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Math Help - Population question

  1. #1
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    Population question

    The population of a city at time t years after 2002 is expressed as
    P(t)=12300(e^0.009t).

    a) Estimate the population of the city in 2012.

    For this one I made t=10 because 2012-2002=10
    the answer was aprox. 13458.3469

    b) Estimate the year the population will be doubled its population in 2002

    I tried to find the initial population by making t=0... also tried t=1 d n both cases the answer was bigger than the answer for a).

    What am I doing wrong?
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  2. #2
    Super Member Quacky's Avatar
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    You could be working at a higher level than I, but here's an attempt at a start of an answer:

    P(t)=12300(e^{0.009t})
    So you substituted in:
    P(0)=12300(e^{(0.009)(0)})=12300
    Then, for the population to be doubled in 'x' years:
    P(x)=12300(e^{0.009x})=24600
    Then e^{0.009x}=2

    By taking logs of both sides, I think you can then get an answer. This will need confirmation, though.
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  3. #3
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    I think what you need to do is finding "t" when you double the population of the year 2002, this means:
    Year 2002 is the year "1", t=1
    So,

    P(t)=12300(e^0.009(1)) = 12411.20

    Then, you double that P(t) and solve for t:

    (2*12411.20)=12300(e^0.009t) =

    24822.40=12300(e^0.009t)=

    Ln(24822.4/12300) = 0.009t

    t = 78 which means the population will be double on year 2080

    Hope this help!
    Best,
    August80
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  4. #4
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    Quote Originally Posted by August80 View Post
    I think what you need to do is finding "t" when you double the population of the year 2002, this means:
    Year 2002 is the year "1", t=1
    No, the problem said "t years after 2002". 2002 was not 1 year after 2002!
    t= 0 for 2002.

    So,

    P(t)=12300(e^0.009(1)) = 12411.20

    Then, you double that P(t) and solve for t:

    (2*12411.20)=12300(e^0.009t) =

    24822.40=12300(e^0.009t)=

    Ln(24822.4/12300) = 0.009t

    t = 78 which means the population will be double on year 2080

    Hope this help!
    Best,
    August80
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