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Exponential functions - converting a linear equation

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Ok, so I have an unknown exponential function:
    y1 = f1(t)

    By measuring the values of t and yi, a linear connection is generated between Yi (=log (yi)) and t:

    Yi = Ait + Bi.
    A1 = -2.12
    B1 = 1.96

    Problem 1: Describe f1 in the following matter:
    y1 = f1(t) = Ceλt


    3. The attempt at a solution
    I've been juggling with numbers and letters without ending up with anything reasonable.
    Ok, let's ramble down some thoughts:
    Adding log function to both sides of the functions gives

    log(y1) = log t + log -2.12 + log 1.96 = log (C) + log (e) λt

    And this is where it all ends. Any help out there? Would be much appreciated!
     
  2. jcsd
  3. Oct 25, 2011 #2

    ehild

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    The red part has no sense...

    The logarithm of y is given: log(y)=At+B=-2.12 t + 1.96.

    And -2.12 t + 1.96=log(e)(λt)+log(C)

    What is the base of the logarithm?

    Compare the terms. The constant terms have to be equal on both sides, and the coefficients of the variable t also have to be equal.

    ehild
     
  4. Oct 25, 2011 #3
    Oh, of course the logarithm of y is given! Why didn't I see that before. Thanks for your reply.

    The base of the logarithm is 10, forgot to say.
    But, how do I solve it from here? This is my first go at a problem like this, so any help is appreciated. How can I solve this when I have no values for neither t, C and λ? and the log gets in the way.
     
  5. Oct 25, 2011 #4

    I like Serena

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    Welcome to PF, LizzieL! :smile:

    ehild's equation must hold for any t.
    This means that -2.12=log(e)(λ) and that 1.96=log(C).

    For instance -2.12=log(e)(λ) can be rewritten as λ=-2.12/log(e).
     
  6. Oct 25, 2011 #5
    Oh my, you guys really are angels sent from up above. Thanks a whole lot! Also for the welcome :smile:

    So, I have a similar problem:

    log(y2) = 1.85t + 1.56 = log(e)μ(t-a)

    I cannot use the same principle as before, right? Or can I?

    If I try again...

    1.85 = log(e)μ
    1.56 = -log(e)μa

    Is this way off?
     
  7. Oct 25, 2011 #6

    I like Serena

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    I'd say it is right on!! :approve:
     
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