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Math Help - decay problem

  1. #1
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    decay problem

    A substance decays at a rate proportional to the amount present, and half of the original quantity Yo is left after 1000 years. What is the percent of the original quantity Yo left after 2000 years?

    basically i use the formula
    Y = Yo*e^(-kt)

    then i solve for k

    \dfrac{1}{2} = 1 * e^{-1000k}

    \dfrac{\ln 1}{\ln 2} = \ln e^{-1000k}

    \dfrac{\ln 1}{\ln 2} = -1000k

    k = \dfrac{\ln \dfrac{1}{2}}{-1000}

    After i get k, then i solve for Y.

    Y = 1e^{(2000)* -(\dfrac{\ln \dfrac{1}{2}}{-1000})}



    is this correct?
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  2. #2
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    You can simplify this down further.

    <br />
\ln\dfrac{1}{2} \equiv \ln 1 - \ln 2<br />

    \ln1 = 0

    So now you have:

    -\ln2 = -1000k

    \ln2 = 1000k

    k = \dfrac{\ln2}{1000}

    This makes working out the second part of the question easier. As you've stated:

    Y = e^{(2000)* -\dfrac{\ln2}{1000}}

    Y = e^{-2\ln2}

    Y = (e^{\ln2})^{-2}

    Y = (2)^{-2}

    Y = \dfrac{1}{4}
    Last edited by Gojinn; March 3rd 2010 at 09:57 PM.
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  3. #3
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    is that all thats different?

    so

    Y = 1e^{(2000)* (\dfrac{-\ln 2}{1000})}
    Y = .25

    After 2000 years, quantity left is 25%.

    correct?
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  4. #4
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    Yeah, I forgot to follow up in my post. lol

    Yeah, .25 is what I got as well.
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  5. #5
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    Lexington, MA (USA)
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    Hello, break!


    A substance decays at a rate proportional to the amount present,
    and half of the original quantity Yo is left after 1000 years.
    What is the percent of the original quantity Yo left after 2000 years?

    Basically i use the formula: . Y \:=\: Y_o\,e^{-kt}

    then i solve for k\!: \;\;\frac{1}{2} \:=\:e^{-1000k}

    . . \frac{\ln 1}{\ln 2} = \ln e^{-1000k}

    . . . {\color{blue}\uparrow}
    . wrong!

    We have: . e^{-1000k} \:=\:\frac{1}{2}

    Then: . \ln\left(e^{-1000k}\right) \:=\:\ln\left(\frac{1}{2}\right) \:=\:\ln\left(2^{-1}\right) \;=\;-\ln(2)

    \text{Hence: }\;-1000k\underbrace{\ln(e)}_{\text{This is 1}} \:=\:-\ln(2) \quad\Rightarrow\quad -1000k \:=\:\-\ln(2) \quad\Rightarrow\quad k \:=\:\frac{\ln(2)}{1000}


    Therefore, the function is: . Y \;=\;Y_o\,e^{-\frac{\ln(2)}{1000}t}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The answer can be further simplified . . .

    . . Y \;=\;Y_o\,e^{-\frac{\ln(2)}{1000}t} \;=\;Y_o\left(e^{\ln(2)}\right)^{-\frac{t}{1000}} \quad\Rightarrow\quad Y \;=\;Y_o\cdot2^{-\frac{t}{1000}}

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  6. #6
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    Red face

    \;-1000k\underbrace{\ln(e)}_{\text{This is 1}} \:=\:-\ln(2) \quad\Rightarrow\quad -1000k \:=\:\-\ln(2) \quad\Rightarrow\quad k \:=\:\frac{\ln(2)}{1000}
    I think you made a mistake. Shouldn't it state:

    \;-1000k\underbrace{\ln(e)}_{\text{This is 1}} \:=\:-\ln(2) \quad\Rightarrow\quad -1000k \:=\:-\ln(2) \quad\Rightarrow\quad k \:=\:\frac{\ln(2)}{1000}<br />

    Only a small thing, but just makes it clearer. lol.
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