If ${l o g}_{12} 27 = a$ , then ${l o g}_{6} 16$ is :

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

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

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

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

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Solution

The correct option is A $\frac{4 (3 - a)}{(3 + a)}$
${l o g}_{12} 27 = a \Rightarrow \frac{l o g 27}{l o g 12} = a$
$\Rightarrow 3 l o g 3 = (2 l o g 2 + l o g 3) a \Rightarrow 3 l o g 3 - a l o g 3 = 2 a log 2$
$\Rightarrow log 3 = \frac{2 a l o g 2}{3 - a}$
${l o g}_{6} 16 = \frac{l o g 16}{l o g 6} \Rightarrow \frac{4 l o g 2}{l o g 2 + l o g 3}$
$= \frac{4 l o g 2}{l o g 2 + \frac{2 a l o g 2}{3 - a}}$
$= \frac{4 (3 - a)}{3 + a}$

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Similar questions

{log}_{12} 27 = a .

{log}_{6} 16 = \frac{4 (3 - a)}{3 + a} .

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