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Math Help - Population of Elk

  1. #1
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    Population of Elk

    I am having trouble getting this one started...

    The population of a herd of elkwas 220 in 1970 and 530 in 1990. What is the population today?
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  2. #2
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    Hello, jigger1!

    The population of a herd of elk was 220 in 1970 and 530 in 1990.
    What is the population today?
    With a population problem, we assume an exponential function: . P \;=\;Ae^{Bt}


    In 1970 (t=0),\;P = 220
    . . We have:. 220 \:=\:Ae^0\quad\Rightarrow\quad A \,=\,220

    The function (so far) is: . P \;=\;220e^{Bt}


    In 1990 (t = 20),\;P = 530
    . . We have: . 530 \:=\:220e^{20B} \quad\Rightarrow\quad e^{20B} \:=\:\frac{53}{22}

    Take logs: . \ln\left(e^{20B}\right) \:=\:\ln\left(\frac{53}{22}\right) \quad\Rightarrow\quad 20B\!\cdot\ln e \:=\:\ln\left(\frac{53}{22}\right)

    . . 20B \:=\:\ln\left(\frac{53}{22}\right)\quad\Rightarrow  \quad B \:=\:\frac{1}{20}\ln\left(\frac{53}{22}\right) \;\approx\; 0.044

    Hence, the function is: . P \;=\;220e^{0.044t}


    Today (t = 38), we have: . P \;=\;220e^{(0.044)(38)} \;=\;1171.016608 \;\approx\;\boxed{1171\text{ elk}}

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  3. #3
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    Thanks Soroban, how did you know that it would be an exponential function equation since it was a population problem? Can you explain?

    I think that is half my problem...
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