Question

An artifact was found and tested for its carbon-14 content. If 83% of the original carbon-14 was still present, what is its probable age (to the nearest 100 years)? Use that carbon-14 has a half-life of 5,730 years.

Answer 1

solve
\[.83=(\frac{1}{2})^{\frac{t}{5700}}\] for t

Answer 2

\[\ln(.83)=\frac{t}{5700}\ln(\frac{1}{2})\]

Answer 3

giving
\[t=5700\frac{\ln(.83)}{\ln(.5)}\]

Answer 4

Let's say you have an initial amount A_0, and a final amount A. We have
\[A=A_0e^{-kt}\]Our first step is almost always to solve for k. We know that after 5730 years, we will have 1/2 of the original amount of carbon. This gives us
\[0.5A_0=A_0e^{-5730k}\]Divide both sides by the constant term A_0.
\[0.5=e^{-5730k}\]Take the natural log of both sides to get the k out of the exponent.
\[\ln(0.5)=\ln(e^{-5730k})\]\[\ln(0.5)=-5730k\]\[k=-\frac{\ln(0.5)}{5730}\approx0.00012097\]
Substitute this value of k back into the original equation to get \[A=A_0 e^{0.00012097t}\]
We want to know what t is when A=0.83*(initial amount)
\[0.83A_0=A_0e^{0.00012097t}\]Again, divide both sides by A_0
\[0.83=e^{0.00012097t}\]Take the natural log of both sides.
\[\ln(0.83)=0.00012097t\]
\[t=\frac{\ln(0.83)}{0.00012097}\]

Answer 5

@jabberwok this method will surely work, but it requires solving twice. if you are given the half life rather than using
\[A=A_0 e^{kt}\] and having to solve for k and going back to solve again for t, it is quicker to use
\[A=A_0(\frac{1}{2})^{\frac{t}{h}}\] where h is the half life. that way you don't have to solve twice, only once.

just thought i would mention it

Answer 6

Huh, I had never seen that formula before. Conceptually, that definitely makes sense when you consider what a half-life is. Thanks!

Answer 7

yw. doesn't seem to matter anyway since mmadern has left.