Which functions have the property $f(ab)=f(a)+f(b)$?

Question

I know that the logarithmic functions satisfy this, as $$\log_x{ab}=\log_x{a}+\log_x{b}$$ but do any other functions satisfy this? Thank you.

I have seen this SE question (A function that verifies the property $f(ab) = f(a) + f(b)$), but I don't really undertand its notation and it hasn't actually given any examples of other fucntions that satisfy the stated property.

https://math.stackexchange.com/questions/98673/examples-of-functions-where-fab-fafb — scoopfaze, Sep 28 '20 at 21:57
@scoopfaze I don't understand how it gets from $f(e^x e^y)$ to $f(e^x)+f(e^y)$. — A-Level Student, Sep 28 '20 at 21:59
@A-levelStudent $f$ is assumed to be a function that has this property — alphaomega, Sep 28 '20 at 22:05
@A-levelStudent. Take $a=e^x$ and $b=e^y$ in $f(ab)=f(a)+f(b).$ — md2perpe, Sep 28 '20 at 22:07

Robert Shore · Answer 1 · 2020-09-28T22:37:49.600

There are a bunch: $2^{2^{\aleph_0}}$ many. Choose a value for $f(1)$. That will force a value of $f(x)$ for all $x \in \Bbb Q$.

Now choose your favorite irrational number. (I'm partial to $\sqrt 2$ myself.) Assign it a value arbitrarily. That forces a value for any number of the form $a+b \sqrt 2$, where $a, b \in \Bbb Q$.

Now choose another irrational number that's not a (finite) linear combination (over $\Bbb Q$) of $1$ and $\sqrt 2$. Lather, rinse, repeat. You'll have $2^{\aleph_0}$ choices to make before you run out of numbers that avoid previous linear combinations, and you can assign a value for each of those choices arbitrarily.

Which functions have the property $f(ab)=f(a)+f(b)$?

1 Answers1