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Solve-the-equation-2-x-8x-4-




Question Number 100879 by Tony6400 last updated on 29/Jun/20
Solve the equation 2^x +8x=4
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2}^{\mathrm{x}} +\mathrm{8x}=\mathrm{4} \\ $$
Commented by bemath last updated on 29/Jun/20
approximation x = 0.342
$$\mathrm{approximation}\:\mathrm{x}\:=\:\mathrm{0}.\mathrm{342} \\ $$
Answered by 1549442205 last updated on 07/Jul/20
⇔2^(x−2) +2x−1=0.Putting f(x)=2^(x−2) +2x−1  f ′(x)=2^(x−2) ln2+2>0∀x∈(−∞;+∞)  f(x)=0⇔x=0.3416043768  Since f(x) is increasing ,the given equation has  only root x=0.3416043768
$$\Leftrightarrow\mathrm{2}^{\mathrm{x}−\mathrm{2}} +\mathrm{2x}−\mathrm{1}=\mathrm{0}.\mathrm{Putting}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2}^{\mathrm{x}−\mathrm{2}} +\mathrm{2x}−\mathrm{1} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{2}^{\mathrm{x}−\mathrm{2}} \mathrm{ln2}+\mathrm{2}>\mathrm{0}\forall\mathrm{x}\in\left(−\infty;+\infty\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0}\Leftrightarrow\mathrm{x}=\mathrm{0}.\mathrm{3416043768} \\ $$$$\mathrm{Since}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{increasing}\:,\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{has} \\ $$$$\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{root}}\:\boldsymbol{\mathrm{x}}=\mathrm{0}.\mathrm{3416043768} \\ $$
Answered by mr W last updated on 29/Jun/20
2^x =8((1/2)−x)  e^(xln 2) =8((1/2)−x)  (((√2) ln 2)/8) =[((1/2)−x)ln 2]e^(((1/2)−x)ln 2)   ((1/2)−x)ln 2=W((((√2) ln 2)/8))  ⇒x=(1/2)−(1/(ln 2))W((((√2) ln 2)/8))  ≈(1/2)−((0.109791)/(ln 2))=0.341605
$$\mathrm{2}^{{x}} =\mathrm{8}\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right) \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{8}\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right) \\ $$$$\frac{\sqrt{\mathrm{2}}\:\mathrm{ln}\:\mathrm{2}}{\mathrm{8}}\:=\left[\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right)\mathrm{ln}\:\mathrm{2}\right]{e}^{\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right)\mathrm{ln}\:\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right)\mathrm{ln}\:\mathrm{2}={W}\left(\frac{\sqrt{\mathrm{2}}\:\mathrm{ln}\:\mathrm{2}}{\mathrm{8}}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}{W}\left(\frac{\sqrt{\mathrm{2}}\:\mathrm{ln}\:\mathrm{2}}{\mathrm{8}}\right) \\ $$$$\approx\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{0}.\mathrm{109791}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{0}.\mathrm{341605} \\ $$

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