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Math Help - Application with natural log

  1. #1
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    Application with natural log

    in my book, they gave formula: y = y1e^kt (y1 base amount).

    the problem is: "The population of the US was 3.9 million in 1790 and 178 million in 1960. If the rate of growth is assumed proportion to the number present, what estimate would you give for the population in 2000?"

    i got 3,900,000^(170k) = 178,000,000 to try to solve for the rate but i dont think thats right.

    the other one is: A population is growing at a rate proportion to its size. After 5 years, the population size was 164,000. After 12 years, the population size was 235,000. What was the original population?

    i tried to set up the base amount with an x and try different approaches but it didn't work. please help.
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  2. #2
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    [QUOTE=Arez;303430]in my book, they gave formula: y = y1e^kt (y1 base amount).

    the problem is: "The population of the US was 3.9 million in 1790 and 178 million in 1960. If the rate of growth is assumed proportion to the number present, what estimate would you give for the population in 2000?"

    i got 3,900,000^(170k) = 178,000,000 to try to solve for the rate but i dont think thats right.[quote]

    Looks good to me :]

    the other one is: A population is growing at a rate proportion to its size. After 5 years, the population size was 164,000. After 12 years, the population size was 235,000. What was the original population?

    i tried to set up the base amount with an x and try different approaches but it didn't work. please help.
    Your question says that: dN/dt = kN

    you can then integrate that and work out what you need (I'm running late so I'll edit it when I get back)
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  3. #3
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    Hello, Arez!

    You're doing great . . .


    they gave formula: . P \:=\:P_oe^{kt}

    The population of the US was 3.9 million in 1790 and 178 million in 1960.
    If the rate of growth is assumed proportion to the number present,
    what estimate would you give for the population in 2000?

    i got: . 3,\!900,\!000\,e^{170k} \:=\: 179,\!000,\!000 to try to solve for the rate . . . . Right!
    Keep going . . .

    . . e^{170k} \:=\:\frac{1790}{39} \quad\Rightarrow\quad 170 k \:=\:\ln\left(\tfrac{1790}{39}\right) \quad\Rightarrow\quad k \:=\:\tfrac{1}{170}\ln\left(\tfrac{1780}{39}\right  ) \;\approx\;0.0225

    We have: . P \;=\;3,900,000\,e^{0.225t}

    Now evaluate at t = 210.




    A population is growing at a rate proportional to its size.
    After 5 years, the population size was 164,000.
    After 12 years, the population size was 235,000.
    What was the original population?
    I set it up like this . . .


    When t = 0\!:\;P = 164k
    When t = 7\!:\;P = 235k
    . . What was the population when t = -5 ?

    And we can solve it with the same procedure as the first problem.

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