3. Let $a, b$ be positive real numbers, and
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}, \\
(a-b)^{2}=4(a b)^{3} .
\end{array}
$$

Then $\log _{a} b=$ $\qquad$
3. -1 .

From $\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}$, we get $a+b \leqslant 2 \sqrt{2} a b$.
$$
\begin{array}{l}
\text { Also, }(a+b)^{2}=4 a b+(a-b)^{2} \\
=4 a b+4(a b)^{3} \\
\geqslant 4 \times 2 \sqrt{a b(a b)^{3}}=8(a b)^{2},
\end{array}
$$

which means $a+b \geqslant 2 \sqrt{2} a b$.
Thus, $a+b=2 \sqrt{2} a b$.
From the condition for equality in equation (1), we get $a b=1$.
Solving this together with equation (2), we have
Therefore, $\log _{a} b=-1$.