\documentclass{article}
\usepackage{amsmath}
\begin{document}
\title{Solving $x^{x^x}=2^{-\sqrt2}$}
\author{Amit Yaron}
\date{Apr 16, 2024}
\maketitle
That's a nice equation I once found on YouTube; You don't need a numeric
method to solve it.\\
The first thing to do is to find how many positive solutions exist and where they lie.\\
Now, $2>0\land-\sqrt2<0\Rightarrow 2^{-\sqrt2}<1\Rightarrow 0<x<1$
Let us now check how many solution exists by checking where LHS decreases
and where increases. We don't have to derive $x^{x^x}$ for that purpose,
just take $ln$.
Deriving $x^x$ is nice, suppose:
\begin{equation*}y=x^x\end{equation*}
Then,
\begin{align}
&\ln(y)=x\ln(x)\Rightarrow\nonumber\\
\Rightarrow&\frac{y'}y=1+\ln(x)\Rightarrow\nonumber\\
\Rightarrow& y'=x^x\ln(x)
\end{align}
Let us define
\begin{equation*}f(x)=x^x\ln(x)\end{equation*}
Then, from (1) with the product rule:
\begin{align}f'(x)&=x^x(\frac1x)+x^x[\ln(x)+1]\ln(x)\nonumber\\
&=x^x[\ln^2(x)+\ln(x)+\frac1x]
\end{align}
From (2), let us find a necessary condition for $f'(x)$ to be zero/
\begin{align*}
&\ln^2(x)+\ln(x)+\frac1x=0\Rightarrow\\
\Rightarrow & \ln^2(x)+\ln(x)+\frac14+\frac1x-\frac14=0\Rightarrow\\
\Rightarrow & [\ln(x)+\frac12]^2+\frac1x-\frac14=0\Rightarrow\\
\Rightarrow & [\ln(x)+\frac12]^2=\frac14-\frac1x\Rightarrow\\
\Rightarrow & \frac14-\frac1x\geq0\Rightarrow\\
\Rightarrow & x-4\geq0\Rightarrow\\
\Rightarrow &x\geq4>1
\end{align*}
Actually, that never happens. Thus $f(x)$ is strictly increasing, which means
that only one solution exists. Let us find it:
\begin{align*}
2^{-\sqrt2}&=0.5^{\sqrt2}\\
&=0.5^{2^{0.5}}\\
&=0.25^{0.5\cdot2^{0.5}}\\
&=0.25^{0.25^{0.5}\cdot0.5^{-0.5}}\\
&=0.25^{0.25^{0.5\cdot0.25^{0.5}\cdot(-0.5)}}\\
&=0.25^{0.25^{0.5}\cdot0.25^{-0.25}}\\
&=0.25^{0.25^{0.5-0.25}}\\
&=0.25^{0.25^{0.25}}=x^{x^x}
\end{align*}
And only one solution exists, thus:
\begin{equation*}\boxed{x=0.25}\end{equation*}
\end{document}
