Chapter 5 - 5.5 - Exponential and Logarithmic Models - 5.5 Exercises - Page 405: 11

$6376.28$ and $15.403$

We have $A=Pe^{rt} $ and $P=Ae^{-rt}=10000 e^{-0.45 \times 10} \approx 6376.28$ Now, we have $A=Pe^{rt} \implies 2P=Pe^{rt}$ $\implies 2 =e^{rt}$ Take the $\log$ on each side. $\log 2 = \log e^{rt} \implies t = \dfrac{\ln 2}{0.045} \approx 15.403$ Our results are: $6376.28$ and $15.403$