left| {begin{array}{*{20}{c}}{{{log }_3}512}& | vedclass

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(B) Let $D_1 = \left| {\begin{array}{*{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\{{{\log }_3}8}&{{{\log }_4}9}\end{array}} ight|$ and $D_2 = \left| {\

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$10$

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(B) Let $D_1 = \left| {\begin{array}{*{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\{{{\log }_3}8}&{{{\log }_4}9}\end{array}} ight|$ and $D_2 = \left| {\begin{array}{*{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\{{{\log }_3}4}&{{{\log }_3}4}\end{array}} ight|$.For $D_1$: $D_1 = (\log_3 512)(\log_4 9) - (\log_4 3)(\log_3 8)$$= (\frac{\log 512}{\log 3} 	imes \frac{\log 9}{\log 4}) - (\frac{\log 3}{\log 4} 	imes \frac{\log 8}{\log 3})$$= (\frac{9 \log 2}{\log 3} 	imes \frac{2 \log 3}{2 \log 2}) - (\frac{\log 8}{\log 4}) = 9 - \frac{3 \log 2}{2 \log 2} = 9 - 1.5 = 7.5$.For $D_2$: $D_2 = (\log_2 3)(\log_3 4) - (\log_8 3)(\log_3 4)$$= (\frac{\log 3}{\log 2} 	imes \frac{\log 4}{\log 3}) - (\frac{\log 3}{\log 8} 	imes \frac{\log 4}{\log 3})$$= (\frac{\log 4}{\log 2}) - (\frac{\log 4}{\log 8}) = 2 - \frac{2 \log 2}{3 \log 2} = 2 - \frac{2}{3} = \frac{4}{3}$.Therefore,$D_1 	imes D_2 = 7.5 	imes \frac{4}{3} = \frac{15}{2} 	imes \frac{4}{3} = 10$.

$\left| {\begin{array}{{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}} \right| \times \left| {\begin{array}{{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}} \right| = $

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$\left| {\begin{array}{*{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}} \right| \times \left| {\begin{array}{*{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}} \right| = $