We know that

$$
\frac{8^{x}}{2^{x+y}}=64 \text{ and } \frac{9^{x+y}}{3^{4 y}}=243
$$

Determine the value of $2 x y$.

#
Solution

Since $8=2^{3}$ and $9=3^{2}$, we have

$$
\begin{aligned}
64 & =\frac{8^{x}}{2^{x+y}} \\
2^{6} & =2^{3 x-(x+y)} \\
& =2^{2 x-y}
\end{aligned}
$$

$$
\begin{aligned}
243 & =\frac{9^{x+y}}{3^{4 y}} \\
3^{5} & =3^{(2 x+2 y)-4 y} \\
& =3^{2 x-2 y}
\end{aligned}
$$

Thus, we have the following system:

$$
\left\{\begin{array}{l}
2 x-y=6 \\
2 x-2 y=5
\end{array}\right.
$$

Subtracting the second equation from the first, we get $y=1$. Substituting this value into the first equation, we get $x=7 / 2$. Therefore, $2 x y=2 \cdot \frac{7}{2} \cdot 1=7$.