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Math Help - Exponential Models and Application help

  1. #1
    Junior Member 22upon7's Avatar
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    Exponential Models and Application help

    I am provided with an example for an Exponential mode, but I can't understand parts of it, I'd appreciate if some one could explain them a bit more.

    Question:

    There are apporximately ten times as many red kangaroos as grey kangatoos in a certain area. If the population of grey kangaroos increases at a rate of 11% per annum while that of the red kangaroos decreases at 5% per annum, find how many years must elapse before the proportions are reversed, assuming the same rates continue to apply.

    Solution

    Let P = population of grey kangaroos at the start.

    Therefore the number of grey kangaroos after n years = P(1.11)^n

    The number of red kangaroos after n years = 10P(0.95)^n

    Question 1 : Why is the percentage raised to the power of n (+n years later)?

    When Proportions are reversed:

    P(1.11)^n = 10* [10P(0.95^n)]
    (1.11)^n = 100(0.95)^n

    Question 2 : Why is [10P(0.95^n)] multiplied by 10?
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  2. #2
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    Quote Originally Posted by 22upon7 View Post
    I am provided with an example for an Exponential mode, but I can't understand parts of it, I'd appreciate if some one could explain them a bit more.

    Question:

    There are apporximately ten times as many red kangaroos as grey kangatoos <<<<< see question 2 in a certain area. If the population of grey kangaroos increases at a rate of 11% per annum while that of the red kangaroos decreases at 5% per annum, find how many years must elapse before the proportions are reversed, assuming the same rates continue to apply.

    Solution

    Let P = population of grey kangaroos at the start.

    Therefore the number of grey kangaroos after n years = P(1.11)^n

    The number of red kangaroos after n years = 10P(0.95)^n

    Question 1 : Why is the percentage raised to the power of n (+n years later)?

    When Proportions are reversed:

    P(1.11)^n = 10* [10P(0.95^n)]
    (1.11)^n = 100(0.95)^n

    Question 2 : Why is [10P(0.95^n)] multiplied by 10?
    I'll take the 2nd line of your calculation:

    (1.11)^n = 100(0.95)^n~\implies~\dfrac{(1.11)^n}{(0.95)^n}=10  0~\implies~\left(\dfrac{1.11}{0.95}\right)^n=100

    Use logarithms to determine n:

    n = \log_{\frac{111}{95}}(100)=\dfrac{\ln(100)}{\ln(11  1)-\ln(95)} \approx 29.6\ years
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  3. #3
    Junior Member 22upon7's Avatar
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    Thanks, I understand the final Log part of the question.

    What I don't understand is why the percentage is raised to the power of n/number of years?

    Also doesn't the equation for red kangaroos [10P(0.95^n)] already reflect the fact that it is 10 times more than the number of Grey Kangaroos?

    Thanks again,

    Dru
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  4. #4
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    Quote Originally Posted by 22upon7 View Post
    Thanks, I understand the final Log part of the question.

    What I don't understand is why the percentage is raised to the power of n/number of years?

    Also doesn't the equation for red kangaroos [10P(0.95^n)] already reflect the fact that it is 10 times more than the number of Grey Kangaroos? <<<<< Yes

    Thanks again,

    Dru
    Let's start at the beginning:

    At the beginning you have a population of P animals and the population increases by 11% per year:

    <br />
year\ 0 : P
    year\ 1: P + P*11\% = P(1+11\%) = P(1+0.11)
    year\ 2: P(1+0.11) + P(1+0.11) * 0.11 = P(1+0.11)(1 + 0.11) = P(1+0.11)^2
    year\ 3: P(1+0.11)^2 + P(1+0.11)^2 * 0.11 = P(1+0.11)^2(1 + 0.11) = P(1+0.11)^3

    ....

    year\ n: P(1+0.11)^{n-1} + P(1+0.11)^{n-1} * 0.11 = P(1+0.11)^{n-1}(1 + 0.11)  = P(1+0.11)^n
    Last edited by earboth; November 9th 2009 at 11:57 AM. Reason: typo
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