Question

The least integer greater than ${\log }_{2}15\cdot {\log }_{1/6}2\cdot {\log }_3\frac{1}{6}$ is

Answer 1

${\log }_{2}15\cdot {\log }_{\frac{1}{6}}2\cdot {\log }_3\frac{1}{6}$
$={\log }_{2}15\cdot \frac{{\log }_{\frac{1}{6}}2}{{\log }_{\frac{1}{6}}3}$
$={\log }_{2}15\cdot {\log }_{3}2$
$=\frac{{\log }_{2}15}{{\log }_{2}3}$
$={\log }_{3}15$

Now, $9<15<27$
$\Rightarrow {\log }_{3}9<{\log }_{3}15<{\log }_{3}27$
$\Rightarrow 2<{\log }_{3}15<3$