A) \[1\]
B) \[6\]
C) \[{{\log }_{5}}\,9\]
D) \[{{\log }_{3}}5.\,{{\log }_{5}}\,81\,\]
Correct Answer: D
Solution :
\[\left| \begin{matrix} {{\log }_{5}}729 & {{\log }_{3}}5 \\ {{\log }_{5}}\,27 & {{\log }_{9}}25 \\ \end{matrix} \right|=\left| \begin{matrix} {{\log }_{5}}\,{{3}^{6}} & {{\log }_{3}}5 \\ {{\log }_{5}}\,{{3}^{3}} & {{\log }_{{{3}^{2}}}}{{5}^{2}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 6\,{{\log }_{5}}\,3 & {{\log }_{3}}5 \\ 3\,{{\log }_{5}}\,3 & \frac{2}{2}\,{{\log }_{3}}5 \\ \end{matrix} \right|\] \[=6\,{{\log }_{5}}\,3{{\log }_{3}}\,5-3\,{{\log }_{5}}\,3{{\log }_{3}}\,5\] \[=6-3\] \[=3\] and \[\left| \begin{matrix} {{\log }_{3}}5 & {{\log }_{27}}5 \\ {{\log }_{5}}9 & {{\log }_{5}}9 \\ \end{matrix} \right|=\left| \begin{matrix} {{\log }_{3}}5 & {{\log }_{{{3}^{3}}}}5 \\ {{\log }_{5}}\,{{3}^{2}} & {{\log }_{5}}{{3}^{2}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} {{\log }_{3}}5 & \frac{1}{3}{{\log }_{3}}5 \\ 2{{\log }_{5}}3 & 2\,{{\log }_{5}}3 \\ \end{matrix} \right|\] \[=2{{\log }_{5}}\,3{{\log }_{3}}5-\frac{2}{3}\,{{\log }_{5}}\,3{{\log }_{3}}5\] \[=2-\frac{2}{3}=\frac{4}{3}\] Now, \[\left| \begin{matrix} {{\log }_{5}}\,729 & {{\log }_{3}}5 \\ {{\log }_{5}}27 & {{\log }_{9}}25 \\ \end{matrix} \right|\left| \begin{matrix} {{\log }_{3}}5 & {{\log }_{27}}5 \\ {{\log }_{5}}9 & {{\log }_{5}}9 \\ \end{matrix} \right|=3.\frac{4}{3}=4\] Take option [d], \[{{\log }_{3}}5.{{\log }_{5}}81={{\log }_{3}}81={{\log }_{3}}{{3}^{4}}=4\]
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