Question

Find the value of ${\log }_{8}128,{\log }_6\frac{1}{216},{\log }_{27}\frac{1}{81},{\log }_{343}49$.

Answer 1

${\log }_{8}128={\log }_{2^3}{(2}^7)=7\left(\frac{1}{3}\right){\log }_{2}2=\frac{7}{3}$

${\log }_6\frac{1}{216}={\log }_6{216}^{-1}={\log }_6{6}^{-3}=-3$

${\log }_{27}\frac{1}{81}={\log }_{3^3}{81}^{-1}={\log }_{3^3}{3}^{-4}=-4\left(\frac{1}{3}\right)=\frac{-4}{3}$

${\log }_{343}49={\log }_{7^3}{7}^2=2\left(\frac{1}{3}\right)=\frac{2}{3}$