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Math Help - Equation involving sum of exponentials

  1. #1
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    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!
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  2. #2
    MHF Contributor ebaines's Avatar
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    There may be an easier way, but here goes:

    Starting with

    <br />
-12.5 e^{-1500t} +16 e^{-2400t} + e^{-600t} = 0<br />

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

    <br />
16 e^{-1800t} -12.5 e^{-900t} +1 =0<br />

    This factors out to the form:
    <br />
(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:

    <br />
A^2 - 12.5A +16 = 0. <br />

    Use the quadratic formula to solve for A:

    <br />
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):

    <br />
(1.448e^{-900t} - 1) (11.052e^{-900t} -1) = 0.<br />

    So:
    <br />
e^{-900t} = \frac 1 {1.448}<br />
    <br />
t = \frac 1 {900} ln(1.448) = 0.000411<br />

    or:
    <br />
e^{-900t} = \frac 1 {11.052}<br />
    <br />
t = \frac 1 {900} ln(11.052) = 0.00267<br />
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  3. #3
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    awesome
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  4. #4
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    Hello, listonroute!

    Another approach . . .


    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)

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  5. #5
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    very nice as well, thanks. I should've seen this...been out of practice too long I suppose.
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