If ${a^2} + 4{b^2} = 12ab,$ then $\log \left( {a + 2b} \right) = $

$\frac{1}{2}\left( {\log a + \log b - \log 2} \right)$

$\log \frac{a}{2} + \log \frac{b}{2} + \log 2$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

{a^2} + 4{b^2} = 12ab,

then

\log \left( {a + 2b} \right) =

View answer and explanation

Correct option: C

\frac{1}{2}\left( {\log a + \log b + 4\log 2} \right)

Option C (

\frac{1}{2}\left( {\log a + \log b + 4\log 2} \right)

) is correct because by completing the square and applying logarithm properties, we can derive the given expression.

Practice All Chapter MCQs →

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\frac{3}{4}