hello welcome back once again today we

have this interesting quadratic equation

so we're going to solve for the values

of X so we have from the left we have X

to the^ 4 this is equal to from the

right we have

81 right so on the left hand side we can

rewrite it as X to the^ 2 then squared

since we know that 2 * 2 is 4 4 then

this will be equal to 81 is same as 9

s now subtract 9 squ from both sides so

we get x ^ 2 2 - 9^ 2 this is equal to Z

from here we're going to make use of the

difference of two squares formula

remember that a

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

Aus B so from here this is the

difference of 2 squares formula so we

going to apply this rule here so the

base here is X2 so we're going to have

X2 plus the base here is 9 we have 9

then multiply by the subtraction that is

the difference of the two bases that

would be

X2 - 9 then this is equal to0 so from

here we have these two products which is

equal to Z so this implies from here

that either X2 + 9 is = 0 or x² - 9 is =

to

0 so from here this 9 is positive if it

is crossing over it will become negative

so we get x² is = to

9 take the square root of both sides you

get X will be equal to the plus or minus

the root

of9 so here we get X is equal to of1 is

Iota so we get plus or minus aota then

sare root of 9 is 3 so we can see this

solution is

imaginary then over to this here we have

x² is equal to this is negative crossing

over it becomes POS 9 take the square

root of both sides we get X is equal to+

or minus the < TK of 9 then from here we

get X is = + or - 3 so here we have the

solutions X is = to plus or minus 3 is

the only two R Solutions and X is equal

to plus or minus Iota root of nine which

is Iota 3 this is the two imaginary

Solutions and the two real solutions

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Interesting quartic equation | Can you solve ?


good evening viewers welcome back to my

Channel today we have this simple

problem if a is equal to negative 1 and

B is equal to negative 2 then a to the

power 3 minus B to the power 3 over a

minus b equals Dutch

please if today is the first day of

visiting this Channel please subscribe

to this Channel and turn on the

notification Bell so that you'll be

notified anytime we upload new videos

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

solve this problem now we're going to

use the following identity

that a raised power 3 minus B raise

power 3 is the same as a minus B

times a squared

plus b squared

plus a b okay now

we have a raise power 3 minus B raise

power 3 all over a minus B to be equal

to

a minus B times a squared plus b squared

plus a b close bracket then all over

a minus B okay

now we're going to cancel out a minus B

from the numerator and from the

denominator here so we are left with

what a squared plus b squared

plus a b okay now recall that a is equal

to negative 1 and B is equal to negative

two so we have a raised to power 3 minus

B raise power 3 all over a minus b equal

to you know a squared plus b squared

plus a b and this is equal to what

negative 1 squared plus negative 2

squared

plus negative 1 times negative two okay

we're going to evaluate it from here so

we have this you know negative 1 squared

2 is even so this is one plus four then

plus you know negative times negative is

positive then one times two is what two

and this is equal to seven and this is

the right answer to the problem thank

you for viewing and please make sure you

subscribe to my YouTube channel for more

interesting videos

Algebra trick


hello good viewers welcome back once

again a lot of student find this very

difficult to evaluate some say that the

answer is one thinking that 2^ 2 to^ 2^

2 is the base here and therefore this ra

to the^ of 0 is one beter will tell you

the truth the answer to this is not

one now let's get into this

video solution here we're going to start

like this we have 2^ 2^ 2^ 2^ 0 this

will be equal

to here we have to take this Force 2

the^ of 0 so that will become 2^ 2^ of 2

now 2^ of 0 here is 1 right okay next we

take this 2 the^ of 1 so here we get 2^

of 2 then 2^ of 1 is 2 right

then next we take this 2 to the^ of 2

and therefore we get two ra the power of

you know 2^ 2 is 4 and therefore 2^ 4 is

2 * 2 * 2 * 2 four places right and that

will give us 16 and therefore the right

answer to this problem is 16 thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye I

An easy way of evaluating tower of powers of numbers


hello everyone welcome back once again

please and please if you have not

subscribed to our Channel please kindly

subscribe subscribe to this Channel and

share our videos today we're going to

evaluate or simplify this expression

here 49 of 2 - 1 all over 50 so let us

try to rewrite the numerator over here

so here we have this is equal to 49 2 -

1M as 1

2ide by what 50 right okay now we can

see that we can make use of difference

of two squares

formula namely a s minus b 2 is equal to

a + b * Aus B right so now making use of

this property here so we have up here 49

that is the base here + 1 then 49 - 1

all IDE

50 so this is equal to 49 + 1 is 50 * 49

- 1 is

48 oride by 50 see 50 cancel 50 here and

we are left with 48 and this is the

right answer please if you're enjoying

this exciting videos please please

kindly subscribe to our Channel and

share our videos

bye-bye

Difference of two squares is the best way out. (49²-1)/50 = ?


hello good viewers here we have an

interesting exponential problem here to

solve we are given that the fifth root

of three of nus 5 is = to

729 here we ask to find the value of n

please if you have not subscribe to my

channel kindly subscribe to this channel

like comment and share here we have the

fifth

route of 3 of N - 5 is equal to

729 let us recall this property when we

have the end root of a of B this is same

as a of B all over n right so from here

this is going to be equal to 3 of N - 5

all over 5 this is equal to 729 is the

same as 3 ra to the^ of 6 so from here

we can see that we have equal bases

right so we're going to equate the

powers or the exponents so here we get N

- 5 all over 5 is equal to 6 so cross

multiply from here we get N - 5 is equal

to 5 * 6 that is 30 add 5 to both sides

we get n is equal to

35 and this is the right answer thank

you for watching please do where to like

comment and share and don't forget to

subscribe to my channel bye-bye

lnteresting exponential equation.


hello good viewers welcome back once

again today we have a very interesting

equation here and this is an infinite

Tower right we have x^ x^ x^ x^ x^ X and

so on that this is equal to 2 then we're

going to figure out the value of x but

before we solve this equation there's

something we need to take note of when

we have x x x x X and so on let's assume

that this is equal to n now there's a

condition that this infinite Tower here

will converge exactly to the right hand

side of the equation and the condition

is that 1 a is less than n and n is less

than a right so if n exceeds the value

of e then this infinite Tower cannot

converge

now let's talk about the right hand side

of that equation let's bring the

equation down right over here we have x^

x^ x to the power x and so on that this

is equal to 2 now we all know that e is

approximately

28 and we can say that e is greater than

two right or we can say 2 is less than a

so which actually mean that this

infinite Tower will converge now we can

go ahead and and solve our

question so here we have x^ x^ x^ x^ X

and so on that this is equal to two so

now we can write this as X the power of

then we can put this in this form then

we can put the powers x to the power of

x to the power of x to the power of X

and so on then this is equal to two now

think about this power here this power

here is exactly

what is at the left hand side of this

equation which is exactly equal to two

so meaning we can replace this with two

right therefore we get x^ of 2 then this

is equal to 2 now we take root of both

sides then we get X is equal to the sare

< TK of 2 and this is the right answer

think about it check your calculator and

continue the infinite Tower < TK of 2 <

TK 2 < TK 2 to the^ two to and so on

this

will converge to two and this is the

right answer thank you for watching

please cly subscribe to my YouTube

channel also like comment and share

bye-bye

An infinite Tower | Exponentiating Exponents 😍


good day welcome back once again today

we have these nice systems of equations

here we have a cub minus B Cub which

isal to 3 we have Aus b = 1 and here

we'll be solving for a * B right okay

now from the second equation here we

have a - b which is equal to 1 so let's

go ahead and CU both sides of this

equation now from the left hand side let

us right as a - b * a - b * a -

b and this is equal to 1 Cub which is 1

so from here let us take this product

first right so here we have a - b and

into bracket a - b * Aus B that will

give us a

s + b 2 - 2 * a * B right

then this is equal to

one okay now from here let us take this

product here so from here a * A 2 will

give us a cubed then plus a * b 2 will

give us a b

2 then minus a * 2 a here will give us 2

a 2 B okay now we have here B * a s will

give us a² B

then B * POS b 2 will give us B Cub B *

2 a will give us positive 2 a b 2 then

this is equal to 1 so from here we have

a

cubed we have here minus B

Cubed then we have

here we have here 2 a² B minus sorry

that is -2 a s b - a b a s b that will

give us -3 a² B then plus a b 2 + 2 a b

2 will give us 3 a b 2 then this is

equal to 1 so from here we have a cued

minus B Cubed then here we have minus 3

a which is common here so we're going to

have here a then minus B then this is

equal to

1 okay now let's go back to the original

problem remember a cubus b Cub is = 3

and we have a minus b = 1 of which we

are solving for a * B right okay so from

here a CU - B Cub is 3 so we have 3 - 3

a b multi by a - b which is 1 then this

is equal to 1 so from here we have 3 - 3

a b

which is equal to 1 okay now from here

we have 3 a b is = to 3 - 1 which is 2

so remember we're solving for a so let's

divide both by coefficient of a which is

three so cancel this here then we have

here a * B is = 2/ 3 and this is the

right answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Interesting algebraic problem | Can you solve ?


hello good day viewers welcome back once

again today we have a very nice Olympiad

math problem we're given x + 1 /x = 7

and here we're going to evaluate this

expression x² ID by this quatic

expression here x^ 4 + X2 +

1 now let's start with this expression

here x^ 2 / x ^ 4 + X2 + 1 1 now let's

divide the numerator and the denominator

by x ^ 2 so here we have x^ 2 / x^ 2

then the whole of this x^ 4 + X2 + 1 all

/ x^ 2 from the numerator this will

cancel this and we left with one then

divided by let's split this denominator

here that will be x^ 4 / X X2 + X2 / X2

+ 1 /

X2 so from here we have 1 / here x^ 4 /

X2 will give us

X2 plus here this will cancel this we

have 1 and here plus 1 / x² of which we

can actually rewrite as 1 / x ^ 2 + 1/

x^ 2 then + 1 right now let's go back to

the original equation we have x + 1 /x =

7 now let's try squaring both sides of

this equation and from the left we shall

make use of this formula remember this

equal a 2 + b 2 + 2 * a * B for this

here our a is going to be X and our B is

going to be 1 / X and with this we have

x 2 plus we have here 1 /x 2 + 2 * x * 1

/x and this equal

49 so from here we have x^ 2 plus here

we're going to make use of this a over

b^ n thisal a^ n / b^ n so from here 1 2

is 1 / x 2 right and here plus this

cancel this we have 2 here this equals

49 subtract two from both sides we have

X2 + 1 / X2 =

47 now remember this is the expression

we are evaluating and if you take very

good look at the bottom here we actually

have x² + 1 / x² right over here so let

us add one to both sides of this

equation to get the exact value of this

denominator right so add one to both

side from here we get x2 + 1 X2 + 1 this

will be equal to 47 + 1 which is 48 so

therefore the expression we asked to

evaluate equals 1 all over 48 and here

is the right answer thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Interesting Olympiad Math | Here is the trick


hello good viewers welcome back once

again in our today's video we're going

to solve the following given simple

exponential equation I made this video

to let some people who still find this

difficult to understand how we can make

use of the logarithm when solving this

kind of exponential

equation so from here we have 5 to^ of

xal 10 so let let us use the logarithm

on both sides so here we have the log

log of 5^ of X this will be equal to the

log of 10 now let us record this

property of the

logm which states that the log of a to

the power of B is same as B MTI the log

of a where this is in base 10 and this

is also in base 10 or they can be in any

base so

here moving this x to the back here so

that will become X X MTI the log of 5

then this is equal to the log of 10 now

since we're solving for x let's divide

both side by the coefficient of x the

coefficient of x is a number here which

is the log of 5 so divide here by the

log of five and equally divide this side

by the log of five so this will cancel

this from here and we can see that our X

is fre X = log of 10 / the log

of five which is approximately equal

to

[Music]

