate x ^ of 1 1999 + 1 /x ^ of 1

right okay well this question looks

crazy but an easy

problem let's see how we can find the

value of this expression

here okay now from the equation given x

1999 - 1 / x 1

1999 is equal to 11 let's Square both

sides of this equation so if you square

both sides

here we have 11 squ from the left we're

going to make use of the

property a - b s this is equal to a + b

s

minus 4 a b right okay cool so if we

make use of this property at the left

hand side we going to have

x 1

1999 + 1 x 1

1999 then here

squared right then minus here we have 4

* x 1 1999 * 1 / x 1 1999 you know this

we cancel out we are left with four and

here this is equal to 11 s which is 12 1

right so from here we have x^

1999 + 1 / X for

1999 squ this is equal to you know if

you add four to both side we get

125 so from here we're going to take

root of both sides and there we find the

value of this expression and the value

is going to be X for

199 + 1 / X is for 1 1999

this is equal to plus or minus < TK 125

so simplifying this s we get plus or

minus 5 < TK 5 and this is the final

answer to the problem thank you for

watching please do wait to subscribe to

my channel also like comment and share

bye-bye

The crazy looking Olympiad Math. How to evaluate x^1999 + ¹/x^1999.


hello welcome back once again today we

have another interesting problem here

we're going to give a step by step

solution to this radical equation so

here we have square root of square root

of 2x + 1 which is equal to 3 so here

we'll be solving for the value of x

right so let's start the first thing

we'll do here is to square both sides of

this equation right right so we going to

square this and equally Square the right

hand side as we know for example Square

< TK of

a^ of 2 is equal to a because Square we

cancel the square root right but here we

have two square root signs so this

Square will cancel the first square root

so here we are left with square OT of 2x

1 right which is equal to from the right

hand side 3 S is 9 okay again we have

another square root so let's go ahead

and square both sides okay so here

Square this

side and equally Square the right hand

side we know this Square will cancel the

square roots and here we have 2X then +

1 which is equal to 9^ 2 is

81 okay so look at this left hand side

of this equation we have positive one so

what happens when we subtract one from

both sides we see that the positive one

here and the negative one will vanish

and that will be left with 2x right so

let's do that so here we have

2X then + 1 - 1 this will be equal to 81

- 1 so from here we can see that + 1 - 1

this is zero so here we have

2X which should be equal to 81 - 1 will

give us 80

right okay so from here the next thing

to do is to divide both sides by the

coefficient of x and the coefficient of

x in this case is two so divide this by

two and divide here by two so this two

we cancel this two and here we left with

X so our answer now we get X is equal to

80 divid by 2 which should give us 40 so

the right answer to this problem is 40

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Very nice radical equation | Step by step solution


hello good day viewers welcome back once

again please cly support my channel by

subscribing also like comment and share

in this video we are going to solve for

x in the following Olympia math

problem here we are given that x ^ of X

is = to 2^ of

248 in this video we are going to find

the solution for X

right let's get started we have here x

to the^ of X this is equal to 2^ of

248 right so from here we have X to the^

of X this is equal to 2 power of you

know

2048 is the same as 2 multi by 1 24

right now let's record this

property a of B * C this is same as a

power of B then in bracket power of C so

from here we have X the^ of X this is

equal

to you know in bracket 2

2 then ra power of 1 to 4 and from here

this is equal to 4 ra power of 1 to

4 okay now from here we have x to the

power of X this is equal to 4 power of

1024 is the same as 2

512 right so we make use of this

property once again from here we get x^

of X this is equal to 4^ 2 then ra power

of

512 and from here the bracket here we

have 16 to^ 51 2 now from here we have

x^ X this is equal to 16 power of 2 *

512 is 2 *

256 and from here we have 16 2 ra the^

of

256 and from here we have x ^ of X this

is equal to 16 s which is

256 to the power of

256 isn't this

so from here this implies that X is

equal to

256 and this is the solution thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Japanese | Nice Olympiad exponential equation | #maths #mathematics


good if he was today shidex is showing

his concern on this problem

this has been a trending problem online

huh the negative 3 squared equals Dutch

okay

now before we go into this problem

please

subscribe to my channel for more

interesting videos please subscribe you

can share

you can comment

and so

you can like now

let's say we have

Negative X

part 2 like this in Brackets okay so

here you know any variable or number in

Brackets where it means the number

multiply itself twice so here we have

negative x times negative X okay so we

get this equals you know minus here

so here we have minus times minus is

plus okay here we have x times x so that

is X2

so this is equal to X raised part two

okay now

let's say we have Negative X

you raise poetry so this is equal to

Negative X

times negative x times negative X so

this is giving us what so you know here

minus times minus here so that is plus

then times this minus here so we get

minus so we get minus x times x times x

that is SQ so we have a negative sign so

this means that whenever we have

Negative X

even numbers

so this will give us answer in positive

form so

in positive form and also when we have

Negative X

raised by odd numbers

so this will give us answer in negative

form okay

so this is what we have now

let's say we have Negative X raised part

two in this one so this is equal to

negative one times x raised power 2. so

most of us

has been assuming that negative x

squared is the same as Negative X raised

by two like this then answer becoming

positive but this is absolutely wrong

okay so here we have negative 1 times x

squared so

in this case we have negative three

respirator so this is equal to negative

one

times negative sorry times three

squared okay so here we have minus one

then times three squared is nine so we

get our answer as negative nine and this

is the right answer to the problem thank

you for viewing and please make sure you

subscribe to our channel for more

interesting videos

My take on this.


hello good viewers in this video we're

going to solve for the value of x in the

following exponential

equation we have here 2 to ^ of x + 1

you know we can move this to the right

hand side that will give us 3 to ^ of x

+ 2 now let us record this property M of

n + P this is same as M of n * m power

of P so from here we have 2^ X Multi 2 1

which is 2 this is equal to 3 x * 3 2

that will give us

9 so from here let's go ahead and divide

both sides by

two this will cancel this we have here

2^ x this is = 3 x * 9 /

2 from here we can go ahead and divide

both sides by 3 to the^ of x

this should cancel this and here we have

you know we have the same power so we

can write it as 2 over 3 all X this will

give us 9 all 2 now let us recall that

the logarithm of M to the power of B to

the base of M is exactly equal to B so

from here we're going to log both sides

but the logarithm must be to the base 2/

3 so we have the

logarithm of 2/ 3 of x x to the base of

2 / 3 this is equal to the

logarithm of 9 / 2 to the base of 2 over

3 so we are left with just X here

because the whole of this will cancel

out we have X is equal to the

logarithm of 9 / 2 to the base of 2/ 3

which is approximately equal

to -3 point this is three

-3 709

513 okay 113 I mean thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Exponential equation | #maths #mathematics


hello good day viewers let's find a

solution to the following given

exponential

equation we have 4^ of otk X this is

equal to 9 now we're going to make use

of the natural

logarithm so from the both side we'll be

taking the natural logarithm so here we

have the Ln of 4^ ofun X this is equal

to the Ln of

9 so from here you know the property of

the logarithm this power we cross over

so here we have the root x * the Ln of 4

this is equal to the Ln of 9 so from

here remember we're solving for x so let

us go ahead and divide both side by the

Ln of 4 so we have otk x * the Ln of 4 /

by the Ln of 4 this is equal to the Ln

of 9 / the Ln of four so from here this

cancel this so we have root X this is

equal to the Ln of 9 is the same as 3 2

/ 4 is the same as 2

2 so from here we have rootx this is

equal to with the property of logarithm

this power is crossing over so we have 2

MTI the Ln of 3 divided by the same

thing happens here so we have two M the

Ln of 2 so this two and this two we

divide out so here we have root X this

is equal to the Ln of 3 / the Ln of

2 now we're solving for x so we're going

to square both side to get rid of this

root sign so here we Square this and

from here we Square this so the square

and the root sign we cancel out and here

we have X is equal to the Ln of 3 / the

Ln of 2 s and this is approximately

equal to 2

613 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 | #maths #mathematics


hello welcome back once again let us

solve this exponential equation from the

left we have 3 of root of x this will be

equal to here 9 is same as 3 S right

then power of the CU root of x now take

note of this when we have a^ B to the

power C this is same as a^ B * C so from

here we get 3^ ofun X this equals here

will be 3^ 2 MTI the cube root of

x okay now from here we have the same

bases right so we're going to equate the

powers so we get root of x this equal 2

* the cube root of

x okay now how do we solve this equation

from

[Music]

here so how do we solve this equation

from here so for my solution let us

first reite the

square root and the cube root sign so

this is same as x^ 1 / 2 then this is

equal to 2 * X to the^ 1 / 3 now let x

equals remember the LCM of 2 and 3 is 6

so we're going to let it equals y^ of 6

right so from here we can see that when

we raise both sides to the^ of 1 / 2 we

get that x^ 1 / 2 2 will be y to the^ of

3 right because 6 * 1 2 is 3 and raise

both sides to the power of 1/3 we get X

of 1/3 = y^

2 okay so from here otk x which is x^ 1

/ 2 will become y^ 3 = 2 * by x^ 1 / 3

which is y^ 2 move this term to the left

so we get y Cub - 2 y^ 2 this = 0 so we

have something common here and that is

y^ 2 we factor that out and we left with

y - 2 then the whole of this equals z

right okay now from here we have that

y^2 = 0 we have from the bracket y - 2 =

0 so from here you know this is l y = 0

and from here Y = 2 now let us recall

again that x = y^ 6 so we have the first

value of x let's call it X1 let's call

This y1 And here Y 2 right so for my

first solution for x x = 0 and is a

valid solution right and for the second

solution which is X2 this

equals y ^ 6 which is 2^ 6 and 2 6 will

give us 64 and here we found the

solutions thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting exponential equation


