5. Let $x y z+y+z=12$. Then the maximum value of $\log _{4} x+\log _{2} y+\log _{2} z$ is $\qquad$ .
5.3.

From the given conditions, we have
$$
\begin{array}{l}
12=x y z+y+z \geqslant 3 \sqrt[3]{x y^{2} z^{2}} \\
\Rightarrow x y^{2} z^{2} \leqslant 64 .
\end{array}
$$

Then $\log _{4} x+\log _{2} y+\log _{2} z=\log _{4} x y^{2} z^{2}$
$$
\leqslant \log _{4} 64=3 \text {, }
$$

with equality holding if and only if $x=\frac{1}{4}, y=z=4$, in which case the maximum value of 3 is achieved.