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Math Help - more exponential growth!

  1. #1
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    more exponential growth!

    ok, so, the population of a city is growing exponentially with the function P(t)=P0e^kt

    the population doubled in the first 30 years.

    a) Firstly, find k.

    My answer was 0.023 but I believe that value is incorrect as I wasn't sure whether to do 2P0 (as the population doubled) or leave P0 as it is when working it out.

    b.) By what factor will the original population have grown in the initial 120 years

    c) When will the original population size have grown by a factor of 10?


    how do I do factors? cheers
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  2. #2
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    Quote Originally Posted by meesukj View Post
    ok, so, the population of a city is growing exponentially with the function P(t)=P0e^kt

    the population doubled in the first 30 years.

    a) Firstly, find k.

    My answer was 0.023 but I believe that value is incorrect as I wasn't sure whether to do 2P0 (as the population doubled) or leave P0 as it is when working it out.

    b.) By what factor will the original population have grown in the initial 120 years

    c) When will the original population size have grown by a factor of 10?


    how do I do factors? cheers
    a) P(t) = P_0e^{kt}.

    You also know that in 30 years, the population doubled.

    So P(30) = 2P(0)

    P_0e^{30k} = 2P_0e^{0k}

    P_0e^{30k} = 2P_0

    e^{30k} = 2

    30k = \ln{2}

    k = \frac{\ln{2}}{30}.


    Therefore P(t) = P_0 e^{\frac{t\ln{2}}{30}}.


    b) In 120 years:

    P(120) = P_0 e^{\frac{120\ln{2}}{30}}

     = P_0 e^{40\ln{2}}

     = P_0 e^{\ln{2^{40}}}

     = 2^{40}P_0.

    So the initial population has grown by a factor of 2^{40}.


    c) To have grown by a factor of 10:

    P_0e^{\frac{t\ln{2}}{30}} = 10P_0

    e^{\frac{t\ln{2}}{30}} = 10

    \frac{t\ln{2}}{30} = \ln{10}

    t\ln{2} = 30\ln{10}

    t = \frac{30\ln{10}}{\ln{2}}.
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