Question

Log 2/3 + log1/3 x=2

Answer 1

$\log_3^2x+\log_{\tfrac{1}{3}}x=2\qquad (x\ \textgreater \ 0)\\ \log_3^2x+\dfrac{\log_3x}{\log_3\dfrac{1}{3}}=2\\ \log_3^2x+\dfrac{\log_3x}{-1}=2\\ \log_3^2x-\log_3x=2\\ \log_3^2x-\log_3x-2=0\\ \log_3^2x+\log_3x-2\log_3x-2=0\\ \log_3x(\log_3x+1)-2(\log_3x+1)=0\\ (\log_3x-2)(\log_3x+1)=0\\\\ \log_3x-2=0\\ \log_3x=2\\ x=3^2=9\\\\ \log_3x+1=0\\ \log_3x=-1\\ x=3^{-1}=\dfrac{1}{3}\\\\ \boxed{x=\left\{\dfrac{1}{3},9\right\}}$