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Question Number 75933 by Rio Michael last updated on 21/Dec/19

solve the inequality a. ln(2x−e) >1 b. (lnx)^2 −lnx−6<0 c. ∣x∣ + ∣x+2∣ ≥ 2 d. ∣2x−5∣ + ∣x +2∣ > 7

solvetheinequality a.ln(2x−e)>1 b.(lnx)2−lnx−6<0 c.∣x∣+∣x+2∣⩾2 d.∣2x−5∣+∣x+2∣>7

Commented bymathmax by abdo last updated on 21/Dec/19

for x>(e/2) we have ln(2x−e)>1 ⇒ln(2x−e)>ln(e) ⇒ 2x−e>e ⇒2x>2e ⇒x>e ⇒S=]e,+∞[ b) let lnx=t with x>0 (e) ⇔ t^2 −t−6<0 Δ=1−4(−6) =25 ⇒t_1 =((1+5)/2) =3 and t_2 =((1−5)/2) =−2 t^2 −t−6<0 ⇔ −2<t<3 ⇒−2<lnx<3 ⇒e^(−2) <x<e^3 c) ∣x∣ +∣x+2∣≥2 ⇔ ∣x∣+∣x+2∣−2≥0 let A(x)=∣x∣+∣x+2∣−2 x −2 0 +∞ ∣x∣ −x −x x ∣x+2∣ −x−2 x+2 x+2 A(x) −2x−4 0 2x case 1 x≤−2 (in) ⇔ /−2x−4 ≥0 ⇔ −2x≥4 ⇔ 2x≤−4 ⇔x≤−2 ⇒S_1 =]−∞,−2] case 2 −2≤x≤0 (in) ⇔ 0≥0 (true)⇒s_2 =[−2,0] case 3 x≥0 (in) ⇔ 2x≥0 ⇒x ≥0 ⇒S_3 =[0,+∞[ ⇒ S =∪ Si=R .

forx>e2wehaveln(2x−e)>1⇒ln(2x−e)>ln(e)⇒ 2x−e>e⇒2x>2e⇒x>e⇒S=]e,+∞[ b)letlnx=twithx>0(e)⇔t2−t−6<0 Δ=1−4(−6)=25⇒t1=1+52=3andt2=1−52=−2 t2−t−6<0⇔−2<t<3⇒−2<lnx<3⇒e−2<x<e3 c)∣x∣+∣x+2∣⩾2⇔∣x∣+∣x+2∣−2⩾0 letA(x)=∣x∣+∣x+2∣−2 x−20+∞ ∣x∣−x−xx ∣x+2∣−x−2x+2x+2 A(x)−2x−402x case1x⩽−2(in)⇔/−2x−4⩾0⇔−2x⩾4⇔ 2x⩽−4⇔x⩽−2⇒S1=]−∞,−2] case2−2⩽x⩽0(in)⇔0⩾0(true)⇒s2=[−2,0] case3x⩾0(in)⇔2x⩾0⇒x⩾0⇒S3=[0,+∞[⇒ S=∪Si=R.

Commented byRio Michael last updated on 21/Dec/19

thanks

thanks

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