Question

Solve: ${\log }_{x}3+{\log }_{3}x={\log }_{\sqrt{x}}3+{\log }_3\sqrt{x}+\frac{1}{2}$

Answer 1

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