I was inspired by this question 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 $$f(a^b) = f(a) + h(b), \qquad h(1) = 0. \tag{1}$$ 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 $$f(e^x) = f(e) + k f(x),\qquad f(1) = 0. \tag{2}$$

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

So my question is, is there a natural solution to (1), either by solving (2), or (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 (1) considered in the literature?