Answer
$ 0.11019$ and $6.290$
Work Step by Step
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 $6.290$
