Answer:
Option D
Explanation:
$log_{12}27$ $=a$
$\Rightarrow\frac{log27}{log12}$ $=a$
$\Rightarrow\frac{log3^{3}}{log(3\times 2^{2})}$ $=a$
$\Rightarrow\frac{3log3}{log3+2log2}$ $=a$
$\Rightarrow\frac{log3+2log2}{3log3}$ $=\frac{1}{a}$
$\Rightarrow$ $\frac{log3}{3log3}$ $+\frac{2log2}{3log3}$ $=\frac{1}{a}$
$\Rightarrow\frac{2log2}{3log3}$ $=\frac{1}{a}-\frac{1}{3}$ $=\left(\frac{3-a}{3a}\right)$
$\Rightarrow\frac{log2}{log3}$ $=\left(\frac{3-a}{2a}\right)$ $\Rightarrow log3$ $=\left(\frac{2a}{3-a}\right)$$log2$.
$log_{6}16$ $=\frac{log16}{log6}$ $=\frac{log_{2}^{4}}{log(2\times 3)}$ $=\frac{4log2}{log2+log3}$
$=\frac{4log2}{log2\left[1+\left(\frac{2a}{3-a}\right)\right]}$
$=\frac{4}{\left(\frac{3+a}{3-a}\right)}$ $=\frac{4(3-a)}{(3+a)}$
