log_5 ((3−x)(x^2 +2)) ≥ log_5 (x^2 −7x+12)+log


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Question Number 89451 by jagoll last updated on 17/Apr/20

$\log_{5} ((3 - x) (x^{2} + 2)) ⩾ \log_{5} (x^{2} - 7 x + 12) + \log_{5} (5 - x)$
	Answered by john santu last updated on 17/Apr/20
	$(1) (3−x)(x^2 +2) > 0 ⇒ x < 3 (2) x^2 −7x+12 > 0 (x−3)(x−4) > 0 ⇒x < 3 ∨ x > 4 (3) 5 −x > 0 ⇒ x < 5 (4) (3−x)(x^2 +2) ≥ (x−3)(x−4)(5−x) ⇒(3−x)(x^2 +2)+(3−x)(x−4)(5−x)≥0 (3−x) {x^2 +2−x^2 +9x−20} ≥ 0 (3−x)(9x−18)≥ 0 ⇒ 2 ≤ x ≤ 3 solution : 2 ≤ x < 3$
	$(1) (3 - x) (x^{2} + 2) > 0$ $\Rightarrow x < 3$ $(2) x^{2} - 7 x + 12 > 0$ $(x - 3) (x - 4) > 0$ $\Rightarrow x < 3 \lor x > 4$ $(3) 5 - x > 0 \Rightarrow x < 5$ $(4) (3 - x) (x^{2} + 2) ⩾ (x - 3) (x - 4) (5 - x)$ $\Rightarrow (3 - x) (x^{2} + 2) + (3 - x) (x - 4) (5 - x) ⩾ 0$ $(3 - x) {x^{2} + 2 - x^{2} + 9 x - 20} ⩾ 0$ $(3 - x) (9 x - 18) ⩾ 0$ $\Rightarrow 2 ⩽ x ⩽ 3$ $s o l u t i o n : 2 ⩽ x < 3$
	Commented by jagoll last updated on 17/Apr/20

	$t h a n k y o u$
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