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$\left| {\,\begin{array}{*{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}\,} \right| \times \left| {\,\begin{array}{*{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|=$
$7$
$10$
$13$
$17$
Solution
(b) $\left| {\,\begin{array}{*{20}{c}}{{{\log }_2}512}&{{{\log }_8}3}\\{{{\log }_3}8}&{{{\log }_3}9}\end{array}\,} \right|$ $×$ $\left| {\,\begin{array}{*{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|$
$ = \left( {\frac{{\log 512}}{{\log 3}} \times \frac{{\log 9}}{{\log 4}} – \frac{{\log 3}}{{\log 4}} \times \frac{{\log 8}}{{\log 3}}} \right)$
$×$ $\left( {\frac{{\log 3}}{{\log 2}} \times \frac{{\log 4}}{{\log 3}} – \frac{{\log 3}}{{\log 8}} \times \frac{{\log 4}}{{\log 3}}} \right)$
$= \left( {\frac{{\log {2^9}}}{{\log 3}} \times \frac{{\log {3^2}}}{{\log {2^2}}} – \frac{{\log {2^3}}}{{\log {2^2}}}} \right)$ $×$ $\left( {\frac{{\log {2^2}}}{{\log 2}} – \frac{{\log {2^2}}}{{\log {2^3}}}} \right)$
$= \left( {\frac{{9 \times 2}}{2} – \frac{3}{2}} \right)\,\left( {2 – \frac{2}{3}} \right) = \frac{{15}}{2} \times \frac{4}{3} = 10$.
