If log 12 27 = a , then log 6 16 =

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Solution

Q.
If $\log_{12} 27 = a$ , then $\log_{6} 16 =$

1 $2 (\frac{3 - a}{3 + a})$

2 $3 (\frac{3 - a}{3 + a})$

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

4 $4 (\frac{2 - a}{2 + a})$

Ans.
(3)

$a = \frac{3}{1 + 2 \log_{3} 2} \Rightarrow \log_{3} 2 = \frac{3 - a}{2 a};$ $\log_{6} 16 = \frac{4 \log_{3} 2}{1 + \log_{3} 2}$

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