If $2^{x+y} = 4^{x-y} = \sqrt{32}$, then $xy =$ ___.

Write everything as a power of $2$:

$4^{x - y} = (2^2)^{x - y} = 2^{2(x - y)}$

$\sqrt{32} = (2^5)^{\frac{1}{2}} = 2^{\frac{5}{2}}$

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When the bases are equal, the powers are equal:

$x + y = \dfrac{5}{2}$

$2(x - y) = \dfrac{5}{2}$

$x - y = \dfrac{5}{4}$

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Adding the two equations:

$2x = \dfrac{5}{2} + \dfrac{5}{4}$

$2x = \dfrac{15}{4}$

$x = \dfrac{15}{8}$

$y = \dfrac{5}{2} - \dfrac{15}{8}$

$= \dfrac{20 - 15}{8}$

$= \dfrac{5}{8}$

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$xy = \dfrac{15}{8} \times \dfrac{5}{8}$

$= \dfrac{75}{64}$

Therefore, $xy = \dfrac{75}{64}$.