Answer: 49

Solution:

Write $x = 2^a,$ $y = 2^b,$ $z = 2^c.$ Then $\log_x(2y) = \frac{b + 1}{a},$ $\log_{2x}(4z) = \frac{c + 2}{a + 1},$ and $\log_{2x^4}(8yz) = \frac{b + c + 3}{4a + 1},$ so the condition is $$\begin{aligned} \frac{2(b + 1)}{a} &= \frac{2(c + 2)}{a + 1} \\ &= \frac{b + c + 3}{4a + 1} \ne 0. \end{aligned}$$

From the first two, $\frac{b + 1}{a} = \frac{c + 2}{a + 1},$ and equal ratios also equal their mediant $\frac{b + c + 3}{2a + 1}.$ Comparing with the third expression gives $\frac{2(b + c + 3)}{2a + 1} = \frac{b + c + 3}{4a + 1}.$ The common value is nonzero, so $b + c + 3 \ne 0,$ and thus $2(4a + 1) = 2a + 1,$ giving $a = -\frac{1}{6}.$ Then $\frac{b + 1}{-1/6} = \frac{c + 2}{5/6}$ yields $c + 2 = -5(b + 1),$ that is, $5b + c = -7.$

Therefore $xy^5z = 2^{a + 5b + c} = 2^{-1/6 - 7} = \frac{1}{2^{43/6}},$ so $p + q = 43 + 6 = 49.$