\[
\begin{aligned}
A & =P e^{r} \\
0.5 & =e^{5730 r} \\
\ln (0.5) & =\ln \left(e^{5730 r}\right) \\
\frac{\ln (0.5)}{5730} & =r \\
r & \approx-3.2676659890376
\end{aligned}
\]
Now use the growth rate to determine the fossil's age.
\[
\begin{aligned}
A & =P e^{r t} \\
0.08 & \approx 2.1 e^{-3.2676659890376 t} \\
0.038095238095238 & \approx e^{-3.2676659890376 t} \\
\ln (0.038095238095238) & \approx \ln \left(e^{-3.2676659890376 t}\right) \\
\ln (0.038095238095238) & \approx-3.2676659890376 t \\
\frac{\ln (0.038095238095238)}{-3.2676659890376} & \approx t \\
t & \approx 27,012.626816226
\end{aligned}
\]
Answer
\( t = \frac{\ln(0.0381)}{-3.2677} \approx 27012.6268 \)
Step 1 ：\( A = Pe^{rt} \)
Step 2 ：\( 0.5 = e^{5730r} \)
Step 3 ：\( r = \frac{\ln(0.5)}{5730} \approx -3.2677 \)
Step 4 ：\( 0.08 = 2.1 e^{-3.2677t} \)
Step 5 ：\( e^{-3.2677t} \approx 0.0381 \)
Step 6 ：\( -3.2677t \approx \ln(0.0381) \)
Step 7 ：\( t = \frac{\ln(0.0381)}{-3.2677} \approx 27012.6268 \)
