Solving Equations with the Lambert W Function: A* (e^2x - e^x) + b*x = c

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In summary, the conversation is discussing the possibility of solving the equation a*(e^(2x)-e^x)+b*x=c in terms of a, b, and c. It is suggested that this can be done by finding the intersection of a line with a quadratic in e^x, represented by the equation e^x*(e^x-1)=mx+k, where m=-b/a and k=c/a. It is also mentioned that there may be a special function that can solve this problem without needing to look at a graph.
  • #1
jya
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I know we can solve e^x=x by the Lambert W function, but is it possible to solve the following equation:

a*(e^(2x)-e^x)+b*x=c

in terms of a, b, and c.
 
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  • #2
Welcome to PF;

solve: a*(e^(2x)-e^x)+b*x=c ... in terms of a, b, and c.
Solving for x given a,b,c you mean? That would be the intersection of a line with a quadratic in e^x... that is $$ e^{x}\left ( e^x - 1\right ) = mx+k$$ ...where ##m=-b/a## and ##k=c/a##.
And you want to find x given m and k.
That help?
 
Last edited:
  • #3
Thank you Simon. Yes. I am wondering if there is some special function can solve this problem, i.e. given the values of a,b,c (or m,k in your equation) I can find out the value of x without looking at the graph.
Simon Bridge said:
Welcome to PF;

Solving for x given a,b,c you mean? That would be the intersection of a line with a quadratic in e^x... that is $$ e^{x}\left ( e^x - 1\right ) = mx+k$$ ...where ##m=-b/a## and ##k=c/a##.
And you want to find x given m and k.
That help?
 

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