, 2015

Solution:
Answer: $\log_{2} 2015-1$
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 $x=e^{x'}$, $y=e^{y'}$, which is possible because $x, y \in \mathbb{R}^{+}$. Then set $h(x)=g\left(e^{x}\right)$. This gives us that
$$g\left(e^{x'+y'}\right)=g\left(e^{x'}\right)+g\left(e^{y'}\right) \Longrightarrow h\left(x'+y'\right)=h\left(x'\right)+h\left(y'\right)$$
for all $x', y' \in \mathbb{R}$. 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, $f(2015)=g(2015)-1=\log_{2} 2015-1$.