hello welcome back once again today we

have this nice radical equation we have

root of root of x = 1 / root of x and we

are given that X is not equal to Z

without even letting us know we all know

that obviously X must not be equal to

zero right so now let's get into this

video so we have here the square root of

root of x this is equal to 1 / root of x

right so now let's go ahead and square

both sides because this is involving the

root sign we all know that the square

and the root signs are enemies right

they always get cancel out so here let's

Square both

sides so we Square both

sides we Square both sides and for the

right hand side here let us take note of

this a /

B to the power of n this is equal to a^

n / p the^ n so but from the left hand

side we know this square and this root

sign they are enemies so they will

cancel out and we are left with one root

sign under this X or X is under 8 I mean

so here this will be equal to 1 s which

is 1 / root of x² remember the square

will cancel this root so we have 1 / X

so in our next step let's go ahead and

square both sides again because of this

root sign here so Square both sides and

here Square both sides so this cancel

this so we have here x is = to 1 2 is 1

/ X2 which is x^ 2 so in our next step

Let's cross multiply so this we multiply

this and this we multiply this right so

here x² * X will give us x^ 3 this will

be equal to 1 * 1 which is 1 so let's

subtract one from both side we get x^ 3

- 1 this is equal to 0 so from here x

Cub - 1 is same as 1 Cub this is equal

to Z so from here let's go ahead and

make use of the difference of two cubes

formula a cub minus B Cub this is = to a

- b into braet a 2 + a b + b 2 so make

use of that we have here x -1 into braet

X2 + 1 * X which is x + 1 s which is one

then this is equal to Z so whenever we

have these two products which is equal

to Zer so we all think about this that

either this is equal to Z or this right

over here is equal to Z so when X - 1 is

= 0 we found our real solution that X is

= 1 and this is the only real solution

and for this quadratic expression we're

going to equate it equal zero right so

here we form this quadratic equation

which we're going to solve with the

general quadratic formulative B here

will be1 plus or minus b s b s is 1 - 4

a which is -3 of -3 is Iota root of

three then divid by two here we have the

two complex

Solutions thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting radical equation | Can you solve ?


hello good day viewers welcome back once

again today we have this nice cubic

equation and here we'll be solving for

the values of X now let's get started

from here we have X cubed plus

x² + x this is equal to Z okay now from

here many student may think that the

best way to approach this problem from

here is to divide through by X but if

you do that you will forget a solution

that X is equal to Z and this solution

you're definitely not going to have this

solution again so what do we do from

here look at the left hand side of this

equation x is common so we Factor it out

so here x Cub / X will give us

x2+ x² / X will give us

X+ x / X that is 1 then this is equal to

Z okay now here we have this product to

be equal to Z then we have either X is

equal to0 so we can see we have this

solution here or the bracket x² + x + 1

is = 0 so from here we're going to make

use of the general quadratic formula so

here we get X isal to B in this case

will be -1 plus or minus Square Ro TK of

b s will be 1 s which is 1us 4 * a * c

will give us 4 then ID 2 so from here

this is equal to1 + or

minus < TK of -3 / 2 now simplifying we

get X is = -1 + or minus now here s of1

is Iota and we have root of 3 then / 2

therefore we have the following

solutions for X so we have x is = 0 and

x is also = to -1 + or minus Iota root

of 3 / by 2 thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Interesting cubic equation


hello good viewers welcome back once

again today we have this interesting

Olympiad mathematics problem involving

the sub factor of 4 here we have 6^ of <

TK of x^ X this is equal to 4 * the subf

factoral of four here we'll be solving

for the R value of

x what is this subf factor of four if

you check my videos I've made video

about this particular topic the sub

factoral but I'm going to talk less

about this here we

have the property here we have the sub

factoral of n or N Sub factoral is equal

to n factorial multip by this sum as K

goes from 0 to n of -1 to the power of k

/ K

factorial we also have the rough

approximation for the sub factorial of n

which is approximately equal to n

factorial /

a now the sub factoral of four using

this s of approximation here is

approximately equal to the four

factorial ID a and this is approximately

equal to 9 but when we use this we get

exactly 9 so therefore this right over

here is equal to 9 therefore 4 * 9 will

give us 36 therefore the left hand side

is equal to

36 now let's go ahead and solve the

problem so here we have 6^ of theare <

TK of x x this is equal to

36 now from here we have 6^ of the < TK

of x^ X this is equal to 36 is the same

as 6

s now take note of this when we have a^

B = a^ C since they have equal base we

can easily say that b is equal to C so

therefore we have equal bases here so we

can equate the power so here we have the

square root of x of X this is equal to 2

to get rid of this root sign we need to

square both sides of this

equation so this Square will cancel the

square roots so we have here x of X this

is equal to 2 to the^ 2

so from here we can see that this imply

that X is = 2 and therefore this is the

solution we have X as the base from the

right hand side we have two at the base

the same X which is the power here the

right hand side we have the same two

which is the power here therefore this

simply implies that xal 2 and here is

the solution thank you for watching

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Interesting exponential equation involving subfactorial.


hello welcome back once again today

we'll be finding the following given

limits we have limits as X is

approaching for of 4 - x / 2 - < TK of

X when we try to make direct

substitution here we get 4 - 4 / 2 - <

TK4 and here we get this 0 all over 0

which is in indetermined

form in this video the most interesting

part is that when we are going to make

use of the litus rule so therefore we're

going to solve this

algebraically so let us start by making

the following substitution so we're

going to

let this square root of x at the bottom

here equals y now Square both sides we

get X will be equal to Y 2 so therefore

our limit will be limit now y will be

approaching remember if X is approaching

4 y will be approaching theare root of 4

which is 2 so here we have 4us instead

of X so we have y^ 2 then / 2us inste of

root of x we replace with Y so from here

this will be equal to limit as Y is

approaching 2 so here 4 is as 2 2 then -

y^ 2 / 2 - y

y so at this point here let us record

the difference of two squares formula

remember a 2us b 2 is equal to a + b *

Aus B so from here we have limit as Y is

approaching 2 so up here we have 2 + y

MTI 2 - Y and down below there we have 2

- y so we can see we have this nice

cancellation here so this is equal to

limit as Y is approaching to of 2 + y so

here we can make our direct substitution

now to get 2 + y here which is 2 which

is equal to 4 and this is the right

answer thank you for watching please

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Very nice Limit | No L'hospital's rule !


hello good day viewers today I'll be

solving this nice Junior Olympiad matth

so we have here 2 power of 5 ra power of

< TK of 2 power of 0 is equal to D we

asked to evaluate this right please if

you have not subscribed to my YouTube

Channel please kindly support this

channel by subscribing and also share my

videos right how do we evaluate this so

I'm going to start with square < TK of

2^ of 0 right over here right so let us

recall that X power of 0 is always 1

provided that X is not equal to 0 right

and here we have sare < TK 2^ of 0 here

we see that square root of 2 is not

equal to zero right so this is going to

be equal to one right so the expression

2 5 otk of 2 0 is going to be the same

thing as 2^ of 5 then power of

1 so this is going to be equal to now

take this power from here we only know

that 5 power of 1 is equal to 5 right so

this going to be equal to

2 5 and 2 5 is the same as 2 * 2 * 2 2 *

2 * 2 that is in five places and this

resulted to 32 and this is the right

answer thank you for watching please do

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Junior Olympiad Math. How to deal? #maths #olympics


hello welcome back once again today

we're going to solve this interesting

equation so we have X - 1 to^ 2 all 3

this is equal to 25 so in our video

we're going to solve for the value of

values of X so let's get started from

the left hand side of this equation we

can rewrite it as follow we have X - 1

then 2 over 3 is same as 2 * 1/ 3 then

from the right hand side this is equal

to

25 now let us make use of this property

we know that a to the power B * C this

is equal to a^ B in bracket to the power

of C this is also equal to a to the

power C in bracket to the power of B so

using this property the left hand side

of this equation we can write it as X X

- 1 to the^ 1 / 3 in bracket to the^ 2

this is equal to

25 now in our next step let's take the

square root of both sides so the left

hand side will become x -1 to^ 1 / 3

this is equal to plus or minus the

square root of

25 from here we have x - 1 to the^ 1 / 3

this is equal to plus or minus >

25 is five right so from here we have

two

equations we have the first one x -1 to^

1 / 3 is = to POS 5 and we have the

second one here x -

1^ 1 / 3 is = to

5 now in our next step let's CU both

sides so we

have X - 1 to the^ 1 / 3 cubed this is

equal to 5 cubed and from

here x -1 to^ of 1/ 3 cubed this is

equal

to

-5 cubed so we know 1 3 * 3 is 1 right

so 3 will cancel 3 so we have X - 1 is =

5 cubed is

125 from here we know three we cancel

three here so we have X - 1 is equal to

-5 cubed will give

us15 because the power is

odd then from here add one to both sides

so we get X is = to 125 + 1 which is

equal to

126 so this is a

solution then from here add one to both

sides so we get X is = to -125 + 1 we

give

us14 and this is our second solution

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Interesting equation with fractional power | Can you solve ?


