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Question Number 80639 by jagoll last updated on 05/Feb/20

((log_2 ^2  (x−4)−log_2 (4−x)^8 +16)/(30−3x−(4−x)^2 )) ≥ 0

$$\frac{\mathrm{log}_{\mathrm{2}} ^{\mathrm{2}} \:\left({x}−\mathrm{4}\right)−\mathrm{log}_{\mathrm{2}} \left(\mathrm{4}−{x}\right)^{\mathrm{8}} +\mathrm{16}}{\mathrm{30}−\mathrm{3}{x}−\left(\mathrm{4}−{x}\right)^{\mathrm{2}} }\:\geqslant\:\mathrm{0} \\ $$

Commented by john santu last updated on 05/Feb/20

((log_2 ^2 (x−4)−8log_2 (x−4)+16)/(30−3x−16+8x−x^2 ))≥0  (1) x>4  (2)(({log_2 (x−4)−4}^2 )/(−x^2 +5x+14))≥0  (({log_2 (x−4)−4}^2 )/((x+2)(x−7)))≤0  ∴ 4<x<7 ∧x=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

$${thank}\:{you} \\ $$

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