55807 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

How to solve simple exponential equation using logarithm


hello good viewers welcome back once

again today we're going to solve for the

values of X in the following given

exponential equation we have e^ x = 3X

now let's get started we have here e to^

x is = to 3 x right okay now from here

let's go ahead and divide both sides of

this equation by 3 so here we have a^ x

/ 3 this is equal to 3x divided by

3 okay from the right this three we

cancel this three and from the right we

have X = from the left here we have e^ X

multip by we can write it as 1 /

3 now from here let's go ahead and

divide both by e^ of X so here x / e^ X

this equals e^ x * 1 / 3 / e^ X

so from here this will cancel this and

here we have x * 1 over e^ X this = 1 /

3 now let us record this property 1 a^ n

= a n so from here this is x * e^ of --x

this = 1 / 3 now we'll be making use of

the Lamb W function remember that the

lamb W function with the following

argument x * e^ x =

x this means that the power of E here

must be exactly the coefficient right so

the power needs to be equal with the

coefficient and here the power of E here

ISX but the coefficient is POS X so

let's go ahead and multiply both by1 and

by so doing we get X x *

e^x this will Beal to -1 / 3 now we can

make use of the Lamb function so let's

lze both sides so we

getx *

e^x then this is equal to the lambat

function of-1 / 3 so from here this will

evaluate tox then this is equal to W of

-1 / 3 from here multiply by1 we get X

is = to W with a subscr n then -1 / 3

okay now this is the most interesting

part of this the d w function has some

Branch points and some points usually

give us the real solutions to the

equation right so and that is at n = 0

and at n = -1 so here we have the first

solution X1 = W with Subs 0 of -1 all

over 3 and this is approximately equal

to

0613 right and we have the other Branch

C which is W which sub -1 then -1 / 3

which is approximately equal to 1.

512 13 4 5 5 and these are the real

values of X in the given equation thank

to my channel also like comment and

share bye-bye

Oxford University Entrance Exam | Can you find the two real roots ? | Lambert W Function application


hello welcome back once again today we

have this interesting equation and we're

going to solve for the value of x that

will satisfy this equation it's

obviously one can clearly say that one

is a solution yes indeed because 1^ 1 is

= to 1 so therefore we can see that X is

equal to 1 now what about X is equal

to0 can that be a solution no because 0

^ of 0 is not equal to 1 rather 0^ 0 is

undefined but how can we prove that X is

equal to 1 is the only

solution now let's see now let's start

by introducing the natural log on both

sides so remember our equation is X to

the^ X is equal to 1 now let do natural

log both sides so we have the Ln of x to

the^ of X this is equal to the Ln of 1

now with the with the property of

logarithm this power of X will come down

here so that will become X MTI the Ln of

X this is equal to the natural log of 1

is z okay now we can see we have this

product x * natural log of x which is

equal to Z so this implies from here

that either X is equal to0 or the

natural log of x is equal to0 and we

know for the natural log of x to be

equal to0 x has to be equal to 1 right

so from here we have X will be equal to

e to^ of 0 right and here we get X is =

1 but 0 is not a solution right

therefore the only solution here is X

isal to 1 thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Oxford University 🎓 Entrance Exam | Can you solve ?


hello good viewers welcome back once

again today we have an interesting

exponential equation problem we are

giving that 2^ of otk X this is equal to

8^ of X now let's go ahead and solve for

the values of X we have here 2 of root X

this is equal to in bracket it8 is can

be expressed as 2^ 3 then power of X now

let's call this

property M power of n ra the power of K

this is equal to m^ of n *

K so from here we have 2^ ofun X this

will be equal to

2 of 3 * X which is 3 x so from here let

us recall this when we have b r power of

let's say m and this is equal to let's

say we have B ra the power of um let's

say we have one or we have K so let's

let's use that so since we have equal

bases from both sides we can easily

equate the powers that M will be equal

to K right okay so from here we can see

that we have equal bases right so we can

equate the powers that is this square

root of x this will be equal to 3 x now

let's Square both sides to get r of this

root sign so we have root

x s this will be equal to

squared now from the left hand side here

the square and the root will cancel out

we have X will be equal to Here let us

record this

MN the power of B this will be same as

m^ of B * n^ of B so here we have 3

2 * x 2 so from the right we have 32

which is 9 * X2 this will be equal to

X now let's subract X from both sides we

obtain 9 x^ 2 - x this will be equal 0

factor out X we have X into braet 9x - 1

this is equal Z so from here this

implies that x is = 0 or 9 x - 1 this is

equal 0 so from here we get X is equal

to0 or we have here x will be equal to 1

all over 9 right so here we have come to

the end of today's video that in the

above exponential equation x is = to

0 and x is = to 1 over9 and these are

the solutions thank you for watching

please kindly subscribe to my YouTube

channel also like comment and share

bye-bye

Interesting exponential equation | Not just one value of x ! | #maths #olympiad


hello welcome back once again let us

solve for the value of x in the given

radical equation here we have otk of 2x

this is equal to < TK 2 * < tkx now in

our first step here we're going to

square both sides of this equation to

get rid of this root signs so here we

have root of

2x 2 then this will be equal to

root of 2 * root X all 2 as we know

Square will cancel the square root so

here we have 2X is equal to here Square

we cancel the square root we have 2ot of

X okay now from here let us go ahead and

divide both sides by two so we have 2X /

2 this will be equal to 2un x / 2 so we

know 2 cancel two here 2 can 2 here and

we have X is = to otk X in our next step

let us go ahead and square both sides

again to get rid of this root sign so

here we have

X2 this will be equal to root of

x^2 so here we have X2 is equal to

square cancels the square root so we

have X now subtract X from both side we

get x^2 - x this is equal to0 now X is

is common we factor out X so we have x

-1 then this is equal to Z from here

either X is equal

0 and this is a solution or x -1 is = 0

which when we add one to both side we

get another solution that X is equal to

1 so therefore in the above equation we

have two possible solutions X = to 0 and

x = to 1 thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Easy and interesting radical equation.


hello welcome back once again today we

have this interesting Olympiad

mathematics problem so in this equation

we're going to solve for M M is an

integer we have 3^ m + 2^ m this is

equal to

35 okay now before we proceed let us

take note of the following

functions now let us Define this

function f of

M is equal to

3 to the^ m + 2 to the^ of M now the

graph of this function looks exactly

like

this so the graph of this function looks

exactly like

this if I'm not mistaken so it is an

increasing

function if I'm not mistaken at all so

the gra have somehow look like this

which we can see obviously that this is

an increasing function why the bigger

the value for M the bigger the result

right so if m is equal to 1 if m is

equal to 1 we get our outcome 3^ 1 + 2^

1 which is 5 then if m is equal to 2 you

can see it is an increasing function

right okay okay now if you look at the

right hand side we have just

35 which we know that that is just a

straight line so if you check along this

Y axis here this vertical Y axis so we

going to have 35 at some points here

right so the graph for this one will

pass through like this which shows that

there's only one intersection point so

therefore for integer M there will only

be one real

solution now let's go back to the

original problem so we have 3 to the^ m

+ 2 to the^ M this is equal to

35 so here we have 3 to the^ m + 2 to

the^ M this is equal to now 35 is same

as

27 + 8

now we have here 3 to the^ m + 2 to the^

M this is equal to 277 3 to the^ 3 then

Plus 8 as 2 to the^ 3 so I think from

here we can compare and from here this

implies that our solution is out and we

have this only real I mean is that not

integer yes of course so the only

integer solution for m is three and this

is the correct answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

India | Nice Olympiad Math problem | Can you solve ?


good day viewers welcome back to my

channel again today we have our second

class on this problem so this is problem

number eight the S mathematical problem

here we have if a all Bal 2 What is the

value of 4 B all over a options are here

okay let's get started

solution now let us bring down this

expression 4 b a so you remember this

expression is the same as 4 * B all a

okay so this is what will be

evaluating okay now remember the

equation we have here is a/ b = 2 okay

now I'm going to take the reciprocal of

both sides so here I have a/ B close

bracket1 is = to 2

are1 so now remember if I want to

rewrite this according to this negative

power the power will become positive

okay so reciprocal of numbers for

example the reciprocal of 1 over 2 is

equal to 2 the reciprocal of 3 is equal

to 1 over 3 and so on so here I'm going

to have B all over a in bracket

to the positive power is equal to 1 all

2 okay now remember B all a to the first

power is still the same so here we have

B all a is = 1 2 now I'm going to

multiply both sides by four in order to

obtain this expression which we are

asked to evaluate okay so here I have 4

* B all a is = to 4 4 * 1 all over 2 so

from here this is over one so 4 * B is 4

B all over a * 1 here is a and here we

have this over one two here one and two

here is two so we get 4 B all a is equal

to two and this is the right answer so

we go to the

option and we see that the option

C is correct thank you for viewing and

please make sure you subscribe to my

YouTube channel for more interesting

videos

SAT math Problem 8.


