The amount of carbon-14 present in animal bones after $t$ years is given by $\mathrm{P}(\mathrm{t})=\mathrm{P}_{0} e^{-0.00012097 t}$. $\mathrm{A}$ bone has lost $23 \%$ of its carbon-14. How old is the bone?
Answer
Our final answer is \(\boxed{3030}\) years.
Step 1 ：Setting \(P(t)\) to \(0.77P_0\) (since the bone has lost 23% of its carbon-14, it retains 77% or 0.77 of its original amount), we find the following: \[0.77P_0 = P_0 e^{-0.00012097t}\]
Step 2 ：Divide both sides by \(P_0\) to get: \[0.77 = e^{-0.00012097t}\]
Step 3 ：Take the natural logarithm of both sides to isolate the exponent: \[\ln(0.77) = -0.00012097t\]
Step 4 ：Divide both sides by \(-0.00012097\) to solve for \(t\): \[t = \frac{\ln(0.77)}{-0.00012097}\]
Step 5 ：Calculate the value of \(t\) to find the age of the bone: \[t \approx 3030\] years
Step 6 ：Our final answer is \(\boxed{3030}\) years.
