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I was trying to solve

$$ f(a^x)=a^{f(x)} $$ for any $a,x$ .

with $a \in \mathbb{R}$, $x \in \mathbb{R}$ and $f(\cdot)$ ranging where the math does not break down on real line (for example it breaks at $f(x)<0$ and $a=2n+1$ with $n$ as a natural number).

This below is what I've tried:

Take $x=1$ so $f(a)=a^{f(1)}$ as $a$ is real as $x$ does so $f(x)=x^{f(1)}$, then $f(x)=x^{c}$ but $f(a^x)=(a^x)^c$ not $f(a^{x})=a^{x^c}$ (by $f(a^x)=a^{f(x)}$). We might think that $a^{x^{c}}=a^{xc}$ holds for some $c$ so we let know what $c$ is, we have $a^{x^c}$ must be $a^{xc}$ so $a^{x^c}=a^{xc}$ implies to $x^{c}\ln a=xc\ln a$ then $x^c=xc$ so $c$ as constant may be $c=1$ then $f(x)=x$. When I made $a^{x^c}=a^{xc}$ in order to find $c$ have I got loss of generality? Is $f = \operatorname{id}$ the only solution?

Davi Américo
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  • I took the liberty to format your question. Hope this is to your liking, and if otherwise, please feel free to make any further changes :) – Sangchul Lee Jun 06 '22 at 19:42
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    Your quantifiers are not clear. So you mean $\forall a;\forall x; f(a^x)=a^{f(x)}$ ... if not, say what you do mean. – GEdgar Jun 06 '22 at 19:45

2 Answers2

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No! Try $f(x)=a^x$. (Assuming $a$ is fixed and $x$ is variable, see the comments.)

For that matter, the functions $f(x)=a^{a^x}$ and $f(x)=a^{a^{a^x}}$ etc. should also work.

Franklin Pezzuti Dyer
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Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that

$$ f(a^x) = a^{f(x)} $$

holds for any $a > 0$ and $x \in \mathbb{R}$. Then for each $a > 0$,

$$ f(a) = f(a^1) = a^{f(1)} = a^{f(1^0)} = a^{1^{f(0)}} = a. $$

Then plugging $a = e^x$ for $x \in \mathbb{R}$,

$$ e^x = f(e^x) = e^{f(x)}. $$

This implies that $f(x) = x$.

Sangchul Lee
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