Question

Prove $\log \frac{162}{343}+2\log \frac{7}{9}-\log \frac{1}{7}=\log 2$

Answer 1

LHS$=log\frac{162}{343}+2log\frac{7}{9}-log\frac{1}{7}$

$=log\frac{162}{343}+log\frac{7^2}{9^2}+log\frac{7}{1}$

$=log\left(\frac{162\times 49\times 7}{343\times 81}\right)$

$=log\left(\frac{162}{81}\right)=log2=$RHS.