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Math Help - Word problem about Chemical Half-Lives

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    Word problem about Chemical Half-Lives

    Dunno how to write this one, so I used a screenshot.

    http://img111.imageshack.us/img111/8039/mathwordux7.jpg

    The thing that trips me up is the 0 seconds.

    <(I hate Word Problems!)
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    Quote Originally Posted by mankvill View Post
    Dunno how to write this one, so I used a screenshot.

    http://img111.imageshack.us/img111/8039/mathwordux7.jpg

    The thing that trips me up is the 0 seconds.

    <(I hate Word Problems!)
    the half-life is 38 seconds. thus, if you start with A_0, after 38 seconds, you will have \frac 12 A_0. thus, the following equation must hold:

    \frac 12 A_0 = A_0a^{38}

    now just solve for a
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    a = 38?
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    Quote Originally Posted by mankvill View Post
    a = 38?
    ok, so lets say a = 38, that would mean \frac 12 A_0 = A_0 (38)^{38} \implies \frac 12 = (38)^{38}. obviously that makes no sense . how did you come up with that answer? do you have problems solving exponential equations?
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    Quote Originally Posted by Jhevon View Post
    ok, so lets say a = 38, that would mean \frac 12 A_0 = A_0 (38)^{38} \implies \frac 12 = (38)^{38}. obviously that makes no sense . how did you come up with that answer? do you have problems solving exponential equations?
    I thought I understood it. D:

    ...a = 19?

    Or 2?

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    Quote Originally Posted by mankvill View Post
    I thought I understood it. D:

    ...a = 19?

    Or 2?

    haha, stop guessing, and solve it. a would have to be less than 1, of course, otherwise, if you raise it to the 38th power, you would get a number greater than or equal to 1, which is bad, since we have 1/2 on the left hand side

    \frac 12A_0 = A_0a^{38}

    \Rightarrow \frac 12 = a^{38}

    Now, how would you solve for a here. what must you do to get rid of that 38th power, and be left with just a on the right side?
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    Quote Originally Posted by Jhevon View Post
    haha, stop guessing, and solve it. a would have to be less than 1, of course, otherwise, if you raise it to the 38th power, you would get a number greater than or equal to 1, which is bad, since we have 1/2 on the left hand side

    \frac 12A_0 = A_0a^{38}

    \Rightarrow \frac 12 = a^{38}

    Now, how would you solve for a here. what must you do to get rid of that 38th power, and be left with just a on the right side?
    Take the 38th root of \frac 12?
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    Quote Originally Posted by mankvill View Post
    Take the 38th root of \frac 12?
    yes!

    and your answer is? (to 6 decimal places)
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    Quote Originally Posted by Jhevon View Post
    yes!

    and your answer is? (to 6 decimal places)
    I got:

    .981925 after rounding.

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    Quote Originally Posted by mankvill View Post
    I got:

    .981925 after rounding.

    yup
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    2nd part: How much of a 4 gram sample would be left after 10 seconds?
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    Quote Originally Posted by mankvill View Post


    2nd part: How much of a 4 gram sample would be left after 10 seconds?
    Plug in A_0 = 4 and t = 10 and solve for A(t) (remember, you just found the value for a)
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    Quote Originally Posted by Jhevon View Post
    Plug in A_0 = 4 and t = 10 and solve for A(t) (remember, you just found the value for a)
    To set it up, would that be:

    1/2 (4) = (4)a^10

    ?
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    Quote Originally Posted by mankvill View Post
    To set it up, would that be:

    1/2 (4) = (4)a^10

    ?
    no, your original equation is A(t) = A_0a^t now plug in the values you got for A_0, ~a, \mbox{ and }t
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    Quote Originally Posted by Jhevon View Post
    no, your original equation is A(t) = A_0a^t now plug in the values you got for A_0, ~a, \mbox{ and }t
    So A(10) = 4(a)^10?
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