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Solving an equations using LambertW function [on hold]

-1

$begingroup$

I have just started learning the LambertW function, so if my question is very basic I am really sorry but I can't understand how solving for x in the below equation

ln(x) + 2*x = 0

gave

LambertW(2)/2

lambert-w

share|cite|improve this question

asked Mar 27 at 18:23

kunalkunal

1

New contributor

kunal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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put on hold as off-topic by T. Bongers, Javi, YiFan, Thomas, Shailesh 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Javi, YiFan, Thomas, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

-1

$begingroup$

I have just started learning the LambertW function, so if my question is very basic I am really sorry but I can't understand how solving for x in the below equation

ln(x) + 2*x = 0

gave

LambertW(2)/2

lambert-w

share|cite|improve this question

asked Mar 27 at 18:23

kunalkunal

1

New contributor

kunal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by T. Bongers, Javi, YiFan, Thomas, Shailesh 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Javi, YiFan, Thomas, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

I have just started learning the LambertW function, so if my question is very basic I am really sorry but I can't understand how solving for x in the below equation

ln(x) + 2*x = 0

gave

LambertW(2)/2

lambert-w

share|cite|improve this question

asked Mar 27 at 18:23

kunalkunal

1

New contributor

kunal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

ln(x) + 2*x = 0

LambertW(2)/2

share|cite|improve this question

asked Mar 27 at 18:23

kunalkunal

1

New contributor

kunal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 27 at 18:23

kunalkunal

1

New contributor

kunal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 27 at 18:23

kunalkunal

1

New contributor

kunal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 27 at 18:23

kunalkunal

1

1 Answer
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3

$begingroup$

The Lambert-W function is the inverse of the function $f(w) = we^w$, $w ge -1$. That is to say, $W(z)e^W(z) = z$ for $z ge - dfrac 1e$.

Say you wish to solve $ln x + 2x = 0$. To bring the function $we^w$ into the picture, exponentiate the original expression to get $$x e^2x = e^ln x + 2x = e^0 = 1$$ so that $$2x e^2x = 2.$$ With $z = 2$ you obtain $W(2) = 2x$ so that $$x = fracW(2)2.$$

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answered Mar 27 at 18:31

Umberto P.Umberto P.

40.2k13370

$endgroup$

add a comment | 

3

$begingroup$

The Lambert-W function is the inverse of the function $f(w) = we^w$, $w ge -1$. That is to say, $W(z)e^W(z) = z$ for $z ge - dfrac 1e$.

Say you wish to solve $ln x + 2x = 0$. To bring the function $we^w$ into the picture, exponentiate the original expression to get $$x e^2x = e^ln x + 2x = e^0 = 1$$ so that $$2x e^2x = 2.$$ With $z = 2$ you obtain $W(2) = 2x$ so that $$x = fracW(2)2.$$

share|cite|improve this answer

answered Mar 27 at 18:31

Umberto P.Umberto P.

40.2k13370

$endgroup$

add a comment | 

3

$begingroup$

The Lambert-W function is the inverse of the function $f(w) = we^w$, $w ge -1$. That is to say, $W(z)e^W(z) = z$ for $z ge - dfrac 1e$.

Say you wish to solve $ln x + 2x = 0$. To bring the function $we^w$ into the picture, exponentiate the original expression to get $$x e^2x = e^ln x + 2x = e^0 = 1$$ so that $$2x e^2x = 2.$$ With $z = 2$ you obtain $W(2) = 2x$ so that $$x = fracW(2)2.$$

share|cite|improve this answer

answered Mar 27 at 18:31

Umberto P.Umberto P.

40.2k13370

$endgroup$

add a comment | 

$begingroup$

The Lambert-W function is the inverse of the function $f(w) = we^w$, $w ge -1$. That is to say, $W(z)e^W(z) = z$ for $z ge - dfrac 1e$.

Say you wish to solve $ln x + 2x = 0$. To bring the function $we^w$ into the picture, exponentiate the original expression to get $$x e^2x = e^ln x + 2x = e^0 = 1$$ so that $$2x e^2x = 2.$$ With $z = 2$ you obtain $W(2) = 2x$ so that $$x = fracW(2)2.$$

share|cite|improve this answer

answered Mar 27 at 18:31

Umberto P.Umberto P.

40.2k13370

$endgroup$

share|cite|improve this answer

answered Mar 27 at 18:31

Umberto P.Umberto P.

40.2k13370

answered Mar 27 at 18:31

Umberto P.Umberto P.

40.2k13370

answered Mar 27 at 18:31

Umberto P.Umberto P.

40.2k13370