A) 7
B) 10
C) 13
D) 17
Correct Answer: B
Solution :
\[n=A{{R}^{r-1}}\Rightarrow \log n=\log A+(r-1)\log R\] \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\left| \,\begin{matrix} {{\log }_{2}}3 & {{\log }_{8}}3 \\ {{\log }_{3}}4 & {{\log }_{3}}4 \\ \end{matrix}\, \right|\] \[=\left( \frac{\log 512}{\log 3}\times \frac{\log 9}{\log 4}-\frac{\log 3}{\log 4}\times \frac{\log 8}{\log 3} \right)\]× \[\left( \frac{\log 3}{\log 2}\times \frac{\log 4}{\log 3}-\frac{\log 3}{\log 8}\times \frac{\log 4}{\log 3} \right)\] = \[\left( \frac{\log {{2}^{9}}}{\log 3}\times \frac{\log {{3}^{2}}}{\log {{2}^{2}}}-\frac{\log {{2}^{3}}}{\log {{2}^{2}}} \right)\] × \[\left( \frac{\log {{2}^{2}}}{\log 2}-\frac{\log {{2}^{2}}}{\log {{2}^{3}}} \right)\] = \[\left( \frac{9\times 2}{2}-\frac{3}{2} \right)\,\left( 2-\frac{2}{3} \right)=\frac{15}{2}\times \frac{4}{3}=10\].
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