Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

$\begin{vmatrix}\log_3\, 512&\log_4\, 3\\\log_3\,8&\log_4\,9\end{vmatrix}×\begin{vmatrix}\log_2\, 3 &\log_8\, 3\\\log_3\,4&\log_3\,4\end{vmatrix}$

Options:

7

10

13

17

Correct Answer:

10

Explanation:

We have,

$\begin{vmatrix}\log_3\, 512&\log_4\, 3\\\log_3\,8&\log_4\,9\end{vmatrix}×\begin{vmatrix}\log_2\, 3 &\log_8\, 3\\\log_3\,4&\log_3\,4\end{vmatrix}$

$=\begin{vmatrix}\log_3\, 2^9&\log_{2^2}\, 3\\\log_3\,2^3&\log_{2^2}\,3^2\end{vmatrix}×\begin{vmatrix}\log_2\, 3 &\log_{2^3}\, 3\\\log_3\,2^2&\log_3\,2^2\end{vmatrix}$

$=\begin{vmatrix}9\log_3\,2&\frac{1}{2}\log_2\,3\\3\log_3\,2&\frac{2}{2}\log_2\,3\end{vmatrix}×\begin{vmatrix}\log_2\,3&\frac{1}{3}\log_2\,3\\2\log_3\,2&2\log_3\,2\end{vmatrix}$

$=(9-\frac{3}{2})×(2-\frac{2}{3})=\frac{15}{2}×\frac{4}{3}=10$  $[∵ \log_2\, 3× \log_3\, 2=1]$