hello good viewers welcome back once

again let's find a solution to this

Olympiad exponential equation problem

we're given that 27 ^ ofun x this is

equal to 1 allun

x before we proceed let us make the

following substitution we're going to

let rootx = Y when you square both sides

we say that X is = to Y 2 right so now

this equation in terms of Y is going to

be 27 ra^ of Y this is equal to 1 / y

now the next thing is to get rid of this

power right over here so what do we do

now what we're going to do is to raise

both sides to the power of the

reciprocal of Y right now from here we

have 27 to^ of

y^ of 1 / y this is equal to 1 / y power

of 1 / y right okay now from the left

hand side we're going to make use of

this property when we have a^ of b c

this is same as a^ of B * C so here we

have

27 of y * 1 / y this is equal to 1 / y^

1 / y now we can see here that Y and Y

we cancel out and here we have 27 ^ 1

which is 27 this is equal 1 / y to the^

of 1 /

y okay now let us recall that 27 is

actually equal to 3 to the power of 3 so

the left hand side we rewrite it in

terms of the powers of 3 and here we

have 3^ of 3 this is equal to 1 / y^ of

1 / y so from here we can see this

pattern here this implies that 1 / Y is

equal to 3 we have the bases here 1 / y

= 3 and from here we have the powers 1/

y = 3 so with this pattern this implies

that 1 / Y is = 3 and when we cross

multiply we get 3 y this is equal to 1

right that is when you cross multiply

that is y * 3 3 y this is equal to 1 * 1

which is 1 so when we divide both sides

by

3 3 cancel 3 and from here we get Y is =

1 / 3 okay now let us recall

again recall that X is = y^ 2 and from

here remember Y is = 1/ 3 so here x is =

to 1/ 3 S and this will give us 1 / 9

and this is the solution thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Indians | Interesting Olympiad Math | #maths #mathematics


hello good viewers welcome back once

again today we're going to find the

square root of a sodic

expression we have theare otk of 6 + <

tk32 now this is equal to the square

root of 6 + otk of 16 * 2 right with the

following property root a * B this is

equal to root

ATI root B right so this is going to be

equal to the square root of first of all

we are going to rewrite six as the sum

of two numbers that will be 4 and

2 then plus the root of 16 which is 4

then multiply by the square < TK of 2

right so from here this is equal to the

square root of 4 is the same as 2 2 plus

here two we can rewrite it rewrite two

as the square of a number and that will

be the square of <

tk2 then plus 4 is the same as 2 multi

by 2 then * <

tk2 okay now let's recall the following

algebraic

property that has to do with a B

expansion right so we have here A + B 2

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

so now let's us compare the right hand

side of this equation from here with

what we have under this root sign so if

we compare this together we have a s and

here we have something squar plus b s we

have another something squared plus 2

MTI by a * B and here we have + 2 MTI by

something time another something right

so if we compare this we find out that a

is equal to 2 y b isal to < tk2 right so

from here this is equal to the square

root

of 2 + < tk2 to the^ of

two and from here this is equal to the

absolute value of 2 + < tk2 right which

is exactly equal to 2 + < tk2 and this

is the right answer thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Chinese junior Olympiad Math | The square root surdic expression. #olympiad #exam


hello welcome back once again today we

have this interesting radical equation

and here we'll be solving for the values

of

X so let's get started from the left

hand side we

have the square root of

x then

plus the < TK of x - 9 this is equal to

9 now

this term right over here let's move it

to the right hand

side so we are left with square < TK of

x - 9 then this will be equal to 9 - the

sare root of

x in our next step let's Square both

sides so here we

have the S < TK of x - 9 squared this is

equal to 9 - < TK

x

2 okay so from the right hand side let

us make use of this identity we know

that a minus b

s this is equal to a

2 + b 2 - 2 * a * B so we're going to

make use of this identity at the right

hand side but here we know Square we

cancel square

roots and we have x - 9 this will be

equal to we have 9

2 + otk x

2 - 2 * 9 * otk of

X so from here we have x - 9 this is = 9

s is 81 Square cancel square root we get

+ x - 18 root of

x we can see that X is positive at the

left hand side and X is also positive at

the right hand side so we need to cancel

them out we all know that if we subtract

X from both sides X will be eliminated

from both

sides so what do we have next so we have

from the left hand side -9 this is = to

81 - 18 otk of

X so from here let us make the right

hand side the left hand side and the

left hand side the right hand side so

move this term to the left and take this

term to the

right so we're going to have 81 don't

change any sign - 18K X this is equal

[Music]

okay so we have this minus this to give

us this so This actually mean that this

is also equal to this minus this so we

have 18 < tkx is = 81 -

9 from here we have 18 tkx is equal to

that will give us 90 divide by 18 so

divide here by 18 and divide here by 18

18 will cancel 18 so we have otk X is =

90 18 is 5 Square both

sides so Square we cancel square root

and here we get our solution X is = 25

and this is the only real solution that

satisfy the equation thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Germany | Nice Olympiad Math | Interesting radical equation | #maths #education #puzzle

#maths #education #mathematics #puzzle #learning


hello good day viewers welcome back once

again today we're going to solve for the

value of x in the following given

exponential equation we have square Ro

TK of 8^ of X this is equal to

16 okay now we're going to get rid of

this root sign here and how do we do

this we're going to square both sides of

the equation

from the left we know the square and the

root sign we cancel out and we have 8 to

the^ of X this is equal to in the

bracket there 16 is the same as 2 to

the^ of 4 then rest power of 2 now from

the left again we need to at least

obtain the same basis from both sides of

this equation right

so here we have 8 to be written as 2^ 3

then X this is equal to 2^ 4 squar now

we're going to make use of this

property a of B of C this is equal to a

of B * C so from here this equals 2^ 3 *

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

is

8 so since we have equal base from both

sides we can equate the power so this

implies that

3x is equal to 8 from here if you divide

both sides by

three then we come up with a solution

that X is =

8/ 3 and this is right answer thank you

for watching please kindly subscribe to

my YouTube channel also like comment and

share bye-bye

Exponential equation | #maths #mathematics


hello good viewers welcome back once

again we're given that a = -8 and here

we're going to evaluate the square root

of a squ I have a reason for saying that

99% fails this please forgive me if my

English is wrong because my main

language is not

English okay now let's get started we

have here a = 8 if we try evaluating

this like this square root of a s = the

square < TK of -8 s and from here this

square and this square root will cancel

out and we still arrive at -8 obviously

this is very

wrong so what do we do let us take note

that square < TK of x to the^ of 2 like

this is not equal to the square root of

x all

squar for instance assuming the equation

was the square root of a all squared

then that will be a different case this

will be -8 I agree but here we are

giving a s under this root sign so let's

take note that the square root of x to

the^ of two in this

form signifies the absolute value of

x and and we know the absolute value of

x equal --x when X is less

than 0 right and the absolute value of x

equal positive X when X is greater than

or equals 0

amazing so with from the given problem

the square root of a squar

equals the absolute value of a and

remember a = -8 right so we have that

this will be equal to the absolute value

of 8 and we know the absolute value of

negative number result to the positive

of that number so the right solution is

eight thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

99% keeps failing this !


hello welcome back once again let's

solve for the real value of x in the

following given equation we have X Cub +

x Cub = 162 here we'll be solving for

only real value of

x now from the left hand side of this

equation we can see that this term here

and this term here are exactly the

same let us take note of this when we

have a + a this is same as 2 * a which

is 2 a right and here we have X Cub + x

Cub which should be same as 2 * X cubed

right then this

equals

162 from here let's divide both sides by

two this we cancel this we have a x

cubed equals here if you divide this by

two that will give us

81 now from here remember we solving for

the real value of x therefore we're

going to take the cube root on both

sides so from here we get X = the cube

root of

81 which obviously we can simplify right

remember 81 is the same as 27 *

3 so from here we can see that when we

have the N

root of M * K this is same as the n root

of

M the N root of K so therefore here will

be same as the cube root of 27 MTI the

cube root of 3 from here we get X = the

cube root of 27 is 3 then multiply by

the cube root of three and this is the

right answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Junior Olympiad Math


hello good viewers welcome back once

again today we have this tricky math

problem we have x^ 60 + 1 x^ 60 = 47 and

right over here we're going to evaluate

x ^ 30 + 1 / x ^ 30 from the given

equation we can actually write it as x

to the power of 30 * 2 + 1 all X to the^

of 30 * 2 and this is equal to

47 from the left hand side here let us

take note of this when we have a^ of B *

C we can actually Express this as a to

the power of B then ra power of C so

from here we get X the power of 30 2 + 1

x ^ 30 2 this is equal to

47 now from the left side let us take

note of this a 2 + b s can be written as

a + b all squ then - 2 * a * B so from

the left making use of this so we have

x^ 30 right then plus 1 / x ^ of 30 you

know why remember we have everything

here squared but here we have one/

something squared what happens to one

here actually this we can express this

as 1 X all to the^ of 30 right so here

we have this 2 then - 2 * x ^ 30 * 1 / X

the^ of 30 then this will be equal to

47 from here this cancel this right so

we have here x ^ 30 + 1 / X 30 or 2 - 2

this equal

47 add two to both side we get x^ 30 + 1

/ x^ 30 2 this equal

49 take root of both sides we get the

right answer to the question that x^ 30

+ 1 x^ 30 = plus orus < 49 which is 7

and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Tricky Algebra Math


[संगीत]

स्ट पी

[संगीत]

[संगीत]

[संगीत]

स्लिट पीटस

[संगीत]

[संगीत]

ब

[संगीत]

Can you solve for x without logarithms ?   #maths #education


hello welcome back once again today we

have this interesting equation and here

we'll be solving for the values of x

from the left hand side we have x to^ 2

to^ 3 this is equal to from the right

hand side x^

2^ 2 right now we're going to simplify

each each exponents from both sides so

we start like this 2 to^ 3 from here

is equal to 8 right and here 2^ 2

here is equal to 4 so from the left hand

side we have x to^ 8 this is equal to x

to^ 4 now move this term to the left we

get x^ 8 - x^ 4 this is equal to 0 we

can see x^ 4 is common so let's Factor

it out so get x^ 4 into bracket X 8 x^ 4

will give us x^ 4 - x^ 4 x^ 4 will give

us 1 and this is equal to0 so this

implies from here that either x^ 4 is

equal 0 or x ^ 4 - 1 is = 0 from here

take the square take the fourth root of

both sides so we get X = fourth root of

0 is still zero right so this is our

first solution and here we're going to

make use of the difference of two

squares but we need to rewrite the

equation as follow x 2 2 we know 2 * 2

is 4 - 1^ 2 this is equal to 0 so making

use of the difference of 2

squar we know this is equal to a + b *

Aus B so that will become X2 + 1 * X2 -

1 this is equal to Z so from here we

have two products which is equal to Z so

this implies that either x² + 1 is = 0

or X2 - 1 is = 0 so from

here subtract one from both sides so we

get x2 is = to -1 and you add one to

both side we get x2 is = 1 taking the

square root of both sides so we get X =

Plus orus theun

of1 and here X = plus orus the root of 1

so we know according TOA theot of1 is I

which is Iota so this is equal plus or

minus I and here is equal to plus or

minus 1 so in the above equation we have

the following solutions for X so x is =

0 x is = + or minus one and for

Imaginary Solutions X = plus or minus

Iota thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Harvard University 🎓 Entrance Exam | Can you solve ?


