I was inspired by [Does there exist a function which converts exponentiation into addition?](https://math.stackexchange.com/questions/4475785/does-there-exist-a-function-which-converts-exponentiation-into-addition) to think about mapping exponentiation onto addition.
The question asks whether there exists $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(a^b) = f(a) + f(b).$$ Given that $f(a) + f(b)$ is symmetric in $a$ and $b$ but $f(a^b)$ is not, we conclude that $f = 0$ is the only solution.


Let's broaden our scope then, to consider $$f(a^b) = g(a) + h(b).$$ Firstly we note that $f(a^1) = g(a) + h(1)$, so without loss of generality restrict to $$f(a^b) = f(a) + h(b), \qquad h(1) = 0 \tag{1}\label{1}$$ for $a>0$. Letting $a=1$ or $b=0$ implies that $h$ need be constant if $f$ is defined for all positive reals, so restrict to $f : (0,1)\cup(1,\infty)\rightarrow \mathbb{R}$.


So, then we want to have a natural way to write $h$ in terms of $f$, and then try to solve for $f$.


My first approach was to restrict to $h = k f$ for $k\in\mathbb{R}$ giving $f(a^b) = f(a) + k f(b)$. Then we could look for $f$ with $$f(x^x) = (1 + k) f(x),\qquad f(1) = 0. \tag{2}\label{2}$$


A second idea I had was to let $h(b) = f(b^b) - f(b)$, motivated by letting $a = b$ in \eqref{1}, giving $f(a^b) = f(a) + f(b^b) - f(b)$. Then we would try to solve $$f(e^x) = f(e) + f(x^x) - f(x). \tag{3}\label{3}$$


So my question is, is there a natural solution to \eqref{1}, either by solving \eqref{2}, or \eqref{3}, or otherwise?


Also, logarithms are widely used and it seems that if such functions exist then they could potentially also be useful. Are functions which satisfy \eqref{1} considered in the literature?



Mapping exponentiation onto addition
$\def\abs#1{\lvert#1\rvert}$Though it’s not clearly stated in the question, I take it $f$ is supposed to be defined only for $a>0$, as otherwise $a^b$ has no sensible definition for non-integer $b$.


For $a>0$, $a\ne1$ and $b\ne0$, a simple solution is
$$\log\abs{\log a^b}=\log\abs{\log a}+\log\abs b.$$


The domain restrictions $a\ne1$ and $b\ne0$ are necessary: as noted in LSpice’s answer, there is no nontrivial choice of $f$, $g$, and $h$ that works for $a=1$, as this forces $h$ to be constant. Likewise, no nontrivial choice works for $b=0$, as it forces $g$ to be constant.


