Let $$f(x)=\frac{e^x-1}{x}=$$
Looking at the graph in comparison to the exponential function $g(x)=e^x$, it is clear that $f^{-1}(x)$ grows in a similar fashion to $g^{-1}(x)=\ln(x)$. However, I am pretty sure that there is not a closed form for $f^{-1}$. Certainly it can not be found by the elementary methods taught in courses such as precalculus, etc. Consider switching $x$ and $y$ and solving for $y$;
$$x=\frac{e^y-1}{y}\Rightarrow xy+1=e^y$$
I mean, I suppose this is a valid expression, it is just not expressable in function notation.
Why can we not find an expression for $f^{-1}$ in terms of the logarithmic function, since $f$ and $g$ are clearly closely related?