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Math Help - half life word problem using logs

  1. #1
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    half life word problem using logs

    a 50-mg sample of cobalt-60 decays to 40mg after 1.6min. determine the half-life of the cobalt-60.

    this is as far as i got before getting stuck

    M = C (\frac{1}{2})^\frac{t}{h}

    40 = 50 (0.5)^\frac{1.6}{h}

    0.8 = (0.5)^\frac{1.6}{h}

    \frac{1.6}{h} = \frac{log0.8}{log0.5}

    then right here i get stuck and can't get the correct answer. can somone tell me what i'm doing wrong?

    i'm getting the answer 0.52 mins, when its susposed to be 5 mins
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by extraordinarymachine View Post
    a 50-mg sample of cobalt-60 decays to 40mg after 1.6min. determine the half-life of the cobalt-60.

    this is as far as i got before getting stuck

    M = C (\frac{1}{2})^\frac{t}{h}

    40 = 50 (0.5)^\frac{1.6}{h}

    0.8 = (0.5)^\frac{1.6}{h}

    \frac{1.6}{h} = \frac{log0.8}{log0.5}

    then right here i get stuck and can't get the correct answer. can somone tell me what i'm doing wrong?

    Edit: your working is actually fine - you just need to use some arithmetic to make h the subject. I would take the reciprocal and multiply by 1.6

    \frac{h}{1.6} = \frac{log(0.5)}{log(0.8)} \: \: \rightarrow \: \: h = 1.6 \, \frac{log(0.5)}{log(0.8)} = 5.0 (2sf)



    Use the exponential decay formula

    A=A_0e^{-kt}

    40=50e^{-1.6t}

    -1.6k = ln(0.8)

    k = -\frac{1}{1.6}\, ln(0.8)

    t_{1/2}=\frac{ln(2)}{k} = -\frac{1.6\, ln(2)}{ln(0.8)} = 5.0 (2sf)

    (see spoiler for why this is so)

    Spoiler:
    Take the exponential decay equation: A=A_0e^{-kt}

    Half life ( t_{1/2}) is defined as the time it takes for half the original substance to decay (ie: A=0.5A_0)

    \frac{0.5A_0}{A_0} = e^{-kt_{1/2}}

    A_0 cancels and then take the log. As 0.5=2^{-1} use the log power law to say that ln(0.5) = -ln(2)

    -ln(2)=-kt_{1/2}

    Divide by -k

    t_{1/2}=\frac{ln2}{k}
    Last edited by e^(i*pi); December 1st 2009 at 12:15 PM. Reason: see post
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