Here's the problem:

7e^(-2t) - 5e^(-5t) + 4e^(-6t) = 0

Solve for t.

I've tried to solve it a few ways (taking the natural log) but each leads to a dead end.

Please help!

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- Jul 14th 2007, 07:47 PMjutexponents and e problem
Here's the problem:

7e^(-2t) - 5e^(-5t) + 4e^(-6t) = 0

Solve for t.

I've tried to solve it a few ways (taking the natural log) but each leads to a dead end.

Please help! - Jul 14th 2007, 08:13 PMKrizalid
Factorise by $\displaystyle e^{-6t}$

Then you tell us... - Jul 14th 2007, 09:22 PMjut
I factored out a $\displaystyle e^{-6t}$ but that didn't seem to help.

I am left with :

$\displaystyle e^{-6t}*(7e^{4t}-e^{t}-4)=0$

- Jul 14th 2007, 10:52 PMCaptainBlack
Well as $\displaystyle e^z \ne 0 \mbox{ for any z } $, this reduces to:

$\displaystyle 7e^{4t}-5 e^{t}-4=0$

putting $\displaystyle y=e^t$ gives:

$\displaystyle 7 y^4-5 y-4=0$

which by Descartes rule of signs we know has exactly one positive root, which is what

we want if we are looking for a real solution to the original equation.

RonL