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.
A video course for students taking the GCSE Maths exam
👉👉👉 bit.ly/3vtDbtW
Disclaimar:
Affiliate Links:
The writing Tablet I use for my videos
Veikk A15 Pro
Buy Now👇👇👇
amzn.to/3Jdkv7B
Wacom Writing Tablet
amzn.to/3b7fS2c
BOOKS
Math Olympiad
amzn.to/3OuGA2F
Awesome Electronics Course
8692fflf220x3b...
If you are new to my channel, please subscribe, like, comment and share.
Playlist for other videos
/ @mathwithmershscore

[@mathwithmershscore]

I have solved this using different approach so check it out: kzbin.info/www/bejne/oIO5pKN_hLarnJY

[@garimasharma2384]

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

[@kafka-x7u]

Yeah literally

[@2002THEBOY]

Yep things can be obvious but the proof is something else

[@kafka-x7u]

@@2002THEBOY nah even the proof is not that tricky

[@antoine2571]

@@2002THEBOY jordan's theoreme is a good example

[@SebastienPatriote]

@@2002THEBOY just natural log on both sides, bring down the x in the power and divide both sides by ln x. You'll end up with x=1.

[@onepiecefan87]

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

[@mathwithmershscore]

Ok thanks being honest with us. You know this channel was created to provide answers to the 'How' to questions. So please share it among your friends too so that we can build a very large family. Thanks for subscribing

[@kananishion1253]

agree with you

[@SatnamGugar]

I’ve never seen anyone say hit or trial, only trial and error lol

[@onepiecefan87]

@@SatnamGugar well my teacher used the term 'hit and trial' when we were first taught about finding roots of a cubic polynomial back in 8 or 9 grade It's just stuck since then, both the terms are correct because the meaning is same

[@SatnamGugar]

@@onepiecefan87 that’s cool but Idk why you would need to use hit and trial finding roots of cubic polynomials

[@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

[@awildscrub]

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

[@gauravghodinde2949]

Or u can diff w r t x

[@ГеоргийПлодущев-с2н]

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

[@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.

[@koenth2359]

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

[@ravirajamadan]

An easier way is : x^x=x which also means X^1/x = x . Multiplying LHIS with LHS and RHS and RHS we get x^2=1 which means x=+-1

[@soumyadeepmukherjee5091]

@Math with Merschscore your assumption that (-1)^(-y-1) cannot be equal to -1 for +ve y is incorrect. Check for y=2 it does not hold true.

[@TazPessle]

It's nice seeing a methodical proof for so.wthing that simple. The mond can find the answers, but proving they are the only answers is something else

[@johnny_eth]

As trivial as x=x. X=1 or x=-1. You can apply the logarithm to both sides, so ln(x) cancels out, and you are left with the first solution. Then, whenever you cancel out anything, you should always consider side effects. Logs can't take negative numbers, so for negative x, do a substitution t=-x , and you get the second solution x=-1.

[@mathwithmershscore]

This method is also nice, check it out: kzbin.info/www/bejne/oIO5pKN_hLarnJY

[@aychinger]

Sorry, but the solution for x (-2)^(-3) = 1 / (-2)^3 = -1/8. As others have mentioned, there are much more elegant and compact solutions (aside from this one being plain false).

[@mathwithmershscore]

Check this out: kzbin.info/www/bejne/oIO5pKN_hLarnJY

[@renukanandini9401]

My first thought- x=1 and it was correct

[@Eucheumeuneu870]

X is equal zero, one or minus one between parentheses, x can have any value.

[@sirmonke8946]

People here are like "I know the solutions just looking at the problem". But how do you know that those are the only solutions? You need to look for EVERY solution.

