Finding inverse of a function. Could someone please help me finding the inverse of the following function: $$f(x)=
\frac{x-1}{\ln(x)}$$ where $x>0$? 
Thank you.
 A: There is no solution in elementary functions but the inverse function is given by
$$ x = -y \mathcal{W}\left( -\frac{e^{-1/y}}{y} \right) $$
where $\mathcal{W}$ is the product log function.

step by step solution:
$$ y= \frac{x-1}{\ln{x}} \Longrightarrow -\ln{x} = - \frac{x-1}{y} \Longrightarrow \exp{\left(-\frac{x}{y}\right)} = \exp{\left(-\frac{1}{y}-\ln{x} \right)}$$
Now multiply both sides with $\frac{x}{y}$ and get:
$$\frac{x}{y}\exp{\left(-\frac{x}{y}\right)} = \frac{\exp{\left(-1/y\right)}}{y}\Longleftrightarrow -\frac{x}{y}\exp{\left(-\frac{x}{y}\right)} = -\frac{\exp{\left(-1/y\right)}}{y}$$
Now apply the definition of $\mathcal{W}$ and my answer mentioned before follows.
A: if you want to  find inverse of function $f(x)=y$  ,try to replace places for $x$ and $y$ or use $f(y)=x$ and solve it to back to $y$ ,in your case  set
$x=(y-1)/(ln(y)$  ,now  interpret $y$  by $x$
now  we have following ine
$x*ln(y)=(y-1)$
or
$ln(y)/(y-1)=1/x$
now please take  on exponent form and also please note that $e^{ln(x)}=x$
