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Math Help - Exponential Function Word Problem

  1. #1
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    Exponential Function Word Problem

    Use the decay eqtn for polonium-218, at A(t)=A_o(0.5)^(t/3.1), where A is the amount remaining after t min. and A_o is the initial amount.

    a) How much will remain after 90 sec from an initial sample of 50 mg?

    b) How long will it take for this sample to decay to 10% of its initial amount of 50 mg?

    c) Would your answer in part b change if the size of the initial sample where changed? Explain why or why not.

    My answer for part a was 35.75 mg, which is correct.

    But I do not understand how to do part b and c. For part b, what I did was sub in 35.75 for A(t) and subbed in 5 (10% of 50, which is 5) for the initial amount A_o. Then, I proceeded on to solve for t.

    I ended up with -9.04, but the answer is supposed to be approx 10.3 min. Could someone explain to me what I did wrong and how I can fix this?

    Also, for part c, the answer in the back says "no" but I think that (even though I don't have part b done), from the answer that they give, the answer would in fact be "yes" because from part a we used 50 mg as the initial sample and that is how we calculated part b. So they must be the same, otherwise the answer would change in part b. Am I wrong on this too? Please explain why.

    Thank you very much to whoever helps!
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  2. #2
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    exponential decay solution

    A_{0} is the initial amount in the equation, so you should have plugged in 50 for A_{0}

    Also, you are right that we have to find out when it decays to 5, but the problem says A is the amount remaining after time t. They in fact mean A(t) is the amount remaining after time t, as the amount remaining is a function of how much you started with. So we need to solve 5=50(.5)^\frac{t}{3.1} and you will get approx 10.3.

    For part c) the time it takes to decay to 10% of the original amount actually does not depend on the size of the initial amount. You can see this buy saying let X be the original amount. Then we have to see when it decays to .10*X.... so we solve

    (.1)X=X(.5)^\frac{t}{3.1} and you will get approx. 10.3
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  3. #3
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    Quote Originally Posted by skeske1234 View Post
    Use the decay eqtn for polonium-218, at A(t)=A_o(0.5)^(t/3.1), where A is the amount remaining after t min. and A_o is the initial amount.

    a) How much will remain after 90 sec from an initial sample of 50 mg?

    b) How long will it take for this sample to decay to 10% of its initial amount of 50 mg?

    c) Would your answer in part b change if the size of the initial sample where changed? Explain why or why not.

    My answer for part a was 35.75 mg, which is correct.

    But I do not understand how to do part b and c. For part b, what I did was sub in 35.75 for A(t) e^(i*pi): Why did you do this? It's unnecessary, you can follow the following step and solve for t
    and subbed in 5 (10% of 50, which is 5) for the initial amount A_o. Then, I proceeded on to solve for t.

    I ended up with -9.04, but the answer is supposed to be approx 10.3 min. Could someone explain to me what I did wrong and how I can fix this?

    Also, for part c, the answer in the back says "no" but I think that (even though I don't have part b done), from the answer that they give, the answer would in fact be "yes" because from part a we used 50 mg as the initial sample and that is how we calculated part b. So they must be the same, otherwise the answer would change in part b. Am I wrong on this too? Please explain why.

    Thank you very much to whoever helps!
    A(t) = A_0(0.5)^{\frac{t}{3.1}}

    b) Let A(t) = 0.1A_0

    0.1A_0 = A_0(0.5)^{\frac{t}{3.1}}

    0.1 = (0.5)^{\frac{t}{3.1}}

    take logs to base 10: -1 = \frac{t}{3.1}log(0.5)

    note that log(0.5) = log(2^{-1}) = -log(2)

    \frac{1}{log(2)} = \frac{t}{3.1}

    t = \frac{3.1}{log(2)} = 10.30

    c) No it wouldn't. Since this is a ratio in question then it would not change. If we set A(t) = 0.1A_0 which is saying 10% of the original sample.

    Divide that through by A_0 and you A_0 will cancel thus making it independent of the initial amount
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