3.4.

From the operation of logarithms, we get
$$
\begin{array}{l}
\log _{a} 16+2 \log _{4} a=4 \log _{a} 2+\log _{2} a \\
=\frac{4}{\log _{2} a}+\log _{2} a .
\end{array}
$$

Since $a>1$, it follows that $\log _{2} a>0$.
By the AM-GM inequality, we have
$$
\frac{4}{\log _{2} a}+\log _{2} a \geqslant 2 \sqrt{\frac{4}{\log _{2} a} \log _{2} a}=4,
$$

with equality holding if and only if $a=4$.
3. Given $a>1$. Then the minimum value of $\log _{a} 16+2 \log _{4} a$ is . $\qquad$