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Math Help - 9th Grade Problem Solving Question

  1. #1
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    9th Grade Problem Solving Question

    I need help with a homework question I was given. I have no idea if Iím meant to use a graph, equation or something else.

    I have to find the age (in carbon-14 half lives) of a sample with a rate of disintegration of C-14 of .11 disintegrations/min/gram.



    Basically I have to find x when y is .11


    Half Life (x) Disintegration rate (y)
    0................... 16
    1................... 8
    2................... 4
    3................... 2
    4................... 1
    5................... 0.5
    6................... 0.25
    7................... 0.125
    8................... 0.0625


    (Note 1 C-14 Half-Life = 5730 Years before present)
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  2. #2
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    Quote Originally Posted by asaver View Post
    I need help with a homework question I was given. I have no idea if Iím meant to use a graph, equation or something else.

    I have to find the age (in carbon-14 half lives) of a sample with a rate of disintegration of C-14 of .11 disintegrations/min/gram.



    Basically I have to find x when y is .11


    Half Life (x) Disintegration rate (y)
    0................... 16
    1................... 8
    2................... 4
    3................... 2
    4................... 1
    5................... 0.5
    6................... 0.25
    7................... 0.125
    8................... 0.0625



    (Note 1 C-14 Half-Life = 5730 Years before present)
    Hi ..

    You can look at its pattern

    (0,2^4) , (1,2^3) , (2,2^2) , (3,2^1) , ...

    (0,2^{4-0}) , (1,2^{4-1}) , ...

    y=2 to the power of 4 minus its 'x' .

    y=2^{4-x}

    so when y is 0.11 , 0.11=2^{4-x}

    Then calculate \ln 0.11 = (4-x)\ln 2

    Can you continue ?
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  3. #3
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    x = 4 - .11/2
    idk
    ???
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  4. #4
    MHF Contributor
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    Quote Originally Posted by asaver View Post
    x = 4 - .11/2
    idk
    ???

    i don understand what you did ..

    4-x=\frac{\ln 0.11}{\ln 2}

    4-x=-3.184

     <br />
x=7.184 <br />
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