Question

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

Answer 1

let $y={\log }_{2}15\cdot {\log }_{\frac{1}{6}}2\cdot {\log }_3\frac{1}{6}$
$\Rightarrow y={\log }_{2}15\cdot (-1){\log }_{6}2\cdot (-1){\log }_{3}6[\because {\log }_{1/a}b=-{\log }_{a}b\&{\log }_a\frac{1}{b}=-{\log }_{a}b]$
$=\frac{\log 15}{\log 2}\times \frac{\log 2}{\log 6}\times \frac{\log 6}{\log 3}[\because {\log }_{b}a=\frac{\log a}{\log b}]$

$={\log }_{3}15[\because {\log }_{b}a=\frac{\log a}{\log b}]={\log }_3(3\times 5)={\log }_{3}3+{\log }_{3}5[\because \log (ab)=\log a+\log b]$
$=1+{\log }_{3}5[\because {\log }_{a}a=1]$
Since, $\log 5>\log 3$
Therefore, $y=1+{\log }_{3}5>2$
and the smallest possible integer is $3$
Ans: 3