Solving for t in Lambert W Function: a=bt+e^(ct)

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The discussion focuses on solving for the variable t in the equation a = bt + e^(ct) using the Lambert W function. The transformation involves rewriting the equation to isolate the exponential term, leading to the expression e^(a/b) = (b/c)xe^x. This ultimately allows for the solution x = W(c/b * e^(a/b)), where W represents the Lambert W function. The process requires a solid understanding of algebraic manipulation and the properties of the Lambert W function.

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How can I use the W function to solve for t?

a = bt + e^(ct),

where a,b,c,e are known constants. e is Euler's number. t is the unknown?

I have only seen examples where a = 0. Thanks.
 
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Do a little thinking and a little algebra to get it into the right form.
Lambert's W function is the inverse function to f(x)= xex so you have to change variables to get your equation into that form. From
a= bt+ ect, a- bt= ect. Okay, let u= a- bt. Then t= -(u-a)/b so
[tex]e^{ct}= e^{\frac{-cu}{b}}e^{\frac{a}{b}}= u[/tex]
so
[tex]e^{\frac{a}{b}}= ue^{\frac{cu}{b}}[/tex]
Now let [itex]x= \frac{cu}{b}[/itex] so [itex]u= \frac{b}{c}x[/itex] and
[tex]e^{\frac{a}{b}}= \frac{b}{c}xe^x[/tex]
Finally,
[tex]\frac{c}{b}e^{\frac{a}{b}}= xe^x[/tex]
so that
[tex]x= W(\frac{c}{b}e^{\rac{a}{b}})[/tex]
 
Thanks a lot. That's cool!
 

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