Exponential equation-Lambert function?

  • Thread starter giorgos
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  • #1
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Hallo

I am trying to solve for x the equation

(b-x)exp^[(a-x)]+c-x=0

where a,b and c are constants.

Could anyone tell me if this equation can be solved using the Lambert function?
Thanks
 

Answers and Replies

  • #2
uart
Science Advisor
2,776
9
Can the solution to that equation be expressed in terms of the Labert W function?

Yes.
 
  • #3
uart
Science Advisor
2,776
9
Hmmm. At first glance I thought yes, but now I'm not so sure.

Certianly anything that can be expressed (with suitable change of variable) as a linear combination of log(x), x and 1 or as a linear combination of exp(x), x and 1, can be solved pretty easily in terms of W. I'm not 100% sure about linear combinations of x*exp(x), exp(x), x and 1 however.
 
  • #4
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I don't know if the solution is accurate or not but here is the method.

I have used a couple of approximations and assumptions.

Assuming a > 0 and for x << a

we can rewrite the equation as

exp(a-x) = (x-c)/(b-x) ;

adding one two both sides :

1 + exp(a-x) = (b-c)/(b-x);

and if a>> x then we can neglect 1 w.rt. exp(a-x) and so,

(b-x)exp(a-x) = b - c;

which yields x = b - W((b-c)exp(b-a));


Can someone check this though? Also, does neone know if we can solve

y = a*exp(b(x - y)) + c*exp(d(x-y));

in terms of W?
 
  • #5
2
0
Hi,

I have a simpler but related question.

How do we solve the following equation:

exp(ax)=bx+c

Thanks.
 
  • #6
2
0
exp(ax) = bx + c;

I think this might be the solution:

let bx + c = y/a;

a*x = (y - c*a)/b;

exp(y/b)*exp(-c*a/b) = y/a;

or

-(a/b)*exp(-c*a/b) = (-y/b)*exp(-y/b);

or -y/b = W(-(a/b)*exp(-c*a/b));

or y = -b*W(-(a/b)*exp(-c*a/b));

where y = a*(bx+c);

You can check for mistakes
 
  • #7
2
0
Thanks a lot. Romain
 

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