Exponential equation-Lambert function?

In summary, the conversation is about solving the equation (b-x)exp^[(a-x)]+c-x=0 using the Lambert W function. The participants discuss the possibility of solving it and provide a method for solving similar equations using approximations and assumptions. They also mention another equation (y = a*exp(b(x - y)) + c*exp(d(x-y))) and provide a potential solution for it. The conversation ends with one participant thanking the others and mentioning his name.
  • #1
3
0
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
 
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  • #2
Can the solution to that equation be expressed in terms of the Labert W function?

Yes.
 
  • #3
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
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
Hi,

I have a simpler but related question.

How do we solve the following equation:

exp(ax)=bx+c

Thanks.
 
  • #6
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
Thanks a lot. Romain
 

1. What is an exponential equation?

An exponential equation is an equation in the form y = ab^x, where a and b are constants and x is a variable. It represents a relationship between a changing quantity and a constant ratio.

2. What is the Lambert function?

The Lambert function, also known as the omega function (ω), is a special function that is defined as the inverse of the function y = xe^x. It is typically denoted as W(x) and is useful in solving exponential equations that cannot be solved algebraically.

3. How is the Lambert function used in solving exponential equations?

The Lambert function is used in solving exponential equations by allowing us to express the variable in terms of the constants and the Lambert function. This makes it possible to solve for the variable using numerical methods, such as iteration.

4. What is the relationship between the Lambert function and the exponential equation?

The Lambert function and the exponential equation are closely related because the Lambert function is the inverse of the exponential function. This means that for any given value of x, the value of W(x) will satisfy the equation y = xe^x.

5. What are some real-world applications of exponential equations and the Lambert function?

Exponential equations and the Lambert function have many real-world applications, including in finance, population growth, and radioactive decay. They are also used in various fields of science, such as physics, chemistry, and biology, to model exponential processes and predict outcomes.

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