Question Number 75933 by Rio Michael last updated on 21/Dec/19
$${solve}\:{the}\:{inequality} \\ $$$${a}.\:\:{ln}\left(\mathrm{2}{x}−{e}\right)\:>\mathrm{1} \\ $$$${b}.\:\left({lnx}\right)^{\mathrm{2}} −{lnx}−\mathrm{6}<\mathrm{0} \\ $$$${c}.\:\mid{x}\mid\:+\:\mid{x}+\mathrm{2}\mid\:\geqslant\:\mathrm{2} \\ $$$${d}.\:\mid\mathrm{2}{x}−\mathrm{5}\mid\:+\:\mid{x}\:+\mathrm{2}\mid\:>\:\mathrm{7} \\ $$
Commented by mathmax 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 .](https://www.tinkutara.com/question/Q75995.png)
$${for}\:{x}>\frac{{e}}{\mathrm{2}}\:\:\:{we}\:{have}\:{ln}\left(\mathrm{2}{x}−{e}\right)>\mathrm{1}\:\Rightarrow{ln}\left(\mathrm{2}{x}−{e}\right)>{ln}\left({e}\right)\:\Rightarrow \\ $$$$\left.\mathrm{2}{x}−{e}>{e}\:\Rightarrow\mathrm{2}{x}>\mathrm{2}{e}\:\Rightarrow{x}>{e}\:\:\:\Rightarrow{S}=\right]{e},+\infty\left[\right. \\ $$$$\left.{b}\right)\:{let}\:{lnx}={t}\:\:{with}\:{x}>\mathrm{0}\:\:\left({e}\right)\:\Leftrightarrow\:{t}^{\mathrm{2}} −{t}−\mathrm{6}<\mathrm{0} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}\left(−\mathrm{6}\right)\:=\mathrm{25}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{5}}{\mathrm{2}}\:=\mathrm{3}\:\:{and}\:{t}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{5}}{\mathrm{2}}\:=−\mathrm{2} \\ $$$${t}^{\mathrm{2}} −{t}−\mathrm{6}<\mathrm{0}\:\Leftrightarrow\:−\mathrm{2}<{t}<\mathrm{3}\:\:\Rightarrow−\mathrm{2}<{lnx}<\mathrm{3}\:\Rightarrow{e}^{−\mathrm{2}} <{x}<{e}^{\mathrm{3}} \\ $$$$\left.{c}\right)\:\:\mid{x}\mid\:+\mid{x}+\mathrm{2}\mid\geqslant\mathrm{2}\:\:\Leftrightarrow\:\mid{x}\mid+\mid{x}+\mathrm{2}\mid−\mathrm{2}\geqslant\mathrm{0} \\ $$$${let}\:{A}\left({x}\right)=\mid{x}\mid+\mid{x}+\mathrm{2}\mid−\mathrm{2} \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$$\mid{x}\mid\:\:\:\:\:−{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x} \\ $$$$\mid{x}+\mathrm{2}\mid\:−{x}−\mathrm{2}\:\:\:\:\:{x}+\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+\mathrm{2} \\ $$$${A}\left({x}\right)\:\:−\mathrm{2}{x}−\mathrm{4}\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{x} \\ $$$${case}\:\mathrm{1}\:\:\:\:\:{x}\leqslant−\mathrm{2}\:\:\:\:\:\:\left({in}\right)\:\Leftrightarrow\:/−\mathrm{2}{x}−\mathrm{4}\:\geqslant\mathrm{0}\:\:\Leftrightarrow\:−\mathrm{2}{x}\geqslant\mathrm{4}\:\Leftrightarrow \\ $$$$\left.\mathrm{2}\left.{x}\leqslant−\mathrm{4}\:\Leftrightarrow{x}\leqslant−\mathrm{2}\:\:\:\:\Rightarrow{S}_{\mathrm{1}} =\right]−\infty,−\mathrm{2}\right] \\ $$$${case}\:\mathrm{2}\:\:\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{0}\:\:\:\:\:\left({in}\right)\:\Leftrightarrow\:\:\mathrm{0}\geqslant\mathrm{0}\:\:\left({true}\right)\Rightarrow{s}_{\mathrm{2}} =\left[−\mathrm{2},\mathrm{0}\right] \\ $$$${case}\:\mathrm{3}\:\:\:\:{x}\geqslant\mathrm{0}\:\:\:\:\:\left({in}\right)\:\Leftrightarrow\:\mathrm{2}{x}\geqslant\mathrm{0}\:\Rightarrow{x}\:\geqslant\mathrm{0}\:\Rightarrow{S}_{\mathrm{3}} =\left[\mathrm{0},+\infty\left[\:\Rightarrow\right.\right. \\ $$$${S}\:=\cup\:{Si}={R}\:. \\ $$$$ \\ $$
Commented by Rio Michael last updated on 21/Dec/19
$${thanks} \\ $$