Question Number 10774 by j.masanja06@gmail.com last updated on 24/Feb/17
Answered by sandy_suhendra last updated on 24/Feb/17
$$=\begin{vmatrix}{\mathrm{log}_{\mathrm{3}} \mathrm{1024}\:\:\:\:\mathrm{1}}\\{\mathrm{log}_{\mathrm{3}} \mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}×\begin{vmatrix}{\mathrm{log}_{\mathrm{2}} \mathrm{3}\:\:\:\:\mathrm{log}_{\mathrm{4}} \mathrm{3}}\\{\mathrm{log}_{\mathrm{3}} \mathrm{4}\:\:\:\:\mathrm{log}_{\mathrm{3}} \mathrm{4}}\end{vmatrix} \\ $$$$=\left(\mathrm{2log}_{\mathrm{3}} \mathrm{1024}−\mathrm{log}_{\mathrm{3}} \mathrm{8}\right)×\left(\mathrm{log}_{\mathrm{2}} \mathrm{3}.\mathrm{log}_{\mathrm{3}} \mathrm{4}−\mathrm{log}_{\mathrm{4}} \mathrm{3}.\mathrm{log}_{\mathrm{3}} \mathrm{4}\right)\:\:\:\: \\ $$$$=\left(\mathrm{log}_{\mathrm{3}} \frac{\mathrm{1024}^{\mathrm{2}} }{\mathrm{8}}\right)×\left(\mathrm{log}_{\mathrm{2}} \mathrm{4}−\mathrm{1}\right) \\ $$$$=\left(\mathrm{log}_{\mathrm{3}} \mathrm{2}^{\mathrm{17}} \right)×\left(\mathrm{2}−\mathrm{1}\right)\:=\:\mathrm{17}\:\mathrm{log}_{\mathrm{3}} \mathrm{2} \\ $$