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

[flash]

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

 

[Cyosis]

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

 

[jackmell]

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?

 

[flash]

Thanks for the help guys, I've got it now :)