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Q.

If $\log_{12} 81 = x, then \log_{6} 3 =$

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a

$\frac{2 x}{x + 4}$

b

$\frac{x + 4}{2 x}$

c

$\frac{x}{x + 4}$

d

$\frac{x + 4}{x}$

answer is C.

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Detailed Solution

$\begin{array}{l} 4 \log_{12}^{3} = x \Rightarrow \frac{4}{x} = \log_{3}^{12} \\ \Rightarrow 1 + \log_{3}^{12} = \frac{4}{x} + 1 = \log_{3}^{36} \\ \Rightarrow \frac{4 + x}{x} = 2 \log_{3}^{6} \Rightarrow \log_{6}^{3} = \frac{2 x}{4 + x} \end{array}$

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If log12⁡81=x, then log6⁡3=