hello welcome back once

again in the following equation we're

going to solve for the value of x and

here we're given that 0 is less than x

and x is less than 1 so going to solve

for the value of x that we satisfy this

equation within this range 0 and one so

let's get started from the left hand

side you have

256 to the^ of X this is equal to from

the right 1

X now let's raise both sides of this

equation to the power of the reciprocal

of the power of

256 so we know the power of 256 is X so

its reciprocal is 1 / X so raise both

sides to the power of 1 / X and raas

this side to the^ of 1 / X we know x * 1

/ X is 1 right so here we have just

256 and this is equal to 1 /

X ra the^ of 1

/x fortunately

256 can be expressing index power of 4

as 4^ 4 so from the left hand side from

the right I mean we have 1

/x we have to the ^ 1 /x this will be

equal to 4 to^ 4 so now we can compy as

follow so we have the base is equal to

the base or the power is equal to the

power so from here this implies that 1

/x is = 4 put this over one and cross

multiply so here we

have x * 4 we give us

4X this is equal to 1 * 1 will give us 1

divide by the coefficient of x which is

four four we cancel four so from here we

have the solution X is is equal to 1 / 4

or

0.25 and here we have the

solution thank you for watching please

comment and share bye-bye

India | Nice Olympiad Math | Can you solve ?


hello welcome back once again today

we're going to solve this equation we

have from the left hand side x * x * X

which is X cubed right so this is equal

to x + x + x which is 3 x so move this

time to the left so we get X Cub -

3x this is equal to Z we can see from

here x is common so we have X into

bracket X2 - 3

then this is equal to Z from here this

implies that either X is equal to Z and

indeed this is a solution or x^2 - 3 is

= to 0 add 3 to both side we get x² is =

3 take root of both sides so here we get

X is = to plus or minus < TK of 3 so

here we have the solutions so three real

solutions exist for equation x = 0 and x

= plus orus otk of 3 thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Interesting Olympiad Math | Can you solve ?


hello welcome back once again today we

have another interesting systems of

equations we have a + b equal to 1 we

have the product of A and B is equal to

2 and here we'll be evaluating a 2 - b 2

interesting right now let's get

started from equation one we have a + b

this is equal to 1 now let's go ahead

and square both sides of this equation

okay now from the left hand side let us

make use of this

property that a + b s this is equal to

in terms of aus B that will be a - b 2 +

4 * a * B amazing property right so at

the left hand side we be making use of

this property and that will become a - b

2 + 4 * a * B this is equal to 1 2 is 1

okay from the original problem remember

a * B is equal 2 so we replace here with

2 that will become a - b 2 + 4 * 2 which

is 8 this is equal to 1 subtract it8

from both sides we get a - b 2 this is =

to 1 - 8 is -7 from here take root of

both side we get a minus B is equal to

plus or minus < TK of 7 which is equal

to plus or minus remember of1 is I then

we have otk 7 okay okay now the we ask

evaluate a s minus b 2 is simply the

difference of two squares right which is

factorized as a + b * a minus B so you

can see the reason why we solve for a

minus B already from the problem we

given that A + B is equal to 1 and from

here we have found Aus B to be equal to

plus orus IUN 7 so we can easily

evaluate that expression Now by

multiplying the value for this and the

value for this together so we have here

a 2us b 2 will be equal to a + b that is

1 * a minus B that is plus or minus IUN

7 and from here we get the right answer

that this equals plus orus I root of 7

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Interesting Olympiad Math | Algebra problem


hello welcome back once again today I

have cooked another nice problem here

and today we're going to find the sum of

the six powers of A and B we give you

that a + b = 2 a 2 + b² = 3 and here

we're going to evaluate a to ^ 6 + b^ of

6 okay the first thing to do here is to

find the product of A and B I think that

will be very helpful right so from

equation one we have a + b = 2 so we cut

to square both sides then from the left

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

which is 4 from the given equation a s +

b² = 3 so we get 3 + 2 a b this = 4

subract three from both side we get 2 a

= 1 and from here we get A8 * b = 1 / 2

amazing so what do we do

next I will adice us to First find the

sum of the thir power of A and B right

that is a cub + B Cub so here a cubed +

B Cubed is given by the following

property or formula that this = A + B

Cub - 3 * a * B into bracket A + B so

you can see the reason why a * B's value

is very helpful right so you can see it

has come up here then this equals from

the given equation A + B is 2 so 2

cubus 3 * 1 over 2 then times a + b

which is 2 from here we get 2 cubed

which is 8 this cancel this and we left

with minus 3 this equals 5

beautiful then what of a ^ of 6 + b ^ 6

remember we can write this as a cub

2+ B Cubed 2 but before we proceed take

note a 2 + b 2 = A + B is a wonderful

Property 2 - 2 * a * B with this

property

here we call a Cub a we call B Cub B and

from there we derive that this equal a

cub + B Cub then squar - 2 a cub B Cub

the same as 2 I mean a * B all

cubed so from here this equals a cub +

BB is 5 right so we have 5^ 2 - 2 into

bracket here A = 1 /2 then we Cube it

right so from here this = 5^ 2 which is

25 - 2 * this will be 1 / 8 right this

equal 25

1us that would be 1/

4 remember 25 is the same as 100 / 4

we're trying to make equal denominator

and from here we can subtract Den

numerator 100 - 1 is 99 / 4 and here is

the right answer

amazing thank you for watching please if

you really enjoy this content please

kindly subscribe turn on the

notification Bell so that you be

notified whenever I upload the next

video thank you bye-bye

Can you solve this Oxford University 🎓 Math problem ?


hello good viewers welcome back once

again please kindly support my channel

by subscribing also like comment and

share today we're going to simplify this

sodic fraction here we have ro3 + 1 over

the conjugate plus ro3 - 1/ conjugates

so we asked to simplify this right so

let's get started now the first step

we're going to find the L let me let me

put it that way of this and this and

that is actually multiplying them

together

right and then the next step the product

of this if you divide it by this

denominator here we're going to have <

tk3 + 1 CU this and this will cancel out

and then we multiply this numerator

byun3 + 1 which is actually the square

of < tk3 + 1 right here we have a plus

sign you know if you divide this by this

we left with3 -1 and if you multiply by

this that is actually the square of <

tk3 - 1 right so from here we have < tk3

+ 1 if you square this we are going to

have 3 + 1 + 2 < tk3 then plus if you

square this we're going to have 3 + 1 -

2 < tk3 then all over do not forget that

down here we have to make use of the

difference of two squares right so here

this is going to be the < tk3 2 - 1 2

I'm very sure we're very familiar with

this when we have a + b * a minus B this

is same as a s - b 2 right so from here

we have this is equal to you know this

and this will cancel out and here 4 + 4

which is 8 then all over you know this

and this will cancel out we have 3 -

1 so from here we have 8 all over two

and that is equal to 4 and this is the

right answer thank you for watching

please do wa to subscribe to my channel

also like comment and share bye-bye

simplification of surds. |Rationalization #maths #education #mathematics


hello good viewers welcome back once

again today we're going to solve for the

values of X in the following given

exponential equation we have x^ of X

this is = to 4 -

2x^ of X many may think that the best

way to solve this problem is to equate

the bases since we have equal

powers that is letting x = 4 - 2x

yes of course luckily a

solution will come up that is 4 all over

3 and that is IR rational

solution what if there are other

Solutions how do we figure them out now

when I try plotting the graph of the two

different functions of

X and G of x equal 4 - 2x^ X I found out

that these two graphs

have two intersection points and one is

an integral solution the other is a

rational solution just as we found here

how do we solve for all real values of X

in this given equation now let's get

started here we have X the^ of X this is

equal to 4 -

2x of X let go ahead and divide both

side by X X power of X and here by x^ of

x from the left hand side this way

cancel out to be one we have here now

let's use this property a^ B over M the

power B this is = a/ m to the power of B

so for here we have 4 -

2x over X then to the power of

X so here we have 1 is equal so when you

split this denominator here we have 4

/x - 2 then power of X okay now we have

some important things to discuss

here if x is even then 4 /x - 2 can be

equal to 1 right so this is our case

one because 1 to the power of an even

number is equal to one then the second

case to be considered

here

when the base here is odd that is if the

base here is

-1 that is 4 /x - 2 to be equal to -1

then X can also be even right because -1

to the power of an even number will give

us one right so if x is even the the

base here can be equal to 1 or

-1 so from here when you add two to both

side we get 4 /x is = to1 sorry 1 + 2

that is 3 right and from here cross

multiply we get 4 is = to 3x and from

here when you divide both by 3 we get X

is = to 4 / 3 which is the rational

solution and from here when you add 2 to

both sides we get 4 /x is = to1 + 2

which is 1 when you cross multiply we

get X is = 4 and here we find the

solutions X is = to 4 and also X is = to

4 over 3 and these are the solutions

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Germany | Olympiad Math | Interesting exponential equation | #maths #mathematics


hello good day everyone welcome back

once again to my YouTube Channel please

kindly support my channel by

subscribing please subscribe to this

YouTube channel and share my videos

subscribe to this Channel please and

share my videos here we are giving this

radical equation right we have the

square root of otk of X +qu root of X is

equal to otk of X right so we are asked

to find the value or values of X so here

we have the equation the sare root of <

TK x

+ < TK X is = to < TK of X right so

obviously we can see our tral solution

that X is equal to 0 because Zer satisfy

this equation right but let us go ahead

and work it out okay now the first thing

I'm going to do here I'm going to

let the sare root of x to be equals K

right so right here I will have X = kqu

of two if you square both sides right so

here I have the square root of you know

K plus K is equal to K right so here we

have the square root root of 2 K is

equal to K so I'm going to go ahead and

square both sides of this equation so

here I have this squ is equal to this

squ so we have to eliminate this root

sign and this way so here we have 2 K

isal to K of 2 so you know we can make

the right hand side to be the left hand

side and left hand side to be the right

hand side so here we have K2 is exactly

equal 2 K so now we're going to subtract

2 K from both sides so here we have k^ 2

- 2 K is equal 0 so the common factor

here is K so we Factor it out K Open

Bracket K - 2 is equal to 0 right

so here we have k isal 0 or k - 2 = 0 so

we have here K = 0 or k isal 2 right so

from here we can go ahead and find the

values of X so here we have when

K is equal to0 x = to k 2 which is 0 S

and that is 0 so this is our false

solution

solution so

when k = 2 x = k 2 which is equal to 2 2

and this is equal to 4 and this is our

second solution so we can see that the

equation has two real solutions Z and

four thank you for watching do wa to

subscribe and share our videos

An interesting radical equation.


