Question

$\left|\begin{array}{cc}{\log }_{3}512& {\log }_{4}3\\ {\log }_{3}8& {\log }_{4}9\end{array}\right|\times \left|\begin{array}{cc}{\log }_{2}3& {\log }_{8}3\\ {\log }_{3}4& {\log }_{3}4\end{array}\right|$

(a) 7
(b) 10
(c) 1
(d) 17

Answer 1

(b) 10

$$\begin{gathered} \left|{\log }_{3}512{\log }_{4}3 \\ {\log }_{3}8{\log }_{4}9\right|\times \left|{\log }_{2}3{\log }_{8}3 \\ {\log }_{3}4{\log }_{3}4\right| \\ =\left|{\log }_3{2}^9{\log }_{2^2}3 \\ {\log }_3{2}^3{\log }_{2^2}{3}^3\right|\times \left|{\log }_{2}3{\log }_{2^3}3 \\ {\log }_3{2}^2{\log }_3{2}^2\right| \\ =\left|9{\log }_{3}2\frac{1}{2}{\log }_{2}3 \\ 3{\log }_{3}2\frac{1}{2}\times 2{\log }_{2}3\right|\times \left|{\log }_{2}3\frac{1}{3}{\log }_{2}3 \\ 2{\log }_{3}22{\log }_{3}2\right|\left[\because \log {}_{b^m}{a}^n=\frac{n}{m}{\log }_{b}a\right] \\ =\left(\left(9{\log }_{3}2\times {\log }_{2}3\right)-\left(3{\log }_{3}2\times \frac{1}{2}{\log }_{2}3\right)\right)\times \left(\left({\log }_{2}3\times 2{\log }_{3}2\right)-\left(\frac{1}{3}{\log }_{2}3\times 2{\log }_{3}2\right)\right)\left[\because {\log }_{m}n\times {\log }_{n}m=1\right] \\ =\left(9-\frac{3}{2}\right)\times \left(2-\frac{2}{3}\right) \\ =\frac{15}{2}\times \frac{4}{3}=10 \end{gathered}$$