MathematicsEasy110×since 2002Q3882The inverse of y=5logxy = {5^{\log x}}y=5logx is :Ax=5logyx = {5^{\log y}}x=5logyBx=y1log5x = {y^{{1 \over {\log 5}}}}x=ylog51Cx=51logyx = {5^{{1 \over {\log y}}}}x=5logy1Dx=ylog5x = {y^{\log 5}}x=ylog5Check answerSkip