A rigid exponential question This question was given by my friend

$$x^{x^{20}}=\sqrt[\sqrt 2]{2}$$

even after applying log this question dosen't simplifies

$x^{20}$ $ln$ $x$ $=$ $\frac{1}{\sqrt{2}}$ $ln$ $2$

if this question is put in wolfram alpha it gives and if I do a approx form it gives 1.067 how does this came

$x=e^{\frac{1}{20} W\left(10 \sqrt{2} \ln (2)\right)}$

How to solve this question
 A: Using by standard techniques, we have
$$\color{red}{\left(x^{x^{20}}\right)^{20}=\left(\sqrt[\sqrt 2]{2}\right)^{20}} \\\begin{align}&\implies x^{20 x^{20}}=\left(\sqrt[\sqrt 2]{2}\right)^{20} \\
&\implies \left({x^{20}}\right)^{ x^{20}}=\left(\sqrt[\sqrt 2]{2}\right)^{20}\end{align}$$
$$\color {blue}{ u(x)^{u(x)}=\left(\sqrt[\sqrt 2]{2}\right)^{20}}$$
$$\begin{align}&\implies \ln u(x)^{u(x)}=\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\\
&\implies u(x) \ln u(x)=\ln \left(\sqrt[\sqrt 2]{2}\right)^{20} \\
&\implies \ln u(x) e^{\ln u(x)}=\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\\
&\implies W \left(\ln u(x) e^{\ln u(x)}\right) =W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right) \\
&\implies \ln u(x)=W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right) \\
&\implies u(x)= e^{W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right)}\\
&\implies x^{20}=e^{W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right)}\end{align}$$
$$\begin{align}\color {gold}{\boxed {\color{black} {{x=e^{\frac {1}{20} W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right)}}}}}\end{align}$$
Note that ,
$$\left(\sqrt[\sqrt 2]{2}\right)^{20}=2^{10\sqrt 2}$$
$$\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}=\ln 2^{10\sqrt 2}=10\sqrt 2 \ln 2.$$

Finally, the real root of our equation is equal to:
$$\large \color {gold}{\boxed {\color{black} {{{x=e^{1/20 ~W\left (10 \sqrt 2 \ln 2\right)}}}}}}$$
A: It happens that this expression
can also be presented in a simple closed form, $x=\sqrt[8]2$:
\begin{align} 
(2^{1/8})^{(2^{1/8})^{20}}
&=
(2^{1/8})^{2^{20/8}}
\\&=
(2^{1/8})^{2^{2+1/2}}
\\&=
(2^{1/8})^{4\sqrt2}
\\&=
2^{{4\sqrt2}/8}
\\&=
2^{1/\sqrt2}
\\&=
\sqrt[\sqrt2]2
.
\end{align}
A: Chances are this problem doesn't have an explicit solution.
Equations of the form
$$x\cdot e^x=c$$
$c\in\mathbb{R}\setminus\{0\}$
cannot be solved by simple algebra, but there is a function called Lambert W which can give us approximations.
It is used to solve such and similar equations (like $x\cdot\sin(x)=2$, $x\cdot\coth(x)=-4$).
