Question

Solve for x & y $\frac{7}{3^{{}^x}}-\frac{6}{2^y}=15$ & $\frac{8}{3^x}=\frac{9}{2^y}$

• A. -3, -2
• B. -2, -3
• C. $\frac{-1}{2},\frac{-1}{3}$
• D. $\frac{-1}{3},\frac{-1}{2}$

Answer 1

The correct option is B -2, -3

Given,$\frac{7}{3^x}-\frac{6}{2^y}=15$ ...(1) and

$\frac{8}{3^x}=\frac{9}{2^y}$ ...(2)

from (2)

$\frac{1}{3^x}=\frac{9}{2^{y+3}}$ since($8=2^3$)

$\frac{7}{3^x}=\frac{63}{2^{y+3}}$ ..(3)

substituting value in (1)

$\frac{63-48}{2^{y+3}}=15$

$15=15\left(2^{y+3}\right)$

$2^{y+3}=1$

$2^{y+3}=2^0$

Taking ${\log }_2\left(\right)$ on both sides

$y+3{\log }_{2}2=0{\log }_{2}2$

$y+3=0$

$y=-3$

On substituting $y=-3$ value in (2)

$\frac{8}{3^x}=\frac{9}{2^{-3}}$

$8\times 2^{-3}=9\times 3^x$

$x=-2$

Therefore $x=-2,y=-3$