Question 1062183
<font face="Times New Roman" size="+2">


If interest rate *[tex \Large r] (expressed as a decimal) is compounded monthly, then when the principal has doubled, *[tex \Large A\ =\ 2P].


So if the investment takes *[tex \Large t] years to double:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2P\ =\ P\left(1\ +\ \frac{r}{12}\right)^{12t}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ +\ \frac{r}{12}\right)^{12t}\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\left(1\ +\ \frac{r}{12}\right)^{12t}\right)\ =\ \ln(2)]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12t\,\cdot\,\ln\left(1\ +\ \frac{r}{12}\right)\ =\ \ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(2)}{12\,\cdot\,\ln\left(1\ +\ \frac{r}{12}\right)]


If compounded continuously:


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


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


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


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


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


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \