Prove or disprove the validity of a statement and find all solutions

Question

Find all possible values of $x$ if $$\sqrt[x]{\frac{2}{x-1}}=(x-1)^{(x-2)}$$

My attempt:

$$\frac{2}{x-1}=(x-1)^{(x-2)x}$$ $$\implies2=(x-1)^{(x^2-2x+1)}$$ $$\implies2=(x-1)^{(x-1)^2}$$ Let $x-1=a$

Therefore $$2=a^{a^2}$$ By observation, $a=\pm\sqrt2$ satisfies the equation. $\implies$ $$x=\sqrt2+1 \:\:\textrm{or} \:\:x=1-\sqrt2$$ My doubts:

$1.$ Is an expression like $$\sqrt[{\sqrt2+1}]{y}$$ or $$\sqrt[{-\sqrt2+1}]{y}$$ even valid$?$

$2.$ How do we find all solutions to the equation $$2=a^{a^2}$$ assuming that the domain of $a$ is both $\mathbb{R}$ and $\mathbb{C}$

Answer 1

Note that to determine real solutions you should also check that we only have one solution, in this case since for $1<x\le 2$ we have $f(x)=(x-1)^{(x-1)^2}\le 1$ and $f(x)$ is continuous and increasing for $x>2$, by IVT, we can claim that $x=\sqrt2+1$ is the unique real solution for $x>1$.

By inspection we find that also the symmetric value $x=-\sqrt2+1$ is also a solution.

This kind of expression $\sqrt[{\sqrt2+1}]{y}$ and $\sqrt[{-\sqrt2+1}]{y}$ are valid and equivalent to

$$\sqrt[{\sqrt2+1}]{y}={y}^{\frac1{\sqrt2+1}}$$ $$\sqrt[{-\sqrt2+1}]{y}={y}^{\frac1{-\sqrt2+1}}=\frac1{{y}^{\frac1{\sqrt2-1}}}$$

with $y>0$ if we are dealing with real expressions.

For a general solution we can use Lambert W function as discusse for example here

• Lambert Function as a solution