Question 465812
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The formula for compound interest is:


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


where *[tex \Large A] is the future value, *[tex \Large P] is the present value, *[tex \Large r] is the annual interest rate expressed as a decimal, *[tex \Large n] is the number of compounding periods per year, and *[tex \Large t] is the number of years.


In this case we really don't care what the present or future values are so long as the future value is three times the present value, so we can re-write the formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ +\ \frac{r}{n}\right)^{nt}\ =\ \frac{A}{P}\ =\ 3]


We are given both *[tex \Large r\ =\ 0.052] and *[tex \Large n\ =\ 1], so substitute the values:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1.052\right)^{t}\ =\ 3]


The only thing left is to take the logarithm of both sides (any base, it doesn't matter in the end) and solve for *[tex \Large t]


Write back and let me know what you get for an answer.


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