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If a function $f(x)$ is proportional to $\ln x$, then we know $$ f(xy) = f(x) + f(y). $$

My question is, is the converse true? If we know that, for an unknown function f, $$ f(xy) = f(x) + f(y), $$ can we conclude that the function must be proportional to $\ln x$? Why?

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    $\begingroup$ Put $g(x)=f(e^x)$ and read this. $\endgroup$ – Pp.. Jan 21 '15 at 6:48
  • $\begingroup$ Hi. Does the function $f$ continuous? $\endgroup$ – Farhad Jan 21 '15 at 7:38
  • $\begingroup$ @Pp.. Much obliged. Plus I learned about an interesting new class of pathological functions! $\endgroup$ – thecommexokid Jan 21 '15 at 7:55
  • $\begingroup$ @Farhad, no, it is not explicitly stated to be so. But I deduce from Pp..'s linked article that a sufficient condition on f(x) for the logarithmic solution to be unique is that it be continuous at a single point (or monotonic over any interval or bounded over any interval). Given the context in which the question arose, I feel alright excluding any functions pathological enough not to meet any of those conditions. $\endgroup$ – thecommexokid Jan 21 '15 at 8:00
  • $\begingroup$ In this post you can find a few links some related equations. This question asks about continuous solutions of your (or Cauchy's) equation. $\endgroup$ – Martin Sleziak Jan 21 '15 at 8:10

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