Question

Show that $log\frac{162}{343}+2log\frac{7}{9}-log\frac{1}{7}=log2$.

Answer 1

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

$log\frac{162}{343}+log{\left(\frac{7}{9}\right)}^2-log\frac{1}{7}[\because \text{a log b = log b^a]}$

$\Rightarrow log\frac{162}{343}+log\frac{49}{81}-log\frac{1}{7}$

$\Rightarrow log\left(\frac{\frac{162}{343}\times 4981}{1/7}\right)[\because \text{log mn = log m + log n , log}\frac{m}{n}\text{= log m - log n ]}$

$\Rightarrow log\left(\frac{2/7}{1/7}\right)$

$\Rightarrow log2=RHS$