Let (x_0, y_0) be the solution of the followi | vedclass

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(C) Given equations are $(2x)^{\ln 2} = (3y)^{\ln 3}$ and $3^{\ln x} = 2^{\ln y}$.Taking $\ln$ on both sides of the first equation:$(\ln 2)(\ln 2 + \l

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$\frac{1}{2}$

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(C) Given equations are $(2x)^{\ln 2} = (3y)^{\ln 3}$ and $3^{\ln x} = 2^{\ln y}$.Taking $\ln$ on both sides of the first equation:$(\ln 2)(\ln 2 + \ln x) = (\ln 3)(\ln 3 + \ln y) \quad (1)$Taking $\ln$ on both sides of the second equation:$(\ln x)(\ln 3) = (\ln y)(\ln 2) \Rightarrow \ln y = \frac{(\ln x)(\ln 3)}{\ln 2}$.Substituting $\ln y$ into $(1)$:$(\ln 2)^2 + (\ln 2)(\ln x) = (\ln 3)^2 + (\ln 3)\left(\frac{(\ln x)(\ln 3)}{\ln 2}\right)$$(\ln x) \left(\ln 2 - \frac{(\ln 3)^2}{\ln 2}\right) = (\ln 3)^2 - (\ln 2)^2$$(\ln x) \left(\frac{(\ln 2)^2 - (\ln 3)^2}{\ln 2}\right) = (\ln 3)^2 - (\ln 2)^2$Dividing both sides by $((\ln 2)^2 - (\ln 3)^2)$,we get:$\frac{\ln x}{\ln 2} = -1$$\ln x = -\ln 2 = \ln(2^{-1})$$x = 2^{-1} = \frac{1}{2}$.Thus,$x_0 = \frac{1}{2}$.

Let $(x_0, y_0)$ be the solution of the following equations: $(2x)^{\ln 2} = (3y)^{\ln 3}$ and $3^{\ln x} = 2^{\ln y}$. Then $x_0$ is

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