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Assume that $f:(0,\infty)\to\mathbb{R}$ is a differentiable monotone function satisfying $$f^{-1}(f'(x))=e^{1/x},\ \forall\ x\in (0,\infty)\tag{1}$$

If $f(x)=\log_a{x}$ for $a>0$ and $a\neq 1$ then, $f$ is a solution of $(1)$. My question is: Is logarithm the only solution of $(1)$?

I was trying to prove that $f(xy)=f(x)+f(y)$ because this would implies (with the monotonocity) that $f$ is a logarithm, however I could not prove it until now.

Tomás
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  • I think you will not be able to prove this, due to the inverse function in there. Clearly any constant (except 0) multiplied with the logarithm is a solution. This is clear from your solution of $\log_a x = \frac{\log x}{\log a}$ with $a \neq 1$. I tried assuming another function $g$ that satisfies (1) and showing $h(x)=f(x)+g(x)$ would have to satisfy as well. However sums of invertible functions need not be invertible. If we assume that $f$ is not only monotonic but strictly monotonic we can show that $h$ is invertible, but not necessarily expresible in terms of $f+g$. – Jan Aug 13 '18 at 09:42

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