[@gonzalotapia1250]

Take logx in both sides. You have xlogx(x) = logx(x) Since log x (x) =1 x*1 = 1 x=1

[@fahdfarachi6232]

We should suppose that x>0 to enter the log function on both sides. However we should also study the case where x

[@ChavoMysterio]

x^x=x ln(x^x)=ln(x) x[ln(x)]=ln(x) x[ln(x)]-ln(x)=0 [ln(x)](x-1)=0 ln(x)=0 x=e^0 x=1 x-1=0 x=1 Therefore, x=1 with multiplicity of 2 🏳️‍🌈

[@mathwithmershscore]

It has two solutions

[@johnyriosrosales7674]

Yo aplique otro logaritmo de X a ambos miembros dé la ecuación X^X=X LOGX(X^X)=LOGX(X) XLOGX(X)=1 X=1

[@堀勇作-l5p]

x=1

[@Vishnu-dk2rc]

We can take log on both sides and solve it in two steps

[@fahdfarachi6232]

We can do that if we suppose x > 0, because ln is defined when x > 0. But we should also study the case where x < 0.

[@mirianaquino9548]

Too easy :)

[@falahalfadhel185]

In one step x^x =x^1 , then x=1

[@mathwithmershscore]

yeah but -1 is also answer to this question

[@falahalfadhel185]

log(x)^x =logx xlogx =logx xlogx - logx=0 logx ( x-1)=0 logx=0 or x-1=0 if logx=0 , then x = 10^0 s0 x=1 or x-1 =0 ,then x=1

[@Nohant2]

It,s quite obvious that the only value that matches such criteria is +-1, no calculations needed here.

[@ganeshdas3174]

Go for x=1

[@Thriftyknight]

literaly just looked at the problem and got 1, Anyone esle here?

[@adgf1x]

X=1(taking logarithm)

[@shader6857]

1

[@Nikioko]

For any number, x¹ = x, therefore x = 1. And for any number x ≠ 0, x⁻¹ = 1 / x, therefore x = -1.

[@rarebeeph1783]

but you have yet to prove that those are the only solutions.

[@josephmathews6096]

There is another easy method

[@kirbycottoncandy807]

1 one

[@LordErnie]

1, -1, or 0

[@timmy1729]

X = 2.86295 + 3.22327 i

[@neverdie0001]

answer: 1 or -1

[@JoseAntonio-pk2nq]

1^1=1 X=1

[@ronaldnoll3247]

you can calculate in your head

[@geektoys370]

S=1…

[@thelordz33]

Isn't x just equal to 1?

[@isonzo686]

Why overcomplicate the problem?

[@ИгорьАрсоу]

Х=1

[@khangphuc778]

x^x = x^1 means x = 1

[@mathwithmershscore]

You're right but -1 is also solution

[@parithiilamaaran.h9829]

Bruh easiest sum ever. Don't complicate with Logs. x^x = x is nothing but x^x = x^1. The bases are same, then exponents are same

[@keesvp]

please stick to teaching 1st-grade arithmetic

[@-mrgd1-384]

If an 11 yr old can solve it anyone can

[@mncubing8160]

Why you keep saying these are Olympiad problems? Olympiad problems are a million times harder than these.

[@matteocalo5975]

x=-1 is not a solution to the equation. For x^x to be defined over R, you need to exclude all negative values of x, therefore x>0.

[@isaacbaruccruzdasilva7701]

If x=-1 (-1)^(-1)=-1 first invert the base there so the exponent is positive (1/-1)^(1)=-1 (1/-1)=-1 That's right!

[@matteocalo5975]

@@isaacbaruccruzdasilva7701 the calculation is indeed right, but x^x, being in the form f(x)^g(x), is only defined for f(x)>0 => x>0

[@isaacbaruccruzdasilva7701]

@@matteocalo5975 but no matter function, the equation is "x^x=x" and that's it. unless you're talking about the way he solved it, then it's up to you, but -1 is a solution to this equation 👍

[@pillgod5043]

No in case you don't know anything to the power - 1 is just that number in the denominator And fractions come under real numbers so - 1^-1 is feasible as it gives the end result 1/-1 which is the exact same thing as - 1

[@matteocalo5975]

if you had 1/x=x, wouldn't you set x≠0, since it's in the denominator? x^x is a continuous function in |N and Z, but not in Q or |R. For example, (-1/2)^(-1/2) would be 1/√(-1/2), which is undefined in |R. Negative values are excluded from |R because it makes the function continuous over its domain. You can also check the domain and graph on Wolfram Alpha

[@josuejacques6309]

wtf

[@isaacbaruccruzdasilva7701]

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

[@PatriceFERLET]

That's exactly what I wonder too...

