$F(xy) = F(x)+F(y)$ Proof

Question

Suppose $F$ is differentiable $\forall x>0$ and $F(xy) = F(x)+F(y)$, $ \forall x,y>0$. Prove that if $F$ is not the zero function, then $\exists$ $ a>0$ such that: $F(x)=\log_a(x)$, $\forall x>0$. I seem to be doing fine except on getting the base $a$ for the log. So far I have that $F'(x)=\frac{F'(1)}{x}$. I know from calculus that $\int\frac{1}{t}dt=\ln(t)$. How can I get the base to be $F'(1)$ instead of $e$?

Answer 1

$$a^x=\left(e^{\text{log}(a)}\right)^x = e^{\text{log}(a)x}$$ $$\text{log}(a^x)=x\text{log}(a)$$ $$\text{log}_a(a^x)=x$$

Extrapolating from here, we have

$$\text{log}_a(x) = \frac {\text{log}(x)}{\text{log}(a)}$$

It shouldn't be too hard to get your answer from this.