hello welcome back once again today we

have to solve this problem right when

students always make mistake when

solving this

problem most students think that the

best way to solve this problem is to

First div B by x² right but if you do so

you miss some

solutions so this is how we're going to

solve this

problem we have x to^ 6 which is equal

to 2 x^ 2 this time let's take it to the

left hand side we have x ^ 6 - 2

x² then this is equal to Z from the left

hand side X2 is common so let's Factor

it out so we have x 2 into braet x ^ 4 -

2 this is equal to 0 from here we have

two products which is equal to zero so

this implies from here that either x 2

is = 0 or x^ 4 - 2 is = 0 from here take

root of both side we get X isal to 0

right so this is one of the solution we

would have missed if we started by

dividing both sides by x² right so here

we're going to manipulate the left hand

side so let us reite this as x^2 to the^

2 we know 2 * 2 is 4 then minus rewrite

this one as < tk2 2 we know Square

cancel square root then this is equal to

Zer you know why now we can make use of

this formula a s - b 2 = A + B * Aus B

called the difference of two squares

right so without formula we get

x2 + <

tk2 * X

- < tk2 then this is equal Z again this

implies from here that either X2 + < tk2

is = 0 or X2 - < tk2 is = 0 now subtract

root2 from both side here we get X = <

tk2 and here add root2 to both side we

get x2 isal

tok2 take theare root of both side we

get xal plus or minus theun of < tk2

again here take the of both get X Plus

or theun ofun of two from here

simplifying we get xal plus or minus of1

is aota and of root2 is same as the 4

root of 2 please do do not forget and

here we get xal plus or minus the 4th

root of 2 remember 0 is a repeated

solution right so here we have the

following solutions for x x = 0 x = 4un

2 x = 4ot of 2 then imaginary Solutions

X = iot posi 4ot of 2 and X = Iota 4

root of 2 and these are the solutions

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Interesting hexic equation | You will probably miss a solution !!!


let us find a solution to this problem

please if today is your first day of

visiting this channel kindly support us

by subscribing to this YouTube Channel

please subscribe to this Channel and

share this video here we are giving this

nice exponential

problem 5^ 2 n^ 2 - 1 is =

to

125 okay here we know that

125 is the same as 5^ 3 so let us go

ahead and replace that with 5^ 3 so here

we have 5^ of 2 n^ 2 - 1 = 5^ 3 so we

have equal bases right from both Sid so

we can easily equate the powers right so

here we have 2 n^ 2 - 1 is = 3 add one

to both side we get 2 n² is = to 4

divide both side by 2

here over two and here over two we get

this canceled out we left with n^2 is =

4 2 is 2 so let us take the root of both

sides so here we have n = plus or minus

the sare RO TK of 2 right and this is

the right

answer thank you for watching do well to

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An exponential equation.


hello good viewers welcome back once

again let's find a solution to this

sodic

equation root of x - root of 6 this is

equal to root of 24 now we can easily

rewrite root of 24 in terms of root of 6

if we recall that 24 is same as 4 * 6

therefore root of 24 will be same as <

TK 4 * 6 which is same as otk of 4

multip by root of 6 from here we get 2 <

TK 6 so from here our equation will

become < TK x - otk 6 this is = 2un 6

now we add root of 6 to both sides we

get otk X is = 2un 6 +un 6 from here we

get root of x is = 3 < TK 6 from here we

go and and square both side to get rid

of this root

sign the square and the root will cancel

out we get X is equal to here 3un 6 2 is

same as 3un 6 * 3un 6 now simplifying 3

* 3 here is 9 then * root of 6 * root of

6 will give us root of 36 from here this

is 9 * root of 36 which is 6 and and

from here oh let me use

this so from here 9 * 6 is 54 right so

here we come up with the solution X isal

to 54 and this is the right answer thank

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to my channel also like comment and

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Easy Surdic Equation | #maths #mathematics


hello good viewers welcome back once

again today we're given the following

exponential equation x^ x^ 4 = 4 here

we're going to evaluate this expression

x^ x^ 8 + x^ X the power of

2 now there's a way I want to solve this

equation and I think that would be very

cool right this equals 4 I'm going to

let X to the^ 4 = Y and obviously X = Y

to^ 1 / 4 so from here remember this

base here has the power of x^ 4 so X

here will be y^ 1/ 4 then all the^ of x^

4 which is y then this equals

4 from here with this property a^ b^ c =

a^ b * C now we have y^ 1 4 * Y is the

same as y / 4 then this is equal

to

4 from here let's raise both side to the

power of four I have reason for

that whatever I do I have good raising

so ra 4 and from here we can multiply

this and four we cancel for we left it

y^ y = 4^ 4 nice reason so from here

this implies that y = 4 and that implies

also that x = y^ 1 4 which is 4^ 1 4

remember 4 is as 2 2 then to^ 1 / 4

multiplying these Powers together we get

2^ 2 / 4 which is equal to 2^ 1 / 2

which is equal to the < TK of 2 amazing

so in the given equation here x = the <

TK of 2 now let's go back to the

expression we ask to

evaluate right so this will be equal to

the < TK of 2 of < TK of 2 of 8 plus the

< TK of

2 < TK 2 of 8 of two I mean sorry for

that okay so from

here first of all we know that the < TK

of 2 2 = 2 remember this8 we can Factor

it as 2 * 4 then the root of 2 of the

two will give us two

which implies that of 2^ 8 is the same

as 2^ of

4 so here we have the < TK of 2 2^ 4

then plus the < TK of 2un of 2^ 2 here

will be two so from here again 2^ 4 is

16 right so let's put it well 2 16 +un

2^ 2 from here remember of 2 2 is 2 16

is same as 2 * 8 so this root of 2 of

the 2 will give us two so this will same

as 2 to the^ of that 8 then plus here

this is two right so this equals

256 + 2 which is equal to

258 and this is the right answer thank

to my channel also like comment and

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Oxford University Entrance Exam


hello good viewers welcome back once

again let's solve this exponential

equation very quickly we have 2^ of 2^

of X this is equal to 3 now what we're

going to do is to make use of the

natural logarithm so we're going to Ln

both sides of this equation so we have

here the Ln of 2 of 2 x this is equal to

the Ln of 3 now we call this property of

logarithm that the Ln of a to the^ of B

this is equal to B * the Ln of a right

so here the power of 2 here is 2 to the^

of X so this power here is crossing over

so here we have 2 to the^ of X MTI the

Ln of 2 this is equal to the Ln of

3 so from here we're going to divide

both side by the Ln of two so here we

divide by the Ln of two and from this

side we divide by the Ln of 2 so we can

see that this Ln of 2 and this Ln of two

divide out so here we have 2 ^ of X this

is equal to the Ln of 3 / the Ln of 2

now let us recall another property of

this logarithm again recall that when we

have the log

of a to the power of B to the base a

this simplifies to B right so now from

both sides we're going to take the

logarithm and this log must be in base 2

which is the base of this x here so here

we have to log both side to the base

two and from the right we have the log

of the Ln of 3 / the Ln of 2

to the base

2 so from here with this property here

so this simplifies to X so X is equal to

the log of the Ln of 3 over the L of 2

to the base 2 and this is approximately

equal to

0.66 4 4 4 right then 8 7

one and this is the right answer thank

you for watching please kindly subscribe

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China | Interesting exponential equation | can you solve ? | #maths #mathematics


hello good day everyone welcome back

once again today we're going to solve

this equation here we have the fourth

root of x = 12 / 7 - 4 root of x from

here we're going to make a substitution

we're going to let the fourth root of x

equals another variable let's call it y

right so from here if we raise both

sides to the^ of 4 we get X is = y ^ 4

right so so from here we're going to

write this equation in terms of Y so

here 4 root of x will be Y which is

equal to 12 / 7 - 4 root of x which is y

now from here we're going to cross

multiply when we cross multiply we have

y

* 7 - y this will be equal to 1 * 12

which is 12 so from here let's open this

bracket y * 7 will give us 7 y y * y

will give us y^ 2 then we can subtract

12 from here and this will be equal to

Zer From Here let us multiply both sides

by-1 so here1 into bracket 7 y - y^ 2 -

12 this is equal to 0 -

-1 from here we have -1 * 7 y y will

give us -7 y -1 * Y 2 will give us + y^

2 and here1 * -12 will give us + 12 and

this will give us 0 from here let's

rewrite this equation as y^ 2 - 7 y + 12

this will be equal to0 now here we have

a quadratic equation which we can

factorize right so this quadratic

equation is factorizable right so we're

going to look for two terms which when

we multiply them we get 2 y^2 and when

we add them we get -7 Y and these two

terms are -4 Y and -3 y so -7 y we can

write it in terms of these two terms and

that will give us y^ 2 - 4 y - 3 y + +

12 = 0 so from here we factor out y we

have y - 4 and here we factor out -3 and

we get y - 4 and the whole of this equal

0 so from here we get y -

4 * y - 3 which is equal 0 and this

implies that y - 4 is = 0 and from here

Y is = 4 also y - 3 3 = 0 so from here Y

is = 3 now let's recall

that 4 root of x = y of which when we

raise both s to the^ of 4 we get X is =

y to the^ 4 now from here we can get the

two real values of X X1 which is 4 ^ of

4 and this will give us

256 and the second value of x which will

be y to ^ 4 which is 3 to ^ 4 and this

will give

us 81 so these are the solutions thank

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Interesting Olympiad Math.


