3. If positive real numbers $a, b$ satisfy $\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2},(a-b)^{2}=4(a b)^{3}$, then $\log _{a} b=$
$$
\begin{array}{l}
(a-b)^{2}=(a+b)^{2}-4 a b=4(a b)^{3} \Rightarrow(a+b)^{2}=4 a b\left[(a b)^{2}+1\right] \geqslant 8(a b)^{2} \\
\Rightarrow\left(\frac{a+b}{a b}\right)^{2} \geqslant 8 \Rightarrow \frac{1}{a}+\frac{1}{b} \geqslant 2 \sqrt{2} \Rightarrow \frac{1}{a}+\frac{1}{b}=2 \sqrt{2} . \text{ When the equality holds, } \\
\left\{\begin{array} { l } 
{ a b = 1 , } \\
{ a + b = 2 \sqrt { 2 } }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ a = \sqrt { 2 } + 1 , } \\
{ b = \sqrt { 2 } - 1 }
\end{array} \text{ or } \left\{\begin{array}{l}
a=\sqrt{2}-1, \\
b=\sqrt{2}+1
\end{array} \Rightarrow \log _{a} b=-1 .\right.\right.\right.
\end{array}
$$