Question

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

• A. 7
• B. 13
• C. 10
• D. 17

Answer 1

The correct option is C 10

$\left|\begin{array}{cc}lo{g}_{3}512& lo{g}_{4}3\\ lo{g}_{3}8& lo{g}_{4}9\end{array}\right|\times \left|\begin{array}{cc}lo{g}_{2}3& lo{g}_{8}3\\ lo{g}_{3}4& lo{g}_{3}4\end{array}\right|=(\frac{log512}{log3}\times \frac{log9}{log4}-\frac{log3}{log4}\times \frac{log8}{log3})\times (\frac{log3}{log2}\times \frac{log4}{log3}-\frac{log3}{log8}\times \frac{log4}{log3})=(\frac{log{2}^9}{log3}\times \frac{log{3}^2}{log{2}^2}-\frac{log{2}^3}{log{2}^2})\times (\frac{log{2}^2}{log2}-\frac{log{2}^2}{log{2}^3})=(\frac{9\times 2}{2}-\frac{3}{2})(2-\frac{2}{3})=\frac{15}{2}\times \frac{4}{3}=10.$