TITLE: Solving $x^x=\frac{1}{\sqrt 2}$
QUESTION [10 upvotes]: The equation $$x^x=\frac{1}{\sqrt 2},x\in \mathbb R$$
has two obvious solutions $0.5$ and $0.25$
One can easily prove they are the only ones using differential calculus.
Is there any natural algebraic manipulation that would lead to finding these solutions ?

REPLY [5 votes]: In fact, it is a special case of the equation $x^x = y^y$ with $x \neq y$, showing (1/2,1/4) is a solution. To find other rational solutions of this equation, one does not need differential calculus.
Set $x = y^s$, so 
$x^x = (y^s)^{y^s} = y^{s.y^s} = y^y$ iff 
$s.y^s = y$ iff 
$y^{(s-1)} = (1/s)$ iff 
$y = (1/s)^{1/(s-1)}$
Note that y is rational iff $1/(s-1) =  N$ is a natural number, hence solving for $s$ gives, 
$s = 1 + 1/N = (N+1)/N$, giving  
$y = [{N \over N+1}]^N$ and
$x = y^s = [{N \over N+1}]^{(N+1)}$
Note that with these $x$ and $y$, $x^x = y^y = [{N \over N+1}]^{N^{(N+1)} \over (N+1)^N}$
For $N=1$ we get $(x(1),y(1)) = (1/2, 1/4)$, hence $(1/2)^{(1/2)} = (1/4)^{(1/4)} = \sqrt 2$
For $N=2$ we get $(x(2), y(2)) = (4/9, 8/27)$, hence 
$(4/9)^{(4/9)} = (8/27)^{(8/27)} = (2/3)^{(8/9)} = ({256 \over 6561})^{(1/9)} = 0.69738... $
etc.