A) \[2\left( \frac{3-a}{3+a} \right)\]
B) \[3\left( \frac{3-a}{3+a} \right)\]
C) \[4\left( \frac{3-a}{3+a} \right)\]
D) \[5\left( \frac{3-a}{3+a} \right)\]
Correct Answer: C
Solution :
\[a={{\log }_{12}}27={{\log }_{12}}{{(3)}^{3}}=3{{\log }_{12}}3\] \[=\frac{3}{{{\log }_{3}}\,12}=\frac{3}{1+{{\log }_{3}}4}\] \[=\frac{3}{1+2\,{{\log }_{3}}2}\] \[\therefore \] \[{{\log }_{3}}2=\frac{3-a}{2a}\] ?(i) \[\therefore \] \[{{\log }_{6}}16={{\log }_{6}}{{2}^{4}}=4{{\log }_{6}}2\] \[=\frac{4}{{{\log }_{2}}\,6}=\frac{4}{1+{{\log }_{2}}3}=\frac{4}{1+\frac{1}{{{\log }_{3}}\,2}}\] [from Eq. (i)] \[=\frac{4}{1+\frac{2a}{3-a}}\] \[=4\,\left( \frac{3-a}{3+a} \right)\]
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