hello welcome back once again today

we're going to solve this nice Olympiad

problem we have this exponential

equation 3 to^ < TK of X this is equal

to 27 to the of

X we know that from the right hand side

27 can be expressed in index power of

three we know this is equal to 3 * 3 * 3

in three places which is equal to 3 to

the^ 3 so from the right from the left

hand side we have 3 to the^ < TK X this

is equal to so 27 is 3 cubed then ra of

X so from here we have 3 to the^ sare ro

of X this is equal to we know that 3 * X

is 3 x so we get here 3^ 3 x we can see

we have the same denominators from both

sides so we're going to equate the

powers so this implies from here that

theun of x isal to

3x so Square both sides so we get S of X

is equal to 3x and we're going to square

both sides so from the left hand side

Square will cancel square roots we have

X is = to 3x2 is 9

x² so let's change the sides of this

equation so we have this the same as 9 x

is = to X so here subtract X from both

sides so we get 9

X2 - x this is equal to Z we can see X

is common here so Factor it out and we

have 9x - 1 is equal Z so from here this

implies that X is equal 0 or 9x - 1 is =

0 add one to both sides we get 9x is =

to 1 and divide by the coefficient of

x so 9 cancel 9 and here we have X is =

1 / 9 so here we have our

solution x is = 0 and x is equal to 1 /

9 thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Interesting exponential equation | can you solve ?

#maths #mathematics #education #puzzle #learning


hello good viewers welcome back once

again today we're going to solve for the

real value of x in the following

exponential

equation notice that this equation is an

Olympian math problem we have here x^ of

X the^ of 3 this is equal to 36 here let

us solve for the value of x we have here

x power of x^ 3 this is equal to 3 six

right now the idea from here is to raise

both sides of this equation to the power

of three right now a very smart question

here we come in this form why did I

raise both sides of this equation to the

power of three or why three not any

other number

right now anytime you encounter this

type of

question this is how you will start

solving it and to raise both sides to

the power of any

number we are considering the power of X

right over here for example if we have X

power of X let's say power of n let's

say this is for general and this is

equal to let's say capital letter N we

are going to raise both side to the

power of M that is is the power of this

x right over

here okay now from here we all know this

property is a very nice property by the

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

a of B * C we can still interchange the

powers as a to the power of C then power

of B now from the left hand side here we

have X power of

x^ 3 *

3 and this is equal to in the bracket

here 36 is 6 2 then power of 3 from here

we have here x Cub * 3 as the power of X

right over here now we can interchange

the power we have here x power of 3 then

x^ of 3 this is equal to 6 of 2 * 3

which is

6 so from here this

implies that X Cub is equal to 6 we can

see the base and we can see the powers

the bases and the powers corresponds

right X cubed as the base here six as

the base here x Cub as the power here

and six as the power here so here we

have x 3 this is equal to 6 so if you

take the CU cube root of both sides we

get here x is equal to the cube root of

6 and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Koreans | Olympiad Math | Interesting exponential equation | #maths #mathematics


hello good day everyone welcome back

once again today we're going to solve

the following given Quantic equation we

have m^ 4 = 5 here we're going to find

both the real and complex values of M

what if m is imaginary we can also find

that now let's get started for the left

hand side we can write that as M to the^

of 2 then of two you know why this 2 *

this 2 will give us four and let's

subtract

5 and that will be equal to zero now

take note of this we can write a as the

square root of a in this form then

squared so five here we can rewrite it

as square root of 5 squ right so here we

get

M2 squ then minus here root of 5 then

squar and this is equal to zero the

reason why I'm doing this is so that we

can factorize this equation if we recall

that this is just the difference of two

squares which is equal to a + b * Aus

B with this idea we can factorize this

expression at the left hand side of this

equation which is going to be

m² + < TK 5 * m 2 - < TK 5 and the whole

of this equals z so from here this

implies that we have

M2 + < tk5 = 0 and we have m^2 - < TK 5

= 0 from here take positive root 5 to

the other side we get m^2 =

root of 5 and from here take root 5 to

the other side that will become posun 5

right and take

note when we are looking for the square

root of a square root of a number a this

is same as the fourth root of

a right so from here let's take root of

both side we get M = plus or minus

square root of NE

5 and from here M = plus or minus < TK

of square < TK of

5 from here we can actually rewrite this

as plus or minus < TK of -1 MTI root of

5 from here we already have plus or

minus the fourth root of 5 which is a

real answer

right so from here we get M = plus minor

otk of -1 is I and ofun 5 is 4 root of 5

so here we come up with the solutions we

have the imaginary Solutions Plus I 4ot

of 5 and I 4 root of 5 and here the real

solutions here plus or minus the four

root of 5 and these are the right

answers thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Nice quartic equation


good day

everyone welcome back to my Facebook

page this is chid online M class today

we are giv the following sodic

expression we are asked to evaluate

this okay let's see how to solve this

problem

now this is in bracket we have the first

one square < TK of 448 + < TK of 2024

ultip by < TK of

448 - < TK of

2024 so if we try to expand this like

multiplying whatever that is in the

second

bracket by this plus multiplying Again

by this that will take a little bit time

right okay now let's see the tricky way

to solve this problem now we will be

making use of the difference of two

squares formula right so here we

go so we have that A + B * Aus B is = to

a 2 - b 2 right okay now if we carefully

look at the left hand side of this

equation so we can see that it look very

similar to what we have to evaluate

right

so here we have a + b and here we have

otk of 448 + < TK of

2024 here we have a minus B here we have

square < TK of

448 minus < TK of

2024 so which actually mean that this

expression is the same as here a s so we

have square root

of 4 0 48

squ right here minus b 2 here we

have square otk of

2024 squared right okay now this is

equal to here we all know that square

cancel out square root so we are left

with 4 0 48 minus here we have square

cancel square root so we are left to two

0 2 4 so if we subtract this we will

have

2024 and this is the right answer thank

you for watching please do wa to like

comment and share bye-bye seeing you in

our next video

surds.   Evaluation of surdic expression


hello good day viewers welcome back once

again here I'm going to solve this

problem again using a different method

from the previous video I made

right in this video I'm going to make

use of the demov

theorem the problem here is x + 1 / X =

tok3 we are going to compute the value

of x power of 2025 + 1 /x to^ of 2025

right now now here we have x + 1 /x

right so we have a positive sign in

between these two terms right so which

actually mean that we're going to make

use of the following demov theorem the

one of x^ of n + 1 / x^ of n is = to 2 *

the cosine of n Theta right so remember

that the power of X here is one right so

here we're going to let

N = 1 and we get x + 1 /x is = to 2 *

the cosine of theta right now remember

from this equation here x + 1 / X =

root3 so here we replace this with root3

and that is equal to 2 * the cosine of

theta right so divide both side by two

we get cosine of theta = < tk3 / 2 and

from here we get Theta isal the are COS

of < tk3 / 2 and this is equal to < / 6

right so now the expression here we

asked to evaluate x^ of

2025 + 1 /x the^ of 2025 is going to be

equal to

you know here we already have Theta as

Pi all 6 right which means the general

formula for this problem here we have

gotten as x^ of n + 1 / x^ n is = 2 *

the

cine of n * < all 6 right so here we're

going to let

N = 2020 5 and from here we now

get 2 * the cosine of

2025 * < 6 and this is equal to Z and

that's the answer thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Application of Demoivre's Theorem.   #maths #education


hello good viewers today we're going to

find the square root of this expression

involving X and Y 6.25 x^2 + 5x y + y^ 2

but before we proceed you know I'm a fan

of algebraic identities there's an

identity we're going to make sure we use

here a + b

s this is equal to a

2+ 2 a b + b s so this is actually

telling us that we need to at least get

an expression that will look exactly

like the expression at the right hand

side of this equation so I will know

that the square root of that will be the

sum of A and B

squ corresponding to X and Y with their

coefficients right okay now this is what

we're going to do first of all observe

that

6.25 is equal to

625 all 100 right which reduces

to if you use 25 that is 25 all over 100

25 that is 4 and this is equal to 5 all

2 the whole of this s so for that we're

going to write we have the square root

of here we have 5 / 2 we can put we can

attach X and then we Square

everything then Plus

here we have to at least have the

coefficient of x and y as 5 2 but before

doing that we need to multiply by two

then we put the five as 5/ 2 then we

have X Y then for here we have y^ 2

right okay the reason for putting this

two * 5 over 2 we need to get something

very similar to what we have at the

right hand side and if you carefully

look at this which we have right over

here we have something squared and here

we have something squared plus 2 *

something and here we have plus 2 time

something then plus something squ and

here we have something squar so this is

actually telling us that this will be

the square root of now we pick the one

the first term here that is 5 over 2 x

then plus we pick the last 10 which is

squared which is Y and the whole of this

is squared we know what is going to

happen here so this is the absolute

value of this or we can easily just

cancel this since we have expression

right over there so this is going to

give us 5 all over 2 x + y and this is

the right answer thank you for watching

please kindly subscribe to my YouTube

channel also like comment and share

bye-bye

A square root problem | Can you find the square of this expression? | #maths #olympiad


