If $f(xy) = f(x) + f(y)$, show that $f(.)$ can only be a logarithmic function.

Question

As the question states, show that the property exhibited can only be satisfied by a logarithmic function i.e no other family of functions can satisfy the above property.

Answer 1

Continuity is necessary.

If $F(x+y)=F(x)+F(y)$, for all $x,y$ and $F$ discontinuous (such $F$ exist due to the Axiom of Choice, and in particular, the fact that $\mathbb R$ over $\mathbb Q$ possesses a Hamel basis) and $f(x)=F(\log x)$, then $f(xy)=f(x)+f(y)$, and $f$ is not logarithmic!

Answer 2

With the assumption that $f$ is continuous, compute the partial derivatives; firstly wrt to $x$ then $y$ \begin{align} yf'(xy) &=f'(x) \\ xf'(xy) &=f'(y) \\ \end{align} Equating terms in $f(xy)$ we have $$\frac{1}{y}f'(x)=\frac{1}{x}f'(y)$$ Or $$xf'(x)=yf'(y)$$ Now, if you know anything about these equations you will notice something interesting about both sides of the equation above, allowing you to make the crucial step in obtaining a very simple ODE to solve; of which the solution should be the one you are looking for.