# exponents and e problem

• Jul 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.

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

Then you tell us...
• Jul 14th 2007, 09:22 PM
jut
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 PM
CaptainBlack
Quote:

Originally Posted by jut
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\$

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