# Exponential equation-Lambert function?

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

uart
Can the solution to that equation be expressed in terms of the Labert W function?

Yes.

uart
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.

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?

Hi,

I have a simpler but related question.

How do we solve the following equation:

exp(ax)=bx+c

Thanks.

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

Thanks a lot. Romain