Question

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

• A. $1$
• B. $6$
• C. ${\log }_{5}9$
• D. ${\log }_{3}5.{\log }_{5}9$

Answer 1

Answer: (D)

$\left|\begin{array}{cc}{\log }_{5}729& {\log }_{3}5\\ {\log }_{5}27& {\log }_{9}25\end{array}\right|\left|\begin{array}{cc}{\log }_{3}5& {\log }_{27}5\\ {\log }_{5}9& {\log }_{5}9\end{array}\right|$
= $\left|\begin{array}{cc}{\log }_5{3}^6& {\log }_{3}5\\ {\log }_5{3}^3& {\log }_{32}{5}^2\end{array}\right|\left|\begin{array}{cc}{\log }_{3}5& {\log }_3{3}^5\\ {\log }_5{3}^3& {\log }_{32}{5}^2\end{array}\right|$
$\left|\begin{array}{cc}6{\log }_{5}3& \log 5\\ 3{\log }_{5}3& \frac{2}{2}{\log }_{3}5\end{array}\right|\left|\begin{array}{cc}{\log }_{3}5& \frac{1}{3}{\log }_3{3}^5\\ 2{\log }_{5}3& 2{\log }_{5}3\end{array}\right|$
$\left|\begin{array}{cc}6{\log }_{5}3& \log 5\\ 3{\log }_{5}3& \frac{2}{2}{\log }_{3}5\end{array}\right|\left|\begin{array}{cc}{\log }_{3}5& \frac{1}{3}{\log }_{3}5\\ 2{\log }_{5}3& 2{\log }_{5}3\end{array}\right|$
$\left(6-3\right)\left(2-\frac{2}{3}\right)=\left(3\right)\left(\frac{4}{3}\right)=4$
= ${\log }_{3}5{\log }_{5}81$
$({\log }_{3}5{\log }_{5}81)={\log }_{3}5{\log }_5{3}^4=4lo{g}_{3}5{\log }_{5}3$
= $4(1)=4$