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Homework Help: Simple exponential problem

  1. May 26, 2008 #1
    Question:

    The number of bacteria present in a culture at time t hours is modelled by the
    continuous variable N and the relationship

    N = 2000e^(kt)

    where k is a constant.

    Given that when t = 3, N = 18 000, find

    (a) the value of k to 3 significant figures

    (b) how long it takes for the number of bacteria present to double, giving your
    answer to the nearest minute

    k is found to be 0.73241 to 5s.f

    For the next part, i simply let N=18000*2=36000, and solved to find t=3.94 therefore it takes 3.94-3 = 0.94 hours for the number to double. This obviously yields the correct answer, however the markscheme let N=4000 subsequently solved to give 0.94 immediately (without having to subtract 3). Can someone please explain the logic behind the markschemes method?

    Thanks
     
  2. jcsd
  3. May 26, 2008 #2

    HallsofIvy

    User Avatar
    Science Advisor

    The "time to double" for any exponential function (or "half life" if k is negative) is a constant so it doesn't matter what t you start with. You chose to start with t= 3 while the "markscheme" started with t= 0. When t= 0, N= 2000 and when the number has doubled, N= 2*2000= 4000. Since N= 2000 when t= 0, the "time to double" is just t so that N(t)= 4000. 4000= 2000e^(kt) so e^(kt)= 2.

    You could as easily have said: "Taking T= time to double, N(3)= 18000= 2000e^{3k} and N(T+3)= 36000= 2000e^{k(T+3)}. Dividing the second equation by the first, 2= e^{k(T+3)}/e^{3k}= e^{kT+ 3k- 3k}= e^{kT}, again getting e^{kT)= 2.
     
  4. May 26, 2008 #3
    Thank you!
     
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