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  1. #1
    Dex
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    Help with Algebra

    Please Help me answer this question, I received my exam back but got this question wrong.

    A 100 mg sample of radioactive actinium 227 decays according this equation:

    N=100e^{-0.03194 t}

    When N is the number of milligrams present after t years . Find the half life of 227 Actinium to the nearest 0.1 of a year. [The half life is when half of the original amount is still present]

    Is this right?:

    50=100e^{-0.03194 t}

    0.5=e^{-0.03194 t}

    I know I have to use logs, but I don't know how to finish because of the  e .

    Please help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Dex View Post
    Please Help me answer this question, I received my exam back but got this question wrong.

    A 100 mg sample of radioactive actinium 227 decays according this equation:

    N=100e^{-0.03194 t}

    When N is the number of milligrams present after t years . Find the half life of 227 Actinium to the nearest 0.1 of a year. [The half life is when half of the original amount is still present]

    Is this right?:

    50=100e^{-0.03194 t}

    0.5=e^{-0.03194 t}

    I know I have to use logs, but I don't know how to finish because of the  e .

    Please help!
    0.5=e^{-0.03194 t}

    Take natural logs:

    \ln(0.5)=-0.03194 t

    CB
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  3. #3
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    Quote Originally Posted by Dex View Post
    Please Help me answer this question, I received my exam back but got this question wrong.

    A 100 mg sample of radioactive actinium 227 decays according this equation:

    N=100e^{-0.03194 t}

    When N is the number of milligrams present after t years . Find the half life of 227 Actinium to the nearest 0.1 of a year. [The half life is when half of the original amount is still present]

    Is this right?:

    50=100e^{-0.03194 t}

    0.5=e^{-0.03194 t}

    I know I have to use logs, but I don't know how to finish because of the  e .

    Please help!
    1. Logarithms to the bas e e are called natural logarithms. Your calculator is able to calculate the corresponding values. Use the function ln().

    2. Taking the natural logs on both sides will give:

    0.5=e^{-0.03194 t}~\implies~\ln(0.5)=\ln\left(e^{-0.03194 t}\right)~\implies~\underbrace{-\ln(2)}_{by\ calculator}=-0.03194 t

    3. Solve for t.
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