Question 328011
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The formula for continuous compounding is:


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


If the investment doubles, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ 2P]


That means that, for *[tex \Large t\ =\ 16], 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{16r}\ =\ 2]


Take the natural log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(e^{16r})\ =\ \ln(2)]


Use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16r\ln(e)\ =\ \ln(2)]


Use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(b)\ =\ 1]


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16r\ =\ \ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ \frac{\ln(2)}{16}\ \approx 4.33%]


My calculator said it, I believe it, that settles it.


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