5. Given $x y z+y+z=12$, then the maximum value of $\log _{4} x+\log _{2} y+\log _{2} z$ is $\qquad$
Solve:
$12=x y z+y+z \geqslant 3 \sqrt[3]{x y^{2} z^{2}} \Rightarrow x y^{2} z^{2} \leqslant 64$, equality holds when $x=\frac{1}{4}, y=z=4$. Therefore, $\log _{4} x+\log _{2} y+\log _{2} z=\log _{4}\left(x y^{2} z^{2}\right) \leqslant 3$.