hello welcome back once again today we

have this interesting exponential

equation and involving a fractional

base and we're going to solve for the

value of x right now let's get started

from the left hand side here we have

25 /

16 ra power of X then this is = to 4 / 5

now take note of this 25 here is 5 S and

16 here is 4 S so this will be same as 5

/ 4 all squ then remember we have of X

and this is equal to 4 / 5 okay with one

of the laws of indices this power here

we multiply together right so we have

here 5/

4 2 * X which is 2 X then this is equal

to 4 /

5 okay now from here let us take note of

this when we have a/

B of n if we want to replace a and b

then we have a negative power that means

that will become B / a ra of n right

then remember this 4 5 here they have a

power of 1 so if you try to replace four

and 5 then we have a negative power so

here we have 5 / 4 to the^ 2x this is be

equal to 5 / 4 of1 so we can see here we

have the same base right so we can

equate the power so here this implies

that 2x is = to -1 since we're solving

for x let's divide both by the

coefficient of x so 2 cancel two and

from here we get the right answer that x

= -1 over two thank you for watching

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Interesting exponential equation with fractional bases


hello good day viewers welcome back

today to another video please subscribe

to my channel if you have not subscribed

to this Channel Please Subscribe please

like comment and share here we're going

to evaluate this expression here right

here we have 7^ of < TK 7 / 7 < TK 3 all

ra the power of < TK 7 + < TK 3 right

okay now take a look at this place we

have equal bases here but we have

different powers so there's a law of

indices we're going to make use here we

have a m over a n so we have a basis so

this is same as a power of m- n so this

is the division law right so in the

bracket we're going to have 7^ of < TK

of

7- < TK 3 then in bracket < TK 7

+un 3 okay now there's another law that

when we have a of B of C we can multiply

the powers together so that becomes a b

* C right okay now here we're going to

have this to be equal to 7 ra the power

of < TK 7 - < TK 3 m by < TK 7 + < TK 3

so from here you know this is difference

of two squares right so recall A + B *

Aus B this is equal to a 2 - b 2 right

so from here this is going to be equal

to 7^ of < TK 7

2us < TK 3 2 so this is equal to 7^ of

you know square and square root cancel

out so we have 7 minus square and square

root cancel out we have 3 right so from

here this is equal to 7^ 4 and this is =

to 7 * 7 * 7 * 7 and this equal

241 and this is the right answer thank

you for watching please do wa to

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comment and share bye-bye

Interesting problem for everyone.


hello welcome back once again in this

video let us find the derivative of this

function x / by the natural logarithm of

X we have here a numerator and a

denominator which is a rational function

here who knows whether it is rational or

irrational I think this function is

irrational right because of this natural

logarithm of x

so here we're going to find the

derivative let us recall the quotient

rule because that would be the rule to

be applied here we know the derivative

with respect to X of this function of x

f of x / G of X will be equal to G of x

* frime of

xus f ofx * G Prime of x then the whole

of this / by g^ 2 of X or G of

X2 so therefore we have here y isal x /

the natural logarithm of X take note of

this the derivative of the natural log

of x is = to 1 /x well known so here

differentiating we have Dy DX is equal

to with this property here we first have

the natural log of x MTI by the

derivative of the numerator X then minus

the numerator multip by the derivative

of the denominator right then the all of

this / by the square of

the

denominator so here we have Dy DX isal

to here derivative of x is 1 1 * natural

log of x is natural log of x

minus here the derivative of natural log

of x is 1 / x x * 1 x is 1 then divided

by the square of the natural log of x

and this is the right answer thank you

for watching please kindly subscribe to

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The Quotient Rule | Differentiation


hello welcome back once again today

we're going to solve the following given

exponential equation we have squ root of

xun x = x here clearly X cannot be equal

to Z right because from the left we're

going to have 0 to the^ of 0 which is

undefined so what do we do now for my

solution we're going to natural log both

sides so here we have the Ln ofun X X

this will be equal to the Ln of X now

with the property of logarithm you know

this power here will come over here so

we're going to have the root of x * the

root of x this equals the root of

x what did I call that that is natural L

of root of x by the way okay now from

here we have the squ root of x multiply

by the natural log of you know of X same

as x^ of 1 / 2 then this equals the

natural of X so from here this power

here is coming over so we have of x * 1/

2 * the natural log of x then let's

subtract this minus natural log of x

this equal 0 so from here we have to

factor out the natural L of X and here

we have the otk of x / 2 right that is

what is left here then minus here will

be one then this equals z so from here

this implies is that the natural of x is

= 0 and from here x 2 - 1 = 0 so from

here we have to raise both

sides um to the power of E here we have

to the power of0 right so here this will

cancel this and we're left with x = e

the^ 0 which is 1 and here this is a

solution then from here let's add one to

both side we get of x / 2 = 1

so cross multiply we have the root of x

= 2 now to get rid of this root sign so

let's go ahead and square both

sides so the square we canc square roots

and here we find another solution x = 2

2 which is four and here we have the two

real solutions thank you for watching

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Interesting exponential radical equation


hello welcome back once again today we

have this Olympian ma problem we have x

-1 to^ x =

1024 here we're going to find the

integral value of x for my solution

we're going to start with the

substitution we're going to let the base

which is x -1 = M from here trying to

solve for x in terms of M that would be

x = m + 1 right because to add one to

both sides okay so the

equation will be written in terms of M

as the base which is M the power the

power which is X and that is m + 1 then

this equals 1 to

4 from here we have m^ n + 1 this is

equal to

1024 is same as 4 to the^ of

5 so from here we have m^ n + 1 this is

equal to 4 now this 5 here is same as 4

+ 1 so from here we can obviously say

that m is equal to 4 right here we have

m n + 1 thisal 4^ 4 + 1 so we can say

that m is equal 4 right okay now let us

recall that x = n +1 and therefore we

have our solution x = 4 + 1 which is = 5

and here is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

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Interesting Olympiad Math | Nice exponential equation


good day viewers welcome back to my

channel once again today we have the

following

interesting problem if Y is = the S of

3x + 5 then d y DX equal that okay

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Channel Please

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share my videos thank you

now with we have different methods of

differentiating

functions here we have a trigonometric

function okay so I'll be making a

substitution here so in this case I'll

be using the Shan rule okay so let's get

started here we have y is equal to

S of

3x + 5 okay so this is what we have I'm

going to make a substitution here I will

say let

U let U

equal

3x + 5 okay U is = 3x + 5 so now our y

will be

S of youu okay wow okay so here I'll be

different

both sides

so from here I have

du is equal to you know to differentiate

this with respect to X so we have

three DX okay and right

here I be differentiate both sides so I

have d y is equal to you know cine of U

du okay now from here if I divide both

sides by DX so I will get

du/ DX is equal to

three and to this place if I divide both

by du I will get

dy over du is equal to the cine of U

okay wow now how do we get the Y the x

value okay so let's see what we're going

to do

d y

DX is equal to if you multiply du

u i mean d y/ du

U and

du over DX together so you know this and

this will get cancel out so we left with

this Dy DX and this is what 3 * the

cosine of U so this is equal to 3 cosine

of youu okay so what is our U our U is

3x + 5 so we get d

y DX is = to 3 the cine of

3x + five and that's the answer thank

you for viewing and please make sure you

subscribe to my YouTube channel for more

interesting videos

Differentiation. Chain rule. Differentiation of trigonometric function


hello welcome back once again today

we're going to solve the following given

exponential equation for x so let's get

started from the left hand side we have

4 to^ of X and we know this is equal to

48 now we're going to make use of the

natural logarithm so let's natural log

both sides so we have the Ln of 4^ X

this is equal to the Ln of 48 so we know

the property of the logarithm so this

will come down here right and that will

become x * the natural log of 4 so this

is equal to the natural log of 48 is

same as 16 * 3 right so now this is x *

the natural log of 4 so this is equal to

you know the property of logarithm the

natural of a * B is same as natural log

of a plus the natural log of B that is

the addition rule right so this will be

the natural log of 16 plus the natural

log of 3 so here we have x * the natural

log of 4 this is equal to the natural

log of 4 S right then plus the natural

log of

3 so from here we have x * the natural

log of 4 this is equal to the property

of logarithm this will come down here so

it will become 2 * the natural log of 4

plus the natural log of

three okay so from

here let's divide div both side by the

coefficient of x which is the natural

log of four so divide both side by the

coefficient of x the natural L of 4 this

will cancel this from here so have X is

equal to split the denominator here 2 *

the natural log of 4 / the natural log

of 4 plus natural log of 3 / the natural

log of

4 so from here this will cancel this and

we have just two left so we have X is =

2 plus now this right over here we can

actually rewrite it so remember that the

natural log of a / by the natural log of

B is same as the log of a to the base B

so this right over here will be log of 3

to the base 4 and here we have the

solution thank you for watching please

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Interesting exponential equation | can you solve ?


hello good viewers welcome back once

again today we have the following

interesting problem x^ n + 1 x^ n = 4

right over here we're going to evaluate

x^ N - 1 / x^ n provided that n is not

equal to

zero now we'll be making use of this

property remember that a + b s this

equals a - b s

+ 4 * a * B now from the given equation

x^ of n + 1 / x^ n = 4 let's go ahead

and square both sides of this

equation from the left hand side will'll

be making use of this property so we

have here x^

nus 1 / x^ n then s + 4 MTI by the first

time here x n MTI the second term here 1

/ x^ n this isal 4 which is

16 so from here we have x^ N - 1 / x^ n

2 then

plus here we have just four because this

will cancel this then this is equal to

16 so from here we have x n N - 1 / x n

2 this is = 16 - 4 that will give us 12

from here let's take root of both we get

x n - 1 / x n this is equal plus or

minus < TK 12 which simplifies to plus

or minus 2 < tk3 and this is the right

answer thank you for watching please

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Can you solve ?  Nice algebra problem.


