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Math Help - WORD

  1. #1
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    WORD

    I know you are suppose to use y(t) = e^(kt) but I don't know where to go. Please help me out guys. Thanks

    In 1970 the population of a town was 21,000 and by 1980 that number decreased to 20,000. If the population decreases exponentially, then
    a. find the population in the year 2000
    b. find the population in the year 1960
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  2. #2
    Junior Member
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    This is a hard one but I think you would use the years as t and continue from there
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  3. #3
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    Hello, bigton!

    A piece that formula is missing . . .


    Use: . y(t) \:= \:{\color{red}a}e^{kt}

    In 1970 the population of a town was 21,000
    and by 1980 that number decreased to 20,000.
    If the population decreases exponentially, then

    a) find the population in the year 2000

    Let 1970 be t = 0.

    We are told that: . y(0) \:=\:21,\!000

    . . We have: . 21,\!000 \:=\:ae^0 \quad\Rightarrow\quad a \:=\:21,\!000

    The function is: . y(t) \;=\;21,\!000e^{kt}


    We are told that: . y(10) \:=\:20,\!000

    . . 20,\!000 \:=\:21,\!000e^{10k} \quad\Rightarrow\quad e^{10k} \:=\:\tfrac{20}{21} \quad\Rightarrow\quad 10k \:=\:\ln\left(\tfrac{20}{21}\right)

    . . k \:=\:\tfrac{1}{10}\ln(\tfrac{20}{21}) \:=\:-0.004879016 \;\approx\;-0.005

    The function is: . y(t) \;=\;21,\!000e^{-0.005t}


    In the year 2000, t = 30\!:\;\;y(30) \;=\;21,\!000e^{(\text{-}0.005)(30)} \;=\; 18074.8675 \;\approx\;18,\!075




    b) find the population in the year 1960

    In the year 1960, t = \text{-}10

    y(\text{-}10) \;=\;21,\!000e^{(\text{-}0.05)(\text{-}10)} \;=\;21,\!000e^{0.05} \;=\;22076.69302 \;\approx\;22,\!077


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