8.14 If $\frac{4^{x}}{2^{x+y}}=8$ and $\frac{9^{x+y}}{3^{5 y}}=243$, where $x, y$ are real numbers, then $x y$ equals
(A) $\frac{12}{5}$.
(B) 4 .
(C) 6 .
(D) 12 .
(E) -4 .
(17th American High School Mathematics Examination, 1966)
［Solution］ From the given, we have
$$
\begin{array}{l}
\frac{4^{x}}{2^{x+y}}=\frac{2^{2 x}}{2^{x+y}}=2^{x-y}=8=2^{3}, \quad \therefore \quad x-y=3, \\
\frac{9^{x+y}}{3^{5 y}}=\frac{3^{2(x+y)}}{3^{5 y}}=3^{2 x-3 y}=243=3^{5}, \quad \therefore \quad 2 x-3 y=5 .
\end{array}
$$

Solving the system of equations above, we get $x=4, y=1$. Therefore, $x y=4$. Hence, the answer is $(B)$.