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Math Help - Radioactive Logarithim

  1. #1
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    Radioactive Logarithim

    Hey guys, just found this forum after searching for days about this problem I've got. No where seems to have any relevant answers so I thought I might ask a forum for some help.
    Basically I've got a Methods Assignment and I can't seem to work out if what I am doing is correct, any help would be great.
    Here's the question:
    The mass of a radioactive isotope is given by M=Mo(10)^-kt. It takes 200 years for the mass of the isotope to halve.
    Find the initial mass (grams) in terms of M.
    I assume Mo10(10) is the initial mass and a logarithm, doesn't say though.

    I've tried to solve this question like this:
    M=Mo10(10)-kt
    M=(-kt)Mo10(10)
    M/Mo10(10)=-kt
    Mo10(10)/M=1/-kt
    Mo10(10)=-M/kt

    Is this correct or have I made some kind of mistake, just doesn't seem right to me :/
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Radioactive Logarithim

    If it takes 200 years for half of the isotope to decay, then we may state:

    \frac{1}{2}m_0=m_0\cdot10^{-200k}

    \frac{1}{2}=10^{-200k}

    \log(2)=200k

    k=\frac{1}{200}\log(2)

    M=m_0\cdot10^{-\frac{t}{200}\log(2)}=m_0\cdot2^{-\frac{t}{200}}

    m_0=M\cdot2^{\frac{t}{200}}
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  3. #3
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    Re: Radioactive Logarithim

    Well then, I was a little off.
    But thank you so much, that makes more sense and helps me with the next part aswell
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