Question

Solve: ${\log }_{3x}\left(\frac{3}{x}\right)+{\log }_3^{2}x=1$

Answer 1

$\frac{\log 3-\log x}{\log 3+\log x}+{\log }_3^{2}x=1\frac{1-{\log }_{3}x}{1+{\log }_{3}x}+{\log }_3^{2}x=11-{\log }_{3}x+{\log }_3^{2}x+{\log }_3^{3}x=1+{\log }_{3}x{\log }_3^{3}x+{\log }_3^{2}x-2{\log }_{3}x=0{\log }_{3}x({\log }_3^{2}x+{\log }_{3}x-2)=0{\log }_{3}x({\log }_{3}x+2)({\log }_{3}x-1)=0x=3^0,3^{-2},3^{-1}x=1,\frac{1}{9},\frac{1}{3}$