The value of \( \begin{vmatrix}\log_3 16 & \log_5 9\\\log_3 25 & \log_4 27\end{vmatrix}+\begin{vmatrix}\log_3 4 & \log_3 4\\\log_2 3 & \log_2 9\end{vmatrix} \) is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4
The log rules needed are \( \log_b a^n=n\log_b a \), \( \log_{b^m}a=\frac{1}{m}\log_b a \) and \( \log_a b\cdot\log_b a=1 \).
First determinant:
\( \displaystyle \log_3 16\cdot\log_4 27=(4\log_3 2)\left(\frac{3}{2}\log_2 3\right)=6 \)
\( \displaystyle \log_5 9\cdot\log_3 25=(2\log_5 3)(2\log_3 5)=4 \)
\( \displaystyle \Delta_1=6-4=2 \)
Second determinant:
\( \displaystyle \Delta_2=\log_3 4\cdot\log_2 9-\log_3 4\cdot\log_2 3 \)
\( \displaystyle =\log_3 4\,(\log_2 9-\log_2 3) \)
\( \displaystyle =\log_3 4\cdot\log_2 3 \)
\( \displaystyle =2\log_3 2\cdot\log_2 3 \)
\( \displaystyle =2 \)
\( \displaystyle \Delta_1+\Delta_2=2+2=4 \)