# Equation involving sum of exponentials

• February 26th 2010, 09:29 AM
listonroute
Equation involving sum of exponentials
-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!
• February 26th 2010, 11:07 AM
ebaines
There may be an easier way, but here goes:

Starting with

$
-12.5 e^{-1500t} +16 e^{-2400t} + e^{-600t} = 0
$

Divide through by $e^{-600t}$, and rearrange:

$
16 e^{-1800t} -12.5 e^{-900t} +1 =0
$

This factors out to the form:
$
(Ae^{-900t} -1)(Be^{-900t}-1) = 0$
(eqn 1)

Where $AB = 16$ and $A+B = 12.5$. You have two equations in two unknowns, which gets you to:

$
A^2 - 12.5A +16 = 0.
$

Use the quadratic formula to solve for A:

$
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):

$
(1.448e^{-900t} - 1) (11.052e^{-900t} -1) = 0.
$

So:
$
e^{-900t} = \frac 1 {1.448}
$

$
t = \frac 1 {900} ln(1.448) = 0.000411
$

or:
$
e^{-900t} = \frac 1 {11.052}
$

$
t = \frac 1 {900} ln(11.052) = 0.00267
$
• February 26th 2010, 11:45 AM
listonroute
awesome
• February 26th 2010, 01:09 PM
Soroban
Hello, listonroute!

Another approach . . .

Quote:

Solve for $t\!:\;\;-12.5e^{-1500t} + 16e^{-2400t} + e^{-600t}\; =\; 0$

I don't know how to take the natural log of a sum of exponentials.
Nobody does!

We have: . $e^{-600t} - \frac{25}{2}e^{-1500t} + 16e^{-2400t} \;=\;0$

Multiply by $2e^{2400t}\!:\;\;2e^{1800t} - 25e^{900t} + 32 \;=\;0$

Let $x \,=\,e^{900t}\!:\;\;2x^2 - 25x + 32 \;=\;0$

Quadratic Formula: . $x \;=\;\frac{25 \pm3\sqrt{41}}{4}$

Back-substitute: . $e^{900t} \;=\;\frac{25\pm3\sqrt{41}}{4} \quad\Rightarrow\quad 900t \;=\;\ln\left(\frac{25\pm3\sqrt{41}}{4}\right)$

. . Therefore: . $t \;=\;\frac{1}{900}\ln\left(\frac{25\pm3\sqrt{41}}{ 4}\right)$

• February 26th 2010, 01:32 PM
listonroute
very nice as well, thanks. I should've seen this...been out of practice too long I suppose.