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Math Help - exponential growth

  1. #1
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    exponential growth

    An invasive bacterial population doubles in the bloodstream of the invaded organism every 3.0 hours after entry. If the total number of bacteria in the bloodstream reaches 106 before suitable antibiotics are administered, the invaded organism will die. If 100 live bacteria make it into the bloodstream through a cut, how many hours, correct to the nearest whole hour, does it take before the bacterial population reaches a lethal level?


    I thought this problem should have been pretty easy, everytime I do it my answer comes up around 50, but the real answer is 40. Can anyone show me how to do this problem correctly?
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  2. #2
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    Quote Originally Posted by nandokommando View Post
    An invasive bacterial population doubles in the bloodstream of the invaded organism every 3.0 hours after entry. If the total number of bacteria in the bloodstream reaches 106 before suitable antibiotics are administered, the invaded organism will die. If 100 live bacteria make it into the bloodstream through a cut, how many hours, correct to the nearest whole hour, does it take before the bacterial population reaches a lethal level?


    I thought this problem should have been pretty easy, everytime I do it my answer comes up around 50, but the real answer is 40. Can anyone show me how to do this problem correctly?
    Is there a typo involved?

    106 is not that much bigger than 100 so the answer would be 0 hours?

    Either way the equation for exponential growth is

    A(t) = A_0e^{kt}

    Where
    • A(t) = Amount at time t
    • A_0 = Initial Amount
    • k = growth constant
    • t = time



    Standard practice is to use what you're given to find k and then put that into the same equation to find what you want.


    We know that the population doubles in 3 hours. In other words when t=3, A(t) = 2A_0

    k = \frac{\ln(2)}{t}
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  3. #3
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    If the number of bacteria double every three hours, and the initial amount if A_0 you can write the number after t hours as A(t)= A_02^{t/3} since "t/3" measures how many doubling occured.

    As e^(i*pi) said, if there were initially 100 bacteria, passing 106 will happen almost immediately! (And 106 seems a very small number of bacteria.)

    In fact, that will occur when A(t)= 100 2^{t/3}= 106. To solve that, divide both sides by 100 to get 2^{t/3}= 1.06. Take the logarithm of both sides of that: (t/3)log(2)= log(1.06) and, finally, divide both sides by log(2)/3: t= 3\frac{log(1.06)}{log(2)} which, according to my calculator (actually, I'm using the one that comes with Windows) is about .25 of an hour or about 15 minutes!
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