# "tl;dr: I did some reading on related topics, an..."

https://arbital.com/p/4yb

by Kaya Fallenstein Jun 30 2016 updated Jun 30 2016

tl;dr: I did some reading on related topics, and it turns out that (1) may be sufficient to define logarithms if we take as an axiom that every set is Lebesgue measurable (which is incompatible with the axiom of choice). Otherwise, we need to add an additional condition to (1).

(1) states that $f(x\cdot y)=f(x)+f(y)$. Given a function $g$ satisfying this condition, we can generate an additional function satisfying this condition by composing $g$ with a function $h$, where $h(x+y)=h(x)+h(y)$:

$$h(g(x\cdot y))=h(g(x))+h(g(y))$$

$h$, as defined, is a solution to Cauchy's functional equation. The family of functions given by $h(x)=ch(x)$ for some constant $c$ is always a solution, giving the usual logarithm family. The existence of other solutions is independent of ZF. When they do exist they are always pathological and generate non-Lebesgue measurable sets (for more, see this stackexchange link).

We can prove the existence of such solutions in ZFC by noting that the solutions of the Cauchy functional equation are exactly the homomorphisms from the additive group of $\mathbb{R}$ to itself. The real numbers form an infinite dimensional vector space over the field $\mathbb{Q}$. Linear transformations from the vector space to itself translate into homomorphisms from the group to itself. Since the axiom of choice implies that any vector space has a basis, we can, for example, find a non-trivial linear transformation by swapping two basis vectors. This in turn induces a homomorphism from the group to itself. (The Wikipedia page gives the general form of a solution to this functional, which turn out to be all the linear transformations on the vector space.)

(I'm not saying that this article should discuss axiomatizations of set theory, but it doesn't seem good to make statements that are only true if you assume, e.g., an unusual alternative to the axiom of choice.)

Wikipedia proves that the pathological solutions must all be dense in $\mathbb{R}$, so to exclude them, we can adopt any number of conditions. Wikipedia points at "$f$ is continuous", "$f$ is monotonic on any interval", and "$f$ is bounded on any interval". Continuity seems to be most intuitive; once we have defined the value of the function on the rationals (which we can do with basically the arguments already on this page), the rest of its values are determined.