Question 295827
<font face="Garamond" size="+2">


Obvious to whom?


For an investment of *[tex \Large P] at *[tex \Large r] interest per year (as a decimal) compounded *[tex \Large n] times per year for *[tex \Large t] years.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ 1000(1\ +\ \frac{0.1}{12})^{12\cdot1}]


For an investment of *[tex \Large P] at *[tex \Large r] interest per year (as a decimal) compounded continuously for *[tex \Large t] years.  Note: *[tex \LARGE e] is the base of the natural logarithms.  "Inv ln" does the trick quite nicely on the Windows built-in calculator in Scientific mode.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ 1000e^{0.1\cdot1}]


The rest is just calculator work.  Whichever comes out larger is the answer to a) and the difference between the two is the answer to b)


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