hello you are welcome here we're going

to solve this exponential equation we

have here M to the^ of 2 m this is equal

to 1 all over 2 now for us to get rid of

this two right over here we're going to

raise both sides of this equation to the

power of the reciprocal of two and we

know the reciprocal of 2 is 1 all/ 2 so

here we have M to the^ of 2 m ra power

of 1 / 2 this is = to 1 all

2^ of 1 / 2 from here we can multiply

these Powers together with the help of

one of the basic laws of IND thises

right that this is same as a^ B * C so 2

m * 1 / 2 2 we cancel 2 and we're left

with M so we have m m this will be equal

to 1 / 2 of 1 / 2 from here obviously m

= 1 /

2 but that's not all we still have

another

solution now we have M of M which is

equal to from

here 1 / 2 is the same as the square <

TK of 1 / 4

right then we have power of 1 / 2 right

over here now take note the square root

of a is same as a^ of 1 / 2 so from here

we get M to the^ of M this is equal to

here we have 1 / 4 ra power of 1 / 2

then ra power of 1 / 2 from here we get

m to the power of M this is equal to

from here we have 1/ 4 then we can

multiply this power together 1 / 2 * 1 /

2 which is 1 / 4 from here through this

manipulation

we still find out that m = 1 4 and this

is the other real solution thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Tricky Algebra problem | Nice exponential equation


hello good viewers welcome back once

again let's solve the following given

interesting radical equation we have

root x + root ofx this is equal to < tk2

now let's get started from here we can

write this as otk

ofx this is equal to < tk2 - < TK of X

now from here we're going to square both

sides but before we proceed we need to

record this property a - b 2 this is

equal to a 2 + b 2 - 2 * a * B right so

from here we have to square both side

root ofx 2 this is equal to < TK 2 - otk

x^

2 and from here they know the square and

this root cancel out we have ax

is equal to now the first time squ will

give us 2 plus the second time squ will

give us X then - 2 * 2 2 * < tk2 * < TK

X which is same as 2 < tk2 * < TK

X now from here remember we're

subtracting this term from this term so

if you if you wish to color L term from

here 2 <

tk2 * rootx will be also equal = to 2 +

x

--x from the left from here 2 < tk2 we

all know this is same Asun 8 right then

multi by root X this is equal to from

here 2 then x - - x this minus and minus

here will be plus as they are

multiplying so we have here 2 + 2 x from

here let's go ahead and square both

sides again so here we have to square

both

sides so from here we have root 8 squ

will give us 8 multip root x^2 will give

us X right then this is equal to from

here there's another binomial expansion

we need to recall a + b 2 is = a 2 + b 2

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

squ which is 4 plus the Second Time

Square give us 4 x 2 + 2 * the first

term * the second term that will give us

8 x right so here we have 2 * 2 4 * 2 x

which is 8

x so from here from the right hand side

we have 4 x^2 + 8 x + 4 this is = to 8

x now we're going to collect like terms

so we have here 4 x2+ plus from here 8 x

- 8 x that is 0 and here we are left

with 4 then this is equal to 0 Let's go

ahead and divide both sides of this

equation by 4 so here we have 4 X2 / 4 +

4 / 4 this is equal 0 / 4 so from here

this will divide out we have

x2+ here 4 / 4 which is 1 this is equal

so from here we have X2 is = -1 from

here take root of both side we get X is

= to plus orus root of1 which is equal

to plus or minus I and here we have the

solution that X is equal to plus or

minus I amazing thank you for watching

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Japanese | Nice Olympiad Math | #maths #mathematics


good day viewers today we have the

following interesting problem a 2 - b 2

= 56 Aus b = 2 then a + b = that please

if you have not subscribed to my Channel

Please

Subscribe like comment and share thank

you now we have a^ 2us b^ 2 =

56 remember we have the difference of

two squares formula here so that is A +

B * a - b is = to

56 here a - b = 2 so here we have a + b

here * 2 is equal to 56 here we'll be

dividing both side by two so this is

over two and here over two so 2 cancel

two here we get a + b = 56 divide by two

so that is equal to 28 and this is the

right answer thank you for viewing and

please make sure you subscribe to my

YouTube channel for more interesting

videos

A simple Algebra trick.


hello welcome back once again let's find

this limit so we have limit as X is

approaching two of x^ 4 - 16 / x -

2 we have limit as X is approaching 2 so

here x^ 4 we going to re write it as x^

2 2 right you know 2 * 2 here will give

us 4 then - 16 is 4 to the^ 2 2 then the

whole of this / x - 2 okay now from here

we're going to make use of the

difference of two squares remember a 2 -

b 2 is same as a + b * a minus B so make

use of that to get

limit as X is approaching

2 so we have here X2 - 4 * X2 + 4 then

ided by x -

2 okay now from here we get

limit as X is approaching 2 so now in

the first bracket here we're still going

to make use of difference of two squares

right so we get x2 minus 4 which is same

as 2^ 2 then here we have X2 + 4 then

ided by X x - 2 so in that first bracket

let us make use of the difference of two

squares formula right so there we're

going to have x - 2 * x + 2 right then

in the next bracket we have x 2 + 4 the

whole of this ided by x - 2 so from here

this we cancel this right then we get

limit as X is approaching

2 of x + 2 * x 2 + 4 right so here we

can make our direct substitution and

here we get 2 + 2 * 2 2 + 4 which is

equal to here 2 + 2 is 4 * by here 2 2 2

2 is 4 4 + 4 which is 8 and from here we

get the right answer which is

32 thank you for watching please kindly

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Interesting limit.


hello welcome back once again today

we're going to solve the following given

exponential equation we have x + < tk3

of x + < tk3 = <

tk3 here we're going to solve for the

real value of x first of all let us

natural log both sides of this equation

so here we have the natural log

of x + < tk3

of x + <

tk3 this will be equal to the natural

log of < tk3 now with the property of

natural logarithm you know this power

here is coming over here right that is

from here to here and that will

become x + < tk3 * by the natural log of

x + < tk3 and from the right this equals

the natural log of

ot3 now from here remember a can be

written as e to the power of natural log

of a right so from here x +3 we can

actually write it as e of the natural

log of x + < tk3 then multiply by the

natural log of x + < tk3 from the right

this equals the natural log of root3

from here we're going to make use of of

the lambar W function recall that the

lamb W function of x * e^ of x = x since

the power of E here is exactly equal to

the coefficient and from here we have

the same argument here right so now

making use of the Lamb W function on

both sides natural log of x + <

tk3 e the natural log of x + < tk3

will be equal to W of the natural log of

tk3 so from here this left side

simplifies to just the natural log of x

+ < tk3 and this equals the W of the

natural log of Ro <

tk3 from here we have x + < tk3 you know

the power rule here will be e to the

power of the W of the natural log of ot3

subtract ro3 from both sides and here

comes the value of x which is e to the

power of the Lamb function of the

natural log of < tk3 minus <

tk3 and this is approximately equal

[Music]

381 6 4 7 and this is a right right

answer thank you for watching please

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Interesting exponential equation.


hello good viewers welcome back once

again let's find a solution to this

Olympiad Mass problem we given that X

the first Power Plus y the 1 power is =

to 1 x the power of 2 that is the second

power + y the power of 2 is = 2 then

we're going to evaluate x^ 4 + y^ 4 many

may assume that four is the right answer

but unfortunately four is not the right

answer what do we do the first idea is

to solve for or x * y cuz that would be

very helpful here so from equation 1 we

have x + y this is equal to 1 Let's

Square both sides from the left we have

X2 + Y 2 + 2 x y this will give us 1 X2

+ y^2 is already 2 so we replace we have

2 + 2 x y this is equal to 1 when we

subtract two from both side and divide

by two we find out that X Y = 1 - 2 / 2

which is a -1 /

2 okay now from equation 2 X2 + y^2 is =

2 let's Square both

sides from the left here we have x² 2

which is x ^ 4 + y ^ 4 then + 2 into

braet x y^ 2 if you expand this this is

exactly what you're going to have and

this is equal to 2 which is 4 remember

XY is =

to1 / 2 and it is squ think about it

when we Square this we find out that x

y^ 2 is = to 1 1/ 2 2 will give us 1 4

and we multiply by

2 we find out that 2 * x y^ 2 will be 1

/ 2 so from here this is 1 /2 and we're

subtracting 1/2 from both sides so we

get x^ 4 + y ^ 4 will be = to 4 - 1 2

right okay 4 is the same as 8 / 2 then

we are subtracting 1 / two from here we

get um 7 / 2 and this is the right

answer thank you for watching please

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Olympiad Math problem | Interesting algebraic problem | #maths #mathematics


hello welcome back once again today

we're going to solve for the value of x

in the following given exponential

equation we have the cube

root of 12 to the^ of X this is equal to

2 from here we're going to cube both

sides of this equation to get rid of

this cube root sign so here we Cube this

and from the right we Cube as well so we

know the cube and the cube root will

cancel out we are left with 12 to the^

of X this is equal to 2 cubed which is

8 from here we're going to make use of

log reading we going to take the log of

both sides but the log must be in Bas 12

so here we have the log of 12 to^ of x x

to the base 12 this is equal to the log

of 8 to the base 12 but what happens to

the left hand side let us record this

property the log of a ra of B to the

base a this simplifies to be the power

of a here so and here we can see we have

similar TX here we have log a to the

base a log of a b to the base a this

simplifies to the power of a right over

here so here we have the log of 12 x the

base 12 so this will cancel out and we

are left with X so from here we get X is

equal to the log of 8 to the base 12 and

this is approximately equal to

0.836 8

to 9 and this is the correct answer

thank you for watching please kindly

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Exponential equation 2 | #maths #mathematics


