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Using e to determine population growth?

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A population started at 10,000 in 1900. The population doubles every 50 years. What was the population in 2000?


    2. Relevant equations

    Q(time) = Q(initial)*e(rate)(time)

    3. The attempt at a solution
    I was able to come up with the answer using the standard growth equation, but got the wrong answer with the given equation. Am I doing anything wrong or is the equation not suitable for this problem?
     
  2. jcsd
  3. Jun 19, 2012 #2

    Mark44

    Staff: Mentor

    I don't think you're supposed to use the exponential equation. If the population was 10,000 in 1900 and it doubles every 50 years, what was it in 1950? (No calculator allowed:tongue:)
    What was it in 2000?
     
  4. Jun 19, 2012 #3
    I wouldn't use e. you could simply use. Q=10000(2)^(x/50)
     
  5. Jun 19, 2012 #4

    HallsofIvy

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    Staff Emeritus
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    In fact, because 2000= 1900+ 50+ 50, no "formula" is necessary. In 50 years the population doubles. If it was 10000 in 1900, what was it in 1950, 50 years later? What was it in 2000, another 50 years later?

    (Should have read the other posts more carefully. I see now that Mark44 already said that.)

    If you really want to use the exponential formula you give, you have to be determine what "rate" is. Since the population doubles in 50 years, you must have [itex]e^{(rate)(50)}= 2[/itex] so that (rate)(50)= ln(2) and rate= ln(2)/50. The population after t years is [itex]10000e^{ln(2)t/50}[/itex].

    Note that this us the same as [itex]10000(e^{ln(2)})^{t/50}[/itex] and since [itex]e^{ln(2)}= 2[/itex] that is the same as [itex]10000(2^{t/50}[/itex].
     
    Last edited: Jun 21, 2012
  6. Jun 21, 2012 #5
    e is not involved. e is only involved during continuous growth.
     
  7. Jun 21, 2012 #6

    HallsofIvy

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    It is very common to approximate discrete changes with continuous models.
     
  8. Jun 21, 2012 #7
    But in this case the approximation is innaccurate. It is also harder, when one can just multiply by ##2^n##
     
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