It's a well known result that, with some additional assumptions (often continuity or monotonicity), all functions $f : \mathbb{R}_{> 0} \to \mathbb{R}$ satisfying $f(ab) = f(a) + f(b)$ are constant multiples of $\log$.
It occurs to me that I've never seen a function which shows the necessity of the additional assumptions. Is there a function which satisfies $f(ab) = f(a) + f(b)$ but is not a constant multiple of $\log$?
There are (many) exotic solutions $\alpha\colon\mathbb{R}\to\mathbb{R}$ of the functional equation $\alpha(x+y)=\alpha(x)+\alpha(y)$. Indeed, $\alpha$ is a solutions iff it is linear with respect to the structure of a rational vector space. The exotic ones are exactly those sich are not multiples of the identity.
So $f:=\alpha\circ\log$ is an exotic logarithmic function if $\alpha$ is as above.
Edit: It can be seen quite easily that $f$ satisfies the equation (1) $f(ab)=f(a)+f(b)$ if $\alpha:=f\circ\exp$ satisfies (2) $\alpha(x+y)=\alpha(x)+\alpha(y)$. Thus $f$ solves (1) iff $f=\alpha\circ\log$ with some $\alpha$ satisfying (2).