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!
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