HMMT November

Let $g(x)=f(x)+1$. Substituting $g$ into the functional equation, we get that $$\begin{gathered} g(x y)-1=g(x)-1+g(y)-1+1 \\ g(x y)=g(x)+g(y) \end{gathered}$$ Also, $g(2)=1$. Now substitute $x=e^{x^{\prime}}, y=e^{y^{\prime}}$, which is possible because $x, y \in \mathbb{R}^{+}$. Then set $h(x)=g\left(e^{x}\right)$. This gives us that $$g\left(e^{x^{\prime}+y^{\prime}}\right)=g\left(e^{x^{\prime}}\right)+g\left(e^{y^{\prime}}\right) \Longrightarrow h\left(x^{\prime}+y^{\prime}\right)=h\left(x^{\prime}\right)+h\left(y^{\prime}\right)$$ for al $x^{\prime}, y^{\prime} \in \mathbb{R}$. Also $h$ is continuous. Therefore, by Cauchy's functional equation, $h(x)=c x$ for a real number c. Going all the way back to $g$, we can get that $g(x)=c \log x$. Since $g(2)=1, c=\frac{1}{\log 2}$. Therefore, $g(2015)=c \log 2015=\frac{\log 2015}{\log 2}=\log _{2} 2015$. Finally, $f(2015)=g(2015)-1=\log _{2} 2015-1$.