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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 $$f(e^x) = f(e) + k f(x),\qquad f(1) = 0. \tag{2}\label{2}$$

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?

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?

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 $a>0$. 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 $$f(e^x) = f(e) + k f(x),\qquad f(1) = 0. \tag{2}\label{2}$$

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?

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 $$f(e^x) = f(e) + k f(x),\qquad f(1) = 0. \tag{2}\label{2}$$

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?

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 $$f(a^b) = f(a) + h(b), \qquad h(1) = 0. \tag{1}\label{1}$$ 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 $$f(e^x) = f(e) + k f(x),\qquad f(1) = 0. \tag{2}\label{2}$$

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?

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 $a>0$. 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 $$f(e^x) = f(e) + k f(x),\qquad f(1) = 0. \tag{2}\label{2}$$

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?