First all, the decay formula is [tex]P(t)=Ae^{-kt} [/tex]
where:
[tex]P(t)[/tex] is the remaining quantity after [tex]t[/tex] years
[tex]A[/tex] is the initial sample
[tex]t[/tex] is the time in years
[tex]k[/tex] is the decay constant
From the problem we know that [tex]A=33[/tex] and [tex]P=100[/tex], but we don't have the time [tex]t[/tex]; to find it we will take advantage of the half-life of the Carbon-14. If you have a sample of 100 mg and Carbon-14 has a half-life of 5730, after 5730 years you will have half of your original sample i.e. 50 mg. We also know that after [tex]t[/tex] years we have a remaining sample of 33mg, so the amount of the sample that decayed is [tex]100mg-33mg=67mg[/tex]. Knowing all of this we can set up a rule 3 and solve it to find [tex]t[/tex]:
[tex] \frac{100mg--\ \textgreater \ 5730years--\ \textgreater \ 50mg}{100mg--\ \textgreater \ (t)years---\ \textgreater \ 67mg} [/tex]
[tex] \frac{5730years--\ \textgreater \ 50mg}{(t)years---\ \textgreater \ 67mg} [/tex]
[tex] \frac{5730}{t} = \frac{50}{67} [/tex]
[tex]t= \frac{(5730)(67)}{50} [/tex]
[tex]t=7678.2[/tex]
Now that we know our time [tex]t[/tex] lets replace all the values into our decay formula:
[tex]33=100e^{-7678.2k} [/tex]
Notice that the constant [tex]k[/tex] we need to find is the exponent; we must use logarithms to bring it down, but first lets isolate the exponential expression:
[tex] \frac{33}{100} = e^{-7678.2k} [/tex]
[tex]e^{-7678.2k} = \frac{33}{100} [/tex]
[tex]ln(e^{-7678.2k} )=ln( \frac{33}{100} )[/tex]
[tex]-7678.2k=ln( \frac{33}{100} )[/tex]
[tex]k= \frac{ln( \frac{33}{100}) }{-7678.2} [/tex]
[tex]k=-0.000144[/tex]
We can conclude that the decay constant [tex]k[/tex] is approximately -0.000144
