Question 434424
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It depends on how often the interest is compounded.


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


A is the future amount -- for your problem $2340.


P is the present amount -- for your problem $1800.


r is given as 5% which is to say 0.05.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ +\ \frac{0.05}{n}\right)^{nt}\ =\ \frac{2340}{1800}\ =\ 1.3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(1\ +\ \frac{r}{n}\right)^{nt}\ =\ \ln(1.3)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ nt\ln\left(1\ +\ \frac{r}{n}\right)\ =\ \ln(1.3)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(1.3)}{n\cdot\ln\left(1\ +\  \frac{r}{n}\right)]


So now all you have to do is decided how often to compound, plug in an appropriate value for n and then do the arithmetic.  Use your calculator.


Annual compounding: n = 1, Semi-annual, n = 2, quarterly, n = 4, monthly, n = 12, and so on.


UNLESS you want to compound continuously, then


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


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{0.05t}\ =\ 1.3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(1.3)}{0.05}]


The fastest you can get there is 5 years 3 months, give or take a day or two.


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