Answer
$(\frac{1}{2},\frac{1}{16})$
Work Step by Step
1. Take the difference between the two equations, we have $log_x(2y)-log_x(4y)=1$ or $log_x(\frac{2y}{4y})=1$ or $log_x(\frac{1}{2})=1$, thus $x=\frac{1}{2}$
2. Use the 1st equation to get: $2y=(\frac{1}{2})^3$, thus $y=(\frac{1}{2})^4=\frac{1}{16}$
3. Check the solution $(\frac{1}{2},\frac{1}{16})$, it fits the original equations.
