Let $a,b$ be positive reals such that $\frac{1}{a}+\frac{1}{b}\leq2\sqrt2$ and $(a-b)^2=4(ab)^3$. Find $\log_a b$.
1. Given the equations:
   \[
   \frac{1}{a} + \frac{1}{b} \leq 2\sqrt{2}
   \]
   and
   \[
   (a - b)^2 = 4(ab)^3
   \]

2. We start by rewriting the second equation. Let \( a = x \) and \( b = y \). Then:
   \[
   (x - y)^2 = 4(xy)^3
   \]

3. We introduce new variables \( s = x + y \) and \( p = xy \). Then:
   \[
   (x - y)^2 = s^2 - 4p
   \]
   Substituting this into the given equation:
   \[
   s^2 - 4p = 4p^3
   \]
   Therefore:
   \[
   s^2 = 4p^3 + 4p
   \]

4. Next, we consider the inequality:
   \[
   \frac{1}{x} + \frac{1}{y} \leq 2\sqrt{2}
   \]
   This can be rewritten using \( s \) and \( p \):
   \[
   \frac{x + y}{xy} \leq 2\sqrt{2}
   \]
   which simplifies to:
   \[
   \frac{s}{p} \leq 2\sqrt{2}
   \]
   Therefore:
   \[
   s \leq 2\sqrt{2}p
   \]

5. Substituting \( s \leq 2\sqrt{2}p \) into the equation \( s^2 = 4p^3 + 4p \):
   \[
   (2\sqrt{2}p)^2 \geq 4p^3 + 4p
   \]
   Simplifying:
   \[
   8p^2 \geq 4p^3 + 4p
   \]
   Dividing by 4:
   \[
   2p^2 \geq p^3 + p
   \]
   Rearranging:
   \[
   p^3 - 2p^2 + p \leq 0
   \]
   Factoring out \( p \):
   \[
   p(p^2 - 2p + 1) \leq 0
   \]
   Factoring the quadratic:
   \[
   p(p - 1)^2 \leq 0
   \]

6. Since \( p = xy > 0 \) and \( p(p - 1)^2 \leq 0 \), the only solution is \( p = 1 \). Therefore:
   \[
   xy = 1
   \]

7. Given \( xy = 1 \), we have \( y = \frac{1}{x} \). We need to find \( \log_x y \):
   \[
   y = \frac{1}{x} = x^{-1}
   \]
   Therefore:
   \[
   \log_x y = \log_x (x^{-1}) = -1
   \]

The final answer is \(\boxed{-1}\).