# exponents and e problem

• July 14th 2007, 07:47 PM
jut
exponents 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.

• July 14th 2007, 08:13 PM
Krizalid
Factorise by $e^{-6t}$

Then you tell us...
• July 14th 2007, 09:22 PM
jut
I factored out a $e^{-6t}$ but that didn't seem to help.

I am left with :

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

• July 14th 2007, 10:52 PM
CaptainBlack
Quote:

Originally Posted by jut
I factored out a $e^{-6t}$ but that didn't seem to help.

I am left with :

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

Well as $e^z \ne 0 \mbox{ for any z }$, this reduces to:

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

putting $y=e^t$ gives:

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

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