Annual Harvard-MIT November Tournament

Solution:
We can re-write the equation as $2^{x}\left(2 \cdot (2^{x})^{2} - 17 \cdot 2^{x} + 8\right) = 0$, or $2 \cdot (2^{x})^{2} - 17 \cdot 2^{x} + 8 = 0$. Make the substitution $y = 2^{x}$. Then we have $2y^{2} - 17y + 8 = 0$, which has solutions (by the quadratic formula) $y = \frac{17 \pm \sqrt{289 - 64}}{4} = \frac{17 \pm 15}{4} = 8, \frac{1}{2}$, so $2^{x} = 8, \frac{1}{2}$ and $x = 3, -1$. The product of these numbers is $-3$.