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Question Number 133936 by liberty last updated on 25/Feb/21
log _(x+8) (x^2 −3x−4) < 2.log _((4−x)^2 ) (∣x−4∣)
logx+8(x2−3x−4)<2.log(4−x)2(∣x−4∣)
Answered by EDWIN88 last updated on 25/Feb/21
log _(x+8) (x^2 −3x−4)<2.log _((4−x)^2 ) (∣x−4∣) { ((x^2 −3x−4>0)),((x+8>0 ; x+8≠1 )),(((4−x)^2 ≠1 , x≠4 )) :} { ((x<−1 ; x>4)),((x>−8)),((x≠4; x≠3; x≠5)) :} ⇔ log _(x+8) (x^2 −3x−4)<2.log _((x−4)^2 ) ∣x−4∣ ⇔log _(x+8) (x^2 −3x−4)<2.(1/2)log _(∣x−4∣) ∣x−4∣ ⇔ log _(x+8) (x^2 −3x−4)<1 ; ((ln (x^2 −3x−4))/(ln (x+8))) <1 ((ln (x^2 −3x−4)−ln (x+8))/(ln (x+8))) <0 numerator :ln (x^2 −3x−4) = ln (x+8) ; x^2 −4x−12=0 x_1 =6 ; x_2 =−2 denumerator : ln (x+8)=0 ; x=−7 solution set is (−8;−7)∪(−2;−1)∪(4;5)∪(5;6)
logx+8(x2−3x−4)<2.log(4−x)2(∣x−4∣){x2−3x−4>0x+8>0;x+8≠1(4−x)2≠1,x≠4{x<−1;x>4x>−8x≠4;x≠3;x≠5⇔logx+8(x2−3x−4)<2.log(x−4)2∣x−4∣⇔logx+8(x2−3x−4)<2.12log∣x−4∣∣x−4∣⇔logx+8(x2−3x−4)<1;ln(x2−3x−4)ln(x+8)<1ln(x2−3x−4)−ln(x+8)ln(x+8)<0numerator:ln(x2−3x−4)=ln(x+8);x2−4x−12=0x1=6;x2=−2denumerator:ln(x+8)=0;x=−7solutionsetis(−8;−7)∪(−2;−1)∪(4;5)∪(5;6)
Answered by bobhans last updated on 25/Feb/21
⇔ log _(x+8) (x^2 −3x−4) < 2.log _((4−x)^2 ) (√((4−x)^2 )) log _(x+8) (x^2 −3x−4) < 1 log _(x+8) (x^2 −3x−4) < log _(x+8) (x+8) ⇔ (x+8−1)(x^2 −3x−4−x−8)< 0 (x+7)(x^2 −4x−12) < 0 (x+7)(x−6)(x+2) < 0 (i) x < −7 ∪ −2 < x < 6 (ii) { (((4−x)^2 ≠ 0 ⇒x≠ 4)),(((4−x)^2 ≠ 1 ⇒ x≠ 3; x≠ 5)),((x+8 > 0 ; x > −8)) :} (iii) x^2 −3x−4 > 0 ; (x−4)(x+1)>0 x<−1 ∪ x > 4 solution : (i)∩(ii)∩(iii)
⇔logx+8(x2−3x−4)<2.log(4−x)2(4−x)2logx+8(x2−3x−4)<1logx+8(x2−3x−4)<logx+8(x+8)⇔(x+8−1)(x2−3x−4−x−8)<0(x+7)(x2−4x−12)<0(x+7)(x−6)(x+2)<0(i)x<−7∪−2<x<6(ii){(4−x)2≠0⇒x≠4(4−x)2≠1⇒x≠3;x≠5x+8>0;x>−8(iii)x2−3x−4>0;(x−4)(x+1)>0x<−1∪x>4solution:(i)∩(ii)∩(iii)

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