## Algebra and Trigonometry 10th Edition

$0.11019$ and $6.290$
We have $A=Pe^{rt}$ Take the $\log$ on each side. $\log (A/P) = \log e^{rt}$ or, $r = \dfrac{1}{t } \log (A/P)=(1/10) \log \dfrac{1505.00}{500} \approx 0.11019$ Now, we have $A=Pe^{rt} \implies 2P=Pe^{rt}$ $\implies 2 =e^{rt}$ Take the $\log$ on each side. $\log 2 = \log e^{rt} \implies t = \dfrac{\ln 2}{0.11019} \approx 6.290$ Our results are: $0.11019$ and $6.290$