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Thread: Radio-active decay problem.

  1. #1
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    Radio-active decay problem.

    I have the following equation.

    $\displaystyle N/6= N*exp(-.0001216t)$

    Then I want to solve to t, and the solution is $\displaystyle ln(6)/.00001216$

    I must be missing something....why is $\displaystyle ln(6)$ and not $\displaystyle ln(1/6)$?
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  2. #2
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    Quote Originally Posted by tactical View Post
    I have the following equation.

    $\displaystyle N/6= N*exp(-.0001216t)$

    Then I want to solve to t, and the solution is $\displaystyle ln(6)/.00001216$

    I must be missing something....why is $\displaystyle ln(6)$ and not $\displaystyle ln(1/6)$?
    $\displaystyle N/6= N*exp(-.0001216t)$

    $\displaystyle 1/6= exp(-.0001216t)$

    $\displaystyle 6= exp(0.0001216t)$

    $\displaystyle \ln 6= (0.0001216t)$

    $\displaystyle t = \frac{\ln 6}{0.0001216}$
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  3. #3
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    Quote Originally Posted by tactical View Post
    I have the following equation.

    $\displaystyle N/6= N*exp(-.0001216t)$

    Then I want to solve to t, and the solution is $\displaystyle ln(6)/.00001216$

    I must be missing something....why is $\displaystyle ln(6)$ and not $\displaystyle ln(1/6)$?
    You could also write the answer as t= ln(1/6)/(-.0001216). Now ln(1/6)= -ln(6).
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