hello good day everyone welcome back

once again today we're going to solve

this exponential equation involving root

of

x so from the left we have X to the^ of

root of x this equals now this base

right over here which is root of x we

can write it as x^ 1 / 2 then we have to

the^ of X so here we have X of root of x

then this will be equal to X remember we

can multiply this power together right

so we have 1 / 2 * X that will give us X

all over two okay now for my solution

here I'm going to natural log both sides

of this equation remember X cannot be

equal to Z right okay so here if you

natural log both side we get natural log

of x to the^ root X this equals the

natural log of x

x / 2 now with the property of natural

logarithm remember this is crossing over

here and this is crossing over here so

that will become root X the natural log

of x = here x 2 m natural log of

x okay from here we have root x m natur

log of x - x 2 m natural X the whole of

this equal 0 so you know natural log of

x is common so we factor that out and we

left with root of

xus x 2 then this equals 0 okay so from

here this implies that the natural log

of x = 0 or from the bracket we have <

TK x - x 2 = 0 and from here we have x =

e^ 0 which is 1 right so here we find a

solution and from here take this time to

the right hand side we get root x = 0 +

x / 2 is X /2 this is over one so we

cross multiply from there

right so when you cross multiply we get

2 * otk X is 2un X then this equal x * 1

is X now for us to get rid of this root

sign we need to square both sides of

this equation

so here first of all 2 2 is 4 root x 2

remember Square we cancel the root sign

we get just X then this equals

x² from the right hand side we can write

this as X2 then subtracting for X then

this equal 0 factor out X and here we

have x - 4 = 0 so from here we get x is

= 0 or from the bracket x - 4 = 0

implies that x = 4 but X is not equal to

Z that is not a solution here we find

another solution that x = 4 and

therefore the two real values of X that

satisfy the above equation is only one

and four and these are the right answers

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comment and share bye-bye


okay here we have theun of x + 1 is =

root3 so let's go ahead and square both

sides since we can see we have root sign

everywhere right so Square both sides

then we can see that this Square will

cancel the square root and we can see

here Square we cancel the square root so

we have here just

X2 + 1 which is equal to 3 subtract one

from both sides so we get

x2 is equal to 3 - 1 which is 2 now you

can take the square root of both sides

so here we get X is equal to plus or

minus theare < TK of 2 and here we have

the solution please subscribe to my

channel

Interesting solution  |  #maths #exam #mathematics


hello guys welcome back once again today

is circle theor right so we're going to

find the value of x here Y and Z as well

so let's see how to solve this problem

right so now let us recall this theorem

that states that angles in opposite

segments are

supplementary what I mean by two angles

are supplementary so let's assume we

have angle a and we have angle B so if

these two angles are supplementary so

this means that a + b is equal to

180° right so now here we have the

opposite angles in this segment here

here we have X and we have 102° so this

actually means that X and 102 are

supplementary angles right so we have

that X Plus

102° is equal to

180° right

so according to this theorem right so

right here we going to find the value of

x here so X here is equal to what

remember 102° is positive here so if

it's crossing over it will change to

negative so here we have

180° minus

102° right and this gives the value of x

as

78 degrees wow wonderful okay now let us

find the value of Y here so if you take

a very good look at this place you can

see that we have a straight line here so

here we have this angle 81° here we have

y so this actually mean that this y+ 81°

should be 180° because this is sum of

angles on a straight line right so here

we have

y plus

81° is equal to what 180° that is sum of

angles on a straight line so here we

have y is equal to what

180° minus 81° right and this gives

99° for y right okay now let us find the

value of Z here right so let us record

this theorem the same theorem we apply

in the first one that angles in opposite

segments are equal so the angles in

opposite M here is z and y right so

since they are supplementary so this

means that y + z isal 180° so here we

have Z Plus the value of y here is 99°

right so 99° is equal to

180° right so here we have Z is equal to

what

180° -

99° and this is giving us what 81

° and that's the answer so here we have

X is equal to what

78° here Y is equal to what Y is equal

to

99° and Z is equal to what

81° and these are the answers thank you

for watching please do well to like

comment and share my videos thank you

bye-bye

Circle Theorem 😍🤩


hello you're welcome today we're going

to solve this nice interesting equation

here we have the fourth root of x plus

the root of x and this is equal to 1 so

let's go ahead and solve for the value

of x that will satisfy this beautiful

looking equation now let's get started

let us take note of this that the end

root of a is equal to a^ 1 / n so the

fourth root of x will be X to 1 / 4 plus

theun of X will be x^ 1 / 2 and this is

equal to 1 so from here we have x^ 1 / 4

plus so here we can rewrite this as x^ 1

/ 4 in bracket power of 2 we know this

property that a^ m n is = a^ M * n we

know 1 4 * 2 we give + 2 4 which is 1 /

2 and that is the power of X here then

this is equal to 1 so it is very clear

to make the following substitution

letting 1 x^ 1 4 = K so if you if you

raise both sides to the^ of 4 we get

that X will be equal to K to the^ 4 all

right so here we have here started from

here we have k 2 then plus k subtracting

one from both sides then this will be

equal to zero right okay so from here

let us solve this quadratic equation

using the general quadratic formula so

this will be K isal to -1 that is B plus

or minus so this will be square root of

B2 is 1 - 4 a c which is -4 1 - - 4 is 5

so we have here 5 / 2 so we can see here

that K1 is = -1 + < TK 5 / 2 and K2 is =

to -1 - < TK 5 / 2 okay so here we're

going to solve for x and x is K to^ 4 so

here we have our X1 is = to K1 ^ 4 which

is -1 + otk 5 /

2^ of 4 so I'm not going to show you

guys how to evaluate this here if you

watch my previous video on how to

evaluate -1 /2 +un 5 / 2 to the^ 9 you

will find how the fourth power of this

is evaluated right so by doing it here

we have here 7 / 2 - 3 < TK 5 all 2 and

when you plug this in the given equation

this satisfies the

equation so here our X2 will be K2 2

which is -1

sorry so that would be -1 - < TK 5 / 2^

4 which should be the conjugate of this

which is 7 / 2 + 3un 5 / 2 but this root

here is an extranos root so this root

does not satisfy the equation and

therefore X is = 7 /2 - 3un 5 / 2 which

is the only satisfying root of the

equation thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Olympiad Math problem | Can you solve ?


hello welcome back once again today we

have this nice olymp mathematics problem

we're given that X to the^ X is equal to

9 so in this video we're going to solve

for x so let's get

started we know that 2 to the^ of 2 is

equal 4 and obviously 4 is less than 9

and again 3 to the^ 3 is equal to 27 and

obviously 27 is greater than 9 so we can

see that the value of x lies between 2

and 3 right yes okay so now let's get

into the main video from the left hand

side we have X to the^ X this is equal

to 9 so we're going to Len both sides of

this equation so here we have the

natural log of x to the^ X this will be

equal to the natural log of 9 right so

we know this property of logm the log of

a to the power B is equal to B the log

of a so here this power here will come

down here right so this will become x *

the log of x so this is equal to the log

of

9 now X here we can actually write it

recall that X is same as e to the^ Ln of

X we know that the base of the natural

logarithm is e so with that base here

this will cancel out leaving us with X

so here x is same as e to the power of

Ln of X so we have multip by this Ln of

X then this is equal to the Ln of 9

right okay now let us recall the lambar

W function we know the lambar W function

with the following argument or E input x

* e to X is = to X right since the power

of E and the coefficient are exactly

equal so here we can see the power of E

is the is exactly with the coefficient

natural log of x and natural log of x so

we're going to take the lamb W on both

sides so here we get the

W of e to^ Ln X the Ln X this is equal

to the W of Ln of 9 so the left hand

side with this property we simplify to

something very nice so that would be the

natural log of x which is equal to here

the W of the natural log of 9 right so

from here we're going to make a simple

base of e right we know that e to the^

of Ln of anything let's use a is just

equal to a so if you make this base of e

so the left hand side we have e to the^

of the Ln of X so this will be equal the

right hand side e to the power of the W

of the natural log of 9 so we can see

this here simplifies to X which we are

looking for and this is equal to e to

the power of the W of the Ln of 9 and

with the help of my calculator this is

finally approximated as

953 9

2802 and this is the correct answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

[Music]

Oxford University Entrance Exam | Can you solve ?


hello welcome back once again please

kindly support this channel by

subscribing to this

Channel please subscribe to this Channel

and share our videos please support this

channel by subscribing to the channel

thank you here we have this function of

X here we have f of x is = X all over x

+ 1 so we are asked to find the first

derivative of this function at x = 1

right okay now let's see how to do this

okay now here we have this function of X

here f of x = x 1 + x so the first thing

to do is to find the derivative of this

function with respect to X so we can see

that this is a rational function right x

1 + x so we call the quotient rule

so we have these functions of X U of X

and V of X so if we are given y equal to

U of X over V of X so these are both

functions of X right then the derivative

is going to be the bottom V of x * the

derivative of the

numerator minus the numerator time the

derivative of the denominator all divide

by the denominator rais to the^ of two

right

so here we have the derivative of this

function FR Prime of X is going to be

making use of this property so the

denominator 1 +

x time the derivative of the numerator

so the numerator here is X so the

derivative of x is just one right so

minus here we can see the numerator that

is X

time the derivative of the denominator

so we have it 1 + x so the derivative of

1 + x with respect to X is just one so

all divide by the square of the

denominator so 1 + x ra^ of 2

right so we have

here F Prime of X that is to simplify

this well so here we open up this

brackets 1 * 1 is 1 + 1 * X is x - 1 * X

is X all over 1 + x of 2 so this is

simplified to you know this X and this x

cancel out so we left with just 1 all

over 1 + x of 2 so this is the

derivative of this function of X here so

now we are here to evaluate the frime of

1 so here we have fime of X so we're

going to let X =

1 then we get fime of 1 is equal to 1

all over 1 + 1^ of 2 so this is equal to

1 all over you know 1 + 1 is 2 then of 2

and this is exactly equal to 1 over 4

and this is the right answer thank you

for watching way to subscribe to this

YouTube channel and share our

videos

A nice function of x. f(x) = x/(1+x), f'(1)=?


