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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

$${Solve} \\ $$ $$\left.\mathrm{a}\right)\:{e}^{\mathrm{2}{x}} −{e}^{{x}+\mathrm{1}} −{e}^{{x}} +{e}<\mathrm{0} \\ $$ $$\left.\mathrm{b}\right)\mathrm{4}.\mathrm{2}^{\mathrm{2}{x}} −\mathrm{9}.\mathrm{2}^{{x}} <−\mathrm{2} \\ $$ $$\left.{c}\right)\mathrm{9}^{{x}} −\mathrm{4}.\mathrm{3}^{{x}+\mathrm{1}} +\mathrm{27}>\mathrm{0} \\ $$ $$ \\ $$

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}

$$\left.{b}\right)\:\:\mathrm{4}\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} \:−\mathrm{9}\left(\mathrm{2}^{{x}} \right)\:+\:\mathrm{2}\:<\:\mathrm{0} \\ $$ $$\:\:\:\:\:\mathrm{4}\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{2}^{{x}} \right)\:−\left(\mathrm{2}^{{x}} \right)\:+\:\mathrm{2}\:<\mathrm{0} \\ $$ $$\:\:\:\:\mathrm{4}\left(\mathrm{2}^{{x}} \right)\left(\mathrm{2}^{{x}} −\mathrm{2}\right)\:−\mathrm{1}\left(\mathrm{2}^{{x}} −\mathrm{2}\right)<\mathrm{0} \\ $$ $$\:\:\:\:\left(\mathrm{2}^{{x}} −\mathrm{2}\right)\left[\mathrm{4}\left(\mathrm{2}^{{x}} \right)−\mathrm{1}\right]\:<\mathrm{0} \\ $$ $${the}\:{zeros}\:{are}\:\:\mathrm{2}^{{x}} =\:\mathrm{2}\:{or}\:\:\mathrm{2}^{{x}} =\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{x}\:=\:\mathrm{1}\:{or}\:{x}\:=\:−\mathrm{2} \\ $$ $$\:\mathrm{2}^{{x}} <\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}<\mathrm{2}^{{x}} <\:\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{2}^{{x}} >\mathrm{2} \\ $$ $$++++\:\:\:\:\:\:\:−−−−−\:\:\:\:\:\:++++ \\ $$ $${solution}\:{set}\:{S}\:=\:\left\{\mathrm{2}^{{x}} :\:\frac{\mathrm{1}}{\mathrm{4}}<\mathrm{2}^{{x}} <\mathrm{2}\right\}\:\:\:{or}\:\:{S}\:=\:\left\{{x}:\:−\mathrm{2}\:<\:{x}\:<\:\mathrm{1}\right\} \\ $$ $$ \\ $$

Commented byMaclaurin Stickker last updated on 01/Oct/19

Great!

$${Great}! \\ $$ $$ \\ $$

Commented byRio 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

$$\left.{a}\right) \\ $$ $$\mathrm{e}^{\mathrm{2}{x}} −\left(\mathrm{e}+\mathrm{1}\right)\mathrm{e}^{{x}} +\mathrm{e}=\mathrm{0} \\ $$ $$\Rightarrow\:\mathrm{e}^{{x}} =\mathrm{1}\vee\mathrm{e}^{{x}} =\mathrm{e}\:\Rightarrow\:{x}=\mathrm{0}\vee{x}=\mathrm{1} \\ $$ $$\Rightarrow\:\mathrm{0}<{x}<\mathrm{1} \\ $$ $$\left.{b}\right) \\ $$ $$\mathrm{2}^{\mathrm{2}{x}} −\frac{\mathrm{9}}{\mathrm{4}}\mathrm{2}^{{x}} +\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$ $$\Rightarrow\:\mathrm{2}^{{x}} =\frac{\mathrm{1}}{\mathrm{4}}\vee\mathrm{2}^{{x}} =\mathrm{2}\:\Rightarrow\:{x}=−\mathrm{2}\vee{x}=\mathrm{1} \\ $$ $$\Rightarrow\:−\mathrm{2}<{x}<\mathrm{1} \\ $$ $$\left.{c}\right) \\ $$ $$\mathrm{3}^{\mathrm{2}{x}} −\mathrm{12}×\mathrm{3}^{{x}} +\mathrm{27}=\mathrm{0} \\ $$ $$\Rightarrow\:\mathrm{3}^{{x}} =\mathrm{3}\vee\mathrm{3}^{{x}} =\mathrm{9}\:\Rightarrow\:{x}=\mathrm{1}\vee{x}=\mathrm{2} \\ $$ $$\Rightarrow\:{x}<\mathrm{1}\vee{x}>\mathrm{2} \\ $$

Commented byMr. K last updated on 01/Oct/19

Amazing

$${Amazing} \\ $$

Commented byMaclaurin Stickker last updated on 01/Oct/19

Great!

$${Great}! \\ $$

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