Question

The value of $\begin{vmatrix} \log _{5} 729 & \log _{3} 5 \\ \log _{5} 27 & \log _{9} 25 \end{vmatrix} \times\begin{vmatrix} \log _{3} 5 & \log _{27} 5 \\ \log _{5} 9 & \log _{5} 9 \end{vmatrix}$ is

• A. $1$
• B. $6$
• C. $\log _{5} 9$
• D. $\left(\log _{3} 5\right) \times\left(\log _{5} 81\right)$

Answer 1

Answer: (D)

$\begin{vmatrix}\log _{5} 729 & \log _{3} 5 \\ \log _{5} 27 & \log _{9} 25\end{vmatrix} \times\begin{vmatrix}\log _{3} 5 & \log _{27} 5 \\ \log _{5} 9 & \log _{5} 9\end{vmatrix}$
$=\begin{vmatrix}6 \log _{5} 3 & \log _{3} 5 \\ 3 \log _{5} 3 & \log _{3} 5\end{vmatrix} \times\begin{vmatrix}\log _{3} 5 & \frac{1}{3} \log _{3} 5 \\ 2 \log _{5} 3 & 2 \log _{5} 3\end{vmatrix}$
$=3 \log _{5} 3 \log _{3} 5\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \times 2 \log _{3} 5 \log _{5} 3\begin{vmatrix} 1 & 1 / 3 \\ 1 & 1
\end{vmatrix}$
$=6[2-1] \times\left[1-\frac{1}{3}\right]=4$
$\because$ Hence the options
$\log _{3} 5 \times \log _{5} 81=\left(\log _{3} 5\right) \times\left(4 \log _{5} 3\right)=4$