Question 313164
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So far, so good, except for one tiny detail.  The formula you want is:


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


Where *[tex \Large A] is the amount at the end, *[tex \Large P] is what you start with, *[tex \Large r] is the rate as decimal, *[tex \Large t] is the time in years, and *[tex \Large e] is the base of the natural logarithms.


Since you want your money to double,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{A}{P}\ =\ 2]


Therefore:


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


You don't need to substitute the rate value just yet...patience, Grasshopper.


Take the natural logarithm of both sides:


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


Next apply the following rule:


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


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


Next use the fact that *[tex \Large \ln(e)\ =\ 1]


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


Since you want to know *[tex \Large t], multiply both sides by *[tex \Large \frac{1}{r}]


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


NOW substitute the value of *[tex \Large r]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(2)}{0.05}]


And finally get all happy with your calculator


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ \approx\ 13.9] years.


By the way, this same formula will work for "How long will it take my money to multiply by *[tex \Large m] at *[tex \Large r] interest rate compounded continuously.


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


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