Mapping exponentiation onto addition

Question

I was inspired by 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?

Answer 1

$\def\abs#1{\lvert#1\rvert}$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.

Answer 2

This answer is to the original version of the question, where $f$ was assumed to be defined on all of $\mathbb R$, and the equation $f(a^b) = g(a) + h(b)$ was apparently demanded for all $a \in \mathbb R$.

Since $f(1^b) = f(1)$ equals $g(1) + h(b)$ for all $b$, we have that $h$ is constant, so that $f$ is constant on powers, in the sense that $b \mapsto f(a^b)$ is constant for every $a$. In particular, $f$ is constant on $\mathbb R_{> 0}$.

I am not sure how to make sense of your proposed equation $f(a^b) = g(a) + h(b)$ for, for example, $a = 0$ and $b = -1$, or $a = -1$ and $b = 1/2$.