Question

If ${72}^x.{48}^y=6^{xy}$, where $x$ and $y$ are nonzero rational numbers, then $x+y$ equals.

• A. $3$
• B. $\frac{10}{3}$
• C. $-3$
• D. $-\frac{10}{3}$

Answer 1

The correct option is D $-\frac{10}{3}$

${72}^x\cdot {48}^y=6^{xy}\Rightarrow (2^3\cdot 3^2{)}^x\cdot (2^4\cdot 3{)}^y=2^{xy}\cdot 3^{xy}$

Equating powers of $2$ and $3$, we get

$\Rightarrow 3x+4y=xy\dots eqn\left(1\right)$

$\Rightarrow 2x+3y=xy\dots eqn\left(2\right)$

Subtract eqn(1) from eqn(2), we get

$x=-3y$ Put this value of $x$ in eqn (2).

$-6y+y=-3y^2\Rightarrow y=\frac{5}{3}$

and $x=-5$

$x+y=-5+\frac{5}{3}=\frac{-10}{3}$