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log-5-3-x-x-2-2-log-5-x-2-7x-12-log-5-5-x-

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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)
log5((3−x)(x2+2))⩾log5(x2−7x+12)+log5(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)(x2+2)>0⇒x<3(2)x2−7x+12>0(x−3)(x−4)>0⇒x<3∨x>4(3)5−x>0⇒x<5(4)(3−x)(x2+2)⩾(x−3)(x−4)(5−x)⇒(3−x)(x2+2)+(3−x)(x−4)(5−x)⩾0(3−x){x2+2−x2+9x−20}⩾0(3−x)(9x−18)⩾0⇒2⩽x⩽3solution:2⩽x<3
Commented by jagoll last updated on 17/Apr/20
thank you
thankyou

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