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Question Number 90806 by john santu last updated on 26/Apr/20

log_((x+4)) (x^2 −8x+12) < (1/2)log_(∣x−2∣) (2−x)^2

log(x+4)(x2−8x+12)<12log∣x−2∣(2−x)2

Commented byjohn santu last updated on 26/Apr/20

⇒(2−x)^2 = ∣2−x∣^2 = ∣x−2∣^2 log_((x+4)) (x^2 −8x+12) < log_(∣x−2∣) ∣x−2∣ log_((x+4)) (x^2 −8x+12) < 1 ⇒ log_((x+4)) (x^2 −8x+12) −log_((x+4)) (x+4)< 0 ((x^2 −8x+12−x−4)/(x+4−1)) < 0 (((x−1)(x−8))/((x+3))) < 0 (ii) x≠2 , x≠3 , x≠1 solution x∈ (−4;−3) ∪ (1;2) ∪ (6;8)

⇒(2−x)2=∣2−x∣2=∣x−2∣2 log(x+4)(x2−8x+12)<log∣x−2∣∣x−2∣ log(x+4)(x2−8x+12)<1 ⇒log(x+4)(x2−8x+12)−log(x+4)(x+4)<0 x2−8x+12−x−4x+4−1<0 (x−1)(x−8)(x+3)<0 (ii)x≠2,x≠3,x≠1 solutionx∈(−4;−3)∪(1;2)∪(6;8)

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