Greatest integer less than or equal to the number $l o g_{2} 15 . l o g_{1 / 6} 2 . l o g_{3} 1 / 6$ is :

4

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3

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2

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1

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Solution

The correct option is D $2$
Given,

${log}_{2} (15) {log}_{\frac{1}{6}} (2) {log}_{3} (\frac{1}{6})$

$= \frac{ln (15)}{ln (2)} \cdot \frac{ln (2)}{ln (\frac{1}{6})} {log}_{3} (\frac{1}{6})$

$= \frac{ln (15)}{ln (\frac{1}{6})} {log}_{3} (\frac{1}{6})$

$= \frac{ln (15) {log}_{3} (\frac{1}{6})}{- ln (6)}$

$= \frac{ln (15) (- {log}_{3} (6))}{- ln (6)}$

$= \frac{ln (15) {log}_{3} (6)}{ln (6)} \approx 2.46497 \dots$

Therefore number greater integer less than or equal to ${log}_{2} (15) {log}_{\frac{1}{6}} (2) {log}_{3} (\frac{1}{6}) = 2$

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