Use the model $Q(t)=Q_{0} e^{-0.000121 t}$ for radiocarbon dating to answer the question.
A sample from a mummified bull was taken from a certain place. The sample shows that $87 \%$ of the carbon-14 still remains. How old is the sample? Round to the nearest year:
The sample is approximately years old.
Answer
Our final answer is \(\boxed{1136}\) years.
Step 1 ：Set \(Q(t)\) to \(0.87Q_0\) (since the sample still retains 87% or 0.87 of its original amount), we find the following: \[0.87Q_0 = Q_0 e^{-0.000121t}\]
Step 2 ：Divide both sides by \(Q_0\) to get: \[0.87 = e^{-0.000121t}\]
Step 3 ：Take the natural logarithm of both sides to isolate the exponent: \[\ln(0.87) = -0.000121t\]
Step 4 ：Divide both sides by \(-0.000121\) to solve for \(t\): \[t = \frac{\ln(0.87)}{-0.000121}\]
Step 5 ：Calculate the value of \(t\) to find the age of the sample: \[t \approx 1136\] years
Step 6 ：Our final answer is \(\boxed{1136}\) years.
