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Is there a way to solve this equation? (maybe with Lambert's W Function?)

Two kind of equations involving natural log and exponentiationHow to solve the exponential inequality $x+3^x<4$A problem in generalizing the Lambert's W functionIs there ANY possible way to solve this equation?Solve analytically an equation with Lambert functionCute problem: Solve this equation involving Lambert-$W$ functionSolving tetration equation with Lambert W FunctionTranscendental Equation.How do I solve the following equation?Solving an equation with Lambert's W function? Or by any other means?

1

3

$begingroup$

I'd like to know if there is a way to solve the equation
$$xln x=alpha+beta x$$
for known constants $alpha,betainmathbbR$.
I know that Lambert's W Function $W$ can be used to solve
$$xln x=alpha$$
because then $x=e^W(alpha )$, but in the upper problem I can't reformulate the equation in a way that lets me use Lambert's W function.

Does anyone know how to do this? Or any way to find a solution for $x$?

Thanks.

lambert-w

share|cite|improve this question

asked Sep 19 '13 at 9:37

echoesechoes

83

$endgroup$

add a comment | 

1

3

$begingroup$

I'd like to know if there is a way to solve the equation
$$xln x=alpha+beta x$$
for known constants $alpha,betainmathbbR$.
I know that Lambert's W Function $W$ can be used to solve
$$xln x=alpha$$
because then $x=e^W(alpha )$, but in the upper problem I can't reformulate the equation in a way that lets me use Lambert's W function.

Does anyone know how to do this? Or any way to find a solution for $x$?

Thanks.

lambert-w

share|cite|improve this question

asked Sep 19 '13 at 9:37

echoesechoes

83

$endgroup$

add a comment | 

$begingroup$

I'd like to know if there is a way to solve the equation
$$xln x=alpha+beta x$$
for known constants $alpha,betainmathbbR$.
I know that Lambert's W Function $W$ can be used to solve
$$xln x=alpha$$
because then $x=e^W(alpha )$, but in the upper problem I can't reformulate the equation in a way that lets me use Lambert's W function.

Does anyone know how to do this? Or any way to find a solution for $x$?

Thanks.

lambert-w

share|cite|improve this question

asked Sep 19 '13 at 9:37

echoesechoes

83

$endgroup$

share|cite|improve this question

asked Sep 19 '13 at 9:37

echoesechoes

83

share|cite|improve this question

asked Sep 19 '13 at 9:37

echoesechoes

83

asked Sep 19 '13 at 9:37

echoesechoes

83

asked Sep 19 '13 at 9:37

echoesechoes

83

3 Answers
3

active

oldest

votes

4

$begingroup$

Note that $xln(x) - xbeta = xln(xe^-beta)$, so we substitute $y=xe^-beta$ and get
$$y ln(y) = alpha e^-beta.$$
Thus $y = exp(W(alpha e^-beta))$ and $x = exp(beta+W(alpha e^-beta))$.

share|cite|improve this answer

answered Sep 19 '13 at 9:49

AbelAbel

6,9011119

$endgroup$

add a comment | 

1

$begingroup$

We have
beginalign*
xlog x &= alpha x +beta\
iff beta &= x(log x - alpha)\
&= xlogbigl(xexp(-alpha)bigr)\
iff exp(-alpha)beta &= exp(-alpha)x logbigl(xexp(-alpha)bigr)\
iff exp(-alpha)x &= expbigl(W(exp(-alpha)beta)bigr)
endalign*

share|cite|improve this answer

answered Sep 19 '13 at 9:49

martinimartini

70.8k45991

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you swapped α and β
  $endgroup$
  – endolith
  Feb 8 '16 at 4:29
  

  
  
  
  

add a comment | 

1

$begingroup$

I will give you a step by step solution.

• First we have $x ln x=alpha +beta x$.


• We rearrange the equation to get $x ln x - beta x = alpha$.


• Factorising gives $x( ln x - beta ) = alpha$.


• Now we substitute $x= exp ( ln x)$ and multiply both sides by $ exp(-beta)$.


• Up to now we have
  $$( ln x - beta)e^ln x - beta= alpha e^-beta$$
  


• Now take $W$ of both sides and remember that $W(xe^x)=x$, so
  $$ln x - beta= W(alpha e^-beta)$$.


• Rearranging and taking exponents of both sides gives you the required result.
  $$x =e^ W(alpha e^-beta)+beta$$.

share|cite|improve this answer

answered Mar 28 at 17:58

Asad MehasiAsad Mehasi

111

New contributor

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

$endgroup$

add a comment | 

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4

$begingroup$

Note that $xln(x) - xbeta = xln(xe^-beta)$, so we substitute $y=xe^-beta$ and get
$$y ln(y) = alpha e^-beta.$$
Thus $y = exp(W(alpha e^-beta))$ and $x = exp(beta+W(alpha e^-beta))$.

share|cite|improve this answer

answered Sep 19 '13 at 9:49

AbelAbel

6,9011119

$endgroup$

add a comment | 

4

$begingroup$

Note that $xln(x) - xbeta = xln(xe^-beta)$, so we substitute $y=xe^-beta$ and get
$$y ln(y) = alpha e^-beta.$$
Thus $y = exp(W(alpha e^-beta))$ and $x = exp(beta+W(alpha e^-beta))$.

