AMC 12 Mock Exam 4 Problem 18

Let $t=\log_2x$; the bases $x,2x,4x$ must differ from $1$, so $t\notin\{0,-1,-2\}$. Since $\log_{2^k x}2=\frac{1}{k+t}$, the equation is $$\frac1t+\frac1{t+1}+\frac1{t+2}=0\iff (t+1)(t+2)+t(t+2)+t(t+1)=0\iff 3t^2+6t+2=0.$$ Its discriminant is $36-24=12>0$, and its roots $t=-1\pm\frac{1}{\sqrt3}$ avoid $0,-1,-2$, so both give solutions. The product of the two values of $x$ is $2^{t_1+t_2}=2^{-2}=\boxed{\tfrac14}$.