Question

If ${\left(\frac{9}{4}\right)}^{x+y}={\left(\frac{3}{2}\right)}^{16}$ and ${\left(\frac{49}{64}\right)}^{x-2y}={\left(\frac{7}{8}\right)}^4$, then $xy$ =

• A. $12$
• B. $14$
• C. $15$
• D. $16$

Answer 1

Answer: (A)

$12$

${\left(\frac{9}{4}\right)}^{x+y}={\left(\frac{3}{2}\right)}^{16}$
$\Rightarrow {\left(\frac{3}{2}\right)}^{2(x+y)}={\left(\frac{3}{2}\right)}^{16}$
$=>2x+2y=16$ --- $\left(1\right)$
${\left(\frac{49}{64}\right)}^{x-2y}={\left(\frac{7}{8}\right)}^4$
${\left(\frac{7}{8}\right)}^{2(x-2y)}={\left(\frac{7}{8}\right)}^4$
$\Rightarrow 2x-4y=4$ --- $\left(2\right)$
Subtracting eqn $\left(2\right)$ from $\left(1\right)$ is
$6y=12$
$\Rightarrow y=2$
So, from eqn $\left(1\right)$

$2x+2\left(2\right)=16$
$\Rightarrow 2x+4=16$
$\Rightarrow 2x=12$
$\Rightarrow x=6$
So, $xy=6\times 2=12$