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Question Number 6021 by sanusihammed last updated on 10/Jun/16

2^x + 2x = 8 find the value of x

$$\mathrm{2}^{{x}} \:+\:\mathrm{2}{x}\:=\:\mathrm{8}\: \\ $$$$ \\ $$$${find}\:{the}\:{value}\:{of}\:{x} \\ $$

Commented by Yozzii last updated on 10/Jun/16

x=2 2^2 +2×2=4+4=8

$${x}=\mathrm{2} \\ $$$$\mathrm{2}^{\mathrm{2}} +\mathrm{2}×\mathrm{2}=\mathrm{4}+\mathrm{4}=\mathrm{8} \\ $$

Commented by sanusihammed last updated on 10/Jun/16

i think that is trial and error

$$ \\ $$$${i}\:{think}\:{that}\:{is}\:{trial}\:{and}\:{error} \\ $$

Commented by Yozzii last updated on 10/Jun/16

Yes it is.

$${Yes}\:{it}\:{is}. \\ $$

Answered by prakash jain last updated on 10/Jun/16

x=u+4 2^(u+4) +2(u+4)=8 2^(u+4) +2u+8=8 8∙2^u +u=0 8∙2^u =−u 8=−(u/2^u )=−(u/e^(uln 2) ) 8=−ue^(−uln 2) 8ln 2=−u∙ln 2∙e^(−uln 2) W(8ln 2)=−uln 2 −u=((W(8ln 2))/(ln 2)) (4−x)=((W(8ln 2))/(ln 2)) x=4−((W(8ln 2))/(ln 2)) W is log product function.

$${x}={u}+\mathrm{4} \\ $$$$\mathrm{2}^{{u}+\mathrm{4}} +\mathrm{2}\left({u}+\mathrm{4}\right)=\mathrm{8} \\ $$$$\mathrm{2}^{{u}+\mathrm{4}} +\mathrm{2}{u}+\mathrm{8}=\mathrm{8} \\ $$$$\mathrm{8}\centerdot\mathrm{2}^{{u}} +{u}=\mathrm{0} \\ $$$$\mathrm{8}\centerdot\mathrm{2}^{{u}} =−{u} \\ $$$$\mathrm{8}=−\frac{{u}}{\mathrm{2}^{{u}} }=−\frac{{u}}{{e}^{{u}\mathrm{ln}\:\mathrm{2}} } \\ $$$$\mathrm{8}=−{ue}^{−{u}\mathrm{ln}\:\mathrm{2}} \\ $$$$\mathrm{8ln}\:\mathrm{2}=−{u}\centerdot\mathrm{ln}\:\mathrm{2}\centerdot{e}^{−{u}\mathrm{ln}\:\mathrm{2}} \\ $$$${W}\left(\mathrm{8ln}\:\mathrm{2}\right)=−{u}\mathrm{ln}\:\mathrm{2} \\ $$$$−{u}=\frac{{W}\left(\mathrm{8ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\left(\mathrm{4}−{x}\right)=\frac{{W}\left(\mathrm{8ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}} \\ $$$${x}=\mathrm{4}−\frac{{W}\left(\mathrm{8ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}} \\ $$$${W}\:\mathrm{is}\:\mathrm{log}\:\mathrm{product}\:\mathrm{function}. \\ $$

Commented by Rasheed Soomro last updated on 10/Jun/16

Can we calculate further x=4−((W(8ln 2))/(ln 2))=2 Because 2 is obvious root of the given equation.

$${Can}\:{we}\:{calculate}\:{further} \\ $$$${x}=\mathrm{4}−\frac{{W}\left(\mathrm{8ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}=\mathrm{2} \\ $$$${Because}\:\mathrm{2}\:{is}\:{obvious}\:{root}\:{of}\:{the}\:{given} \\ $$$${equation}. \\ $$

Commented by sanusihammed last updated on 10/Jun/16

Intdrdsting. please i dont know what is product log..... sorry for asking too much question. i will love to know how the product log is express

$${Intdrdsting}. \\ $$$${please}\:{i}\:{dont}\:{know}\:{what}\:{is}\:{product}\:{log}..... \\ $$$${sorry}\:{for}\:{asking}\:{too}\:{much}\:{question}. \\ $$$${i}\:{will}\:{love}\:{to}\:{know}\:{how}\:{the}\:{product}\:{log}\:{is}\:{express} \\ $$

Commented by FilupSmith last updated on 10/Jun/16

Wikipedia has a lot of information on the product log. So does wolfram alpha a good old google search is your best friend! :)

$$\mathrm{Wikipedia}\:\mathrm{has}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{of}\:\mathrm{information} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{product}\:\mathrm{log}.\:\mathrm{So}\:\mathrm{does}\:\mathrm{wolfram}\:\mathrm{alpha} \\ $$$$ \\ $$$$\mathrm{a}\:\mathrm{good}\:\mathrm{old}\:\mathrm{google}\:\mathrm{search}\:\mathrm{is}\:\mathrm{your}\:\mathrm{best} \\ $$$$\left.\mathrm{friend}!\::\right) \\ $$

Commented by prakash jain last updated on 10/Jun/16

Rasheed′s comment W(8ln 2)=2ln 2 (Wolfram alpha) so x=2

$$\mathrm{Rasheed}'\mathrm{s}\:\mathrm{comment} \\ $$$${W}\left(\mathrm{8ln}\:\mathrm{2}\right)=\mathrm{2ln}\:\mathrm{2}\:\left(\mathrm{Wolfram}\:\mathrm{alpha}\right) \\ $$$$\mathrm{so}\:{x}=\mathrm{2} \\ $$

Commented by Rasheed Soomro last updated on 11/Jun/16

THANKS!

$$\mathfrak{THANKS}! \\ $$

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