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Question Number 89451 by jagoll last updated on 17/Apr/20
log_5  ((3−x)(x^2 +2)) ≥ log_5 (x^2 −7x+12)+log_5 (5−x)
$$\mathrm{log}_{\mathrm{5}} \:\left(\left(\mathrm{3}−{x}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)\right)\:\geqslant\:\mathrm{log}_{\mathrm{5}} \left({x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\right)+\mathrm{log}_{\mathrm{5}} \left(\mathrm{5}−{x}\right) \\ $$
Answered by john santu last updated on 17/Apr/20
(1) (3−x)(x^2 +2) > 0   ⇒ x < 3   (2) x^2 −7x+12 > 0  (x−3)(x−4) > 0   ⇒x < 3 ∨ x > 4  (3) 5 −x > 0 ⇒ x < 5  (4) (3−x)(x^2 +2) ≥ (x−3)(x−4)(5−x)   ⇒(3−x)(x^2 +2)+(3−x)(x−4)(5−x)≥0  (3−x) {x^2 +2−x^2 +9x−20} ≥ 0  (3−x)(9x−18)≥ 0  ⇒ 2 ≤ x ≤ 3  solution : 2 ≤ x < 3
$$\left(\mathrm{1}\right)\:\left(\mathrm{3}−{x}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)\:>\:\mathrm{0}\: \\ $$$$\Rightarrow\:{x}\:<\:\mathrm{3}\: \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\:>\:\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\:>\:\mathrm{0} \\ $$$$\:\Rightarrow{x}\:<\:\mathrm{3}\:\vee\:{x}\:>\:\mathrm{4} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{5}\:−{x}\:>\:\mathrm{0}\:\Rightarrow\:{x}\:<\:\mathrm{5} \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{3}−{x}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)\:\geqslant\:\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left(\mathrm{5}−{x}\right)\: \\ $$$$\Rightarrow\left(\mathrm{3}−{x}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)+\left(\mathrm{3}−{x}\right)\left({x}−\mathrm{4}\right)\left(\mathrm{5}−{x}\right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{3}−{x}\right)\:\left\{{x}^{\mathrm{2}} +\mathrm{2}−{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{20}\right\}\:\geqslant\:\mathrm{0} \\ $$$$\left(\mathrm{3}−{x}\right)\left(\mathrm{9}{x}−\mathrm{18}\right)\geqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}\:\leqslant\:{x}\:\leqslant\:\mathrm{3} \\ $$$${solution}\::\:\mathrm{2}\:\leqslant\:{x}\:<\:\mathrm{3}\: \\ $$
Commented by jagoll last updated on 17/Apr/20
thank you
$${thank}\:{you} \\ $$

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