share|cite|improve this answer

answered Sep 19 '13 at 9:49

AbelAbel

6,9011119

$endgroup$

add a comment | 

$begingroup$

Note that $xln(x) - xbeta = xln(xe^-beta)$, so we substitute $y=xe^-beta$ and get
$$y ln(y) = alpha e^-beta.$$
Thus $y = exp(W(alpha e^-beta))$ and $x = exp(beta+W(alpha e^-beta))$.

share|cite|improve this answer

answered Sep 19 '13 at 9:49

AbelAbel

6,9011119

$endgroup$

share|cite|improve this answer

answered Sep 19 '13 at 9:49

AbelAbel

6,9011119

answered Sep 19 '13 at 9:49

AbelAbel

6,9011119

answered Sep 19 '13 at 9:49

AbelAbel

6,9011119

1

$begingroup$

We have
beginalign*
xlog x &= alpha x +beta\
iff beta &= x(log x - alpha)\
&= xlogbigl(xexp(-alpha)bigr)\
iff exp(-alpha)beta &= exp(-alpha)x logbigl(xexp(-alpha)bigr)\
iff exp(-alpha)x &= expbigl(W(exp(-alpha)beta)bigr)
endalign*

share|cite|improve this answer

answered Sep 19 '13 at 9:49

martinimartini

70.8k45991

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you swapped α and β
  $endgroup$
  – endolith
  Feb 8 '16 at 4:29
  

  
  
  
  

add a comment | 

1

$begingroup$

We have
beginalign*
xlog x &= alpha x +beta\
iff beta &= x(log x - alpha)\
&= xlogbigl(xexp(-alpha)bigr)\
iff exp(-alpha)beta &= exp(-alpha)x logbigl(xexp(-alpha)bigr)\
iff exp(-alpha)x &= expbigl(W(exp(-alpha)beta)bigr)
endalign*

share|cite|improve this answer

answered Sep 19 '13 at 9:49

martinimartini

70.8k45991

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you swapped α and β
  $endgroup$
  – endolith
  Feb 8 '16 at 4:29
  

  
  
  
  

add a comment | 

$begingroup$

We have
beginalign*
xlog x &= alpha x +beta\
iff beta &= x(log x - alpha)\
&= xlogbigl(xexp(-alpha)bigr)\
iff exp(-alpha)beta &= exp(-alpha)x logbigl(xexp(-alpha)bigr)\
iff exp(-alpha)x &= expbigl(W(exp(-alpha)beta)bigr)
endalign*

share|cite|improve this answer

answered Sep 19 '13 at 9:49

martinimartini

70.8k45991

$endgroup$

share|cite|improve this answer

answered Sep 19 '13 at 9:49

martinimartini

70.8k45991

answered Sep 19 '13 at 9:49

martinimartini

70.8k45991

answered Sep 19 '13 at 9:49

martinimartini

70.8k45991

1

$begingroup$

I will give you a step by step solution.

• First we have $x ln x=alpha +beta x$.


• We rearrange the equation to get $x ln x - beta x = alpha$.


• Factorising gives $x( ln x - beta ) = alpha$.


• Now we substitute $x= exp ( ln x)$ and multiply both sides by $ exp(-beta)$.


• Up to now we have
  $$( ln x - beta)e^ln x - beta= alpha e^-beta$$
  


• Now take $W$ of both sides and remember that $W(xe^x)=x$, so
  $$ln x - beta= W(alpha e^-beta)$$.


• Rearranging and taking exponents of both sides gives you the required result.
  $$x =e^ W(alpha e^-beta)+beta$$.

share|cite|improve this answer

answered Mar 28 at 17:58

Asad MehasiAsad Mehasi

111

New contributor

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

$endgroup$

add a comment | 

1

$begingroup$

I will give you a step by step solution.

• First we have $x ln x=alpha +beta x$.


• We rearrange the equation to get $x ln x - beta x = alpha$.


• Factorising gives $x( ln x - beta ) = alpha$.


• Now we substitute $x= exp ( ln x)$ and multiply both sides by $ exp(-beta)$.


• Up to now we have
  $$( ln x - beta)e^ln x - beta= alpha e^-beta$$
  


• Now take $W$ of both sides and remember that $W(xe^x)=x$, so
  $$ln x - beta= W(alpha e^-beta)$$.


• Rearranging and taking exponents of both sides gives you the required result.
  $$x =e^ W(alpha e^-beta)+beta$$.

share|cite|improve this answer

answered Mar 28 at 17:58

Asad MehasiAsad Mehasi

111

New contributor

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

$endgroup$

add a comment | 

$begingroup$

I will give you a step by step solution.

• First we have $x ln x=alpha +beta x$.


• We rearrange the equation to get $x ln x - beta x = alpha$.


• Factorising gives $x( ln x - beta ) = alpha$.


• Now we substitute $x= exp ( ln x)$ and multiply both sides by $ exp(-beta)$.


• Up to now we have
  $$( ln x - beta)e^ln x - beta= alpha e^-beta$$
  


• Now take $W$ of both sides and remember that $W(xe^x)=x$, so
  $$ln x - beta= W(alpha e^-beta)$$.


• Rearranging and taking exponents of both sides gives you the required result.
  $$x =e^ W(alpha e^-beta)+beta$$.

share|cite|improve this answer

answered Mar 28 at 17:58

Asad MehasiAsad Mehasi

111

New contributor

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

$endgroup$

share|cite|improve this answer

answered Mar 28 at 17:58

Asad MehasiAsad Mehasi

111

New contributor

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

answered Mar 28 at 17:58

Asad MehasiAsad Mehasi

111

New contributor

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

answered Mar 28 at 17:58

Asad MehasiAsad Mehasi

111