[@mcmystix]

x=1 and -1

[@pedrosso0]

because -1

[@pedrosso0]

plus what about complex solutions?

[@kilian8250]

Idk and then you missed a solution

[@Hotsk]

1

[@irwandasaputra9315]

x=1

[@mrtrinity143]

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

[@mathwithmershscore]

Yeah that's true but -1 is also a solution to this problem

[@jimmyaesthetics6942]

just for a complex solution - 1 he had to do this long I guess

[@Al-hy2wc]

@@mathwithmershscore sorry, but where is error in solution of MATHatapang?

[@olixx1213]

@@Al-hy2wc he's missing a solution , you can't equate the exponents like that because of this

[@mathisnotforthefaintofheart]

You have to be REALLY careful with your rule is the variable is in both the base and exponent!

[@rarebeeph1783]

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 01/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

[@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 x^(x-1)=1. Since x-1< - 1< 0 => |x|=1 (must be so) => x=-1 (since x try (-1)^2=1 - good to go, So the final two are 1 and -1

[@mathwithmershscore]

Thanks for a lot for sharing with us. I'm so happy because you're making the lesson engaging 💕

[@williamperez-hernandez3968]

True he didn't multiply both sides of the equation, but argument that (-y)^(-y-1) is not negative is valid because it is equal to +1.

[@towalks]

@@williamperez-hernandez3968 try again to get + with y = 2 or 4 or any positive even number. You get +1 assuming that y=1, right? if you know at that stage that y=1, why you are still trying to find y? :)

[@bubuqq9988]

Imo the correct separation should be (-1)^(-y - 1) x (y)^{-y - 1). Only then (y)^{-y - 1) can be confirmed positive.

[@ChaosPod]

Yeah, I was also confused by the statement "(-y)^(-y-1) is never negative because y is positive" that would imply (-y-1) is an even number which is not guaranteed with y being just positive. @TOWalks why do you say that |x| must equal 1 in your line Since x-1< - 1< 0 => |x|=1 (must be so) Is it because x^(x-1) = 1 implies either a^0 = 1 => x-1 = 0 => x = 1 or 1^b = 1 => x = 1 Also, I'd say x=0 is not an answer because in the original equation x^x = x => 0^0 = 0 => 1 = 0 false.

[@renyxadarox]

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

[@aronbucca6777]

The function y=x^x is defined for some negative values even if it's not continuous in the interval (-♾;0). If you take the ln on both sides of the equation you are assuming that they are positive, resulting in the "loss" of the solution x=-1.

[@renyxadarox]

@@aronbucca6777 The function y=x^x is defined for some negative x when the domain of x is Z. The function y=x^x is defined only for positive x when the domain of x is R

[@aronbucca6777]

@@renyxadarox you said it, "-1" is an integer number

[@renyxadarox]

@@aronbucca6777 if someone asks you to solve an equation for unknown x, by default it means the domain of x is real numbers.

[@aronbucca6777]

@@renyxadarox yes, but -1 is also an integer, for wich the function f(x)=x^x IS defined. The fact that it isn't continuous for negative values does not mean that you have to reject them

[@tahititoutou3802]

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.

[@ricoronaldoyt9474]

but how do u know that is only 2

[@tahititoutou3802]

@@ricoronaldoyt9474 Because any number larger or smaller than 1 (absolute value) will give a value greater or smaller than itself if raised to any power other than 1. So then x ^x ≠ x anymore.

[@davidbrown8763]

Exactly. I solved this problem within 20 seconds of seeing it. The solution x = 1 was almost instantaneous. Then I questioned whether there were any other solutions?...and, about 19 seconds later, I realised that the other solution was x = -1. No complicated logarithms needed.

