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Solving an Equation with Three e-Factors

  1. Mar 7, 2007 #1
    I'm trying to solve the following equation:

    20+54e^{-0.0682t}= \frac{50160e^{-0.075t} - 780e^{-0.75t} + 18280}{914}

    With the basic algebra I'm aware of, I can simplify the equation but I cannot get past the fact that there are three terms with e, meaning they are added up or subtracted from one another, and thus I cannot work them into one (excuse my messy choice of words, I'm not too familiar with the English mathematical terms). If anyone could tell me how to go about this, I'd much appreciate it.

    EDIT: I'm messing up the latex, those -0.075t, -0.075t and -0.75t are supposed to be exponents. I'll try and fix it.

    EDIT TWO: My apologies, I inserted the equation wrong. Edited.

    EDIT THREE: For some reason, though I changed the LaTeX (clicking it shows so), it still displays the old equation. Is this just so for me or do the source and the image look different to everyone? (I'm hoping it's just my browser pulling the image from cache and refusing to load it anew)
    Last edited: Mar 7, 2007
  2. jcsd
  3. Mar 7, 2007 #2
    The third "e-term" is just a constant. The two others are raised to the same power. Rearrange terms so that [tex]e^{-0.075t}[/tex] is alone on one side. Then take logarithms and you've got a linear equation.
  4. Mar 7, 2007 #3
    I'm sorry, as my second edit mentions, I inserted the equation incorrectly. Also, I do not believe any of the terms containing an e are constants as they all contain a t. Note my first edit. Still trying to figure that one out, if anyone could tell me what I'm overlooking, I'd appreciate that as well.
  5. Mar 7, 2007 #4


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    Your "first edit says
    I see no t in the third one.
  6. Mar 7, 2007 #5
    I apologize again, maybe I imagined that t there. I thought I'd put it there. I edited my post. For some reason, the LaTeX also looked quite different earlier today, when I was using a public computer. I suppose that added to all the confusion.

    At any rate, I edited my post. It should be correct now. Sorry for all my stumbling.
  7. Mar 7, 2007 #6


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    Okay, you can try to get the same exponents.

    .075= (0.0682)(1.0997...) or (0.0682)(1.1) approximately so
    [tex]e^{0.075t}= (e^{.0682t})^{1.1}[/tex]
    Also 0.75= (0.0682)(10.997...) or (0.0582)(11) approximately so
    [tex]e^{0.75t}= (e^{.0682t})^{11}[/tex]
    Letting x= e0.0682t your equation becomes
    [tex]20+54x= \frac{50160x^{1.1} - 780x^{-11} + 18280}{914}[/tex]
    Those numbers look like approximations to me anyway so a numerical method like Newton's method would probably be sufficient.
  8. Mar 9, 2007 #7
    I don't really see how to solve this equation either. The x in your new equation is still something to the power of t, so I'm still stuck with t as an exponent three times. Could you elaborate on your method?
  9. Mar 11, 2007 #8
    What HallsofIvy did is what you usually do with this kind of equation. If you'd had, for example, [tex]3+4e^{t}-5e^{2t}=0[/tex], you'd set [tex]x=e^t[/tex], so you get a new equation [tex]3+4x-5x^2=0[/tex], which is just a polynomial. Once you solve the polynomial equation, you have [tex]t=\log x[/tex] (if t is real, [tex]x=e^t[/tex] is positive, so throw away negative solutions to the polynomial). Generally, equations of your kind are never easier to solve than polynomials. With my little example, you get a quadratic equation, but the one HallsofIvy got is much harder (it probably has no elementary solution), so you're best off solving it numerically.
  10. Mar 11, 2007 #9
    Ah, right, I understand. Thanks for explaining.
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