Question

Let $(x_0,y_0)$ be thhe solution of the following equations $(2x{)}^{ln2}=(3y{)}^{ln3}$ and $3^{ln3}$ and $3^{lnx}=2^{lny}$. Then $xy=$

• A. $1/6$
• B. $1/3$
• C. $1/2$
• D. $0$

Answer 1

Answer: (A)

$1/6$

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

Taking $\ln$ both side,

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

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

$\Rightarrow \frac{\left(\ln 2\right)}{\left(\ln y\right)}(\ln 2+\ln x)=\left(\ln 3\right)(1+\frac{\ln 3}{\ln y})$

$\Rightarrow \left(\ln 2\right)(\frac{\ln 2}{\ln y}+\frac{\ln x}{\ln y})=\ln 3(1+\frac{\ln 3}{\ln y})$(equation $1$)

And $3^{\ln x}=2^{\ln y}$

Taking $\ln$ both side,

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

$\Rightarrow \frac{\ln x}{\ln y}=\frac{\ln 2}{\ln 3}$(equation $2$)

From $1$ and $2$

$\Rightarrow \left(\ln 2\right)(\frac{\ln 2}{\ln y}+\frac{\ln x}{\ln y})=\ln 3(1+\frac{\ln 3}{\ln y})$

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

$\Rightarrow \ln y=-\ln 3$

$\Rightarrow y=\frac{1}{3}$

And from $1$

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

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