## Precalculus (6th Edition) Blitzer

Step 1. Identify the given quantities: $P=8000, t=3, r=0.08$ Step 2. If the interest is compounded continuously, we have $A_1=Pe^{rt}=8000e^{0.08(3)}\approx10170$ dollars. Step 3. If the interest is compounded monthly, we have $n=12$ and $A_2=P(1+\frac{r}{n})^{nt}=8000(1+\frac{0.08}{12})^{12(3)}\approx10162$ dollars. Step 4. Find the difference: $A_1-A_2=10170-10162=8$ dollars. That is the return is 8 dollars more if the interest is compounded continuously than if it is compounded monthly.