Question

$\begin{array}{l}\log_{3} 512\\\log_{3} 8\end{array}$ $\begin{array}{l}\log_{2} 3\\\log_{4} 3\end{array}$ $\begin{array}{l}\log_{8} 3\\\log_{3} 4\end{array}$ is $\begin{array}{l}\log_{3} 5\\\log_{4} 9\end{array}$ equal to

• A. 5
• B. 7
• C. 10
• D. 12

Answer 1

Answer: (C)

10

Answer 2

The question is interpreted as the product of two determinants. The first determinant is $\begin{vmatrix} \log_{3}512 & \log_{4}3 \\ \log_{3}8 & \log_{4}9 \end{vmatrix}$. Evaluating the terms using logarithm properties: $\log_{3}512 = 9\log_{3}2$, $\log_{4}3 = \frac{1}{2}\log_{2}3$, $\log_{3}8 = 3\log_{3}2$, $\log_{4}9 = \log_{2}3$. The determinant is $(9\log_{3}2)(\log_{2}3) - (\frac{1}{2}\log_{2}3)(3\log_{3}2) = 9 - \frac{3}{2} = \frac{15}{2}$.

The second determinant is $\begin{vmatrix} \log_{2}3 & \log_{8}3 \\ \log_{3}4 & \log_{3}4 \end{vmatrix}$. Evaluating the terms: $\log_{2}3$, $\log_{8}3 = \frac{1}{3}\log_{2}3$, $\log_{3}4 = 2\log_{3}2$. The determinant is $(\log_{2}3)(2\log_{3}2) - (\frac{1}{3}\log_{2}3)(2\log_{3}2) = 2 - \frac{2}{3} = \frac{4}{3}$.

The product is $\frac{15}{2} \times \frac{4}{3} = 10$.