Question

problem: cave paintings were discovered in a cave in france. the paint contained 18% 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: Identify the formula and values

The exponential decay model for carbon - 14 is \(A = A_0e^{-0.000121t}\), where \(A\) is the amount of carbon - 14 remaining, \(A_0\) is the initial amount of carbon - 14, and \(t\) is the time in years. We know that \(A = 0.18A_0\) (since 18% of the original carbon - 14 remains).

Substitute \(A = 0.18A_0\) into the formula:
\(0.18A_0=A_0e^{-0.000121t}\)

Step2: Solve for \(t\)

Divide both sides of the equation by \(A_0\) (assuming \(A_0
eq0\), which makes sense in the context of carbon - 14 dating):
\(0.18 = e^{-0.000121t}\)

Take the natural logarithm of both sides:
\(\ln(0.18)=\ln(e^{-0.000121t})\)

Using the property of logarithms \(\ln(e^x)=x\), we get:
\(\ln(0.18)=- 0.000121t\)

Now, solve for \(t\):
\(t=\frac{\ln(0.18)}{- 0.000121}\)

Calculate \(\ln(0.18)\approx - 1.7147\)

Then \(t=\frac{- 1.7147}{- 0.000121}\approx14171\)

Answer:

14171