Question 277249
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(m)\ =\ P(1\ +\ \frac{r}{n})^{nm}]


Where *[tex \Large P] is the starting principal, *[tex \Large r] is the interest rate expressed as a decimal, that is 5% is expressed as 0.05, *[tex \Large m] is 
the number of years of the investment, and *[tex \Large n] is the number of compounding periods per year.


A function specifically for your given conditions would be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(m)\ =\ 600(1\ +\ \frac{.05}{4})^{4m}]


At 3 years:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(m)\ =\ 600(1\ +\ \frac{.05}{4})^{4(3)}]


You get to do your own calculator work.


Solve your function for *[tex \Large m] after substituting 1400 for A(m)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1400}{600}\ =\ (1\ +\ \frac{.05}{4})^{4m}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\frac{7}{3}\right)\ =\ 4m\ln(1.0125)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{\ln\left(\frac{7}{3}\right)}{4\ln(1.0125)}]


Again, you can do your own calculator button pushing.


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