Question 513103
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The same amount of time it takes any amount of money to double at 5% compounded monthly.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ P\left(1\ +\ \frac{r}{m}\right)^{mt}]


Where *[tex \Large A] represents the Future Value, *[tex \Large P] represents the Present Value, *[tex \Large r] is the annual interest rate expressed as a decimal, *[tex \Large m] is the number of compounding periods in one year, and *[tex \Large t] is the number of years.


You double your money when *[tex \Large \frac{A}{P}\ =\ 2], hence you need to solve for *[tex \Large t] where:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ +\ \frac{0.05}{12}\right)^{12t}\ =\ 2]


Do the arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1.00417\right)^{12t}\ =\ 2]


Take the log of both sides (any base, doesn't matter)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(1.00417\right)^{12t}\ =\ \ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12t\,\cdot\,\ln\left(1.00417\right)\ =\ \ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(2)}{12\,\cdot\,\ln\left(1.00417\right)]


The rest is calculator work.  Enjoy.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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