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The title is self explanatory... How can I prove (or find a counter example) that $f(x)=\log{x^\lambda}$ is the only analytical solution to this functional equation? Assumptions be made on $\alpha$, $f$ and $x$ belonging to real.

Thanks.

gudise
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    $x^\lambda$ is not a solution... – Exodd Sep 21 '22 at 11:50
  • How about $f(x^{a+1})=f(x)+f(x^a)$? – Тyma Gaidash Sep 21 '22 at 11:52
  • @Exodd Very sorry you're right, I forgot the log – gudise Sep 21 '22 at 11:53
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    Let y=xα

    your equation can turn into f(xy)=f(x)+f(y), you can find many article about this Cauchy functional equation.

     – Abel Wong Sep 21 '22 at 12:01
  • @Tyma Gaidash of course both equations are equivalent (only taking logarithms at both sides). I wrote it in the "weird" form of exponentials because to my eye it made it evident that f being the inverse function of exp was a solution. – gudise Sep 21 '22 at 12:02
  • @Abel Wong Funny fact: I made $x^{\alpha}=y$ in an attempt to solve the original problem :-) – gudise Sep 21 '22 at 12:04
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    Could it happen that $e^{f(x^{\alpha+1})}=e^{f(x)}e^{f(x^\alpha)}$ even if $e^{f(xy)}=e^{f(x)}e^{f(y)}$ fails for general $x$ and $y$? Do we have to talk about $x^\alpha$ [or $x^\lambda$] where $x<0$ and $\alpha$ [or $\lambda$] is irrational? – GEdgar Sep 21 '22 at 12:06
  • @gudise. lol, I see. – Abel Wong Sep 21 '22 at 12:12
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    Since there is some dense function solution for g(x+y)=g(x)+g(y), which is not linear( g(x)=ax). Take f(x)=log(g(x)). Then f(xy)=f(x)+f(y) is a solution not in $f(x)=logx$ form. You can see link about the non-continuous solutions. Oh, can I log a dense function?, If I cannot, then your log function is the only solution. – Abel Wong Sep 21 '22 at 12:22

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