Inverseof the function $f(x)=\frac{a^{2x}-1}{a^{2x}+1}$

Question

I have problem to invert this function , Can anyone help me to solve it? $$f(x)=\frac{a^{2x}-1}{a^{2x}+1}$$

My attempt: change $x$ to $y$ and try to solve for $y$, but I could not.

Answer 1

$\begin{align} y &=\dfrac{a^{2x}-1}{a^{2x}+1} \\ \implies ya^{2x}+y&=a^{2x}-1\\\implies a^{2x}(y-1)&=-(y+1)\\\implies a^{2x}&=\dfrac{y+1}{1-y} \\\implies Then\ switch\ X\ and \ Y \\ x&= \frac12 \log_a(\dfrac{y+1}{1-y}) \\\implies y&= \frac12 \log_a(\dfrac{x+1}{1-x})\end{align}$

Answer 2

Hint: Switch $x$ and $y$ first in the equation, then solve for $y$.

$x = \dfrac{a^{2y} - 1}{a^{2y}+1} \to xa^{2y} + x = a^{2y} - 1 \to a^{2y} = \dfrac{1+x}{1-x} \to \log_a\left(\frac{1+x}{1-x}\right) = 2y \to y = ??$