Logarithms | Arithmetic aptitude | Discussion Page

If $log_{12}27$ $=a$ , then $log_{6}16$ is :

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

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

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

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

Option D

$log_{12}27$ $=a$

$\Rightarrow\frac{log27}{log12}$ $=a$

$\Rightarrow\frac{log3^{3}}{log(3\times 2^{2})}$ $=a$

$\Rightarrow\frac{3log3}{log3+2log2}$ $=a$

$\Rightarrow\frac{log3+2log2}{3log3}$ $=\frac{1}{a}$

$\Rightarrow$ $\frac{log3}{3log3}$ $+\frac{2log2}{3log3}$ $=\frac{1}{a}$

$\Rightarrow\frac{2log2}{3log3}$ $=\frac{1}{a}-\frac{1}{3}$ $=\left(\frac{3-a}{3a}\right)$

$\Rightarrow\frac{log2}{log3}$ $=\left(\frac{3-a}{2a}\right)$ $\Rightarrow log3$ $=\left(\frac{2a}{3-a}\right)$ $log2$ .

$log_{6}16$ $=\frac{log16}{log6}$ $=\frac{log_{2}^{4}}{log(2\times 3)}$ $=\frac{4log2}{log2+log3}$

$=\frac{4log2}{log2\left[1+\left(\frac{2a}{3-a}\right)\right]}$

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

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