If ${\log _{12}}27 = a,$ then ${\log _6}16 = $

$3\left( {\frac{{4 - a}}{{4 + a}}} \right)$

$3\left( {\frac{{4 + a}}{{4 - a}}} \right)$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

{\log _{12}}27 = a,

then

{\log _6}16 =

View answer and explanation

Correct option: B

4\left( {\frac{{3 - a}}{{3 + a}}} \right)

Option B (

4\left( {\frac{{3 - a}}{{3 + a}}} \right)

) is correct because it is the result of converting the logarithms to a common base and simplifying the expression.

Practice All Chapter MCQs →

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