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Angela invested $\$15{,}000$ in a savings account. If the interest rate is $4\%\text{,}$ how much will be in the account in $10$ years by each method of compounding? Round each answer to the nearest cent.

compound quarterly: $
compound monthly: $
compound continuously: $

Solution

For compounding quarterly, $n=4\text{.}$ Plugging this and $P=\$15{,}000\text{,}$ $r=0.04\text{,}$ and $t=10$ into the compound interest formula, we have:
$\begin{aligned} A\amp=P\left(1+\frac{r}{n}\right)^{nt}\\ \amp=\substitute{15000}\left(1+\frac{\substitute{0.04}}{\substitute{4}}\right)^{(\substitute{4}\cdot \substitute{10})}\\ \amp=15000\left(1+\frac{0.04}{4}\right)^{40}\\ \amp\approx \$22333 \end{aligned}$

For compounding monthly, $n=12\text{.}$ Plugging this and $P=\$15{,}000\text{,}$ $r=0.04\text{,}$ and $t=10$ into the compound interest formula, we have:
$\begin{aligned} A\amp=P\left(1+\frac{r}{n}\right)^{nt}\\ \amp=\substitute{15000}\left(1+\frac{\substitute{0.04}}{\substitute{12}}\right)^{(\substitute{12}\cdot \substitute{10})}\\ \amp=15000\left(1+\frac{0.04}{12}\right)^{120}\\ \amp\approx \$22362 \end{aligned}$

For compounding continuously, we use the formula $A=Pe^{rt}\text{,}$ and plug in $P=\$15{,}000\text{,}$ $r=0.04\text{,}$ and $t=10\text{.}$ This gives us:
$\begin{aligned} A\amp=Pe^{rt}\\ \amp=\substitute{15000}e^{(\substitute{0.04}\cdot \substitute{10})}\\ \amp\approx \$22377 \end{aligned}$