[@yanything9613]

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.

[@HallStar15]

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

[@gonzalosanchez4572]

Thanks! I thought I was wrong, did it the same way.

[@taylorfinn1496]

But then you are missing a solution

[@scathiebaby]

@@taylorfinn1496 For some reason, this reminds me of the hyperbola y = 1/x

[@C4rnee]

@@taylorfinn1496 can't you just do x^x=x^-1?

[@taylorfinn1496]

@@C4rnee well yes it makes sense and you are getting the correct solution because x=1 and it would make sense to check if negative 1 also works but these methods of solving don't really show for sure that those are the only solutions. Hence why he used cases to prove that these are the only solutions.

[@tachles_math]

B4 seeing the video my answer is: X=-1 / 1 / 0 Edit: Yeah everything to the power of 0 = 1 so 0 is not an answer

[@mathwithmershscore]

Great

[@maths_plus7092]

x=0 is not solution. so for x0 : 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}.

[@insterquiliniisinvenitur4774]

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

[@Amoeba_Podre]

Its 1

[@joeanderson2024]

You claim (-y)^( -y-1) will never be negative because y is positive. Why? -y-1 will never be negative but that means (-y)^(-y-1)=-1/[(-y)^(y+1)] which will be negative if y is odd

[@cebbolla4629]

i was searching for this comment, it needs more upvotes

[@fahdfarachi6232]

He made a mistake, in fact it's y**(-y-1) and not (-y)**(-y-1), because he already wrote (-1)**(-y-1) next to it. And y**(-y-1) > 0 since y>0, meaning that (-1)**(-y-1) is positive because we have the positive value "1" on the right side, and since (-1)**(-y-1) only takes 1 and -1, so it's equal to 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.

[@mathwithmershscore]

So nice

[@youtubevirus2307]

Oh man

[@Yandel_sw]

Brilliant

[@sirmonke8946]

Why are you able to take the absolute value on both sides? Can't x^x still be positive when x

[@GirishManjunathMusic]

@@sirmonke8946 no it can't be. for a > 0, -a↑-a = 1/(-a↑a) = -(/a↑a) < 0.

[@taito404]

My dumb ass would have tried w lambert function and fail miserably

[@DeJay7]

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

[@mouinhourani6956]

???

[@DeJay7]

@@mouinhourani6956 the exponential equation b^x = a can be re-written as x = log_b(a), and in this case x^x = x => x = log_x(x) but if the base and argument are equal then it's equal to 1 (log_a(a) = 1)

[@mouinhourani6956]

@@DeJay7 Thanks, log_x(x) should be x > 0 so this method dosn't cover x

[@mathwithmershscore]

Check this out: kzbin.info/www/bejne/oIO5pKN_hLarnJY

[@ayden41920]

I looked at this and just thought, hmmm 1^1 = 1

[@GillAgainsIsland12]

But the answer is obviously 1 or -1. No need for calculation.

[@armacham]

The point is to prove it. Anyone can say such and such is "obvious."

[@goopyzoomer5798]

It’s same base

[@koenth2359]

Are you sure this has to be solved over the real numbers? looks too easy for a MO question 1. exclude x=0, because 0^0 is indeterminate 2. divide by x , you get x^(x-1)=1 3. notice that either the exponent has to be 0, either the base has to be of norm 1. Over the reals: 4. try x=-1 and x=1: both work Over the complex numbers it's a bit more challenging

[@xristit1]

This is the best answer. The OP is very weak I've seen some of his videos.

[@robertlunderwood]

The reason why -1 works is because of the even exponent that results.

[@stuartgrier5605]

Easy. X=1 or -1

[@adanlopez928]

If x^x = x^1, x=1

[@abbeyroad3657]

X is 1, ready!

[@tinodaperson7174]

x^x=x 1^1=1 x=1

[@basti9935]

Smart bro

[@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!

[@niranjanchakraborty1139]

Ans x=1.

