China Mathematical Competition

From $\frac{1}{a}+\frac{1}{b}\le 2\sqrt{2}$, we have $a+b\le 2\sqrt{2}ab$. On the other hand, $$\begin{gathered} (a+b{)}^2=4ab+(a-b{)}^2=4ab+4(ab{)}^3 \\ \ge 4\cdot 2\sqrt{ab}\cdot (ab{)}^3=8(ab{)}^2, \end{gathered}$$ and that means $$a+b\ge 2\sqrt{2}ab.\stackrel{◯}{1}$$ Therefore, $$a+b=2\sqrt{2}ab.\stackrel{◯}{2}$$ The equality in 1 holds only when $ab=1$. Associating it with 2, we find $$\{\begin{array}{l}a=\sqrt{2}-1,\\ b=\sqrt{2}+1,\end{array}\text{and}\{\begin{array}{l}a=\sqrt{2}+1,\\ b=\sqrt{2}-1.\end{array}$$ So the answer is ${\log }_{a}b=-1$.