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log-x-4-x-2-8x-12-lt-1-2-log-x-2-2-x-2-




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)(x28x+12)<12logx2(2x)2
Commented by john 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)
(2x)2=2x2=x22log(x+4)(x28x+12)<logx2x2log(x+4)(x28x+12)<1log(x+4)(x28x+12)log(x+4)(x+4)<0x28x+12x4x+41<0(x1)(x8)(x+3)<0(ii)x2,x3,x1solutionx(4;3)(1;2)(6;8)

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