Question 219496
<font face="Garamond" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ Pe^{rt}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{rt}\ =\ \frac{A}{P}]


If the principal has doubled, then *[tex \Large \frac{A}{P}\ =\ 2], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{rt}\ =\ 2]


Take the natural logarithm of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln{\left(e^{rt}\right)}\ =\ \ln{(2)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ rt\,\ln{(e)}\ =\ \ln{(2)}]


But *[tex \Large \ln{(e)} = 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ rt\ =\ \ln{(2)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln{(2)}}{r}]


Plug in *[tex \Large r = 0.08] and crank up your calculator.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>