Step 1: Understanding the Concept:
The determinant of a $2 \times 2$ matrix \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} \) is \( ad - bc \). We also use the logarithm property \( \log_a (b^n) = n \log_a b \).
Step 2: Detailed Explanation:
The determinant is \( (\log_3 512 \cdot \log_4 9) - (\log_4 3 \cdot \log_3 8) \).
1. Note that \( 512 = 2^9 \) and \( 8 = 2^3 \), so \( \log_3 512 = 9 \log_3 2 \) and \( \log_3 8 = 3 \log_3 2 \).
2. Note that \( 9 = 3^2 \), so \( \log_4 9 = 2 \log_4 3 \).
Determinant = \( (9 \log_3 2 \cdot 2 \log_4 3) - (\log_4 3 \cdot 3 \log_3 2) \)
Determinant = \( (18 \log_3 2 \cdot \log_4 3) - (3 \log_3 2 \cdot \log_4 3) \)
Determinant = \( 15 (\log_3 2 \cdot \log_4 3) \).
Using the change of base formula \( \log_a b \cdot \log_b c = \log_a c \):
\( \log_3 2 \cdot \log_4 3 = \frac{\ln 2}{\ln 3} \cdot \frac{\ln 3}{\ln 4} = \frac{\ln 2}{\ln 4} = \frac{\ln 2}{2 \ln 2} = \frac{1}{2} \).
Determinant = \( 15 \cdot \frac{1}{2} \). Wait, let's re-calculate:
\( \log_3 2^9 \cdot \log_4 3^2 - \log_4 3 \cdot \log_3 2^3 = 18 (\log_3 2 \cdot \log_4 3) - 3 (\log_4 3 \cdot \log_3 2) = 15 (\log_3 2 \cdot \log_4 3) \).
\( \log_3 2 \cdot \log_4 3 = \frac{\log 2}{\log 3} \cdot \frac{\log 3}{\log 4} = \frac{\log 2}{\log 2^2} = \frac{\log 2}{2 \log 2} = 0.5 \).
Determinant = \( 15 \times 0.5 = 7.5 = 15/2 \).
Correction: Based on the provided options, if the determinant is 15, please verify the input values of the determinant. With the given values, it equals 7.5.
Step 3: Final Answer:
Based on the calculation, the result is 7.5 (15/2).