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Math Help - when the population is doubled

  1. #1
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    when the population is doubled

    In a certain population the annual birth rate and annual death rate are 3.94% and 1.94% respectively.Find the number of years in which the population is doubled assuming that there is no immigration and emigration ?

    now what i've tried is this

    let x be the total population today

    now we want 2x population
    now we know that
    total population in n years = total population born in n years - total population die in n years

    => 2x =x (1.0394)^n -x (1.0194)^n [ i've used the formula of compund interest which is derive by using a geometric series]

    now what ???
    after taking x common and canceling it , what should i do ?
    please help me
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  2. #2
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    You forgot to add x. You start with say 100 people, if you subtracted 1.02 of 100 from 1.04 of 100, you would have lost most of your population, so you need to add on x because that is the starting number, and the rest of your formula is just the increase.

    2x = x + x(1.0394^n) - x(1.0194^n)
    x = x(1.0394^n) - x(1.0194^n)
    1 = 1.0394^n - 1.0194^n
    1 + 1.0194^n = 1.0394^n

    This is as much as I can help you. I hope I have helped a bit, maybe you can take it from here?
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  3. #3
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    Quote Originally Posted by LordMaximus View Post
    In a certain population the annual birth rate and annual death rate are 3.94% and 1.94% respectively.Find the number of years in which the population is doubled assuming that there is no immigration and emigration ?

    now what i've tried is this

    let x be the total population today

    now we want 2x population
    now we know that
    total population in n years = total population born in n years - total population die in n years

    => 2x =x (1.0394)^n -x (1.0194)^n [ i've used the formula of compund interest which is derive by using a geometric series]

    now what ???
    after taking x common and canceling it , what should i do ?
    please help me
    \frac{dP}{dt} = \frac{3.94}{100} P - \frac{1.94}{100} P = \frac{P}{50}.

    Therefore P = P_0 e^{t/50}.

    Solve 2 P_0 = P_0 e^{t/50} \Rightarrow 2 = e^{t/50} for t.
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  4. #4
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    You are making this too difficult! If the annual birth rate is 3.94% and the annual death rate is 1.94% then the population is increasing at 3.94- 1.94= 2% per year (that's where mr. fantastic got his "P/50"). With problems that talk about "doubling" or "half life" I prefer NOT to use "e" like mr. fantastic did. The fact that the rate of increase is a constant tells us there IS a specific time until doubling and we can write the population at year t as P(t)= P(0)2^{t/T} where P(0) is the population at t=0 and T is the time to double. Because the annual increase is 2%, P(1)= P(0)+ 0.2P(0)= 1.02P(0).
    From the previous formula, P(1)= P(0)2^{1/T} and setting those equal and cancelling P(0) on both sides of the equation, [tex]2^{1/T}= 1.02[/itex]. Solve that equation for T. Of course, you would use logarithms to solve that equation just as you would for e^{t/50}= 2 in mr fantastic's method.
    Last edited by HallsofIvy; February 10th 2009 at 06:28 AM.
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