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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Matrices
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$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times \begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}=$

$\begin{array}{1 1}(A)\;7&(B)\;10\\(C)\;13&(D)\;17\end{array}$

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1 Answer

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$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times \begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$
$\Rightarrow \big(\large\frac{\log 512}{\log 3}\times \frac{\log 9}{\log 4}-\frac{\log 3}{\log 4}\times \frac{\log 8}{\log 3}\big)\times \big(\large\frac{\log 3}{\log 2}\times \frac{\log 4}{\log 3}-\frac{\log 3}{\log 8}\times \frac{\log 4}{\log 3}\big)$
$\Rightarrow \big(\large\frac{\log 2^9}{\log 3}\times \frac{\log 3^2}{\log 2^2}-\frac{\log 2^3}{\log 2^2}\big)\times \big(\large\frac{\log 2^2}{\log 2}-\frac{\log 2^2}{\log 2^3}\big)$
$\Rightarrow \big(\large\frac{9\times 2}{2}-\frac{3}{2})$$(2-\large\frac{2}{3}\big)$
$\Rightarrow \large\frac{15}{2}\times \frac{4}{3}$
$\Rightarrow 10$
Hence (B) is the correct answer.
answered Apr 24, 2014 by sreemathi.v
 

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