If {log _{12}}27 = a, then {log _6}16 = | vedclass

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(C) Given $a = \log_{12} 27 = \frac{\log 27}{\log 12} = \frac{3 \log 3}{\log 3 + 2 \log 2}$.Rearranging for $\log 3$: $a(\log 3 + 2 \log 2) = 3 \log 3

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

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(C) Given $a = \log_{12} 27 = \frac{\log 27}{\log 12} = \frac{3 \log 3}{\log 3 + 2 \log 2}$.Rearranging for $\log 3$: $a(\log 3 + 2 \log 2) = 3 \log 3$ $\Rightarrow a \log 3 + 2a \log 2 = 3 \log 3$ $\Rightarrow 2a \log 2 = (3 - a) \log 3$ $\Rightarrow \log 3 = \frac{2a \log 2}{3 - a}$.Now,evaluate $\log_6 16 = \frac{\log 16}{\log 6} = \frac{4 \log 2}{\log 2 + \log 3}$.Substituting $\log 3$: $\frac{4 \log 2}{\log 2 + \frac{2a \log 2}{3 - a}} = \frac{4 \log 2}{\frac{(3 - a) \log 2 + 2a \log 2}{3 - a}} = \frac{4(3 - a) \log 2}{(3 - a + 2a) \log 2} = \frac{4(3 - a)}{3 + a}$.

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

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