Let left(x_0, y_0right) be the solution of th | vedclass

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Question

$(2 x)^{\ln 2}=(3 y)^{\ln 3}$

Taking $\ln$ both sides,

$\Rightarrow(\ln 2)(\ln 2 x)=(\ln 3)(\ln 3 y)$

$\Rightarrow(\ln 2)(\ln 2+\ln x)=(\ln 3

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

$(2 x)^{\ln 2}=(3 y)^{\ln 3}$

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 $x_0$.

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

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