## Algebra and Trigonometry 10th Edition

$5.7 \%$ and $1122.04$
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 $1122.04$.