Question 606820
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I did one very similar to this earlier today.  You should be able to figure this one out from the work I showed in that problem, the only substative difference is that you want to triple rather than double.  You are probably clever enough to figure out where to put a 3 instead of a 2 -- see below:


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


Where *[tex \LARGE A] is the future value, *[tex \LARGE P] is the present value, i.e. the initial investment, *[tex \LARGE e] is the base of the natural logarithms which is the limit of *[tex \LARGE \left(1\ +\ \small{\frac{1}{n}}\LARGE\right)^n] as *[tex \LARGE n] becomes large, *[tex \LARGE r] is the annual interest rate expressed as a decimal, and *[tex \LARGE t] is the number of years.


Regardless of the amount of the initial investment, in order for it to double, *[tex \LARGE \frac{A}{P}] and therefore *[tex \LARGE e^{rt}] must equal 2.


Hence:


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


and solve for *[tex \LARGE t]


Take the natural log of both sides.  Actually, the log of any base will work and, given correct arithmetic, will provide the correct answer.  However, since the equation involves *[tex \LARGE e], the arithmetic is greatly simplified by using the natural log.


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


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(a^n\right)\ =\ n\,\cdot\,\log_b(a)]


To write


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


Then use


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


to write


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


and finally


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


Just use your calculator to divide *[tex \LARGE \ln(2)] by the interest rate.  Don't forget that the interest rate must be expressed as a decimal.


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After that, toss the following into your financial formulas toolkit:


An initial investment of *[tex \LARGE P] at an interest rate of *[tex \LARGE r] compounded continuously will increase by a factor of *[tex \LARGE m] in *[tex \LARGE \frac{\ln(m)}{r}] years.


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