-12.5exp(-1500t) + 16exp(-2400t) + exp(-600t) = 0
solve for t. I don't know how to take the natural log of a sum of exponentials. thanks for your help!
There may be an easier way, but here goes:
Starting with
$\displaystyle
-12.5 e^{-1500t} +16 e^{-2400t} + e^{-600t} = 0
$
Divide through by $\displaystyle e^{-600t}$, and rearrange:
$\displaystyle
16 e^{-1800t} -12.5 e^{-900t} +1 =0
$
This factors out to the form:
$\displaystyle
(Ae^{-900t} -1)(Be^{-900t}-1) = 0 $ (eqn 1)
Where $\displaystyle AB = 16$ and $\displaystyle A+B = 12.5$. You have two equations in two unknowns, which gets you to:
$\displaystyle
A^2 - 12.5A +16 = 0.
$
Use the quadratic formula to solve for A:
$\displaystyle
A = \frac {12.5 \pm \sqrt{12.5^2 -64}} 2 = $ 1.448 or 11.052, B = 12.5-A = 11.052 or 1.448
Put these values back into the equation (1):
$\displaystyle
(1.448e^{-900t} - 1) (11.052e^{-900t} -1) = 0.
$
So:
$\displaystyle
e^{-900t} = \frac 1 {1.448}
$
$\displaystyle
t = \frac 1 {900} ln(1.448) = 0.000411
$
or:
$\displaystyle
e^{-900t} = \frac 1 {11.052}
$
$\displaystyle
t = \frac 1 {900} ln(11.052) = 0.00267
$
Hello, listonroute!
Another approach . . .
Solve for $\displaystyle t\!:\;\;-12.5e^{-1500t} + 16e^{-2400t} + e^{-600t}\; =\; 0$
I don't know how to take the natural log of a sum of exponentials.
Nobody does!
We have: .$\displaystyle e^{-600t} - \frac{25}{2}e^{-1500t} + 16e^{-2400t} \;=\;0$
Multiply by $\displaystyle 2e^{2400t}\!:\;\;2e^{1800t} - 25e^{900t} + 32 \;=\;0$
Let $\displaystyle x \,=\,e^{900t}\!:\;\;2x^2 - 25x + 32 \;=\;0$
Quadratic Formula: .$\displaystyle x \;=\;\frac{25 \pm3\sqrt{41}}{4}$
Back-substitute: .$\displaystyle e^{900t} \;=\;\frac{25\pm3\sqrt{41}}{4} \quad\Rightarrow\quad 900t \;=\;\ln\left(\frac{25\pm3\sqrt{41}}{4}\right) $
. . Therefore: .$\displaystyle t \;=\;\frac{1}{900}\ln\left(\frac{25\pm3\sqrt{41}}{ 4}\right) $