smc

Try solving both equations for $a$. Taking the $x$-th root of both sides in the first equation and the $z$-the root of both sides in the second gives $a=c^{\frac{q}{x}}$ and $a=c^{\frac{y}{z}}$. So $\frac{q}{x}=\frac{y}{z}$. Multiplying both sides by $xz$, $qz=xy$. $\boxed{xy=qz}$.