Question Number 106907 by Study last updated on 07/Aug/20
Answered by Don08q last updated on 08/Aug/20
$$\:\:\:\:\:\:=\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{8}} \left(\mathrm{64}\right)^{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{39}} } \right)\:\underset{} {\:} \\ $$$$\:\:\:\:\:\:=\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{8}} \mathrm{8}^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{38}} } \right)\:\underset{} {\:} \\ $$$$\:\:\:\:\:\:=\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{8}} \mathrm{8}^{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{38}} } \underset{} {\right)} \\ $$$$\:\:\:\:\:\:=\:\mathrm{log}_{\mathrm{2}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}\underset{} {\right)}^{\mathrm{38}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{log}_{\mathrm{2}} \:\mathrm{2}_{} ^{−\mathrm{38}} \\ $$$$\:\:\:\:\:\:=\:−\mathrm{38}\: \\ $$