Question

If $(2x{)}^{\ln 2}=(3y{)}^{\ln 3}$, $3^{\ln x}=2^{\ln y}$ and $(x_0,y_0)$ is the solution of these equations, then $x_0$ is

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

Answer 1

Answer: (C)

$\frac{1}{2}$

Let $(2x{)}^{\ln 2}=(3y{)}^{\ln 3}$ $\Rightarrow$ (i)

$3^{\ln x}=2^{\ln y}$ $\Rightarrow$ (ii)

(i) $\Rightarrow \left(\ln 2\right)\left(\ln 2x\right)=\left(\ln 3\right)\left(\ln 3y\right)$

$\Rightarrow \left(\ln 2\right)(\ln x+\ln 2)=\left(\ln 3\right)(\ln 3+\ln y)$

Let $\ln x=a,\ln y=b$

$\Rightarrow \left(\ln 2\right)(a+\ln 2)=\left(\ln 3\right)(\ln 3+b)$ $\Rightarrow$ (iii)

(ii)$\Rightarrow \left(\ln x\right)\left(\ln 3\right)=\left(\ln 2\right)\left(\ln y\right)$

Substituting $a,b$

$\Rightarrow a\left(\ln 3\right)=b\left(\ln 2\right)$ $\Rightarrow$ (iv)

Substituting the value of b from (iv) in (iii), we get

$a\left(\ln 2\right)+\left(\right(\ln 2{)}^2)=({\ln 3}^2)+a\frac{(\ln 3{)}^2}{\ln 2}$

On simplifying we get,

$a=\ln 2\frac{\left(\right(\ln 3{)}^2-\left(\ln 2{)}^2\right)}{\left(\right(\ln 2{)}^2-\left(\ln 3{)}^2\right)}$

$\Rightarrow a=-\ln 2$

$\Rightarrow \ln x=-\ln 2$

$\Rightarrow x=\frac{1}{2}$