# CAT 2018 Quant Questions

Question:
${\log _{12}}81 = p,then\;3\left( {\frac{{4 - p}}{{4 + p}}} \right)$ is equal to

 log416 log68 log616 log28

Given that: $\log_{12}{81}=p$

$\Rightarrow$ $\log_{81}{12}=\frac{1}{p}$

$\Rightarrow$ $4\log_{3}{3*4}=\frac{1}{p}$

$\Rightarrow$ $1+\log_{3}{4}=\frac{4}{p}$

Using Componendo and Dividendo,

$\Rightarrow$ $\frac{1+\log_{3}{4}-1}{1+\log_{3}{4}+1}=\frac{4-p}{4+p}$

$\Rightarrow$ $\frac{\log_{3}{4}}{2+\log_{3}{4}}=\frac{4-p}{4+p}$

$\Rightarrow$ $\frac{\log_{3}{4}}{\log_{3}{9}+\log_{3}{4}}=\frac{4-p}{4+p}$

$\Rightarrow$ $\frac{\log_{3}{4}}{\log_{3}{36}}=\frac{4-p}{4+p}$

$\Rightarrow$ $3*\frac{4-p}{4+p}=\frac{3\log_{3}{4}}{\log_{3}{36}}$

$\Rightarrow$ $3*\frac{4-p}{4+p}=\frac{\log_{3}{64}}{\log_{3}{36}}$

$\Rightarrow$ $3*\frac{4-p}{4+p}=\log_{36}{64}$

$\Rightarrow$ $3*\frac{4-p}{4+p}=\log_{6^2}{8^2}=\log_{6}{8}$.

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