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Population Growth question

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img14.imageshack.us/img14/1286/99892583.jpg [Broken]


    2. Relevant equations
    in the picture


    3. The attempt at a solution
    so i know that
    p(0)e^2k = 950
    p(0)e^8k = 9500

    how do i find k?
    for p(0) i get 440.950939 is that right?

    btw. ignore the answers in the picture
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 21, 2009 #2

    lanedance

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    how do you find p(0) without knowing k?

    try dividing you 2 equations together, then use logs
     
  4. Oct 21, 2009 #3
    what i did is made both equations equal to k
    so
    k1 = [ln(950/p(0))] / 2
    k2 = [ln(9500/p(0))] / 8

    k1 = k2
    [ln(950/p(0))] / 2 = [ln(9500/p(0))] / 8
    solving for p(0) i get 440.9509 which lets me plug it in the main equation and get the k value..
    but looks like its wrong eh?
     
  5. Oct 21, 2009 #4

    lanedance

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    so dividing the 2 equations gives
    [tex] e^{6k} = 10 [/tex]

    can you solve for k?

    you should get some constant a such that
    [tex]k = \frac{1}{a} ln(10) [/tex]

    the equation then becomes
    [tex]p(t) = p(0).e^{kt} = p(0).e^{(t/a) ln(10)} = p(0).e^{ln(10^{t/a}} = p(0) .10^{t/a}[/tex]
    [tex]p(0) = p(2).10^{-2/a}[/tex]

    which gives me the same p(0) value as you, so your way was fine
     
  6. Oct 21, 2009 #5
    oh so i just did it the long way... kk that makes sense.
    now how do i find the growth rate after 5 hours?
    k = 0.38376418 so growth rate = 38.376%
     
  7. Oct 21, 2009 #6

    lanedance

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    i would take the growth rate to mean p'(t)
     
  8. Oct 22, 2009 #7
    oh i got it!! tnx man
     
    Last edited: Oct 22, 2009
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