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Math Help - Radioacive Decay

  1. #1
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    Radioacive Decay

    Can Someone help me with this?
    Attached Thumbnails Attached Thumbnails Radioacive Decay-m.png  
    Last edited by mr fantastic; October 31st 2010 at 04:20 PM.
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  2. #2
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    Quote Originally Posted by lilwayne View Post
    Can Someone help me with this?
    If the half life is 3.3 days and you have m = 880 mg at t = 0 then you will obviously have 440 mg when t = 3.3 ....
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    If the half life is 3.3 days and you have m = 880 mg at t = 0 then you will obviously have 440 mg when t = 3.3 ....
    yeah i know that but i dont get what to fill in the blanks..
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  4. #4
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    440 = 880e^{3.3k}

    solve for k and finish the problem.
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  5. #5
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    Quote Originally Posted by skeeter View Post
    440 = 880e^{3.3k}

    solve for k and finish the problem.
    yeah i ended up doing that and what i got was 1/2 and 3.3 for the first 2 blanks.

    2 and 3.3 for the second 2 blanks.

    880, 2 and 3.3 for the last 3 blanks.

    but it was wrong, what did i do wrong?
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  6. #6
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    Quote Originally Posted by lilwayne View Post
    yeah i ended up doing that and what i got was 1/2 and 3.3 for the first 2 blanks.

    2 and 3.3 for the second 2 blanks.

    880, 2 and 3.3 for the last 3 blanks.

    but it was wrong, what did i do wrong?
    you forgot to divide the right side by 880 ...

    440 = 880e^{3.3k}

    divide both sides by 880 ...

    \displaystyle \frac{1}{2} = e^{3.3k}<br />

    continue ...
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  7. #7
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    yeah i did that, then i did ln1/2 = lne^3.3k

    which is ln1/2 = 3.3k

    and i was told that ln(1/2) = -ln(2)

    so i had k = -ln(2)/3.3

    is that right so far?
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  8. #8
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    Nvm got it.
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  9. #9
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    Quote Originally Posted by lilwayne View Post
    Nvm got it.
    you're welcome.
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