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

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Homework Help Overview

The discussion revolves around the equation (2t+1)e^{-2t}=5, with participants exploring the possibility of finding an exact solution for t. The subject area includes transcendental equations and the Lambert W-function.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the lack of an exact solution in elementary functions and consider the use of the Lambert W-function. There are attempts to manipulate the equation into a suitable form for applying the Lambert W-function, with questions about the necessary steps to achieve this.

Discussion Status

The discussion includes various attempts to reformulate the original equation, with one participant expressing confidence in their understanding after receiving input from others. However, there is no explicit consensus on the steps to take, indicating ongoing exploration of the topic.

Contextual Notes

Some participants mention the challenge of finding an exact solution and the reliance on numerical methods, highlighting the constraints of the problem.

flash
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Hi,

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

[tex](2t+1)e^{-2t}=5[/tex]

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:

[tex](2t+1)e^{-2t}=5[/tex]

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:

[itex]f(t)e^{f(t)}=k[/itex]

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

[itex]f(t)=W(k)[/itex]

how about then I just multiply by -1:

[itex]-(2t+1)e^{-2t}=-5[/itex]

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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