hello good viewers welcome back once

again today we're going to solve for the

values of X in the following given

exponential equation when we see this

type of equation all we need to think

about is to natural log both sides right

so let's go ahead and do that so from

the left we have the natural log of 5 ^

of X2 this will be equal to the natural

log of 7 to the^ of X now with the

property of logarithm you know the

natural log of a to the^ of B will be b

m the natural log of a right so here the

power of 5 here is X2 and that will

cross over so here we have X2 * the

natural log of 5 this will be equal to x

* the natural log of 7 now the term at

the right hand side let's move it to the

left hand side so we have x^2 * the

natural log of

5us x * the natural log of 7 this is

equal to Z so from the left we have a

common factor and that is X then we

factor that out we left at x * the

natural log of

5us the natural log of 7 and this is

equal to 0 so from here this implies

that X isal to 0 or from the bracket x *

the natural of 5 minus the natural log

of 7 is also equal to Z so add natural

log of 7 to the both sides so we get x *

the natural log of 5 this will be equal

to the natural log of 7 so remember

we're soling for X so we need to divide

both side by the coefficient of x which

is the natural log of five right by the

natural L of five so here this cancel

this and here we're left with xal the

natural log of 7 all over the natural L

of 5 and this is a solution but there's

something we need to recall recall that

when we have the natural log of a all

over the natural log of B this can be

written in in logarithmic form as the

log of a to the base B right so this

actually mean that our answer can be

written as the log of 7 to the base 5

also right so we have solution x = 0 and

x = log 7 to the base 5 thank you for

watching please kindly subscribe to my

channel thank you once again thank you

thank you bye-bye

Olympiad Math | Interesting exponential equation | #mathematics #maths


hello welcome back once again to my

YouTube Channel please and please

support this channel by subscribing to

this Channel please subscribe to my

YouTube channel and share my

videos right here we are given that A +

B = < tk2 A minus B = < tk3 we asked to

find the value of a * B in this solution

I'll making use of this

property we have this property in

algebra that a b is equal to the square

of A +

B minus the square of a minus B all

divide by 4 so if we Square this and

square this when we subtract them and

divide by 4 we are going to have a * B

right so we are going to make of this

identity right over here so in our

solution we have that a b is equal to

the square of a + b here we have a + b

is = < tk2 so here we have <

tk2 2 minus the square of aus B here we

have Aus B is = < tk3 so we have here <

tk3 2 all divide by four so this is

going to be no the square root and the

square cancel out we are left with 2

minus the square root and the square

cancel out we are left with 3 all over 4

you know 2 - 3 is = -1 all over 4 and

this is the right answer thank you for

watching please do well to subscribe to

my YouTube channel and share my

videos

An interesting problem. Watch out the way I solved this problem


hello good viewers welcome back once

again today we're going to simplify this

expression of number here we have 9 with

power of 49 and 48 right okay let's get

started we given 9 ^

49- 9 ^

48 / by 9^ of

48 okay now from here 9 to the^ of 49 we

can write it as 9 ^ 48 + 1 right then -

9 ^ 48 then divided by 9^ of

48 okay now 9^ of 48 + 1 we can actually

rewrite it right then we'll be making

use of this basic algebraic identity

when we have a^ of m + n this is same as

a^ of M

* ^ n right so 9^ 48 + 1 we can rewrite

it as 9^ 48 * 9^ 1 then- 9^ 48 divided

by 9^ 48 so we can see that from this

numerator here we have a common factor

of 9 to the^ of 48 then when we Factor

it out we are left with 9 ^ 1 here which

is 9

minus 9 48 9^ 8 we left with 1 here then

the whole of this ID 9 to the^ of 48 so

from here this and this we cancel right

and we are left

with and we're left

with 9 - 1 and that is equal to 8 and

this is the right answer thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Can you find out the trick ? No calculators please !

#maths #learnhowtosolvetheexponentialequation #learning #mathematics #automobile


hello welcome back once again today

we're given this crazy looking equation

x^ of aota is equal to Iota we know that

Iota according to Ila is equal to the

square root of1 right so here we're

going to solve for the value of x that

we satisfy this

equation Iota is the imaginary unit

right if you're new to it it is equal to

theare root of -1

but do not be surprised X is a real

number is that possible so let's see now

we need to take note of the following we

know Iota in exponential form is equal

to e to the power of Iota Pi / 2 so is a

very nice identity and again we're going

to use this logarithmic rule a to the

power of B in exponential form we can

write it as e to the power of the power

B MTI the Ln of the base a right so the

left hand side of the equation which is

X to the power of aota we can rewrite it

as e to the power of Iota time the

natural log of x and the right hand side

iot we're going to rewrite it as e to

the power of I pi ID 2 so the equation

x^ of I is = I we

become e to^ i * the natural log of x

this is equal to e to the of I Pi / 2 so

we can see we have obtained equal bases

from both sides so we're going to equate

the powers so this implies from here

that I * the natural log of x X is = to

I

Pi / 2 so we can see we can cancel I

from both sides and then we have the

natural of X is = to / 2 so make a

simple base of e we can see that X is

equal to e to the power of Pi / two wow

amazing so X is finally real and that's

is the correct answer so in the above

equation x = e to the power of PK / 2

and this is the correct answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Oxford University 🎓 Entrance Exam | Could you believe that x is real  ?


hello welcome back once again today

we're going to solve the following

radical equation we have here square

root of 3x which we can rewrite as 3x

all the^ 1 / 2 then this equals cube

root of 4X which we can rewrite as 4x^ 1

/ 3 now we're going to raise both sides

to the power of a particular number and

how can we figure out what the number is

the LM of 2 and 3 is 6 right so let's

raise both sides to the power of 6 so

3x^ 1 / 2 all ra to^ of 6 this equal 4X

^ 1 /

3 all power of 6 with this property a^ b

to^ c = a^ b * C we can multiply 1/2

with 6 and from there we get 3x in

bracket then 1/ 2 * 6 will give us 3

right then this equals here 4X in braet

1 3 * 6 that will give us

2 so from here 3x cubed will give us 27

x cubed right then this equals 4x2 will

give us 16

x^2 now let's move this term from the

right to the left so we have here 27 x

Cub - 16 x^2 this equals z from here we

have a common factor of x² right so we

have here

27x -

16 this equal 0 so from here we have X2

= 0 and from here we have 27 x - 16 = 0

from here we take Ro of both side x is

exactly equal to 0 this is a solution

and from here move -6 to the right hand

side will become positive and divide by

27 we get x = 16 / 27 and these are the

solutions thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting radical equation | Algebra problem


hello welcome back to another video

today we're going to evaluate the root

of some irrational numbers right so we

have rooot of 2 + < tk3 + root of 2 - <

tk3 and here we're going to evaluate

this without calculators right now let's

start with this we're going to

let

t equal the expression > 2 + < tk3

+ < TK 2 - < tk3 right okay we need to

be very careful because we're expecting

T to be greater than zero

right okay now from here let's go ahead

and square both sides so we get t^ 2

will be equal to < TK 2 + < tk3

+un 2 - < tk3 all 2 and from the right

hand side let us take note of this a + b

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

right so from here we get t^ 2 will be

equal to the first term here squared

that is otk of 2 + < tk3

squar then plus the second term here

squared which is otk of 2 - < tk3 2 then

we have + 2 2 * the first term < TK 2 +

< tk3 multip by the second term in the

bracket < TK 2 -

root3 okay so from here we have t^2 is

equal to remember this Square we cancel

this Roots this Square will cancel this

root sign so we have 2 + < tk3 plus here

2 - < tk3 then plus 2 multiply by

remember this are under the same r signs

right so we can actually write that as

root of 2 +3 * 2 - < tk3 remember this

is difference of 2 squares right recall

that A + B * a minus B this is equal to

a 2us b 2 right and we can see we have

the only difference here this plus sign

and this negative sign so to multiply

this and this together that would be 2^

2us < tk3

squ okay so from here we have t^2 is

equal to this and this become Z right

root3 - root3 is 0 then we have 2 + 2

which is 4 then + 2 now from here this

is 4 - 3 because this Square will cancel

this root sign and 2 2 is 4 4 - 3 is 1

and root of 1 is 1 2 * 1 is 2 so we can

see here we have 2 2 = 4 + 2 which is =

6 now take root of both side remember T

is greater than Z so no need for plus or

minus we only take the positive square

root so we have t is equal to square otk

of 6 and this is the right answer so we

cannot say that the square < TK of 2 + <

tk3 + S < TK of 2 - < tk3 is = to < TK 6

and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Interesting radical expression | Can you Denest ?


hello good viewers welcome back once

again please kindly subscribe to my

channel also like comment and share

today we are given the following nested

radical equation x is = the < TK 12 +

the < TK of 12 + the < TK of 12+ and so

on here we're going to compute the exact

value of x

right okay let's get started we have X

is = to the square < TK of 12 + this

sare < TK of 12 + the < TK of 12

+ and so on right so here we have X is

equal to the square root of you know

this is

12+ now carefully look at this this is

exactly equal to X since X is equal otk

of 12 +un of 12 + the < TK of 12 and so

on and from here we have the < TK of 12

+ the < TK of 12 plus and so on so we

replace that with X now to get rid of

this root sign we are going to square

both sides of this equation right so we

know from the right the root and the

square will cancel out so here we have

X2 is = 12 + x so here we're going to

move the whole term from the right to

the left so we're going to have X2 here

- x - 12 this is equal to 0

so we need to bear this in mind that X

must be greater than zero because we're

dealing with Ned radical that is root

sign right so X is going to be greater

than zero

right so from

here we are going to use the general

quadratic formula when we have the

general form of any quadratic equation a

X2 +

BX + C is equal to Z then X is given by

B plus or minus theun of b^ 2 - 4 a c

all 2 a right

okay so from here a is equal to coent of

x² which is 1 B is equal to coent of X

which is -1 and C is equal to the

constant which is -12 so X is going to

be

B-1 plus or minus the root of B S1 s

- 4 * a which is 1 * C which is -12

right then over 2 * a which is 2 * 1 so

this is equal to here 1 plus or minus

theare root of 1 1 sare which is 1 here-

* minus we have plus 4 * 1 4 * 12 which

is 48 10 /

2 so from here we get X is = 1 + so

minus you know this is from here this is

equal to the root of 49 which is 7 right

so we get 7 / 2 so this is equal to 1 +

7 / 2 or 1 - 7 / 2 so this is equal to

you know 1 + 7 is 8 / 2 which is 4 or 1

- 7 -6 / 2 which is -3 we all know that

this value is not acceptable right since

X is is greater than Z here we go for

this value so here we can deduce that

the < TK of 12 + the < TK of 12 + the <

TK of 12 plus and so on which was equal

to X is exactly equal to 4 and this is

the right answer thank you for watching

please do wa to subscribe to my channel

also like comment and share bye-bye

An infinite nested radical equation. #maths #education


hello good day viewers welcome back once

again let's solve the following given

quartic equation very quickly so we're

given that x ^ 4 - 9 is = 0 now we're

going to write x ^ 4 as x ^ 2 then s

then - 9 is the same as 3 2 and this is

equal to Z Now from the left hand side

of this equation we're going to make use

of the difference of two squ formula a 2

- b 2 is = A + B * a - b right so from

the left we have here

X2 + 3 then multiply by

X2 - 3 and this is equal to Z so from

here this implies that x² + 3 is = 0 or

we have from here x²

- 3 is = 0 now from here subtract from

both sides we get x2 is = 0 - 3 will

give us -3 and from this side X2 is = 0

+ 3 which is

three from here let's take root of both

side we get X is equal to plus or minus

square otk of -3 and from here take root

of both side we get X isal to plus or

minus Square t of

3 now do not forget that the square root

of1 is I that is the imaginary unit

right so we have here plus or minus I

root of 3 so here we have Solutions we

have X1 equals here let's take the the

real roots for so we have <

tk3 we have X2 = root three these are

the real Roots now let's go to complex

Solutions X1

equals for complex Roots X1 = I root of

3 then X2

IUN of 3 and these are the solutions

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

A nice quartic equation.

