\(\log _{1}^{1}=y=\log _{9}^{2 r-1}\) \(\log _{1}^{\prime}=y\) In index form \(x=3^{\prime} \ldots \ldots \ldots(1)\) also, \(y=\log _{9}^{2 r-1}\) in index form \(2 x-1=9^{\prime} \ldots \ldots \ldots(2)\) substitute equation(1) into (2) \(2\left(3^{\prime}\right)-1=9^{\prime}\) \(2\left(3^{\prime}\right)-1=3^{2 x}\) let \(3^{\prime}=m\) \(2 m-1=\left(3^{\prime}\right)^{2}\) \(2 m-1=m^{2}\) or \(m^{2}-2 m+1=0\) \(m^{2}-m-m+1=0\) \(m(m-1)-1(m-1)=0\) \((m-1)(m-1)=0\) \(m=1\) twice but \(m=3^{\prime}\) \(1=3^{\prime}\) \(3^{y}=1\) \(3^{\prime}=3^{0}, y=0\) from \((1)\) \(x=3^{\prime} \cdot x=3^{0}=1\) therefore, \(x+y=1\)