[@SuperYoonHo]

oh it looks like it's 1 and -1

[@joelfotso7227]

Why all these steps

[@touhami3472]

x^x=x: •if x=0==> 0^0 indefinded: 0 not solution, • if x0 then we can divide both sides by x: x^(x-1)=1 x^p=1 , p=x-1. But x^p=1 • p=0 and x0: x=1 is solution, • or x=1 and p in R: x=1 is solution, • or x=-1 and p is even: -1 is solution.

[@shantanusrivastava9898]

Even if we ignore the previous errors in case 2, even the Last two steps in case 2 are wrong. If -y-1 =1, then -y is 1 , which means -x is 1. Which is the same answer as case 1. How are you getting x is minus 1.

[@paulmartos7730]

This is a problem? Obviously 1^1 = 1, and -1^-1 (or 1 over -1) = -1.

[@-mrgd1-384]

1 right?

[@mathwithmershscore]

-1 is also a solution so you may check it again🙏

[@justabunga1]

Here is another method if you look at it another way. In order to extend more values and since the domain isn't mentioned in this video here, the absolute value must be there in x for the base on the left hand side of the equation and also the right hand side. This can be written as |x|^x=|x|. Again, rewrite this as e^(xln(|x|)=e^ln(|x|). This means xln(|x|)-ln(|x|)=(x-1)ln(|x|)=0, so x=±1 is the answer. However, if you were to graph the equation y=x^x, we have to use calculus, and notice that there is a hole at x=0, to show that it is undefined. The limit comes out to be 1 from the use of l'Hopital's rule since it's in the indeterminate form of 0^0. The only solution to that answer would only be x=1 since the domain of y=x^x is x>0.

[@angelmendez-rivera351]

The domain is not x > 0. (-1)^(-1) = -1, so x = -1 is a solution.

[@justabunga1]

@@angelmendez-rivera351 the domain of y=x^x is x>0. x=-1 is out of the domain. Even though you could plug in negative numbers in the x, some values may be real or imaginary. For this graph to happen, we want everything to be continuous. Therefore, the domain is x>0. However, the video asks for what the value of x is here. It didn't say anything about the domain. Other than x=1, therefore, x=-1 is also a valid solution.

[@eduardionovich4425]

Правильное решение! Всем отечественным придуркам( вроде Трушина и т.п.) разобрать и выучить наизусть.

[@soniaalboresi5488]

x=1

[@mathwithmershscore]

You are right but but -1 is also an answer

[@Poyu_dumb]

When you spent 5min solving the equation while a kid said that it only took him 5 seconds :

[@GodbornNoven]

bruh too easy x^x = x log_x on both sides x=log_x x = 1 LOL

[@WagesOfDestruction]

I think the eqn at 5:04 is wrong on the right side it is not 1 but (-1)^(-y-1)

[@_SANDHIYAS]

We can just compare the powers of x and find its 1

[@jimaanders7527]

Spoiler Alert: X = 1 (or -1) well...duh

[@arunimabera9858]

The answer will be 1 X^x=X X^x=X^1 X=1

[@bhosdibalekalove]

Abby log in both side by base should be logx

[@soundariyasenthil8365]

Taking log xlnx = lnx x=1 if lnx not equal to 0 So x can't be 1 We don't get answer by using log

[@dpage446]

You can rewrite the equation as xlnx-lnx=0. lnx(x-1)=0 Now before taking logarithm we have to add a modulus because otherwise we assume x to be positive. Therefore, ln|x|=0 gives x=+1 or -1.

[@Kavya0704]

Can't you just take log on both sides

[@Justintro]

just look at it

[@roddurde5462]

Me 8th grader 1 to the power 1 = 1 hehe boi

[@202alexwong]

Log x^x = Log x X Log x = Log x X = 1

[@SunilSingh-xx7qp]

Sir answer of question no 1 = 1

[@hans429]

1^1=1, no more solutions...

[@th1v5]

1 because 1 to the 1 is 1

[@mahamaarouf9933]

1^1 =1 Right????

[@jacksonloke2395]

X=+-1