Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$$\mathrm{2}^{{x}} =\mathrm{8}\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right) \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{8}\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right) \\ $$$$\frac{\sqrt{\mathrm{2}}\:\mathrm{ln}\:\mathrm{2}}{\mathrm{8}}\:=\left[\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right)\mathrm{ln}\:\mathrm{2}\right]{e}^{\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right)\mathrm{ln}\:\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right)\mathrm{ln}\:\mathrm{2}={W}\left(\frac{\sqrt{\mathrm{2}}\:\mathrm{ln}\:\mathrm{2}}{\mathrm{8}}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}{W}\left(\frac{\sqrt{\mathrm{2}}\:\mathrm{ln}\:\mathrm{2}}{\mathrm{8}}\right) \\ $$$$\approx\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{0}.\mathrm{109791}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{0}.\mathrm{341605} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com