Is There an Exact Solution to (2t+1)e^{-2t}=5?

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There is no exact solution to the equation (2t+1)e^{-2t}=5 in terms of elementary functions. However, a close-form solution can be derived using the Lambert W-function. The discussion emphasizes transforming the equation into the form f(t)e^{f(t)}=k, allowing the application of the Lambert W-function to solve for f(t). Participants in the forum provided guidance on manipulating the equation to achieve this form.

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Hi,

I'm wondering if there is an exact solution to:

(2t+1)e^{-2t}=5

I've tried and have been unable to solve this for t. I can get an approximate numerical answer using the calculator but that's it.

Cheers
 
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There is no exact solution in terms of elementary functions. You can however find a close formed solution by using the Lambert W-function.
 
flash said:
Hi,

I'm wondering if there is an exact solution to:

(2t+1)e^{-2t}=5

I've tried and have been unable to solve this for t. I can get an approximate numerical answer using the calculator but that's it.

Cheers

How about if I start it for you and see if you can finish it. You want to get it into a form:

f(t)e^{f(t)}=k

so that you can then take the Lambert-W of both sides and write:

f(t)=W(k)

how about then I just multiply by -1:

-(2t+1)e^{-2t}=-5

see how I'm starting to build the exponent to look like it's coefficient. What else then would I have to do to get it into Lambert-W form?
 
Thanks for the help guys, I've got it now :)
 

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