Functional equations with involutions

Question

Having seen that the topic of functional equations is loved by StackExchange, I have constructed this problem hoping that it will please readers.

Solve the functional equation $$ [f(x)]^2f\left(\ln\left(\frac {e^x+1}{e^x-1}\right)\right)=ax $$ where $a$ is a positive constant and $x>0$.

HINT:

Prove first that $g(x)= \ln\left(\frac {e^x+1}{e^x-1}\right)$ is an involution.

Answer 1

Let $g(x)=\log\left(\frac{e^x+1}{e^{x}-1}\right)$: $g$ maps $\mathbb{R}^+$ into $\mathbb{R}^+$, $g(g(x))=x$ and the only fixed point of $g$ is $x_0=\log(1+\sqrt{2})$. For any $x\in\mathbb{R}^+\setminus\{x_0\}$, let $y=g(x)$. We have to fulfill:

$$ f(x)^2\,f(y) = a x,\qquad f(x)\,f(y)^2 = a y $$ hence by dividing the first identity by the square root of the second one$^{(*)}$ we get:

$$ f(x)^{\frac{3}{2}} = \sqrt{a}\frac{x}{\sqrt{y}} $$ hence: $$ f(x)^3 = \frac{a x^2}{y} = \frac{a x^2}{g(x)} $$ and:

$$ f(x) = \sqrt[3]{\frac{ax^2}{\log\left(\frac{e^x+1}{e^{x}-1}\right)}}.$$

$^{(*)}$ $f$ is positive since $f(y)=\frac{ax}{f(x)^2}> 0$ gives $f(x)> 0$.