Question

How old is an ivory tusk that has lost 70% of it's carbon-14? The half life of carbon-14 is 5750 years.

Answer 1

Let \(C_i\) be the initial amount of carbon it has. The equation for the amount of carbon it has with respect to time is:\[\Large C_i\left(\frac{1}{2}\right)^\frac{t}{5750}\]Notice how when \(t=5750\), we have \(\frac{1}{2}C_i\). Now what we also know that the current amount of carbon is \(\frac{70}{100}C_i=0.7C_i\). We want to find the \(t\) such that:\[\Large 0.7C_i=C_i\left(\frac{1}{2}\right)^\frac{t}{5750}\]Make sense?

Answer 2

yes

Answer 3

So can you solve for \(t\) here?

Answer 4

999914?

Answer 5

Um, how did you do it?

Answer 6

k=ln2/5750
k=0.012% a day

0.30Co=Coe^-0.00001205t
t=99914.76

?

Answer 7

Oh right, it should be 0.3

Answer 8

But I'm still getting a different answer.

Answer 9

\[\Large (1-0.7)C_i=C_i\left(\frac{1}{2}\right)^\frac{t}{5750}\]So I first divide byt \(C_i\).\[\Large 0.3=\left(\frac{1}{2}\right)^\frac{t}{5750}\]Then I take the natural log of both sides.

Answer 10

\[\Large \ln(0.3)=\ln\left(\left(\frac{1}{2}\right)^\frac{t}{5750}\right)\]Use an exponent rule:\[\ln(0.3)=\frac{t}{5750}\ln\left(\frac{1}{2}\right)\]

Answer 11

\[\frac{\ln(0.3)}{\ln(0.5)}=\frac{t}{5750}\]\[t=5750\cdot\frac{\ln(0.3)}{\ln(0.5)}\]

Answer 12

This is getting me 9987.55 years.

Answer 13

oh I think I put too many zeros. thanks