Let (x_0, y_0) be the solution of the following equations (2 x)^ 2...

Taking log on both sides, $$\log 2 \cdot \log (2 x)=\log 3(\log 3 y)$$\begin{array}{ll} \Rightarrow & \log 2\{\log 2+\log x\} \\ & =\log 3\{\log 3+\log y\} \\ \text{ and } & \log x \cdot \log 3=\log y \log 2 \\ & \log y=\frac{\log x \cdot \log 3}{\log 2} \end{array} From Eqs. (i) and (ii), we get \begin{aligned} & \log 2\{\log 2+\log x\} \\ &=\log 3 \cdot\left\{\log 3+\frac{\log x \cdot \log 3}{\log 2}\right\} \\ & \Rightarrow \quad(\log 2)^2+\log 2 \cdot \log x \\ &=(\log 3)^2+\frac{(\log 3)^2}{(\log 2)} \cdot \log x \\ & \Rightarrow \quad \log x\left\{\frac{(\log 3)^2}{\log 2}-\log 2\right\} \\ &=(\log 2)^2-(\log 3)^2 \end{aligned}\begin{aligned} & \Rightarrow \quad \log x=-\log 2=\log 2^{-1} \\ & \therefore \quad x=\frac{1}{2} \end{aligned}