Evaluate the determinant `Delta=|(log_3 512,log_4 3),(log_3 8,log_4 9)|`

Options

A. `15/2`
B. 12
C. `14/3`
D. 6

Answer

To evaluate the determinant

\[ \Delta = \begin{vmatrix} \log_3 512 & \log_4 3 \\ \log_3 8 & \log_4 9 \end{vmatrix} \]

we will follow these steps:

### Step 1: Rewrite the logarithms in terms of base 2.
We know that:

\[ \log_a b = \frac{\log_c b}{\log_c a} \]

Using base 2, we can rewrite the logarithms:

\[ \log_3 512 = \frac{\log_2 512}{\log_2 3}, \quad \log_4 3 = \frac{\log_2 3}{\log_2 4}, \quad \log_3 8 = \frac{\log_2 8}{\log_2 3}, \quad \log_4 9 = \frac{\log_2 9}{\log_2 4} \]

Calculating the values:

\[ \log_2 512 = 9 \quad (\text{since } 512 = 2^9), \quad \log_2 4 = 2 \quad (\text{since } 4 = 2^2), \quad \log_2 8 = 3 \quad (\text{since } 8 = 2^3), \quad \log_2 9 = 2 \log_2 3 \]

### Step 2: Substitute back into the determinant.
Now substituting these values back into the determinant:

\[ \Delta = \begin{vmatrix} \frac{9}{\log_2 3} & \frac{\log_2 3}{2} \\ \frac{3}{\log_2 3} & \frac{2 \log_2 3}{2} \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} \frac{9}{\log_2 3} & \frac{\log_2 3}{2} \\ \frac{3}{\log_2 3} & \log_2 3 \end{vmatrix} \]

### Step 3: Calculate the determinant.
Using the formula for the determinant of a 2x2 matrix:

\[ \Delta = ad - bc \]

where \( a = \frac{9}{\log_2 3}, b = \frac{\log_2 3}{2}, c = \frac{3}{\log_2 3}, d = \log_2 3 \):

\[ \Delta = \left(\frac{9}{\log_2 3} \cdot \log_2 3\right) - \left(\frac{\log_2 3}{2} \cdot \frac{3}{\log_2 3}\right) \]

This simplifies to:

\[ \Delta = 9 - \frac{3}{2} = 9 - 1.5 = 7.5 \]

### Step 4: Final answer.
Thus, the value of the determinant \(\Delta\) is:

\[ \Delta = \frac{15}{2} \]