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Logarithm's and Such

  1. Oct 10, 2006 #1
    There are initially 200 bacteria in a culture. After five minutes, the population has grown to 4080 bacteria. Estimate the doubling period.

    I DID THIS:

    [tex]4080=200(k)^5

    4080/200=k^5

    k=1.82[/tex]

    IS THIS CORRECT?
     
  2. jcsd
  3. Oct 10, 2006 #2
    what equation are you using? Exponential growth is N=Ni*e^(k*t). It doen't look like you did that. And simply solving the equation is not answering estimate the doubling period. once you solve for k you will need to then solve for t when the bacteria population is 400.
     
  4. Oct 10, 2006 #3
    ok:
    4080=200(k)^5

    4080/200}=k^5

    k=1.82

    I"M USING N=k(a)^x
     
  5. Oct 11, 2006 #4

    OlderDan

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    You can do it your way, but you actually used (k) in place of (a), or N = a(k)^x. Your k is just a bit off. Check it again. Now you need to find x that satisfies

    400 = 200(k)^x

    It doesn't have to be 400 and 200. All that is required is that the ratio be 2.
     
  6. Oct 11, 2006 #5
    [tex] N = N_{0}e^{kt} [/tex], [tex] N(0) = 200 [/tex].


    [tex] N = 200e^{kt} [/tex]

    [tex] 4080 = 200e^{5k} [/tex]

    [tex] k \doteq 0.603 [/tex]

    doubling time = 1.149 mins
     
    Last edited: Oct 11, 2006
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