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Question: Answered & Verified by Expert
$\left|\begin{array}{cc}\log _2 512 & \log _4 3 \\ \log _3 8 & \log _4 9\end{array}\right| \times\left|\begin{array}{ll}\log _2 3 & \log _8 3 \\ \log _3 4 & \log _3 4\end{array}\right|=$
MathematicsDeterminantsJEE Main
Options:
  • A 7
  • B 10
  • C 13
  • D 17
Solution:
1661 Upvotes Verified Answer
The correct answer is: 10
$\left|\begin{array}{cc}\log _3 512 & \log _4 3 \\ \log _3 8 & \log _4 9\end{array}\right| \times\left|\begin{array}{ll}\log _2 3 & \log _8 3 \\ \log _3 4 & \log _3 4\end{array}\right|$
$=\left(\frac{\log 512}{\log 3} \times \frac{\log 9}{\log 4}-\frac{\log 3}{\log 4} \times \frac{\log 8}{\log 3}\right)$ $\times\left(\frac{\log 3}{\log 2} \times \frac{\log 4}{\log 3}-\frac{\log 3}{\log 8} \times \frac{\log 4}{\log 3}\right)$
$\begin{array}{r}=\left(\frac{\log 2^9}{\log 3} \times \frac{\log 3^2}{\log 2^2}-\frac{\log 2^3}{\log 2^2}\right) \times\left(\frac{\log 2^2}{\log 2}-\frac{\log 2^2}{\log 2^3}\right) \\ =\left(\frac{9 \times 2}{2}=\frac{3}{2}\right)\left(2-\frac{2}{3}\right)=\frac{15}{2} \times \frac{4}{3}-10\end{array}$

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