Question Number 70066 by Maclaurin Stickker last updated on 30/Sep/19
$${Solve} \\ $$$$\left.\mathrm{a}\right)\:{e}^{\mathrm{2}{x}} −{e}^{{x}+\mathrm{1}} −{e}^{{x}} +{e}<\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\mathrm{4}.\mathrm{2}^{\mathrm{2}{x}} −\mathrm{9}.\mathrm{2}^{{x}} <−\mathrm{2} \\ $$$$\left.{c}\right)\mathrm{9}^{{x}} −\mathrm{4}.\mathrm{3}^{{x}+\mathrm{1}} +\mathrm{27}>\mathrm{0} \\ $$$$ \\ $$
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}](https://www.tinkutara.com/question/Q70083.png)
$$\left.{b}\right)\:\:\mathrm{4}\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} \:−\mathrm{9}\left(\mathrm{2}^{{x}} \right)\:+\:\mathrm{2}\:<\:\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{4}\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{2}^{{x}} \right)\:−\left(\mathrm{2}^{{x}} \right)\:+\:\mathrm{2}\:<\mathrm{0} \\ $$$$\:\:\:\:\mathrm{4}\left(\mathrm{2}^{{x}} \right)\left(\mathrm{2}^{{x}} −\mathrm{2}\right)\:−\mathrm{1}\left(\mathrm{2}^{{x}} −\mathrm{2}\right)<\mathrm{0} \\ $$$$\:\:\:\:\left(\mathrm{2}^{{x}} −\mathrm{2}\right)\left[\mathrm{4}\left(\mathrm{2}^{{x}} \right)−\mathrm{1}\right]\:<\mathrm{0} \\ $$$${the}\:{zeros}\:{are}\:\:\mathrm{2}^{{x}} =\:\mathrm{2}\:{or}\:\:\mathrm{2}^{{x}} =\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{x}\:=\:\mathrm{1}\:{or}\:{x}\:=\:−\mathrm{2} \\ $$$$\:\mathrm{2}^{{x}} <\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}<\mathrm{2}^{{x}} <\:\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{2}^{{x}} >\mathrm{2} \\ $$$$++++\:\:\:\:\:\:\:−−−−−\:\:\:\:\:\:++++ \\ $$$${solution}\:{set}\:{S}\:=\:\left\{\mathrm{2}^{{x}} :\:\frac{\mathrm{1}}{\mathrm{4}}<\mathrm{2}^{{x}} <\mathrm{2}\right\}\:\:\:{or}\:\:{S}\:=\:\left\{{x}:\:−\mathrm{2}\:<\:{x}\:<\:\mathrm{1}\right\} \\ $$$$ \\ $$
Commented by Maclaurin Stickker last updated on 01/Oct/19
$${Great}! \\ $$$$ \\ $$
Commented by Rio Michael last updated on 02/Oct/19
$${thanks} \\ $$
Answered by MJS last updated on 30/Sep/19
$$\left.{a}\right) \\ $$$$\mathrm{e}^{\mathrm{2}{x}} −\left(\mathrm{e}+\mathrm{1}\right)\mathrm{e}^{{x}} +\mathrm{e}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{e}^{{x}} =\mathrm{1}\vee\mathrm{e}^{{x}} =\mathrm{e}\:\Rightarrow\:{x}=\mathrm{0}\vee{x}=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{0}<{x}<\mathrm{1} \\ $$$$\left.{b}\right) \\ $$$$\mathrm{2}^{\mathrm{2}{x}} −\frac{\mathrm{9}}{\mathrm{4}}\mathrm{2}^{{x}} +\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}^{{x}} =\frac{\mathrm{1}}{\mathrm{4}}\vee\mathrm{2}^{{x}} =\mathrm{2}\:\Rightarrow\:{x}=−\mathrm{2}\vee{x}=\mathrm{1} \\ $$$$\Rightarrow\:−\mathrm{2}<{x}<\mathrm{1} \\ $$$$\left.{c}\right) \\ $$$$\mathrm{3}^{\mathrm{2}{x}} −\mathrm{12}×\mathrm{3}^{{x}} +\mathrm{27}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{3}^{{x}} =\mathrm{3}\vee\mathrm{3}^{{x}} =\mathrm{9}\:\Rightarrow\:{x}=\mathrm{1}\vee{x}=\mathrm{2} \\ $$$$\Rightarrow\:{x}<\mathrm{1}\vee{x}>\mathrm{2} \\ $$
Commented by Mr. K last updated on 01/Oct/19
$${Amazing} \\ $$
Commented by Maclaurin Stickker last updated on 01/Oct/19
$${Great}! \\ $$