MCQ
$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times\begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$
- A$7$
- ✓$10$
- C$1$
- D$17$
✓
Answer
Correct option: B.
$10$
$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times\begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$
$=\begin{vmatrix}\log_32^9&\log_{2^{2}}3\\\log_32^3&\log_{2^2}3^3\end{vmatrix}\times\begin{vmatrix}\log_23&\log_{2^{3}}3\\\log_32^3&\log_32^2\end{vmatrix}$
$=\begin{vmatrix}9\log_32&\frac{1}{2}\log_23\\3\log_32&\frac{1}{2}\times2\log_23\end{vmatrix}\times\begin{vmatrix}\log_23&\frac{1}{3}\log_23\\2\log_32&2\log_32\end{vmatrix}$
$=\Big(\big(9\log_32\times\log_23\big)-\big(3\log_32\times\frac{1}{2}\log_23\big)\Big)\times\Big(\big(\log_23\times2\log_32\big)-\Big(\frac{1}{3}\log_23\times2\log_32\Big)\Big)$
$=\Big(9-\frac{3}{2}\Big)\times\Big(2-\frac{2}{3}\Big)$
$=\frac{15}{2}\times\frac{4}{3}$
$=10$
$=\begin{vmatrix}\log_32^9&\log_{2^{2}}3\\\log_32^3&\log_{2^2}3^3\end{vmatrix}\times\begin{vmatrix}\log_23&\log_{2^{3}}3\\\log_32^3&\log_32^2\end{vmatrix}$
$=\begin{vmatrix}9\log_32&\frac{1}{2}\log_23\\3\log_32&\frac{1}{2}\times2\log_23\end{vmatrix}\times\begin{vmatrix}\log_23&\frac{1}{3}\log_23\\2\log_32&2\log_32\end{vmatrix}$
$=\Big(\big(9\log_32\times\log_23\big)-\big(3\log_32\times\frac{1}{2}\log_23\big)\Big)\times\Big(\big(\log_23\times2\log_32\big)-\Big(\frac{1}{3}\log_23\times2\log_32\Big)\Big)$
$=\Big(9-\frac{3}{2}\Big)\times\Big(2-\frac{2}{3}\Big)$
$=\frac{15}{2}\times\frac{4}{3}$
$=10$
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