show $f(x)=f^{-1}(x)=x-\ln(e^x-1)$

Question

$f(x)=f^{-1}(x)=x-\ln(e^x-1)$, letting $f^{-1}$ denote the invese of $f$.

I have no idea how to show this algebraically and any Calculus methods I tried failed in horrible differential equations like $y'+e^y=1$ so I'm not sure.

It seems to work though https://www.desmos.com/calculator/yjktqiv1kz

Answer 1

Assuming $f^{-1}$ exists:

$y= x - \ln(\exp(x)-1)$, $x>0.$

Solve for $x$:

$\exp(y) = \exp(x)(\exp(x) -1)^{-1}.$

Set $z=\exp(x):$

$(z-1)\exp(y)=z;$

$z(\exp(y) -1) =\exp(y).$

$z= \exp(y)(\exp(y)-1)^{-1}.$

$x= \ln(z)= y -\ln(\exp(y)-1)$.

Answer 2

Do you just want to verify that $f^{-1}=f$? All this means is that $f(f(x))=x$. But \begin{align} f(f(x))&=f(x)-\ln(\exp(f(x))-1)\\ &=x-\ln(e^x-1)-\ln(e^x/(e^x-1)-1)\\ &=x-\ln(e^x-1)-\ln(1/(e^x-1))\\ &=\cdots \end{align}etc.