$x^x=(1-x)^{(1-x)}$

Question

I have this equation $x^x=(1-x)^{(1-x)}$ and I want to find $x$. The solution $x=1/2$ is clear and from the function plots ploted with, for example, wolfram alpha, is obvious this is the only situation. But how can I prove that this is the only one? I tried logarithmizing the equation and ariving at $x \ln x =(1-x) \ln(1-x)$, but I still couldn't prove it.

Thank you in advance!

Typically 0^0 is defined to equal 1. So x = 0 and x = 1 are also solutions. — Peter Kagey, Nov 14 '16 at 19:26
@PeterKagey 0^0 is not defined. You can eventually find its limit, but it's not defined in the usual sense. — Enrico M., Nov 14 '16 at 19:29
Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik): Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and x^0 have different limiting values when x decreases to 0. But this is a mistake. We must define x^0=1 for all x , if the binomial theorem is to be valid when x=0 , y=0 , and/or x=-y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant. — Peter Kagey, Nov 14 '16 at 19:39
Peter's point (if $0^0=1$, then $x=0$ and $x=1$ are solutions to the equation) is correct. But I wouldn't describe Concrete Mathematics as a “typical” source. — Matthew Leingang, Mar 18 '23 at 19:48

score 6 · Accepted Answer · answered Nov 14 '16 at 19:49

For $\;0<x<1\;$, of course:

$$x^x=(1-x)^{1-x}\iff e^{x\log x}=e^{(1-x)\log(1-x)}\iff x\log x=(1-x)\log(1-x)\iff$$

$$x\left[\log x+\log(1-x)\right]-\log(1-x)=0\;\;\;\color{red}{(**)}$$

Now, we check the function

$$f(x)=x\left[\log x+\log(1-x)\right]-\log(1-x)\implies f'(x)=\log\left(x(1-x)\right)+2$$

Observe that $\;\;0<x<1\implies 0<x(1-x)\le\cfrac12$, and from here, using monotony of the logarithm, we get:

$$f'(x)\le \log\frac12+2=-\log2+2>0$$

so $\;f\;$ is monotone ascending and thus injective, so $\;x=\frac12\;$ is the only solution to $\;\color{red}{(**)}\;$

hamam_Abdallah · Answer 2 · 2016-11-14T20:47:20.770

2

Put $t=\frac 12 -x$

the equation becomes

$$(\frac 12 -t)^{\frac 12-t}=(\frac12+t)^{\frac 12+t}$$

$\iff$

$$(\frac{1-2t}{1+2t})^{\frac 12-t}=(\frac{1+2t}{2})^{2t}=(\frac{1+2t}{1-2t})^{t-\frac 12}$$

$\implies$

$t=0, $ or $t=\frac 12$ or $t=-\frac 12$

$\implies$

$x=\frac 12$ or $x=0$ or $x=1$

edited Nov 14 '16 at 20:47

answered Nov 14 '16 at 19:53

hamam_Abdallah

62,951

Thanks for the answer! But how did you obtain the second egality? Because I am obtaining ${{((1-2t)}/{(1+2t))}}^{1/2} = {{(1+2t)}^t} / {2^t} * {{(1-2t)}^t} / {2^t}$ – Jane Doe Nov 14 '16 at 20:23
@JaneDoe i just used $A^n=(\frac{1}{A})^{-n}$. – hamam_Abdallah Nov 14 '16 at 20:27
Yes, I also used that and I obtain the above equation. – Jane Doe Nov 14 '16 at 20:33
@JaneDoe Yes you are right. i made some correction. is it ok now. – hamam_Abdallah Nov 14 '16 at 20:49

score 0 · Answer 3 · answered Mar 18 '23 at 21:16

We know that $f(x)=x^x$ has one local extremum in $(0,1)$ which is the local minimum $(\frac1e,\frac{1}{\sqrt[e]{e}}).$

Assume that $a\in [0,1]$ is a solution of the equation $a^a=(1-a)^{1-a}$. Then $(a-a^2)^a=1-a$ and hence $(a-a^2)^{a^2}=(1-a)^a.$ Dividing the former by the latter, we get $$(a-a^2)^{a-a^2}=(1-a)^{1-a}=a^a.$$ Since we have one local extremum in the open interval, we deduce that either $a-a^2=1-a\implies a=1$ or $a-a^2=a\implies a=0$ or $a=1-a\implies a=\frac12$.

$x^x=(1-x)^{(1-x)}$

3 Answers3