College Algebra, Chapter 5, 5.4, Section 5.4, Problem 72

Suppose that an amount of $\$ 6500$ is invested in an account that pays $6\%$ interest per year, compounded continuously.

a.) What is the amount after 2 years?

b.) How long will it take for the amount to be $\$ 8000$?


a.) Recall that the formula for interest compounded continuously,



$\begin{equation} \begin{aligned} A(t) =& Pe^{rt} \\ \\ \text{So},& \\ \\ A =& 6500 \left[ e^{(0.06)(2)} \right] \\ \\ A =& \$ 7328.73 \end{aligned} \end{equation}$


b.) If $A = \$ 8000$, then


$\begin{equation} \begin{aligned} 8000 =& 6500 e^{(0.06)t} \\ \\ \frac{8000}{6500} =& e^{(0.06)t} \\ \\ \ln \left( \frac{8000}{6500} \right) =& (0.06)t \\ \\ t =& \frac{\displaystyle \ln \left( \frac{8000}{6500} \right)}{0.06} \\ \\ t =& 3.46 \text{ yrs} \\ \\ t =& 3 \text{ yrs} + 0.46 \text{ yrs} \left( \frac{12 \text{ months}}{1 \text{ yr}} \right) \\ \\ t =& 3 \text{ yrs} + 5.52 \text{ months} \end{aligned} \end{equation}$


It shows that the amount will be $\$8000$ when $t$ is approximately $3$ years and $6$ months.