Answer :
To calculate the product of the two determinants, we first need to evaluate each determinant separately.
- Determinant 1:
[tex]\begin{vmatrix} \log_3 512 & \log_4 3 \\ \log_3 8 & \log_4 9 \end{vmatrix}[/tex]
The determinant of a 2x2 matrix [tex]\begin{vmatrix} a & b \\ c & d \end{vmatrix}[/tex] is calculated as [tex]ad - bc[/tex].
Let's substitute in the values:
- [tex]a = \log_3 512[/tex]
- [tex]b = \log_4 3[/tex]
- [tex]c = \log_3 8[/tex]
- [tex]d = \log_4 9[/tex]
The determinant becomes:
[tex]\log_3 512 \cdot \log_4 9 - \log_4 3 \cdot \log_3 8[/tex]
We can simplify terms using logarithmic identities:
- [tex]\log_3 512 = \log_3 (2^9) = 9 \cdot \log_3 2[/tex]
- [tex]\log_4 9 = \log_4 (3^2) = 2 \cdot \log_4 3[/tex]
- [tex]\log_3 8 = \log_3 (2^3) = 3 \cdot \log_3 2[/tex]
- [tex]\log_4 3[/tex] remains as it is.
Thus, the determinant is:
[tex](9 \cdot \log_3 2)(2 \cdot \log_4 3) - (\log_4 3)(3 \cdot \log_3 2)[/tex]
Simplifying this gives us:
[tex]18(\log_3 2)(\log_4 3) - 3(\log_3 2)(\log_4 3)[/tex]
[tex]= 15(\log_3 2)(\log_4 3)[/tex]
- Determinant 2:
[tex]\begin{vmatrix} \log_2 3 & \log_5 3 \\ \log_3 4 & \log_3 4 \end{vmatrix}[/tex]
We notice that the second column values are identical, meaning the determinant is zero because any column that is repeated in this arrangement results in a determinant of zero. Let's see this step:
[tex]\log_2 3 \cdot \log_3 4 - \log_5 3 \cdot \log_3 4[/tex]
[tex]= \log_3 4(\log_2 3 - \log_5 3)[/tex]
Since the terms become zero:
[tex]\log_3 4 \cdot (\log_2 3 - \log_5 3) = 0[/tex]
Thus, the determinant is 0.
Therefore, the product of the two determinants is:
[tex]15(\log_3 2)(\log_4 3) \cdot 0 = 0[/tex]
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