1. ## 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).

2. ## Re: Half Life of Radon Gas

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

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

Spoiler:
divide both sides by $\displaystyle a$

3. ## Re: Half Life of Radon Gas

Originally Posted by pickslides
For your equation $\displaystyle f(x)=ab^x$ after 30 hours $\displaystyle 0.8\times a=ab^{30} \implies 0.8=b^{30}\implies b= 0.9926$
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?

4. ## Re: Half Life of Radon Gas

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

I used a spredsheet to solve for $\displaystyle b$ , so use a computer or calculator.

5. ## Re: Half Life of Radon Gas

A little mental approximation: $\displaystyle 0.8=b^{30}\Rightarrow 0.8^3=0.512=b^{90}$ so the half-life is a little over 90 hours.

6. ## Re: Half Life of Radon Gas

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