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log-x-2-7x-2-x-3-log-1-x-2-x-2-3x-log-x-2-5-x-

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Question Number 140680 by bemath last updated on 11/May/21
log _((x+2)) (7x^2 −x^3 )−log _(1/(x+2)) (x^2 −3x)≥ log _((√(x+2)) ) ((√(5−x)) )
log(x+2)(7x2−x3)−log1x+2(x2−3x)⩾logx+2(5−x)
Answered by liberty last updated on 11/May/21
(1) x+2> 0 ∧x+2 ≠1 ⇒x>−2 ∧x≠−1 (2)7x^2 −x^3 >0⇒x^2 (7−x)>0 ⇒ x < 7 ∧x≠ 0 (3)x^2 −3x > 0⇒x(x−3)>0 ⇒x<0 ∨ x > 3 (4) 5−x > 0 ⇒x < 5 (5) log _((x+2)) (7x^2 −x^3 )−log _((x+2)) (x^2 −3x) ≥ log _((x+2)) (5−x) ⇒log _((x+2)) (7x^2 −x^3 )−log _((x+2)) (x^2 −3x)−log (5−x)≥0 ⇒log _((x+2)) (7x^2 −x^3 )−[ log _((x+2)) (x^2 −3x)(5−x)]≥0 ⇒ ((7x^2 −x^3 −(x^2 −3x)(5−x))/(x+2−1)) ≥ 0 ⇒((7x^2 −x^3 −(5x^2 −x^3 −15x+3x^2 ))/(x+1)) ≥ 0 ⇒((15x−x^2 )/(x+1)) ≥ 0 ; ((x(15−x))/(x+1)) ≥ 0 ⇒x<−1 ∨ 0≤x≤15 Solution set is ⇒−2<x<−1 ∨ 3<x<5
(1)x+2>0∧x+2≠1⇒x>−2∧x≠−1(2)7x2−x3>0⇒x2(7−x)>0⇒x<7∧x≠0(3)x2−3x>0⇒x(x−3)>0⇒x<0∨x>3(4)5−x>0⇒x<5(5)log(x+2)(7x2−x3)−log(x+2)(x2−3x)⩾log(x+2)(5−x)⇒log(x+2)(7x2−x3)−log(x+2)(x2−3x)−log(5−x)⩾0⇒log(x+2)(7x2−x3)−[log(x+2)(x2−3x)(5−x)]⩾0⇒7x2−x3−(x2−3x)(5−x)x+2−1⩾0⇒7x2−x3−(5x2−x3−15x+3x2)x+1⩾0⇒15x−x2x+1⩾0;x(15−x)x+1⩾0⇒x<−1∨0⩽x⩽15Solutionsetis⇒−2<x<−1∨3<x<5

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