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Expononential Growth

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data
    A bacterial population size N is known to be growing exponentially. If the population triples between noon and 2pm, at what time will N be 100 times the noon population.

    2. Relevant equations
    Firstly. Is this a distribution function??
    If so; f(t)=ue^ut
    where E(t) = 1/u

    3. The attempt at a solution
    I have no idea where to start ...
    t=2 -> 3 times initial population (N)
    I have no idea ...
  2. jcsd
  3. Oct 3, 2007 #2


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    This is an exponential growth problem. pop(t)=P0*exp(k*t) where P0 is the initial population and k is the growth constant with t=0 being noon. If t=2hr then pop(t)=3*P0. Can you find k? Once you've found k, can you say at what value of t is pop(t)=100*P0?
  4. Oct 3, 2007 #3
    @ t=2 -> 3Po=Poe^kt
    therefore ln3=2k and k=0.5493

    @ what time is P(t)=100Po
    therefore t=8.38
    or approxiamtely 8:23pm

    Thanks for your help Dick!

  5. Oct 3, 2007 #4


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    Remember that all exponentials are equivalent: [itex]a^x= e^{x ln(a)}[/itex] so the only difference is a coefficient.
    Since you are told that "the population triples between noon and 2pm", that is that it triples every 2 hours, it is much easier to use 3t/2 where t is in hours. Since there were initially N bacteria, P(t)= N(3t/2)= 100N. Solving 3t/2= 100, (t/2)ln(3)= ln(100) so t= 2ln(100)/ln(3) which gives exactly the answer you got. Of course, you could also use common logs to solve the equation.
  6. Oct 3, 2007 #5
    good thinking HallsofIvy!

    makes perfect sense
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