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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?

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    $\begingroup$ Not necessarily closed form, but usually the inverse of complicated functions involving exponentials and logs result in the Lambert-W function. For this example it would be, $$ x= \frac{-yW(-1/(e^{1/y}y))-1}{y}$$ $\endgroup$ – Dando18 Jul 10 '17 at 3:06
  • $\begingroup$ Yeah, I have seen the lambert function before, though have not studied it at all. $\endgroup$ – AveryJessup Jul 10 '17 at 3:08
  • $\begingroup$ Out of curiosity, is the purpose of representing $f^{-1}$ to prove it is increasing/decreasing or convex/concave over $\mathbb{R}^+$? If so, you may just derive such properties from $f$, that is studied here: math.stackexchange.com/questions/2356878/… $\endgroup$ – Jack D'Aurizio Jul 16 '17 at 0:00
  • $\begingroup$ No, it just seems so strange to me, @JackD'Aurizio, that something so similar to the exponential function does not have an inverse, notationally is really what i mean, that is similar to the natural logarithmic function. In terms of a power series representation, the function differ by shifted coefficients, which seems like a basic transformation, and I'm not sure why we can't have a similar phenomenon for the inverse. $\endgroup$ – AveryJessup Jul 16 '17 at 1:43
  • $\begingroup$ @AveryJessup: $x-\varepsilon\sin(x)$ is very close to the identity function, but en.wikipedia.org/wiki/Kepler%27s_equation $\endgroup$ – Jack D'Aurizio Jul 16 '17 at 1:44

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