Math Olympiad Questions | Find X | X^x=X | challenging Algebra Problem | Olympiad Mathematics

In this video, we will solve this algebraic equation problem trending 2022. The problem is find the value of x in X^x=X . Math Olympiad Questions
we will solve for all possible values of X. This is actually one of the challenging math problems so do not ignore this video.

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[garimasharma]

This is the most obvious problem I have encountered. Literally got the answer after a look on the problem.

[onepiecefan]

To be honest i could not have done this but i knew answer would be 1 or - 1 because of hit and trial method

[isaacbaruccruzdasilva]

but... why all this?
x^x=x
x^x=x^1
Equal bases, equal exponents
x=1

[tahititoutou]

The solution x = 1 is trivial : 1^1 = 1. No log nor ln needed. The solution x = -1 is slightly less obvious but still a college student could find that (-1)^(-1) is the same as 1 / (-1) which equals -1. Thus there are 2 solutions possible : 1 and -1. No logarithms needed.

[yanything]

It would be so much easier just to take the log base x of both sides, getting log_x(x^x) = log_x(x).
Then, since we know x^x is x to the xth power and x is x to the 1st power, we can reduce the equation to x = 1.

[mrtrinity]

they have the same base "X" so their exponents can be equated... x=1

[GirishManjunathMusic]

Before watching the video:
Given:
x↑x = x

To find:
x

Taking log on both sides:
log(x↑x) = log(x)

using log(a↑b) = b·log(a):
x·log(x) = log(x)

moving all terms to LHS:
x·log(x) – log(x) = 0
log(x)·(x – 1) = 0
From here, log(x) = 0 or x – 1 = 0.

both cases give you x = 1.



By taking log of both sides, we implicitly took x > 0.
For x < 0, we can take absolute value on both sides and result in the same process, and by negating the RHS to invert the absolute operator, we get x = –1 as a possible result. Cross checking against the original equation, we find that x = –1 also works (–1↑(–1) = 1/(–1) = –1). Thus, x = ±1 are the only solutions.

[SebastienPatriote]

What an absolutely convoluted way of complexing this problem just gir the sake of a longer video.

Just take natural log on both sides, and use log properties.

You'll get x lnx = lnx

Divide by ln x on both sides since lnx =/= 0 and you end up with x=1

[rarebeeph]

what i did was graph x^x at various points (-2, -1, 1/4, 1/2, 1, 2). -1 and 1 obviously solve the equation, but to prove there are no other options, we take the derivative of x^x, that being (ln(x)+1)x^x, and since x^x is never 0, there must only be turning points where ln(x)+1 is 0, which occurs only at x = 1/e. therefore for 0<x<1, x^x > x. then, since ln(x)+1) and x^x are strictly increasing for x>1/e, we know x^x is greater than x for x>1 as well. for negative values, we model x^x as -(-x)^x. this gives a continuous model, which agrees at all points where x^x is defined and negative, of the lower (closer to x) branch of the negative values of x^x. as it is continuous, we can take its derivative as well: -(ln(-x)+1)(-x)^x. this is only 0 when ln(-x)+1 is 0, i.e., x=-1/e. by similar logic to before, for 0<x<-1, -(-x)^x < x. furthermore, since -1/(-x)^(-x) is a mirror image of -1/x^x, which is strictly greater than -1/x for x>1 since their reciprocals have the reverse relationship, we know that -(-x)^x is strictly greater than 1/x, and thus x, for x<-1, QED.

[uueoztc]

Делим обе части на х
х^х:х=1
х^х-1=1
Значит либо
х-1=0 х≠0
Либо
х=1 х-1-любое число
Значит х=1
Ответ:1
Замечание :если степень с рациональным показателем, то основание может быть только неотрицательным

[towalks]

you messed up everything at 3:22 by multiplying only left side of the equation to (-1)^(-y-1) and your following statement "(-y)^(-y-1) is never negative" is totally wrong (for y=2, (-2)^(-2-1)=(-1/2)^3= -1/8 -negative number) Here is what you had to do:
x<0 => x^(x-1)=1.
Since x-1< - 1< 0 => |x|=1 (must be so) => x=-1 (since x<0) => try (-1)^2=1 - good to go, So the final two are 1 and -1

[robertlunderwood]

I didn't need logs to solve this. From x^(x-1) =1, there are three branches. The first is that the exponent, x-1, is 0. This leads to x =1. The second branch is that the base, x, is 1. The third and less considered branch is that the base is -1 with an even exponent. If x = -1, the exponent is -1-1 = -2 which is even.

[insterquiliniisinvenitur]

0^0 is not indeterminate
It's undefined
Only when the limit of x approaches 0^0 can we say it's indeterminate.

[awildscrub]

You can also just take the log of the absolute value of both sides and you'll get x=-1 or 1

[HallStar]

Bro it's way simpler than that

x^x = x

Therefore x^x = x^1

Since both bases are equal then we can equate the exponents

x^x = x^1

x=1

[koenth]

I believe the reasoning is flawed at 3:46: if y is positive and even, e.g.2, then (-1)^(-2-1) =1/(-1)^3 = -1

Reasoning could go like this: If we are solving for real y, then (-y)^(-y) = -y with y>0 implies that y is an odd integer. Then -1/y^y = -y so we have 1=y/y^y so y=y^y, which is the problem we already solved for x, with 1 as its solution. y=1 is indeed an odd integer, so our solution is x=-y=-1.

Bonus: Solving over the complex numbers we can infer from ||x^x|| =||x|| that ||x||=1 so we can write x=exp(iφ) for real φ. Then we get (exp(iφ))^exp(iφ) = exp(iφ).
Using the power rule (a^b)^c=a^(bc) and the fact that the function exp is periodic mod 2πi to infer that
iφ exp(iφ) = iφ +2kπi for integer values of k. Divide by i to get

φ (exp(iφ)-1) = 2kπ

The RHS is real, and the LHS is real when φ is an integer multiple of π. Setting φ=nπ and divide by nπ, the equation becomes

n((-1)^n -1) = 2k

If n is even this has solution k=0 and we get x=1
If n is odd this has solution k=-n and we get x=-1

So, solving over the complex numbers does not give any new values for x

[renyxadarox]

if you are solving the equation in real numbers, x=-1 must be rejected because f(x)^g(x) => f(x)≥0

[maths_plus]

x=0 is not solution.
so for x<>0 : x^x=x <====> x^(x-1)=1.
But x^p=1 only in 2 cases:
* IxI=1 x=1 or x=-1: 2 solutions 1, -1
* or p=0 <====> x-1=0 <====> x=1.
Therfore, s={ -1; 1}.

[DeJay]

To get x = 1 you could have very easily just said from x^x = x to x = log_x(x) which is 1 so x = 1
For x = -1 I have no idea

[owlsmath]

I guessed 1 and -1 but was too lazy to do all the work. But thanks for doing it for me! And good video!