(3) Let $a, b$ be positive real numbers, $\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}, (a-b)^{2}=4(a b)^{3}$, then $\log _{a} b=$
(3) -1 Hint: From $\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}$, we get $a+b \leqslant 2 \sqrt{2} a b$. Also,
$$
\begin{array}{c}
(a+b)^{2}=4 a b+(a-b)^{2}=4 a b+4(a b)^{3} \\
\geqslant 4 \cdot 2 \sqrt{a b \cdot(a b)^{3}}=8(a b)^{2},
\end{array}
$$

which means
$$
a+b \geqslant 2 \sqrt{2} a b .
$$

Thus,
$$
a+b=2 \sqrt{2} a b .
$$

From the condition for equality in inequality (1), we get $a b=1$. Solving this together with (2), we obtain $\left\{\begin{array}{l}a=\sqrt{2}-1, \\ b=\sqrt{2}+1,\end{array}\right.$ or $\left\{\begin{array}{l}a=\sqrt{2}+1, \\ b=\sqrt{2}-1,\end{array}\right.$ Therefore, $\log _{a} b=-1$.