Question

prehistoric cave paintings were discovered in a cave in france the paint contained 4% of the original carbon - 14. use the exponential decay model for carbon - 14, ( a = a_0e^{-0.000121t} ), to estimate the age of the paintings the paintings are approximately ( square ) years old (round to the nearest integer.)

Explanation:

Step1: Substitute the given values into the formula

Given \(A = 0.04A_0\) (since \(A\) is \(4\%\) of \(A_0\)), and the formula \(A = A_0e^{-0.000121t}\). Substitute \(A = 0.04A_0\) into the formula:
\(0.04A_0=A_0e^{- 0.000121t}\)
Divide both sides by \(A_0\) (since \(A_0
eq0\)):
\(0.04 = e^{-0.000121t}\)

Step2: Take the natural logarithm of both sides

Using the property \(\ln(e^x)=x\), take \(\ln\) of both sides:
\(\ln(0.04)=\ln(e^{-0.000121t})\)
Since \(\ln(e^{-0.000121t})=- 0.000121t\), we have:
\(\ln(0.04)=-0.000121t\)

Step3: Solve for \(t\)

We know that \(\ln(0.04)\approx - 3.218876\). Then:
\(t=\frac{\ln(0.04)}{-0.000121}\)
Substitute \(\ln(0.04)\approx - 3.218876\) into the formula:
\(t=\frac{-3.218876}{-0.000121}\)
\(t\approx26602.28\)

Answer:

\(26602\)