#maths #mathematics #simplification #exponents


hello good viewers welcome back once

again to another video please kindly

support my channel by subscribing also

like comment and share today we have

this nice algebraic problem here we're

given X2 + Y2 is = 34 x - Y is = 2 right

over here we're going to compute the

value of the product of X and Y

right okay in this video we're going to

make use of some nice subject break

identities right I love these identities

because they are really cool right they

helps us in solving all these

complicated problems right here is the

identity X2 + y s can be written as x -

Y 2 right + 2 * x * Y is that not cool

yes of course this is really nice

okay this actually mean that this

equation x^2 y^2 + Y 2 I mean which is

equal to 34 we can write the left hand

side as x - y^ 2 + 2x Y and this is

equal to 34 now there's a nice question

that should be asked here why did I make

use of x - y the reason is because I

have seen that in the second equation

the value of x - + y is given right so

we can still rewrite this in terms of x

+ y but the reason here is clear that x

- Y is given us two right so from here

we can make the substitution from here

so x - Y is 2 so here we have 2^ 2 and

here + 2 x y this is equal to 34 right

so you know 2 2 is 4 so here we have 4 +

2 x y this is equal to 34 so from here

we subtract four from both sides and

here we get 2 * X and Y you know this is

equal to 30 so from here we're going to

divide both sides by two right so here

over two and here over two this two

cancel this two and here we get x * Y is

= 30 / 2 which is 15 and this is the

right answer take thank you for watching

please do wa to subscribe to my channel

also like comment and share bye-bye

Interesting algebraic problem. Can you do this without finding the value of x and y ? #maths


hello welcome back once again today we

have this quatic equation and we're

going to solve for all values of X now

let's get

started now in this video we're not

going to make any substitution so we're

going to solve this problem

straight so let us get started from the

left hand side we have x to^ 4 plus x²

then this is equal to 20 so here we have

x to the 4 + X2 this is = 20 as 16 +

4 so 16 we're going to take it to the

left and also four take it to the left

so they're going to have opposite signs

so we have X to the^ 4 - 16 + X2 - 4

this is equal to

0 so from here we have X to the^ 2 in

bracket to the^ 2 CU 2 * 2 will give us

4 - 16 is same as 4 2 then plus x^2 - 4

this is equal to

0 From Here let us make use of the

difference of two squares formula we

have a 2 - b 2 is = A + B * Aus

B okay now sorry for the noise so here

we're going to make use of this

difference of two squares formula here

so that is going to be X2 + 4 * x^2 - 4

then here we have plus let's put in

bracket x^2 - 4 then this is equal to Z

so we can see that X2 - 4 is Common from

the left so let's Factor it out so we

have x^ - 4 then into bracket here we

have X2 + 4 left then Plus because in

this brackets here when you factor out

this x - 4 we have just one then this is

equal to Z so we have these two products

which is equal to Z so we have either so

this implies that either X2 - 4 is = 0

or X2 + 4 + 1 is 5 is = to 0 from here

take4 to the right so this will become

X2 is = to pos4 here x² is = -5 take

root of both sides so we get X is = plus

or- 4 and here x is = to+ orus < TK

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

so we have only two real solutions and

here x is = to otk of1 is Iota so we

have plus or minus Iota root of 5

and these are the only two imaginary

Solutions thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Stanford University 🎓 Entrance Exam | Can you solve ?


hello good viewers welcome back once

again today we're going to find the

following limit limit as X is

approaching to X CU - 8 all x - 2 when

we make the direct substitution here we

get 2 Cub - 8 2 - 2 and this is equal to

0 0 you know this is an indeterminate

form right so we need to manipulate this

function here I think the function can

further be simplified right now let us

talk about this function right over here

we have X Cub - 8 all x - 2 this is

equal to X Cub - 8 is the same as 2

cubed then all x - 2 now let us recall

the difference of 2 Cub formula a cubus

b Cub this is equal to a minus B into

braet a 2 + a b + b 2 from here this

will be x - 2 into braet X2 + 2 x right

then + 2 2 which is 4 then all over x -

2 so from here we cancel out and here

this simplifies to X2 + 2x + 4 right now

the limit as as X is approaching 2 of X

Cub - 8 / x - 2 this is same as the

limit as X is approaching 2 of X2 + 2x +

4 from here we make the direct

substitution we have 2^ 2 + 2 * 2 + 4 so

from here 2 2 is 4 + 2 * 2 which is 4 +

4 1 again and from here this simplifies

to 12 and this is the right answer thank

you for watching please kindly subscribe

to my YouTube channel also like comment

and share bye-bye

Interesting Limit | #maths #olympiad


hello good viewers welcome back once

again today we are given the following

should I call it optic equation right

and we are asked to evaluate x ^ 4 + 1 *

x^ 4 - one right I know that this

problem looks a little bit we right but

believe me this is a very simple problem

all we need to do is to manipulate this

equation very well right now let's get

into the video we are given that x^ of 8

is equal to 9 right now let us go ahead

and subtract one from both sides of this

equation so from here x^ of 8 - 1 this

is equal to 9 - 1 right so we can see

that the equation still remain balance

okay now from the left here we're going

to rewrite x ^ 8 as x^ 4 in bracket s

right then- 1 is the same as 1 SAR and

from the right 9 - 1 is = to 8 now from

the left hand side we're going to make

use of the difference of two squares

right so recall that when we have a 2

minus b 2 this can be factorized as a +

b * Aus B right

so that is the sum of the

bases then multiply by the difference of

the bases right so if you take a very

look at this one here the base here is x

^ 4 and the other base is one right so

make use of this property here so this

will actually give us x^ 4 + 1 * x^ 4 -1

and that is equal to 8 I think this is

what we're looking for

okay nice we can see that we have found

the solution to the problem thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Interesting algebra problem | Here is the trick | #maths #mathematics


hello welcome back once again today we

have this interesting exponential

equation involving root of x and

reciprocal of

x from the left here the base is four

and four we can write it in index form

as 2 to^ 2 from the right the base here

is 8 and 8 we can also relate it in

index form as 2 to the^ of 3 so the left

hand side we're going to write that for

as 2 to the^ 2 then root X this equals

from the right the eight there will be 2

cubed then of the reciprocal of

X now let us take note of this property

that a^ b^ c is same as a to the^ B *

C so here this will be 2 2 * root X will

give us 2un X then this equals here 2 3

* 1 / X is same as 3 /

X from here we can see that we have the

same base right so we're going to equate

the powers so this implies that 2un x =

3 / X here we're solving for the R value

of x so let's Square both sides to get

rid of this root sign so here this will

be squared and here this will be

squared so from here to square 2 * root

X we're going to square 2 first and that

will give us 4 * root x 2 remember

remember the square will cancel the

roots so we get X then this will be

equal to to square this we're going to

square the numerator and also Square the

denominator when you square the

numerator we get 3 2 which is 9 /

X2 which is X2 right so from here this

is over one Let's cross multiply X2 * 4X

will give us 4 x Cub then this = 1 * 9

will give us 9 so from here divide both

sides by four

this cancel this here we get X Cub = 9 /

4 from here let us take root of I mean

the cube root of both sides I mean so

this give us X = the cube root of 9 /

four and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Interesting exponential equation |


hello good day viewers welcome back once

again let us solve the following given

radical

equation solution we have here the < TK

of 5 ^ x * the < TK of 25 ^ x * the of

125 to^ x this is equal

to

125 let's get started so here we have

the square root

of 5 to^ of x X multip by the square

root of you know here 25 is the same as

5 S right so that would be 5 to^ of 2x

right then multiply by the square root

of 125 is the same as 5 cubed then with

the power of X here that will become 5

to the^ of 3x and this is equal to 1 2 5

is 5 to the power of

3 so from here we have the square root

of 5^ x MTI the square root

of 5 ^ of 2x * by now remember that root

m is = to m^ of 1 / 2 right so from here

this will become 5^ of 3x / 2 right then

this is equal to 5 to the^ of 3 so from

here we have the square root of 5 to^ of

X multip by the square root of remember

from here we have the same basis right

so we need to add the powers so here we

have 5 to^ of 2x + 3 x 2 2x + 3x 2 that

will give us 7 x all over 2 then this is

equal to 5 to the^ of

3 now from here make use of this

property again so we have the square

root

of 5^ of X multi by here will be 5^ 7 x

/ 2 then remember we have root sign so

that would be 2 * 2 which is 4 then this

is equal to 5 to the^ of 3

so from here remember we have the same

basis again so we're going to add the

powers so we have the square root of 5^

of x + 7 x 4 that will give us 11 x

right then all over 4 then this is equal

to 5 to the power of three so from here

making use of that property that is

square root of n m is = m ^ 1 / 2 so

from here this will be 5^ 11x / 4

remember with this root sign that will

be 4 * 2 and that is 8 this is equal to

5 to the power of 3 from here we have

the same basis so we need to equate the

powers right so we have 11 x / 8 this is

equal to

3 so here we cross multiply 11 x * 1

will give us 11 x this is equal to 8 * 3

that will give us 24 so divide by 11

since we are solving for x and here we

come up with the solution that X is =

24/ 11 and that's the right answer thank

you for watching please canly subscribe

to my channel also like comment and

share bye-bye

Interesting radical equation | #maths #mathematics


[Music]

[Music]

[Music]

s spe

[Music]

know

[Music]

oh

[Music]

[Music]

slitz spe

[Music]

Interesting limit.


