Chapter 5 - 5.5 - Exponential and Logarithmic Models - 5.5 Exercises - Page 405: 7

$19.804$ and $1419.07$

We have $A=Pe^{rt} \implies 2P=Pe^{rt}$ $\implies 2 =e^{rt}$ Take the $\log$ on each side. $\log 2 = \log e^{rt} \implies t = \dfrac{\ln 2}{r}=\dfrac{\ln 2}{0.035} \approx 19.804$ Now, $A=Pe^{rt} = 1000 e^{0.035 \times 10}$ or, $A \approx 1419.07$ Our results are: $19.804$ and $1419.07$