good day everyone welcome back once

again please if you have not subscribed

to my Channel please and please support

this my channel by

subscribing thank you today we have this

limit problem here the limit as X is

approaching

Infinity of this function over here 2x -

8 all divide by 6 x -

4 right if you try to make direct

substitution here we have here 2 *

Infinity which is infinity and infinity

minus 8 is infinity over 6 * infinity is

infinity infinity minus 4 is infinity so

we will have Infinity over infinity

right and this is in indeterminate form

this is not allowed in mathematics right

but this function here as X is

approaching Infinity that is as X is

getting larger larger larger and larg

larger must be

approaching a particular value

right okay now let's see what to do the

idea here is to divide the numerator

throughout by X and also the denominator

throughout by X okay let us do that and

see here this is equal to

limit as X is approaching

Infinity here we have

2x -

8 / X then again the denominator 6 x -

4 / X right so this is equal

to limit as X is approaching

Infinity you know we're going to split

here we have

2x/ x - 88 / X and

again 6 x / x - 4 /

X right so let us try to simplify here

we have X cancel X and here x cancel X

so we have this is equal to

limit as X is approaching

Infinity of 2us 8 all

x/ 6 - 4

/x so this is equal to now let's see now

8 x as X is getting larger is

approaching zero right so we can deduce

that the limit of 8 x as X is

approaching Infinity is z right so here

we have 2 - 0 over 6 here 4X as X is

approaching Infinity is approaching zero

so 6 - 0 and this equal to 2 over 6 this

simplifies finally to 1 over three and

this is the final answer thank you for

watching do where to subscribe to this

channel share our videos bye-bye

Fascinating limit


hello good day ladies and gentlemen

welcome back once again today we are

going to solve for the values of X in

the following cubic

equation we have X Cub - 1 this is equal

to 0 now let's get

started for the left hand side we are

going to write it as

follows X cubed - 1 is the same as 1

cubed this is equal to

zero now from the left we are going to

make use of the difference of two cubes

formula we have a

cubus B Cubed this is equal to a - b * a

2 + a b + b

2 so from the left make use of this

formula here we have x

-1 into braet

X2 + x * 1 which is x + 1 2 which is 1

and this is equal

0 so from here we have that X - 1 is = 0

also we have

x² + x + 1

this is equal to

Z for the left hand side here we have

just a linear equation and right hand

side we have a quadratic

equation so for this equation here x is

= to 1 and this is a

solution now for the equation here we

are going to make use of the general

quadratic formula we have

here x is equal to B and here a is equal

to the coefficient of x² which is 1 B is

equal to the coent of X which is 1 and C

which is the constant is also one right

and the formula we are going to make use

of it's B plus or

minus the square root of B 2 - 4 a

c all/ 2

a so from here we have X is = to B which

is 1 plus or minus the root of B 2 is 1

2 which is 1 - 4 a c 4 * a * C is the

same as 4 * 1 * 1 and that is 4 then

over 2 a which is 2 * 1 which is

2 so from here we have X is is = to -1

plus or minus theare < TK of -3 then 2

and from here we have -1 plus or minus

of -3 is the same as I < tk3 then all 2

and from here we come up with the

solutions we have one real solution X is

= to 1 we also have two complex

Solutions x

= -1 + I < tk3 / 2 and the other complex

solution X is = -1 - I < tk3 over two

and these are the solutions thank you

for watching please kindly subscribe to

my channel also like comment and share

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Another well-known cubic equation | Interesting problem | #maths #olympiad


hello welcome today we're going to solve

this simultaneous equation

within some seconds how do we solve it

remember X and Y belongs to real numbers

okay now let us add the equations

together so we have X Plus X here so we

get 2X

and here we have minus y plus Y is zero

so is equal to 2 plus 4 here we get six

so here x is equal to 6 all over 2 and

that is equal to 3 okay how do we solve

for y let's subtract the equations so we

have x minus X is zero minus one minus y

is minus 2y is equal to you know 2 minus

4 here so we get what negative two here

Y is equal to negative 2 over negative 2

and Y is equal to one therefore X is

equal to 3 and

Y is equal to one thank you for viewing

please subscribe to this channel

Best way to solve simultaneous equation within seconds


hello good viewers welcome back once

again today we have an interesting

problem we're given that 72 ra^ of 1 - 2

x = 8 and here we're going to evaluate 3

^ of X let's get started now from the

left hand side of this equation we shall

make use of the following algebraic

property remember when we have a to the

power of m- n this is same as a to the

power of M / a to the^ of n right so

here we have 1 - 2 X so we can write

this as

72 of 1 / 72 of 2 all X then from the

right hand side this is equal to 8 now

from here let us cross multiply remember

this is 8 over 1 so here 72 1 is 72

right then then if you multiply by 1 you

will have 72 and from here

72 2 x * 8 so that will give us 8 *

72 2 all X then from the right this is

equal to 72 now from here let's go ahead

and divide both side by 8 so divide here

by 8 divide here by 8 here8 cancel 8

here then from the left we have here 72

ra the^ of 2 all over X then this is

equal to 72 / 8 that will give us a 9

now from here let us raise both sides to

the power of X right so here to the

power of X and here to the power of

X now notice that from the left hand

side these Powers here can actually

multiply right that is if we recall that

a to the power of B in bracket of

this is same as a^ B * C so when we

multiply 2 x * X you know X will cancel

X then we left with what 72 then ra^ of

2 which is equal to 9 to the^ of X now

from here let us change the sides of

this equation so we have 9^ X which is

equal to

72 of 2 now from here we shall take

square root of both sides so here theare

< TK of 9 ^ x this is equal to the

square < TK of 72 2 right okay now from

the left recall that 9 is the same as 3

S then with this x that will become 2x

right then square root the same as 1 2

that is the power of 1 2 then from here

remember that when we have the square

root of x² in this form this is actually

the absolute value of x right so from

here this is same as the absolute value

of 72 right and that will give us

72 so here from the right we have here

72 now from here notice that this power

can multiply together right so we have 2

x * 1 / 2 2 will cancel 2 and we left

with 3 to the power of X which should be

equal to 72 and this is the right answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Very interesting homemade algebra problem | #maths #mathematics


hello good day viewers once again

welcome back again in today's video

we're going to solve this exponential

equation right 4^ x * 16 3 X - 1 is = to

8 here we ask to solve for the value of

x right okay please kindly subscribe to

my

channel also like comment and share okay

what do we need to do here first the

first thing we need to do here is to

make sure that from both sides the whole

bases are

equal right and then we can equate the

powers and here we have four 4 is the

same as 2^ 2 we have 16 here 16 is the

same as 2 ^ 4 here we have 8 8 is the

same as 2 to^ of 3 and do not forget

that a to the^ of B in bracket of C is

the same as a^ of B * C

okay now let's see how this will help in

this solution here we have to express as

2 ^ 2 then X multip by here 2^ 4^ 3 X -

1 this is equal to 2^ of 3 right okay so

now we make use of this law here that

will give us 2^

2x * here 2^ 4 into braet 3x - 1 this is

equal to 2^ of 3 so you know from here

let us recall this property once again A

to the^ B * a the^ c this is same as

since we have equal bases a^ b + C so

from here we have 2^

2X + 4 into bracket 3x - 1 this is equal

to 2^ 3 right okay from here here we

have equal bases from both sides so we

can equate the powers so here we have 2X

+ 4 into braet 3x - 1 this is equal to 3

so here we have 2X + open this bracket

with four we have here 12x - 4 this is

equal to

3 so from here we have 2X + 12 x that is

14 x - 4 this is equal to 3 so from here

we add four to both sides we get 14 X is

= to 3 + 4 which is 7 so from here we

divide both side by the coefficient of x

which is 14 here

14 so from here we get X is equal to you

know 7 over 14 is the same as 1 over 2

because 7 here 1 and 7 here two and here

we get 1 / 2

or5 and this is the solution thank you

for watching please kindly subscribe to

this channel also do where to like

comment and share bye-bye bye

Interesting exponential equation. #olympiadmathematics #maths #education


hello good viewers welcome back once

again let us find a solution to the

following radical equation we have here

root of x * root 3 this is equal to

2 we have root of x * root of 3 this is

equal to 2 now let's Square both

sides now when you square both sides

here the square and this root sign we

cancel out and we're left with x * root

of 3

right so x * root of 3 then this is

equal to 2 2 which is 4 remember we

solving

for x right so from here we divide both

side by the coefficient of x which is

root3 right so divided by root3 from the

right hand side 4 /

root3 we Sol that is

root3 okay now from here root3 we cancel

root3 and here we come up with the

solution X is = 4 all over < tk3 or we

can rationalize this by multiply byun3

over3 that is 4/ <

tk3 multi

byun3 over <

tk3 now from here we get 4 * < tk3 which

is 4 < tk3 over < tk3 * tk3 isun 9 and

that will give usun of 9 which is 3 so

here we come up with a solution X is =

43 / 3 and that's the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Radical equation | We need to be careful with the root sign. #maths #mathematics


