3. If the positive real numbers $x, y$ satisfy $y=2016 x$, and $x^{y}=y^{x}$, then $\log _{2016} x+\log _{2016} y=$ $\qquad$
3. $\frac{2017}{2015}$.

Notice that,
$$
x^{y}=y^{x} \Leftrightarrow x \log _{2016} y=y \log _{2016} x \text {. }
$$

By $y=2016 x$
$$
\begin{array}{l}
\Rightarrow 2016 x \log _{2016} x=x \log _{2016} 2016 x \\
\Rightarrow \log _{2016} x=\frac{1}{2015} \\
\Rightarrow \log _{2016} x+\log _{2016} y \\
\quad=2 \log _{2016} x+1=\frac{2017}{2015} .
\end{array}
$$