hello welcome back once again today we

have to solve for the value of x that

will satisfy this equation but X must be

greater than

zero I have a very simple way to solve

this

problem first of all let's consider 65

we know 65 is the same as 81 - 16 I

think with this idea this problem is

already

solved let's go back to the original

problem we have 3 ^ of x - 2^ x so this

will be equal to

65 so with this idea so this will become

3 ^ xus 2 to^ x so this will be equal to

81 - 16 so we have 3 ^ x - 2^ x so this

will be equal to 81 is the same as 3 ^

4us 16 is the same as 2 to the^ 4 from

here this implies that X is = to 4 and

this is the solution thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Germany | Interesting Olympiad Math


again please kindly support my channel

by subscribing also like comment and

share today we have this nice problem

here to solve we have here that 2x - 3 Y

is = 5 right over here we're going to

evaluate 4 power of X all 8 to^ of Y

right okay now let's go to this

expression here let us try to manipulate

this expression we have here 4^ x all 8

power of Y you know this isal to 4 is

the same as 2 2 then of X all over 8 is

the same as 2 cubed then of Y so from

here you know this

property a of B of C is equal to a of B

* C right so from here these Powers will

multiply together we have 2

2x all over these Powers multiply

together 2^ 3 y right let us not forget

this property as well when we have a to

the^ of B over a to the power of C with

the same basis we can subtract the

powers right so that will be a power of

B minus C

right okay so from here this is going to

be equal to 2 2^ 2x - 3 y right and do

not forget that 2x - 3 Y is exactly

equal to 5 right so from here this is

equal to 2 power 5 and 2^ of 5 is equal

to 32 and this is the right answer thank

you for watching please do wa to

subscribe to my channel also like

comment and share bye-bye

Easy exponential problem. The trick is really cool. #maths #education


hello welcome back once again today

we're going to find the square root of

an imaginary number here I equals theare

root of1 which is also known as the

imaginary unit right okay for my video

here we're going to take the square root

of I if you recall from the previous

video I made the principle < TK of I = 1

+ I all / the otk of two yes if you

check my videos very well I made a video

on how to find the square root of I and

what it equals right okay so another

thing we need to consider here is when

we have the square root of m n x we can

actually Express this as squ root of M

Munk of n * otk of

X right okay now let's get into the

video here we have the sare OT of -2 I

here we can write this as square root of

-1 right Multi 2 then MTI I and with

this property right over here this

equals

-1 then multi by theun of 2 MTI root of

I just like I stated here the square

root of I that is the principal square

root = 1 + I all < tk2 so going to make

use of it here and here root of -1 = I

right so this equal I * the < TK of 2

multiply by here is going to be 1 + I /

theun 2 remember the all of this there

are products right so here root of two

we cancel root of two so this equal I

into braet 1 + I from here I * 1 is I

then I * I is I which is what -1 so here

we have the solution that the < TK of -2

i = i - one amazing thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Can you find the square of negative two times iota ???


hello good day viewers in this video we

are going to find the following limit

limit as X is approaching Z of e to^ 2x

-1 / e ^ x

-1 now e ^ 2x we can write it as that's

e to ^ of 2x we can write it as e to^ of

X all squared if you recall that a to

the power of

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

then in bracket power of C right so now

the expression here

e to the^ of 2x -1 all e ^ x -1 will be

equal to e ^ x 2 - 1 then all e^ x

-1 now from the numerator here we can

make use of the difference of two

squares right we can rewrite it as e ^

x^ 2 minus here 1 is same as 1 2 then

all e^ X - 1 now make use of the

difference of 2 squares that is a 2 - b

2 this is equal to a + b * a minus B so

from the

numerator we are going to to factorize

that as e^ x + 1 * e ^ x

-1 then all over e^ x -1 so from here

this will cancel out and we have e^ of x

1 okay now the limit we asked to find

limit as X is approaching 0 of e to^ of

-1/ e^ x -1 will be equal to limit

as X is approaching Z of e to the^ of x

+ 1 and from here we make the sub the

direct substition we obtain e to ^ of 0

+ 1 and e^ 0 is 1 so we get 1 + 1 and

that is two and this is the right answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Interesting limit | No L'hospital's rule | #maths #mathematics


good evening viewers here welcome back

to my Channel today we have the

following interesting problem

if x raised by two plus y respirator is

equal to 11 and x times Y is equal to 8

then

X Plus Y in bracket squared is equal to

dash this problem is very simple

so I'll be presenting only one solution

but um very much methods of solving this

problem okay let me go ahead and expand

this so I have here X Plus y raised by 2

is equal to

X raised for two

loss

2 times x times y

plus y squared okay most of us are very

familiar with this expansion okay so

here we write this as follows x squared

nausea y squared then plus 2 times x

times y okay now we're going to make our

substitutions here x squared plus y

squared here we all know that this is

equal to 11

plus 2 times x times y we all know that

is equal to eight okay so here we have

this is equal to 11 plus 16. okay

so we have 11 plus 16 and this is equal

to 27 and this is the right answer to

the problem thank you for viewing and

please make sure you subscribe to my

YouTube channel for more interesting

videos

A secondary student's problem.


hello welcome back once again today

we're going to solve this interesting

equation right and we're going to solve

for the values of a we need to know that

a is not equal to Z if a is equal to Z

the left hand side of the equation will

be

undefined so let's solve for other real

values of a from the left hand side a *

a * a is a cubed so we have a cub

then / a + a + a is 3

a then this is equal to 4 put this over

one and cross multiply so a cub * 1 will

give us a

cub so this is equal to 3 a * 4 will

give us 12 a subtract 128 from both

sides so we get

here a CU - 12 a this is equal to Z from

the left hand side a is common so let's

Factor it out so we get a into bracket a

2 - 12 this is = 0 so we can see we have

two products which is equal to Z so this

implies from here that either a is equal

to Z or a 2 - 12 is = 0 but obviously

this is not a solution right so add 12

to both side we get a 2 is = 12 now take

square root of both sides so we get a

will be equal to plus or minus theare <

TK 12 so simplifying the S A will be

equal to plus or minus the square root

of 12 is as 4 * 3 then from here we get

our solution a will be equal to plus or

minus 4 is 2 then we have here root of 3

so here we have our only two real

solutions that satisfy the given

equation thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting equation | Can you solve ?


hello good day viewers welcome back once

again today we're going to solve for the

values of X in the foll quatic equation

we have x^ 4 + x^2 - 42 this is equal 0

now x^ 4 we can write it as x^ 2^ 2 then

+ X2 - 42 this is equal to Z now we're

going to make the following substit

we're going to let X2 = y so from here

we have y^ 2 + y - 42 this is equal to Z

so for here this left hand side of this

equation can be factorized right and how

do we factorize that first of all this y

we can rewrite y as y² + 7 y - 6 y then

- 42 this is equal to Z from here we

factor out Y and we left with y + 7 and

from this side we factor out -6 and we

left with y + 7 and this is equal to Z

so from here this implies that y -

6 * y + 7 is = 0 so we have here y - 6

is = 0 or y + 7 is equal

0 so from here we find solution for y y

is = 6 or Y is = 0 - 7 which is

-7 now let us recall that x² is equal to

Y so from here we can state that when Y

is = 6 x² is = 6 which implies that X is

= plus or minus otk

6 when y is = to

--7 X2 is = -7 so this implies that X is

= to plus or minus otk of

-7 which is equal to+ - IUN 7 so from

here we find the solutions that X is

equal to plus or minus < TK 6 this is

for R solution

right and we have X isal to plus or

minus I root of 7 this is for Imaginary

Solutions thank you for watching please

kindly subscribe to my YouTube channel

also like comment and share bye-bye

Olympiad Math | A nice quartic equation | Interesting problem | #maths #olympiad


hello welcome back once again let us

solve this interesting equation from the

Oxford

University so from the left hand side we

have this x ra^ otk of X and this is

equal to 3 so let us take the log of

both sides so we have the natural log of

x to the^ otk X this is equal to the

natural log of 3 so we know this

property of logm for example the log of

a the^ b is equal to B * the log of a so

here this will become otk x m the

natural log of x so this is equal to the

natural log of 3 right so now we know

root of x same as x^ 1 / 2 then multi by

the natural log of x this is equal to

the natural log of 3 now we know this

interesting property that a to the power

of B is same as e to the power of B MTI

the Ln of a right so here the power of X

here is 1 / 2 so this is going to be e

to 1 / 2 * the Ln of X then times the Ln

of X this is equal to the Ln of 3 now

we're going to make use of the Lamb W

function but we need to take note of the

following now the lamb function is the

inverse function

for the function x * e^ X right so now

the lamb function of

x * e^ of x simplifies to X provided

that the power of E must be equal with

the coefficient of e so now you can see

here the power of E and the coefficient

here they are not yet equal the only

slight difference here is 1 all/ 2 so

we're going to multiply both sides of

this equation by 1 all/ 2 so stting from

here natur of x * 1/ 2 is as 1 / 2 * the

natural of X then multi e to^ 1 / 2 *

the natural log of x so this will be

equal to 1 / 2 * the natural log of 3

now we can take the lambat on both

sides okay so the left hand side we

simplify to 1 / 2 * the natur of X so

this is equal to the

W of 1 / 2 * the natural of 3 now we can

multiply both side by two so if we

multiply both side by 2 2 * 1/ 2 here is

1 so we are left with natural log of x

so this will be equal to 2 * the W of 1

/ 2 * natur of 3 so now we're going to

make this simple base on both sides so

we're going to write e^ Ln X from the

left hand side will also be equal to e^

2 W * 1 / 2 * the natural L of 3 so we

know e^ Ln anything cancel out to leave

us with the team so here e e and the

power of this Ln we cancel out and we

just have X so this is equal to e to

the^ 2 W * 1 / 2 * the natural of 3 so

with the help of my calculator we get

this nice approximation for x x is

approximately equal to

799 383

02 and this is the correct answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Oxford University Entrance Exam | Can you solve ?


hello welcome back guys here is the

continuation of the previous video you

know in the video I told you guys to

check the next video for the evaluation

of this without calculator right okay

now let's see how to do

that now here we have 2^ 10 - 1 right so

we're going to rewrite as follows 2 ^

5 2 - 1

s right you know these powers can

multiply together so that we can still

have 2^ of 10 so from here we're going

to make use of the difference of two

squares

formula here we go a 2us b squ is equal

to a + b * Aus B so from here this is

going to be equal

to 2 ^ 5 + 1 right * 2 ^ 5 - 1 so this

is equal to you know 2^ 5 is 32 and here

32 + 1 is 33 right then times 2^ 5 is 32

32 - 1 that will give us 31 right so we

need to multiply this together so is

there any other trick on how we can do

this well there are so many tricks but

this one is straightforward right just

have 33 * 31 3 3 99 right and we have

one Z 23 and here comes the right answer

to the problem thank you for watching

please don't wa to subscribe to this

channel also like comment and share

bye-bye

Olympiad Math problem.