Prove that $f(ab) = f(a) + f(b)$ Question : Assume only that $f: (0,\infty)\to{\mathbb{R}}$ is differentiable and that $f'(x) = 1/x$, and $f(1)=0$. Prove that for all $a,b \in(0,\infty)$, $f(ab)=f(a)+f(b)$. [Hint: Let $g(x)=f(ax)$]
My solution : We know that
$$f(x) = \int f'(x)dx = \int\frac{1}{x}dx = \ln(x) + C$$
Also, we are given that $f(1)=0$, so $f(1)=\ln(1)+C=0$, which gives us $C=0$. Therefore, $f(x)= \ln(x)$. 
Now, by the laws of logarithm, we have $\ln(ab)=\ln(a)+ \ln(b)$ for all $a,b\in(0,\infty)$. Hence, $f (ab) = f(a) + f(b)$.
Is this the right approach? I am not really sure how to use the hint of letting $g(x) = f(ax)$.
 A: Usually such a theorem is used to prove the properties of the natural logarithm which is defined as $\int_1^x \frac{1}{t}\ dt$. In that case your proof is invalid simply because $\ln$ is not defined yet!
To prove the theorem directly, consider $\int_1^{ab} \frac{1}{t}\ dt = \int_1^a \frac{1}{t}\ dt + \int_a^{ab} \frac{1}{t}\ dt$ and try to prove that $\int_a^{ab} \frac{1}{t}\ dt = \int_1^b \frac{1}{t}\ dt$.
Note that while egreg's answer does not need integration, it does not prove the existence of the function with the specified properties. This isn't necessary since the question already gave such a function, but one rigorous way of defining $\ln$ is precisely via the integral, so in fact this answer proves more than required, at the expense of using stronger mathematical machinery.
A: You need no integral.
Consider, as the hint tells you, $g(x)=f(ax)$ and compute its derivative:
$$
g'(x)=af'(ax)=a\frac{1}{ax}=\frac{1}{x}
$$
Therefore there exists $k$ such that, for every $x$,
$$
g(x)=k+f(x)
$$
(because differentiable functions defined over an interval that have the same derivative differ by a constant, a well known consequence of the mean value theorem).
Now compute at $x=1$: $g(1)=k+f(1)$, that means $f(a)=k$. Now the above identity, for $x=b$, reads
$$
f(ab)=f(a)+f(b)
$$
A: Shortly,
$$f'(ax)-f'(x)=\frac a{ax}-\frac1x=0.$$ 
By differentiability $f(ax)-f(x)$ is a constant, so that with $x=b$ and $x=1$
$$f(ab)-f(b)=f(a)-f(1)=f(a).$$
