How to prove that if a continuous function satisfies $f(a b)=f(a) + f(b)$, this function must be a log function? How to prove that if a continuous function satisfies $f(ab)=f(a)+f(b)$ and both $a$ and $b$ are positive real numbers, this function must be a log function?  i.e., proof of uniqueness. Thanks
 A: If $f(xy) = f(x) + f(y)$ then taking $g(x) = f(e^x)$ we get $g(x+y) = g(x) + g(y)$ which is Cauchy's functional equation. This equation have been discussed in many questions on this site, see Overview of basic facts about Cauchy functional equation for a very good overview with many links. If $f$ is assumed to be continuous (at a single point) then the only solutions $f:\mathbb{R_{>0}}\to \mathbb{R}$ are given by $f(x) = C \log(x)$ for some constant $C$. If continuity is not assumed there there does not have to be a unique solution (see the link above for how to construct explicit examples).
See also the more directly related questions:


*

*Functional equation $f(xy)=f(x)+f(y)$ and continuity

*Functional equation $f(xy)=f(x)+f(y)$ and differentiability

*Is there another function with a property like the log?
A: Here's a quick proof which neatly sidesteps all the $(1+1/n)^n$ stuff in Euler's original:
Let 
$$
{df\over dx} = g(x)
$$
Applying first principles
$$
{df\over dx} = \lim_{h\to 0} {f(x+h) - f(x) \over h}
$$
$$
= \lim_{h\to 0} {f(1+{h\over x})\over h}
$$
by virtue of the functional equaiton.
Now let $t=h/x$ and rewrite as a limit in $t$:
$$
{df\over dx} = \lim_{t\to 0} {f(1+t) \over tx}
$$
$$
= {g(1)\over x} \quad \text{by de l'Hopital's rule}
$$
This shows that only solutions to the functional equation have derivatives of the form $g(1)/x$.
We have a free choice of $g(1)$, which is equivalent to the choice of base for the log. Setting it to the 'natural choice' of $1$ gives us natural logs.
It's not hard to show that the differentiable solutions span all continuous solutions. Suppose there is a continuous solution $f'$. Select a value $x$ and consider the differentiable $f$ such that $f(x) = f'(x)$. Then $f$ and $f'$ must be equal for all rational powers of $x$. As these are dense, we can apply a limit to any real value and the two functions are therefore equal.
A: We know that:
$f(ab) = f(a) + f(b)$, that give us :
$f(1\cdot a) = f(1) + f(a)$ which give us :
$f(1) = 0$
Now : $f(\frac{1}{a} \cdot a) = f(1) = 0 = f(\frac{1}{a}) + f(a)$
Also : $f(a^{n}) = nf(a)$ , and using some kind of induction we could get: $f(a_{1}^{\alpha_{1}} \dots) = \sum \alpha_{i}f(a_{i})$.
So this function return us zero in $1$. Return us sum of power product with their multiples. And equals negate function in inverse point.
So this is $log_{a}(x)$.
