Cauchy's logarithmic equation

Remarks

Example: $f(x) = \log_2 x$ is a solution, since $\log_2(xy) = \log_2 x + \log_2 y$. Non-example: $f(x) = x - 1$ is not a solution, since $f(2 \cdot 2) = 3$ while $f(2) + f(2) = 2$. The substitution $x = e^s$, $y = e^t$ turns the equation into additivity of $A(s) = f(e^s)$. So the solutions are $f(x) = A(\ln x)$ with $A$ an additive function, and the continuous ones are $f(x) = c \ln x$. On the domain $\R \setminus \set{0}$ the solutions are $f(x) = A(\ln \abs{x})$; on all of $\R$ only $f = 0$ is a solution, since $f(0) = f(0) + f(x)$.