Question

Let $\left(x_0,y_0\right)$ be 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. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. 6

Answer 1

Answer: (C)

$(2x{)}^{\ln 2}=(3y{)}^{\ln 3}$ ...(i)
$3^{\ln x}=2^{lny}$ ...(ii)
$\Rightarrow (\log x)(\log 3)=(\log y)\log 2$
$\Rightarrow \log y=\frac{(\log x)(\log 3)}{\log 2}$ ...(iii)
In (i) taking $\log$ both sides
$(\log 2)\{\log 2+\log \}=\log 3\{\log 3+\log y\}$
$(\log 2{)}^2+(\log 2)(\log x)=(\log 3{)}^2+\frac{(\log 3{)}^2(\log x)}{\log 2}$ from (iii) $\Rightarrow (\log 2{)}^2-(\log 3{)}^2=\frac{(\log 3{)}^2-(\log 2{)}^2}{\log 2}(\log x)$
$\Rightarrow -\log 2=\log x$
$\Rightarrow x=\frac{1}{2}$
$\Rightarrow x_0=\frac{1}{2}$