Question

If $\frac{4^x}{2^{x+y}}=8$ and $\frac{9^{x+y}}{3^{5y}}=243$, $x$ and $y$ are real numbers, then $xy$ equals

• A. $\frac{12}{5}$
• B. $4$
• C. $6$
• D. $12$

Answer 1

Answer: (B)

$4$

$\frac{4^x}{2^{x+y}}=8$
$2^{2x-x-y}=2^3$
$2^{x-y}=2^3$
$x-y=3$ (1)
and $\frac{9^{x+y}}{3^{5y}}=243$
$\frac{3^{2(x+y)}}{3^{5y}}=3^5$
$3^{2x+2y-5y}=3^5$
$3^{2x-3y}=3^5$
$\therefore 2x-3y=5$ (2)
Multiplying equation (1) with 3 and subtracting from equation (2),
$\begin{array}{ccccc}2x& -& 3y& =& 5\\ 3x& -& 3y& =& 9\\ (-)& & (+)& & (-)\\ --& -& --& -& -\\ -x& & & =& -4\end{array}$
$x=4$
We know that
$x-y=3⟹4-y=3⟹y=1$
$\therefore xy=4\times 1=4$