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It looks like there should be a way to do it: $e^{x}$ satisfies $f(x+y)=f(x)f(y)$.

Meanwhile its inverse, the natural logarithm, satisfies a similar looking but inverted equation: $f(xy)=f(x)+f(y)$.

Surely there must be some way to manipulate or derive the second functional equation from the first functional equation, right?

John Joe
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  • I dont know if I understood right, but suppose that $f$ is bijective and satisfies the first equality and let $g$ be its inverse. Hence, $f(g(x)+g(y))=xy$. Then $g(f(g(x)+g(y)))=g(xy)$ and $g(x)+g(y)=g(xy)$. – VJunior Feb 07 '18 at 21:24
  • You can label anything you want as an inverse, I'm certainly not stopping you. – John Joe Feb 07 '18 at 21:25
  • Is there a generalized shortcut for doing this? It seems like you can almost swap each instance of a variable for the inverse function and vice versa? – John Joe Feb 07 '18 at 21:45

1 Answers1

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In general you can't, but if you know that $f$ is invertible, then given $a,b$ you can take $x=f^{-1}(a)$, $y=f^{-1}(b)$, and then the first equality gives you $$ f^{-1}(ab)=f^{-1}(f(x)f(y))=f^{-1}(f(x+y))=x+y=f^{-1}(a)+f^{-1}(b). $$ As you can do this for any $a,b$, you have obtained the second equality.

Martin Argerami
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