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Math Help - Exponential Pattern Problem - Part A

  1. #1
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    Question

    ...
    Last edited by Jaydee; January 7th 2010 at 08:24 AM. Reason: Merged posts (I didn't say to start a new thread for each part of a >single< question!! Go back and read what I said.)
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  2. #2
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    Yeah, another way to set it up is

    \frac{106}{250} = \frac{x}{400}

    Then you do the math

    42,400=250x \rightarrow x=169.6
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  3. #3
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    Oh, I see! That make sense, too. We've been doing a very similar thing in Chemistry lately.

    Thank you for the alternate method, and I am also glad I did this correctly.
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  4. #4
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    Hello, Jaydee!

    You are correct.
    I would write it: . C(t) \:=\:\frac{169.6}{2^{\frac{t}{5.5}}}

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  5. #5
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    Remember the equation...

    A = A_0e^{kt} with A_0 being the original amount and k is the rate of growth/decay and t is time.

    You know half life means after 5.5 hrs there will be half the substance left so you can....

    .5 = e^{k5.5}

    Because if you put in the values before like....

    84.8 = 169.6e^{k5.5} or even for the original in your part A 53 = 106e^{k5.5} they both will reduce to...

    .5 = e^{k5.5}

    now you need to know the rate, k.

    You could make .5 = e^{k5.5} even shorter to begin with, but I think it's important to understand how this came about in case you are given other problems where it's not so cut and dry around half life or exponential growth or decay.

    Oh, I forgot to include that once you find k, you can get the equation, with k filled in...

    f(t) = 169.6e^{kt}
    Last edited by JSB1917; December 19th 2009 at 05:56 PM.
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