Question

If $2^{x+y}=6^y$ and $3^{x-1}=2^{y+1}$, then the value of $\frac{\log 3-\log 2}{x-y}$ will be

• A. $1$
• B. ${\log }_{2}3-{\log }_{3}2$
• C. $\log \frac{3}{2}$
• D. none of these

Answer 1

The correct option is C $\log \frac{3}{2}$

$2^{x+y}=6^y\Rightarrow (x+y)\log 2=y\log 6\Rightarrow (x+y)\log 2=y(\log 2+\log 3)$
$\Rightarrow x\log 2=y\log 3$ ...(1)
$3^{x-1}=2^{y+1}\Rightarrow (x-1)\log 3=(y+1)\log 2$
$\Rightarrow x\log 3-y\log 2=\log 3+\log 2$ ...(2)
From (1) & (2)
$x=\frac{\log 3}{\log 3-\log 2}$
$y=\frac{\log 2}{\log 3-\log 2}$
$\Rightarrow x-y=\frac{\log 3-\log 2}{\log 3-\log 2}\Rightarrow \frac{\log 3-\log 2}{x-y}=\log \frac{3}{2}$