Mathematics - Basic maths Question with Solution | TestHub

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