Proving a property of logarithms

Question

I have been looking online for this proof, but unable to find it (even on Math Stack Exchange).

I want to double check that everything here is correct:

Let $\begin{equation} F:(0,\infty ) \mapsto \mathbb{R}\end{equation}$ be a differentiable function such that $\ F'(x) ={ 1\over{x}}$ and $\;F(1)=0$.
Prove that $\ F(ab) = F(a) + F(b) $.

Here's how I went about it:
Define $h(x) = F(ax) -F(a)$ (*)

$h'(x) = 1/x$, $\;\;h(1)=F(a)-F(a)=0$

Consequently, h(x) must be equal to F(x) (they are equal at a point and derivative is equal for $\mathbb{R}$). Then, $F(x) = F(ax)-F(a)$ from (*) and substituting $x=b$, we have $F(ab) = F(a) +F(b)$.

Am I missing any conditions here? Does it make sense?

Answer 1

I would do

$$h(x)=F(ax)-F(x) \Rightarrow h'(x)=0$$

then $h(x)$ is constant.

$$h(1)=F(a)$$

So $F(ax)-F(x)=F(a)$

now make $x=b$ and get $F(ab)=F(a)+F(b)$.

Answer 2

For $a,x>0$ we define

$g(x)=F(ax)-F(x)-F(a)$.

$g$ is differentiable at $(0,+\infty)$,

and

$$\forall x>0\;\; g'(x)=aF'(ax)-F'(x)$$

$$=a\frac{1}{ax}-\frac{1}{x}=0$$

$$\implies \forall x>0 \;\;g(x)=g(1)=0$$

$$\implies \forall a,x>0\;\; F(ax)=F(a)+F(x)$$