Question

The value of $\left|\begin{array}{cc}{\log }_{5}729& {\log }_{3}5\\ {\log }_{5}27& {\log }_{9}25\end{array}\right|\times \left|\begin{array}{cc}{\log }_{3}5& {\log }_{27}5\\ {\log }_{5}9& {\log }_{5}9\end{array}\right|$ 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)

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