What is $x$ if $x^{x^x} = {(1/2)}^{\sqrt 2}$?

Question

What is $x$ if $x^{x^x} = {(1/2)}^{\sqrt 2}$?

Answer provided more or less like this : \begin{align}{\left(\frac 12\right)}^{\sqrt 2}&=\frac1{2^{\sqrt 2}}\\ 2^{\sqrt 2} &= 2^{(2^{(2/4)})} \\ &= 2^{(4^{(1/4)})}\\ & = 4^{1/2(4^{(1/4)})} \\ &= 4^{2^{-1}(4^{(1/4)})} \\ &= 4^{4^{-1/2}(4^{(1/4)})}\\ &= 4^{(4^{(-1/4)})}\\ &= 4^{({1/4}^{(1/4)})} \\ {\left(\frac 12\right)}^{\sqrt 2} &= \frac1{2^{\sqrt 2}} \\ &= \frac1{4^{({1/4}^{(1/4)})}} \\ &= {\left(\frac14\right)}^{({1/4}^{(1/4)})} \\ x&=\frac 14\end{align}

Is there more elegant way that show $x=\frac 14$ the only answer?

Edit : source https://youtu.be/d-E5isaIDTA

Answer 1

Here is a calculus-based proof that $x=1/4$ is the only positive real solution to the equation $x^{x^x}=(1/2)^\sqrt2$.

It suffices to show that $x^{x^x}$ is increasing for all $x\gt0$. This is clear for $x\ge1$, so it remains to consider what happens for $0\lt x\lt1$. It's convenient to let $x=e^{-u}$ with $u\gt0$, and, after taking logarithms (twice), show that $f(u)=\ln u-ue^{-u}$ is increasing (from $-\infty$ as $u\to0$ to $\infty$ as $u\to\infty$). For this we need to show that

$$f'(u)={1\over u}-(1-u)e^{-u}={1-u(1-u)e^{-u}\over u}\ge0$$

for all $u\gt0$. But this is clear since $u\gt0$ implies $e^{-u}\lt1$, hence

$$u(1-u)e^{-u}\lt u(1-u)\le{1\over4}\lt1$$

(the maximum for $u(1-u)$ occurring at $u=1/2$).

If there is a non-calculus proof that $x^{x^x}$ is increasing, I'd be keen to see it.