Answer
$5.7 \%$ and $1122.04$
Work Step by Step
We have the continuous compound interest formula:
$A=Pe^{rt} \implies 2P=Pe^{12r}$
Take $\log$ on each side.
$ \ln 2= \ln e^{12 r} \implies r= \dfrac{\ln 2}{12} \approx 0.0578$
so, the rate is approximately $5.7 \%$
Further, $A=Pe^{rt} \implies 2000 =Pe^{0.0578(12)t}$
$P= \dfrac{2000}{e^{0.578}} \approx 1122.04$
so, the principal is approximately $1122.04$
Our results are: $5.7 \%$ and $1122.04$.
