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Let $\left(x_0, y_0\right)$ be the solution of the following equations $(2 x)^{\ln 2} =(3 y)^{\ln 3}$ $3^{\ln x} =2^{\ln y}$ . Then $x_0$ is
$\frac{1}{6}$
$\frac{1}{3}$
$\frac{1}{2}$
$6$
Solution
$(2 x)^{\ln 2}=(3 y)^{\ln 3}$
Taking $\ln$ both sides,
$\Rightarrow(\ln 2)(\ln 2 x)=(\ln 3)(\ln 3 y)$
$\Rightarrow(\ln 2)(\ln 2+\ln x)=(\ln 3)(\ln 3+\ln y) \rightarrow(1)$
Now, we will take the second equation,
$3^{\ln x}=2^{\ln y}$
Taking $\ln$ both sides,
$\Rightarrow(\ln x)(\ln 3)=(\ln y)(\ln 2)$
$\Rightarrow(\ln y)=\frac{(\ln x)(\ln 3)}{\ln 2}$
Putting value of $\ln y$ in $(1)$,
$(\ln 2)(\ln 2+\ln x)=(\ln 3)\left(\ln 3+\frac{(\ln x)(\ln 3)}{\ln 2}\right)$
$\Rightarrow(\ln x)\left(\ln 2-\left(\frac{(\ln 3)^2}{\ln 2}\right)\right)=(\ln 3)^2-(\ln 2)^2$
$\Rightarrow \frac{\ln x}{\ln 2}\left((\ln 2)^2-(\ln 3)^2\right)=(\ln 3)^2-(\ln 2)^2$
$\Rightarrow \frac{\ln x}{\ln 2}=-1$
$\Rightarrow(\ln x)=\ln (2)^{-1}$
$\Rightarrow x=2^{-1} \Rightarrow x=\frac{1}{2}$, which is the required value of $x_0$.