Question

Let $(x_0,y_0)$ be the solution of the following equations
$(2x{)}^{\ln 2}=(3y{)}^{\ln 3}$
$3^{\ln x}=2^{\ln y}$
Then $x_0$is

• A. $\frac{1}{6}$
• B. $6$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

We have,

$(2x{)}^{\ln 2}=(3y{)}^{\ln 3}$
$\Rightarrow \left(\ln 2\right)lo{g}_{3}2x=\left(\ln 3\right)lo{g}_{3}3y$
$\Rightarrow lo{g}_{3y}2x=lo{g}_{2}3$
$\Rightarrow \left(2x\right)=3^a,\left(3y\right)=2^a,$ say
Also $3^{\ln x}=2^{\ln y}$
$\Rightarrow \ln x=\ln y\left(lo{g}_{3}2\right)$
$\Rightarrow lo{g}_{y}x=lo{g}_{3}2$
$\Rightarrow x=2^k,y=3^k$
$\Rightarrow 3^a=2^{k+1}$
$\Rightarrow a=0\text{and}k+1=0$
$\Rightarrow 2x=3^a$
$\Rightarrow x=\frac{1}{2}$