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

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The discussion centers on solving the equation a*(e^(2x) - e^x) + b*x = c using the Lambert W function. It is established that the equation can be reformulated to find the intersection of a line with a quadratic in e^x. Specifically, the transformation leads to the equation e^x(e^x - 1) = mx + k, where m and k are derived from the parameters a, b, and c. Participants express interest in determining x directly from the values of a, b, and c without graphing. The conversation emphasizes the potential for special functions to facilitate this solution.
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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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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?
 
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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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