hello good day everyone welcome back

once again today we have this

interesting Olympiad mathematics problem

we're given the following quadratic

equation X2 + 7 x + 49 is = 0 and here

we'll be evaluating x to the thir

power now let's get

started from the original equation we

have that

x² Plus 7

x + 49 is =

to0 now from here let's move this second

term and the constant here to the right

remember both are positive here so if

they are crossing over they will become

negative so here we have

x² is equal to this will become

-7x - 49

right okay so here we have that X2 is

equal

to7 x - 49 this will be very

important now let's proceed so far we

have gotten x² to be equal to

-7x - 49 remember the question asks for

X to^ of 3 so for us to get X to to the

power of 3 at the left hand side we need

to multiply both sides of this equation

by x to

the to the 4 power right so here we have

x^ 1 * X2 this will be equal to x ^ 1

into braet --x -

49 so now let us recall this property a

m a n remember they have the same base

so this will be a^ n + n

right so here we have the same base we

need to add the power so here we get X

to^ of 3 this will be equal to remember

x^ of 1 is just X right so here we have

x * -7 x will give us -7 x 2 then minus

x * 49 will give us

49x so we can see what is going on here

so here we have X cu op this will be

equal to remember this is

-77 right then this x² here check back

this is the value in terms of x -7x - 49

so we replace that

with 7 x - 49 then we have -

49x so from here we get X cubed is = to

-7 * -7 x is that 7 X oh sorry this

is -7 oh okay so -7 * -7 x will give us

POS 49x right then -7 * -49 we give us+

343 then minus remember we have 49x here

so obviously 49x - 49x is Z and here we

have our value for x cubed which is

equal to

343 amazing thank you for watching

please kindly subscribe to my channel

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Olympiad Math problem | Can you solve ?


hello good viewers welcome back once

again let's find solution to the

following Olympiad ma problem we have

here Ro otk X +un Y this is equal to 1

from here we get otk x * y this isal to

1 and right over here we're going to

evaluate this expression here X2 + y^ 2

now from this equation two

here when you square both sides we have

that x * y will also be equal to 1

right okay now let's see how to solve

for X2 + Y 2 from equation 1 we have otk

x + otk y this is = one now let's go

ahead and square both sides from the

left hand side we'll be making use of

this

property a + b 2 this is equal to a 2 +

+ b 2 + 2 * a * B now make use of this

property we have the first 10 squ which

is otk X2 which will give us X plus the

second 10 square which is root y^2 that

will give us y plus 2 * the first term

otk X MTI by the second term here root y

then this is equal to

1 so from here we have have x + y + 2

now otk x * otk Y is the same as root of

x * Y and this is equal to 1 now from

the equation two we have the value for

root X Y which is 1 so here we

substitute we have x + y + 2 * 1 which

is 2 this is equal to 1 so from here we

subtract two from both sides and we get

x + y this is = to 1 - 2 which is a

-1 now from here let's go ahead and

square both sides once again because

we're solving for X2 + Y 2 now when you

square both

sides from the left you know this

property here we we have x² + y^ 2 + 2 *

x * y this is equal to now -1 to the

power of an even number will give us 1

now do not forget from here x * Y is = 1

so let's go ahead and substitute we get

here X2 + Y 2 + 2 * 1 right x * 1 x * Y

is 1 so 2 * 1 will give us 2 and this is

equal to one now subtract two from both

sides we arrive at the solution that x^2

+ y^2 is = 1 - 2 which is -1 and this is

the right answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Chinese | Interesting Olympiad Math | #maths #mathematics


wow hello welcome back once again please

subscribe to my channel if you have not

done that please like comment and

share okay so here we have the ramanan

problem here we given that x - y = 5 and

also the root of x +un of Y is equal to

5 here we're going to find the value of

values of X and Y right okay I'm going

to solve this problem like this so I'm

going to

let this squ < TK of x = a so here squ <

TK of y = b so Square both side we get X

is = a 2 and here Square both side Y is

= b 2 right okay so from equation 1 x -

y this is going to be a 2 - b 2 and this

is equal

to 5 and from here we

have of X which is a + of Y which is B

this is also equal to 5 right so this is

our equation one in terms of A and B and

here equation two okay so from the

equation one from the left hand side we

have difference of two squares right so

that is going to be a + b MTI by a minus

B this is equal to 5 right so but in

equation two we have the value for a + b

which is 5 so right over here we replace

it with 5 then multi a minus B and this

is equal to 5 so divide both sides by 5

and here by

5 okay so here we get a minus B which is

equal to 1

right okay so we can call this equation

three right so we're going to find the

value of a and b using using the

equation two and equation three right

and we're going to solve the two

equations

simultaneously so here we have a + b is

= to 2 is equal to 5 right so this is

equation one equation two I mean and

here A minus B is equal to 1 and this is

equation three okay so we're going to

use the elimination method so let us add

both equations together to eliminate B

Because B has opposite signs right so

here A+ a which is 2 a this is equal to

5 + 1 is 6 so from here we divide both

side by

2 and here we get a is equal to

3 right so we can solve for the value of

B from any of the

equations

so we can make use of equation two or

equation three so let us make use of

equation two a plus b is given us five

here we have a is = 3 so here we have 3

+ B is = 5 so subtract from both side we

get B isal 5 - 3 which is 2 right so let

us recall that X is = a 2 and also B is

X I mean Y is equal to B2 right so here

x is = to a 2 which is 3 S and here Y is

= b 2 which is 2^ 2 so from here we get

X = to 3 S which is 9 and also here Y is

= 4 and this are the solution to the

problem thank you for watching please

kindly subscribe to my YouTube Channel

please like my videos comment and also

share bye-bye

Sir Srinivasa Ramanujan'a like problem


hello welcome back once again today we

have this interesting exponential

equation and this is an Olympia

mathematics problem so we're given from

the left hand side that the square root

of

x root of

x is equal to X right so in this video I

would like to use log rithm right so

we're going to take the log of both

sides so here we have the natural log

ofun

xun X this will be equal to the natural

log of x so we know this property of

log the natural log of a to the power of

B is equal to B MTI the natural log of a

right so we apply this property here so

this will become the root of x * the

natural log of root of x then this is

equal to the natural log of

x okay we know that the root of x is

same as x^ 1 / 2 right so we have here

root of x * the natural log of x^ 1 / 2

square root means to the^ of 1 / 2 then

we have here minus natural log of x this

one here then this is equal to

zero okay so from

here we use this property again so this

will become 1 / 2 * X Asun x /

2 then times the natural log of x and

here minus the natural log of x this

isal

to0 so from here we can see the natural

log of x is common so let's Factor it

out so we have the natural log of x into

bracket so here we are left with square

root of x / 2 then minus here we're left

with one then this is equal to zero so

from here we have these two product s is

equal to Z so which implies from here

that either the natural of X is equal

to0 or the < TK x / 2 - 1 is = 0 from

here we take a common base of a right so

here we have X will be equal to e to

the^ of 0 so here we get our solution

that X is = 1 right okay and from here

take1 to the right hand side to become

positive we getun x / 2 isal to 1 then

cross multiply this is over 1 x * 1 isun

x so this isal to 2 * 1 is 2 so from

here for us to get rid of the root sign

we are going to square both sides of the

equation so this Square will cancel the

square root and here we have our second

solution X is = to 2^ 2 which is 4 so

here we have found the two R solutions

that satisfy the given

equation thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting Olympiad Math | Can you solve this equation ?


hello welcome back to another video here

we have this interesting Olympia matth

problem please if you not subscribed to

my channel kindly support my channel by

subscribing please like comment and

share here we're given that A + B is

equal to 2 a² + B2 is = 5 and right over

here we asked to evaluate a cubed + B

Cubed right okay we're going to make use

of this identity in this video a cubed +

b Cub is equal to a + b Cub - 3 * a * B

A+ B right so we have a problem here and

the only problem we have here is the

value of a * B which is not given in the

original problem but right over here A

plus b and A+ B the value is already

here as two right so how do we figure

out the value of a times B right

okay we're going to do that by also

recording this identity a² + b² = a + b^

2 - 2 a b right so from here we're going

to make substitutions here a s + b s is

5 so here we have 5 this is equal to a +

b 2 that is 2 2 and here we have - 2 a b

so here we have 5 is = to 2 Square which

is 4 - 2 a b so here col terms we're

going to have 2 a b is = 4 - 5 which

is1 so divide both sides by

[Music]

2 so from here we have a * B is equal to

-1 all over 2 cool so we're going to

find the value of a cub + B Cubed from

here right so here we have a cubed + B

Cubed which is = to a + b Cub - 3 * a *

B A + B so this is going to be equal to

a + b that is 2 so we have here 2 cubus

3 * a * B is1 / 2 then times a + b so

that is 2 right so this is going to be

equal to 2 cubed is 8 and here 2 cancel

2 so here we have -3 * 1 which is pos3

and right over here we get this to be

equal to 11 therefore the value of a cub

plus b Cub is 11 thank you for watching

please if you find my video interesting

kindly support my channel by subscribing

also please like comment and share

bye-bye

Interesting Olympiad Math.