Let $(x_0,y_0)$ be the solution of the following equations: $(2x)^{In\,2} = (3y)^{In\,3}, 3^{In\,x} = 2^{In\,y}$

Then, $x_0$ is

$\frac{1}{2}$

We have, 

$(2x)^{In\,2} = (3y)^{In\,3}$ and, $3^{In\,x} = 2^{In\,y}$

$⇒\log(2x)^{In\,2} = \log(3y)^{In\,3}$ and $\log(3^{In\,x}) = \log(2^{In\,y})$

Taking log on both sides

$⇒\log 2 (\log 2+ \log x) = \log 3 (\log 3+ \log y)$ and, $\log 3 \log x = \log 2 \log y$

$⇒(\log 2)^2 + (\log 2) \log x=(\log 3)^2 + (\log 3) (\log y)$ and, $\log 3 \log x = \log 2 \log y$

$⇒(\log 2)^2 + (\log 2) (\log x) = (\log 3)^2 +\frac{(\log 3)^2\log x}{\log 2}$ [On eliminating $\log y$]

$⇒\frac{(\log 2)^2-(\log 3)^2}{\log 2}\log x=(\log 3)^2-(\log 2)^2$

$⇒\log x=-\log 2⇒x=\frac{1}{2}$