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Math Help - Half Life of Radon Gas

  1. #1
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    Half Life of Radon Gas

    I'm working on a study sheet for an upcoming quiz in brief calculus. This section concerns exponential functions.

    A building is found to contain radon gas. Thirty hours later, 80% of the initial amount of gas is still present.


    I need to find an equation to represent the percentage of the initial amount of radon gas present after 't' hours, and then figure out the half-life of radon gas. The problem seems pretty simple, but I keep getting confused on finding the initial condition (a) to fit into f(x)=ab^x. I found the constant multiplier, 1.8, by taking the decimal form of 80% and adding 1 to it. The part about thirty hours is tripping me up. I can't figure out if I should be solving for the time (x) or the initial condition (a).
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    Re: Half Life of Radon Gas

    For your equation f(x)=ab^x after 30 hours 0.8\times a=ab^{30} \implies 0.8=b^{30}\implies  b= 0.9926

    Now for the half life, find x when \frac{a}{2}=a\times 0.9926^x

    Spoiler:
    divide both sides by a
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    Re: Half Life of Radon Gas

    Quote Originally Posted by pickslides View Post
    For your equation f(x)=ab^x after 30 hours 0.8\times a=ab^{30} \implies 0.8=b^{30}\implies  b= 0.9926
    I'm confused at this point, and I think I'm misunderstanding a piece of algebra. Did you take the log of both sides? How were you able to evaluate the b^30?
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    Re: Half Life of Radon Gas

    Well the algebra would suggest taking \log_b to both sides, but that leaves you with 30 = \log_b0.8 which maynot be that helpful.

    I used a spredsheet to solve for b , so use a computer or calculator.
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    Re: Half Life of Radon Gas

    A little mental approximation: 0.8=b^{30}\Rightarrow 0.8^3=0.512=b^{90} so the half-life is a little over 90 hours.
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    Re: Half Life of Radon Gas

    Thanks for the help, I guess I just needed to play around with the numbers a bit.
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