Question

Solve the following equations.
${\log }_{7}2+{\log }_{49}x={\log }_{1/7}\sqrt{3}$

Answer 1

$\frac{\log 2}{\log 7}+\frac{\log x}{2\log 7}=\frac{-\log \sqrt{3}}{\log 7}2\log 2+\log x=-2\log \sqrt{3}\log 4+\log x=\log \left(\frac{1}{3}\right)4x=\frac{1}{3}x=\frac{1}{12}$