If $a^{x}= c^{q}= b$ and $c^{y}= a^{z}= d$, then
$\textbf{(A)}\ xy = qz\qquad\textbf{(B)}\ \frac{x}{y}=\frac{q}{z}\qquad\textbf{(C)}\ x+y = q+z\qquad\textbf{(D)}\ x-y = q-z$
$\textbf{(E)}\ x^{y}= q^{z}$
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 $xz$, $qz=xy$. $\boxed{\textbf{(A)}}$.