Question Number 80639 by jagoll last updated on 05/Feb/20
$$\frac{\mathrm{log}_{\mathrm{2}} ^{\mathrm{2}} \:\left({x}−\mathrm{4}\right)−\mathrm{log}_{\mathrm{2}} \left(\mathrm{4}−{x}\right)^{\mathrm{8}} +\mathrm{16}}{\mathrm{30}−\mathrm{3}{x}−\left(\mathrm{4}−{x}\right)^{\mathrm{2}} }\:\geqslant\:\mathrm{0} \\ $$
Commented by john santu last updated on 05/Feb/20
$$\frac{\mathrm{log}_{\mathrm{2}} ^{\mathrm{2}} \left({x}−\mathrm{4}\right)−\mathrm{8log}_{\mathrm{2}} \left({x}−\mathrm{4}\right)+\mathrm{16}}{\mathrm{30}−\mathrm{3}{x}−\mathrm{16}+\mathrm{8}{x}−{x}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{x}>\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\frac{\left\{\mathrm{log}_{\mathrm{2}} \left({x}−\mathrm{4}\right)−\mathrm{4}\right\}^{\mathrm{2}} }{−{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{14}}\geqslant\mathrm{0} \\ $$$$\frac{\left\{\mathrm{log}_{\mathrm{2}} \left({x}−\mathrm{4}\right)−\mathrm{4}\right\}^{\mathrm{2}} }{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{7}\right)}\leqslant\mathrm{0} \\ $$$$\therefore\:\mathrm{4}<{x}<\mathrm{7}\:\wedge{x}=\mathrm{20} \\ $$
Commented by jagoll last updated on 05/Feb/20
$${thank}\:{you} \\ $$