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Solve-a-e-2x-e-x-1-e-x-e-lt-0-b-4-2-2x-9-2-x-lt-2-c-9-x-4-3-x-1-27-gt-0-

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Question Number 70066 by Maclaurin Stickker last updated on 30/Sep/19
Solve a) e^(2x) −e^(x+1) −e^x +e<0 b)4.2^(2x) −9.2^x <−2 c)9^x −4.3^(x+1) +27>0
Solvea)e2x−ex+1−ex+e<0b)4.22x−9.2x<−2c)9x−4.3x+1+27>0
Answered by Rio Michael last updated on 30/Sep/19
b) 4(2^x )^2 −9(2^x ) + 2 < 0 4(2^x )^2 −8(2^x ) −(2^x ) + 2 <0 4(2^x )(2^x −2) −1(2^x −2)<0 (2^x −2)[4(2^x )−1] <0 the zeros are 2^x = 2 or 2^x = (1/4) ⇒ x = 1 or x = −2 2^x <(1/4) (1/4)<2^x < 2 2^x >2 ++++ −−−−− ++++ solution set S = {2^x : (1/4)<2^x <2} or S = {x: −2 < x < 1}
b)4(2x)2−9(2x)+2<04(2x)2−8(2x)−(2x)+2<04(2x)(2x−2)−1(2x−2)<0(2x−2)[4(2x)−1]<0thezerosare2x=2or2x=14⇒x=1orx=−22x<1414<2x<22x>2++++−−−−−++++solutionsetS={2x:14<2x<2}orS={x:−2<x<1}
Commented by Maclaurin Stickker last updated on 01/Oct/19
Great!
Great!
Commented by Rio Michael last updated on 02/Oct/19
thanks
thanks
Answered by MJS last updated on 30/Sep/19
a) e^(2x) −(e+1)e^x +e=0 ⇒ e^x =1∨e^x =e ⇒ x=0∨x=1 ⇒ 0<x<1 b) 2^(2x) −(9/4)2^x +(1/2)=0 ⇒ 2^x =(1/4)∨2^x =2 ⇒ x=−2∨x=1 ⇒ −2<x<1 c) 3^(2x) −12×3^x +27=0 ⇒ 3^x =3∨3^x =9 ⇒ x=1∨x=2 ⇒ x<1∨x>2
a)e2x−(e+1)ex+e=0⇒ex=1∨ex=e⇒x=0∨x=1⇒0<x<1b)22x−942x+12=0⇒2x=14∨2x=2⇒x=−2∨x=1⇒−2<x<1c)32x−12×3x+27=0⇒3x=3∨3x=9⇒x=1∨x=2⇒x<1∨x>2
Commented by Mr. K last updated on 01/Oct/19
Amazing
Amazing
Commented by Maclaurin Stickker last updated on 01/Oct/19
Great!
Great!

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