A loan program offers an interest rate of 5 percent per year, compounded continuously. Assuming no payments are made, how much would be owed after three years on a loan of 3300 dollars?

Step-by-step solution

1. 1

   Choose the right formula and know where it comes from. Compounding $n$ times a year gives $P\left(1+\tfrac{r}{n}\right)^{nt}$; letting $n\to\infty$ sends this to the continuous-compounding formula

   $A=Pe^{rt}.$

   So the presence of $e$ is not a convention - it is the limit of ever more frequent compounding.

2. 2

   Identify the inputs in consistent units. The principal is $P=3300$, the rate must be a decimal per year, $r=0.05$, and the time must be in the same time unit as the rate, $t=3$ years. Mixing a percent with a decimal, or months with an annual rate, is the usual source of error here.

3. 3

   Form the exponent first. Compute $rt$ before exponentiating:

   $rt=0.05\times 3=0.15,\qquad A=3300\,e^{0.15}.$

4. 4

   Evaluate without rounding early. $e^{0.15}=1.1618342427\ldots$, so

   $A=3300\times 1.1618342427=3834.053001\ldots$

   Rounding only at the very end gives \3834.05$. (Rounding$e^{0.15}$to$1.16$first would have given$$3828.00$, off by six dollars.)

5. 5

   Sanity-check against ordinary annual compounding. With interest added once a year the balance would be 3300(1.05)^3=\3820.16$. Continuous compounding is larger, as it must be, but only by about$$13.89$ - a useful reminder that at moderate rates the compounding frequency matters far less than the rate itself.