Solving $x^x=\frac{1}{\sqrt 2}$

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 ?

Solution 1:

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.

Solution 2:

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.