Set up the equation to solve. You only need to plug in the numbers  not solve for particular values!
How much money would I have after 6 years if I invested $3000 in a bank account that paid 4.5% interest,
 (a) compounded monthly?
 (b) compounded continuously?
Foundations:

The primary purpose of this problem is to demonstrate that you understand the difference between continuous compounding and compounding on an interval of time. When we compound on an interval, say monthly, the value in the account only changes at the end of each interval. In other words, there is no interest accrued for a week or a day. As a result, we use the formula

 $A\,=\,P\left(1+{\frac {r}{n}}\right)^{nt},$

where $A$ is the value of the account, $P$ is the principal (original amount invested), $r$ is the annual rate and $n$ is the number of compoundings per year. The value of $n$ is $365$ for compounding daily, $52$ for compounding weekly, and $12$ for compounding monthly. As a result, the exponent $nt$, where $t$ is the time in years, is the number of compounding periods where we actually earn interest. Similarly, $r/n$ is the rate per compounding period (the annual rate divided by the number of compoundings per year).

For example, if we compound monthly for $7$ years at a $6\%$ rate, we would compound $nt\,=\,12\cdot 7\,=\,84$ times, once per month, at a rate of $0.06/12\,=\,0.005$, or $0.5\%$), per monthly period. Notice that we always use the decimal version for interest rates when using these equations.

On the other hand, interest compounded continuously earns rate in just that way  continuously. I can have any value I want for time $t$, and the total amount in the account will change with each and every second. Therefore, we express the account value as

 $A\,=\,Pe^{rt}.$

The goal in the two parts of this problem is to choose the correct equation for each.

(a):

We are given all the pieces required. We begin with $\$3000$ of principal, and compound monthly, or $12$ times per year. Using the formula in 'Foundations', the equation for the account value is

 $A\,=\,P\left(1+{\frac {r}{n}}\right)^{nt}\,=\,3000\left(1+{\frac {0.045}{12}}\right)^{12\cdot 6}.$

(b):

Again, we need to apply the formula from foundations to find

 $A\,=\,Pe^{rt}\,=\,3000e^{0.045\cdot 6}.$

Final Answer:

 (a) $A\,=\,3000\left(1+{\frac {0.045}{12}}\right)^{12\cdot 6}.$

 (b) $A\,=\,Pe^{rt}\,=\,3000e^{0.045\cdot 6}.$

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