13. Let $a$ be a constant, $0<a \neq 1$. Find all functions $f: \mathrm{R}^{+} \rightarrow \mathrm{R}$, such that for any $\mathrm{x}, \mathrm{y} \in \mathrm{R}^{+}, f(\mathrm{xy})=f(\mathrm{x})+f(\mathrm{y})$ and $f(a)=1$.
Solution: Taking $\mathrm{x}=\mathrm{y}=1$ we get $f(1)=0, f^{\prime}(x)=\lim _{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}=\lim _{y \rightarrow 0} \frac{f\left(x\left(1+\frac{y}{x}\right)\right)-f(x)}{y}=\lim _{y \rightarrow 0} \frac{f\left(1+\frac{y}{x}\right)}{y}$ $=\frac{1}{x} \lim _{\frac{y}{x} \rightarrow 0} \frac{f\left(1+\frac{y}{x}\right)-f(1)}{\frac{y}{x}}=\frac{1}{x} f^{\prime}(1)$, let $\mathrm{C}_{1}=f^{\prime}(1)$, then $f^{\prime}(x)=\frac{C_{1}}{x}, f(x)=C_{1} \ln x+C_{2}$, where $\mathrm{C}_{1}$, $\mathrm{C}_{2}$ are constants, since $f(1)=\mathrm{C}_{2}=0, f(a)=\mathrm{C}_{1} \ln a=1$, so $C_{1}=\frac{1}{\ln a}, f(x)=\frac{\ln x}{\ln a}=\log _{a} x$.

It is also clear that $f(x)=\log _{a} x$ satisfies the equation, hence $f(x)=\log _{a} x$ is the solution.