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2-to-the-power-x-2-to-the-power-x-1-3-lt-0-ease-ans-asap-

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Question Number 5130 by Apoorva last updated on 16/Apr/16
2 to the power x+2 to the power (−x+1)−3<0.ease ans asap
$$\mathrm{2}\:{to}\:{the}\:{power}\:{x}+\mathrm{2}\:{to}\:{the}\:{power}\:\left(−{x}+\mathrm{1}\right)−\mathrm{3}<\mathrm{0}.{ease}\:{ans}\:{asap} \\ $$
Commented by prakash jain last updated on 17/Apr/16
2^((x+2)^((−x+1)−3) ) <0 If x∈R then 2^((x+2)^((−x+1)−3) ) is always >0. There is no x∈R such that 2^((x+2)^((−x+1)−3) ) <0
$$\mathrm{2}^{\left({x}+\mathrm{2}\right)^{\left(−{x}+\mathrm{1}\right)−\mathrm{3}} } <\mathrm{0} \\ $$$$\mathrm{If}\:{x}\in\mathbb{R}\:\mathrm{then}\:\mathrm{2}^{\left({x}+\mathrm{2}\right)^{\left(−{x}+\mathrm{1}\right)−\mathrm{3}} } \:\mathrm{is}\:\mathrm{always}\:>\mathrm{0}. \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:{x}\in\mathbb{R}\:\mathrm{such}\:\mathrm{that}\:\mathrm{2}^{\left({x}+\mathrm{2}\right)^{\left(−{x}+\mathrm{1}\right)−\mathrm{3}} } <\mathrm{0} \\ $$
Commented by Apoorva last updated on 17/Apr/16
Sorry sir I made typing mistake it is 2 to the power (x+2) + 2 to the power (−x−1)−3(3 is not in the power of 2 it is separate)<0(/)
$${Sorry}\:{sir}\:{I}\:{made}\:{typing}\:{mistake}\:{it}\:{is}\:\mathrm{2}\:{to}\:{the}\:{power}\:\left({x}+\mathrm{2}\right)\:+\:\mathrm{2}\:{to}\:{the}\:{power}\:\left(−{x}−\mathrm{1}\right)−\mathrm{3}\left(\mathrm{3}\:{is}\:{not}\:{in}\:{the}\:{power}\:{of}\:\mathrm{2}\:{it}\:{is}\:{separate}\right)<\mathrm{0}\frac{}{} \\ $$
Commented by prakash jain last updated on 17/Apr/16
2^(x+2) +2^((−x−1)) −3<0? Pls confirm.
$$\mathrm{2}^{{x}+\mathrm{2}} +\mathrm{2}^{\left(−{x}−\mathrm{1}\right)} −\mathrm{3}<\mathrm{0}? \\ $$$$\mathrm{Pls}\:\mathrm{confirm}. \\ $$
Commented by Apoorva last updated on 17/Apr/16
Yes sir you got the question exactly correct.Please answer it
$${Yes}\:{sir}\:{you}\:{got}\:{the}\:{question}\:{exactly}\:{correct}.{Please}\:{answer}\:{it} \\ $$
Answered by prakash jain last updated on 18/Apr/16
2^(x+2) +2^(−x−1) −3<0 2^x =a 4a+(1/(2a))−3<0 ((8a^2 +1−6a)/(2a))<0 ((8a^2 −4a−2a+1)/(2a))<0 ((4a(2a−1)−1(2a−1))/(2a))<0 (((2a−1)(4a−1))/(2a))<0 since (1/(2a))=(1/2^(x+1) )>0, we need (2a−1)(4a−1)<0 (2a−1)(4a−1)<0 between the two roots a=(1/2) and a=(1/4) or (1/4)<a<(1/2) (1/4)<2^x <(1/2) or −2<x<−1
$$\mathrm{2}^{{x}+\mathrm{2}} +\mathrm{2}^{−{x}−\mathrm{1}} −\mathrm{3}<\mathrm{0} \\ $$$$\mathrm{2}^{{x}} ={a} \\ $$$$\mathrm{4}{a}+\frac{\mathrm{1}}{\mathrm{2}{a}}−\mathrm{3}<\mathrm{0} \\ $$$$\frac{\mathrm{8}{a}^{\mathrm{2}} +\mathrm{1}−\mathrm{6}{a}}{\mathrm{2}{a}}<\mathrm{0} \\ $$$$\frac{\mathrm{8}{a}^{\mathrm{2}} −\mathrm{4}{a}−\mathrm{2}{a}+\mathrm{1}}{\mathrm{2}{a}}<\mathrm{0} \\ $$$$\frac{\mathrm{4}{a}\left(\mathrm{2}{a}−\mathrm{1}\right)−\mathrm{1}\left(\mathrm{2}{a}−\mathrm{1}\right)}{\mathrm{2}{a}}<\mathrm{0} \\ $$$$\frac{\left(\mathrm{2}{a}−\mathrm{1}\right)\left(\mathrm{4}{a}−\mathrm{1}\right)}{\mathrm{2}{a}}<\mathrm{0} \\ $$$$\mathrm{since}\:\frac{\mathrm{1}}{\mathrm{2}{a}}=\frac{\mathrm{1}}{\mathrm{2}^{{x}+\mathrm{1}} }>\mathrm{0},\:\mathrm{we}\:\mathrm{need} \\ $$$$\left(\mathrm{2}{a}−\mathrm{1}\right)\left(\mathrm{4}{a}−\mathrm{1}\right)<\mathrm{0} \\ $$$$\left(\mathrm{2}{a}−\mathrm{1}\right)\left(\mathrm{4}{a}−\mathrm{1}\right)<\mathrm{0}\:\mathrm{between}\:\mathrm{the}\:\mathrm{two}\:\mathrm{roots} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:{a}=\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{or} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}<{a}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}<\mathrm{2}^{{x}} <\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${or}\:−\mathrm{2}<{x}<−\mathrm{1} \\ $$
Commented by Apoorva last updated on 18/Apr/16
Thanks a lot sir
$${Thanks}\:{a}\:{lot}\:{sir} \\ $$

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