$4 \cdot 234$ If $\log _{8} a+\log _{4} b^{2}=5$, and $\log _{8} b+\log _{4} a^{2}=7$, find $a b$.

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[Solution] Adding the two equations given in the problem, we get
$$
\log _{8} a+\log _{8} b+\log _{4} b^{2}+\log _{4} a^{2}=12,
$$

Using the properties of logarithms, we have
$$
\log _{8}(a b)+\log _{4}(a b)^{2}=12,
$$

Using the change of base formula, we get
$$
\frac{\log _{2}(a b)}{3}+\log _{2}(a b)=12,
$$

Simplifying, we get $\quad \log _{2}(a b)=9$,
Solving, we get $a b=2^{9}=512$.