College Algebra, Chapter 5, Review Exercise, Section Review Exercise, Problem 90

A sum of $\$5000$ is invested at an interest rate of $8 \frac{1}{2} \%$ per year, compounded semiannually.
(a) Determine the amount of the investment after $1 \frac{1}{2}$ years.
(b) On what period of time will the investment amount to $\$7000$
(c) Determine how long would it take for the amount to grow to $\$ 7000$, if interest were compounded continuously instead of semiannually.


a.) Recall that formula for Interest compounded $n$ years
$\displaystyle A(t) = P \left( 1 + \frac{r}{n} \right)^{nt}$
If the interest is compounded semi-anually, then $n = 2$


$
\begin{equation}
\begin{aligned}
A &= 5000 \left( 1 + \frac{0.085}{2} \right)^{2 \left( \frac{3}{2} \right)}\\
\\
A &= \$ 5664.9776 \quad \text{ or } \quad \$5665
\end{aligned}
\end{equation}
$


b.) If $A = \$7000$, then

$
\begin{equation}
\begin{aligned}
7000 &= 5000 \left( 1 + \frac{0.085}{2} \right)^{2(t)}\\
\\
\frac{7000}{5000} &= \left( 1 + \frac{0.085}{2} \right)^{2t} && \text{Take ln of both sides}\\
\\
\ln \frac{7}{5} &= 2t \left[\ln \left( 1 + \frac{0.085}{2} \right) \right] && \text{Solve for } t\\
\\
t &= \frac{\ln \left( \frac{7}{5} \right)}{2 \ln \left( 1+\frac{0.085}{2} \right)}\\
\\
t &= 4.04 \text{ years} \quad \text{ or } \quad 4 \text{ years}
\end{aligned}
\end{equation}
$


c.) Recall that the formula for interest compounded continuously is
$A (t) = Pe^{rt}$
if $A = 7000$, then


$
\begin{equation}
\begin{aligned}
7000 &= 500 e^{(0.085)t} && \text{Divide by 5000}\\
\\
\frac{7000}{5000} &= e^{0.085t} && \text{Take ln of both sides}\\
\\
\ln \frac{7}{5} &= 0.085t && \text{Recall } \ln e = 7\\
\\
t &= \frac{\ln \left( \frac{7}{5} \right)}{0.085} && \text{Solve for } t\\
\\
t &= 3.9585 \text{ years}
\end{aligned}
\end{equation}
$


It shows that if the interest is compounded continuously, it will only take a bit shorter compared to the period in which interest is compounded semi anually.