Question

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Answer 1

$\it a= log_{12}27=\dfrac{log_2 27}{log_2 12}=\dfrac{log_2 3^3}{log_2 4\cdot3}=\dfrac{3log_2 3}{log_2 4+log_23} =\dfrac{3log_2 3}{2+log_2 3} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{a}{3}=\dfrac{log_23}{2+log_2 3} \Rightarrow \dfrac{a}{3-a}=\dfrac{log_2 3}{2+log_23-log_23}\Rightarrow \dfrac{a}{3-a}=\dfrac{log_23}{2}\Rightarrow\\ \\ \\ \Rightarrow log_23=\dfrac{2a}{3-a}\ \ \ \ \ \ (*)$

$\it log_616=\dfrac{log_216}{log_26}=\dfrac{log_22^4}{log_2{2\cdot3}} = \dfrac{4log_22}{log_22+log_23} =\dfrac{4}{1+log_23}\ \stackrel{(*)}{=}\ \dfrac{4}{1+\dfrac{2a}{3-a} }=\\ \\ \\ =\dfrac{4(3-a)}{3-a+2a}=\dfrac{4(3-a)}{3+a}$