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Radioactive decay - log question

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data

    The amount (A) of cesium-137 remaining after t years is given by

    A=A[itex]_{0}[/itex]*2[itex]^{\frac{-t}{30.3}}[/itex]

    where A[itex]_{0}[/itex] is the initial amount. In what year will the cesium-137 be 10% of that which was released at the Chernobyl disaster in 1986?

    2. Relevant equations
    x[itex]^{\frac{m}{n}}[/itex]=(x[itex]^{\frac{1}{n}}[/itex])[itex]^{m}[/itex]


    3. The attempt at a solution
    plug in my 0.1 on the LHS

    0.1A[itex]_{0}[/itex]=A[itex]_{0}[/itex]*2[itex]^{\frac{-t}{30.3}}[/itex]

    A[itex]_{0}[/itex] on both sides cancels out

    0.1=2[itex]^{\frac{-t}{30.3}}[/itex]

    convert decimal to a fraction

    [itex]\frac{1}{10}[/itex]=2[itex]^{\frac{-t}{30.3}}[/itex]

    take the reciprocal of the RHS to convert the negative power to a positive

    [itex]\frac{1}{10}[/itex]=[itex]\frac{1}{2^{\frac{t}{30.3}}}[/itex]

    take the reciprocal of both sides to remove the fractions

    10=2[itex]^{\frac{t}{30.3}}[/itex]

    isolate the variable t and calculate the remaining values

    10=(2[itex]^{\frac{1}{30.3}}[/itex])[itex]^{t}[/itex]

    10=1.02313981[itex]^{t}[/itex]

    then this part I'm unsure if I'm doing right, and I'm not aware of how I should be trying to check my work

    t=[itex]\frac{log 10}{log 1.02313981}[/itex]

    t=100.6544213

    1986+100=2086?
     
    Last edited: Apr 28, 2013
  2. jcsd
  3. Apr 28, 2013 #2

    Dick

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    That looks just fine to me.
     
  4. Apr 28, 2013 #3
    thank you very much for the quick response, my mind has been put at ease :smile:
     
  5. Apr 28, 2013 #4

    Curious3141

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    Homework Helper

    For future reference, there's a neater and less error-prone way to do the manipulation. Basically, we're using a rule of logs: ##\log a^b = b\log a##, where the logarithm can be to any base.

    In this case, use common (base 10 logs) because one side is a power of 10 (actually, 10 itself). So:

    $$10 = 2^{\frac{t}{30.3}}$$

    Take common logs of both side:

    $$1 = \log_{10}2^{\frac{t}{30.3}}$$

    Use that rule mentioned above:

    $$1 = {\frac{t}{30.3}}\log_{10}2$$

    Rearrage:

    $$t = \frac{30.3}{\log_{10}2} = 100.65$$

    and so forth. The calculation can be done in one step right at the end. Most modern scientific calculators can immediately calculate base-10 logs (it's the button marked 'log').
     
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