Inverse transformation for exponential function

Question

Given the function:

$$ f(x)\ :=\ 2^{\frac{x+1}{x-1}} $$

What is the correct approach for $f(x)$ transformation as its inverse?

Kindly regards for the support.

Answer 1

The inverse function:

$$ g(y)\ :=\ \frac{\log_2(y)+1}{\log_2(y)-1} $$

REMARK   Domains:

$$ f : \mathbb R\setminus\{1\}\ \rightarrow\ (0;\infty)\setminus\{2\} $$ and $$ g : (0;\infty)\setminus{2}\ \rightarrow\ \mathbb R\setminus\{1\} $$

 

Explanation:

Function $\ f\ $ is a composition of two functions:

$$ f = \psi\circ \phi $$

where $\ \phi(s)\ :=\ \frac{s+1}{s-1}\ $ and $\ \psi(t)\ :=\ 2^t.\ $ The inverse of $\ \phi\ $ is this function itself. The inverse of the exponential function $\ \psi\ $ is $\ \log_2.\ $ Thus inverse of $\ f\ $ is $\ g\ :=\ \phi\circ\log_2.$

Answer 2

Answer is : $$ x = {{\log(y) + \log(2)} \over {\log(y)-\log(2)}} $$ Demo :
$$ y = f(x) \\ {{x+1} \over {x-1}} = {{\log y} \over {\log 2}} = P \\ x+1=P(x-1) \\ x+1=Px-P \\ x={{P+1} \over {P-1}} = {{{{\log y} \over {\log 2}}+1} \over {{\log y} \over {\log 2}}-1} $$