5. If real numbers $x, y$ satisfy
$$
2^{x}+4 x+12=\log _{2}(y-1)^{3}+3 y+12=0 \text {, }
$$

then $x+y=$ $\qquad$ .
5. -2 .
$$
\begin{array}{l}
\text { Let } s=x-2 \text {, then } \\
2^{x}+4 x+12=0 \\
\Leftrightarrow 2^{s}=-s-5 ; \\
\text { Let } t=y-1 \text {, then } \\
\log _{2}(y-1)^{3}+3 y+12=0 \\
\Leftrightarrow \log _{2} t=-t-5 .
\end{array}
$$

Notice that, the graphs of the functions $y=2^{x}$ and $y=\log _{2} x$ are symmetric with respect to the line $y=x$, and the graph of the function $y=-x-5$ is also symmetric with respect to the line $y=x$. The intersection point of $y=x$ and $y=-x-5$ has an x-coordinate of $-\frac{5}{2}$, so $s+t=-5$.
$$
\text { Therefore, } x+y=s+t+3=-2 \text {. }
$$