The skull of an animal found in an archaeological dig contains about 20% of the original amount of carbon-14. If the half-life of carbon-14 is 5730 years, how long ago did the animal die?

Solution Let $A=A(t)$ be the amount of carbon-14 present in the skull at time $t$. Then $A$ satisfies the differential equation $\dfrac{{\it dA}}{dt}=kA$, whose solution is $$A=A_{0}e^{kt}$$

where $A_{0}$ is the amount of carbon-14 present at time $t=0$. To determine the constant $k$, we use the fact that when $t=5730,$ half of the original amount $A_{0}$ remains. $$\begin{eqnarray*} \dfrac{1}{2}A_{0} &=&A_{0}e^{5730k} \\ \dfrac{1}{2} &=&e^{5730k} \\ 5730k &=&\ln \dfrac{1}{2} = -\ln 2 \\ k &=&-\dfrac{\ln 2}{5730} \end{eqnarray*}$$

The relationship between the amount $A$ of carbon-14 present and the time $t$ is $$A(t) =A_{0}e^{\left( -\ln 2/5730\right) t}$$

In this skull, 20% of the original amount of carbon-14 remains, so $A(t) =0.20A_{0}.$ $$\begin{eqnarray*} 0.20A_{0} &=&A_{0}e^{( -\ln 2/5730) t} \\ 0.20 &=&e^{( -\ln 2/5730) t} \end{eqnarray*}$$

Now, we take the natural logarithm of both sides. $$\begin{eqnarray*} \ln 0.20 &=&-\dfrac{\ln 2}{5730}\cdot t \\ t &=&-5730\cdot \dfrac{\ln 0.20}{\ln 2}\approx 13{,}300 \end{eqnarray*}$$

The animal died approximately 13,300 years ago.