Problem
\[ \begin{aligned} A & =P e^{r} \\ 0.5 & =e^{5730 r} \\ \ln (0.5) & =\ln \left(e^{5730 r}\right) \\ \frac{\ln (0.5)}{5730} & =r \\ r & \approx-3.2676659890376 \end{aligned} \] Now use the growth rate to determine the fossil's age. \[ \begin{aligned} A & =P e^{r t} \\ 0.08 & \approx 2.1 e^{-3.2676659890376 t} \\ 0.038095238095238 & \approx e^{-3.2676659890376 t} \\ \ln (0.038095238095238) & \approx \ln \left(e^{-3.2676659890376 t}\right) \\ \ln (0.038095238095238) & \approx-3.2676659890376 t \\ \frac{\ln (0.038095238095238)}{-3.2676659890376} & \approx t \\ t & \approx 27,012.626816226 \end{aligned} \]
Solution
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 ：\( t = \frac{\ln(0.0381)}{-3.2677} \approx 27012.6268 \)
