Question 1130906
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Compound Interest Formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ F\ =\ P\(1\ +\ \frac{r}{100n}\)^{nt}]


Where *[tex \Large F] is the future value, *[tex \Large P] is the present value, *[tex \Large r] is the annual interest rate as a percent, *[tex \Large n] is the number of compounding periods per year, and *[tex \Large t] is the number of years in the term of the investment.


If the value of the investment doubles, then *[tex \Large \frac{F}{P}\ =\ 2], so solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(1\ +\ \frac{6}{100}\)^{t}\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(1\ +\ \frac{6}{100*12}\)^{12t}\ =\ 2]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(1\ +\ \frac{6}{100*365}\)^{365t}\ =\ 2]


For *[tex \LARGE t]


Hint: Start by taking the logarithm of both sides and then using *[tex \LARGE \log_b(x^n)\ =\ n\log_b(x)]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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