Question

If ${\log }_{9}27-{\log }_{27}9=k$, then the value of $36k$ is equal to

Answer 1

${\log }_{9}27-{\log }_{27}9={\log }_{3^2}{3}^3-{\log }_{3^3}{3}^2$
$=\frac{3}{2}{\log }_{3}3-\frac{2}{3}{\log }_{3}3=\frac{3}{2}-\frac{2}{3}=\frac{5}{6}[\because {\log }_{a^n}{b}^m=\frac{m}{n}{\log }_{a}b,{\log }_{a}a=1]\therefore 36k=30$