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 ?
Using the Lambert W function, which is the inverse of $f(x)=xe^x$, we get $$ \begin{align} x^x&=a\\ x\log(x)&=\log(a)\\ \log(x)e^{\log(x)}&=\log(a)\\ \log(x)&=\mathrm{W}(\log(a))\\ x&=e^{\mathrm{W}(\log(a))} \end{align} $$ There are an infinite set of complex branches of the Lambert W function corresponding to the multiple solutions of $xe^x=y$. There is one real branch for $y\gt0$ and two real branches for $y\lt0$.
In the question at hand, $\log\left(\frac1{\sqrt2}\right)\lt0$, so we get two real branches, giving the answers $\frac12$ and $\frac14$ for $x^x=\frac1{\sqrt2}$ .
I gave an algorithm for computing the real branches of Lambert W in this answer.
One natural thing to try is taking logarithms. This gives
$$x\log x = \log\left({1\over\sqrt2}\right)={1\over2}\log\left({1\over2}\right)$$
from which the solution $x=1/2$ stands out. One would likewise find the solution $x=1/3$ for the equation
$$x^x = {1\over\sqrt[3]3}$$
The other solution, $x=1/4$, however, can be traced to the fact that $2^2=2\times2$, so that
$${1\over4}\log\left({1\over4}\right)={1\over4}\log\left({1\over2^2}\right)={2\over4}\log\left({1\over2}\right)={1\over2}\log\left({1\over2}\right)$$
There is no correspondingly nice relationship when the $2$'s are replaced by $3$'s.
Nope; $x^x$ is a (non-elementary) transcendental function so, in general, an equation involving it has no algebraic solution.
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.
Sometimes you will see the Lambert W-function (or some branch thereof) invoked to handle such questions.
