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Math Help - Logarithms - Half-life

  1. #1
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    Logarithms - Half-life

    A 105-milligram sample of a radioactive substance decays according to the equation

    N=105 * e^-0.038t
    where N is the number of milligrams present after t years.

    (a) Find the half-life of the substance to the nearest tenth of a year.

    The half-life is _________ years.

    (b): After how many years, to the nearest tenth of a year, there will be 105/4 milligrams present?

    After _______ years there will be 26.25=105/4 milligrams of the substance left.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lemontea View Post
    A 105-milligram sample of a radioactive substance decays according to the equation

    N=105 * e^-0.038t
    where N is the number of milligrams present after t years.

    (a) Find the half-life of the substance to the nearest tenth of a year.

    The half-life is _________ years.
    The hard way: solve the equation 105/2 = 105e^{-0.038t} for t. why?

    can you continue?



    The easy way: half-life = (ln(2))/(0.038)

    this is going by the formula: For the radioactive decay equation of the form P = P_0e^{-rt}, the following formula holds:

    rt_h = \ln 2

    where t_h is the half-life and r is the rate of decay

    (b): After how many years, to the nearest tenth of a year, there will be 105/4 milligrams present?

    After _______ years there will be 26.25=105/4 milligrams of the substance left.
    Solve the equation 105/4 = 105e^{-0.038t} for t

    can you continue?
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