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Question Number 140680 by bemath last updated on 11/May/21
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)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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