${\log }_{9}20={\log }_9\left(10\cdot 2\right)={\log }_{9}10+{\log }_{9}2=\frac{1}{\log 9}+\frac{\log 2}{\log 9}=\frac{1+\log 2}{\log 9}=\frac{1+\log 2}{\log 3^2}=\frac{1+\log 2}{2\cdot \log 3}=\frac{1+x}{2y}$