If 3000 dollars invested in a bank account for 8 years, compounded quarterly, amounts to 4571.44 dollars, what is the interest rate paid by the account?

Nov 13, 2017

About 5.3%.

Explanation:

We'll use the compounding interest formula, $A = P {\left(1 + \frac{r}{n}\right)}^{n t}$.
Here, $A$ is the final amount; $P$ is the principal, or original amount invested; $r$ is the interest rate written in decimal form; $n$ is the number of times the money is compounded per time $t$. Usually, $t$ is in years and $n$ is number of compoundings per year. Quarterly means $n = 4$.

Plug in the information you have: $4 , 571.44 = 3 , 000 {\left(1 + \frac{r}{4}\right)}^{4 \cdot 8}$
So $4 , 571.44 = 3 , 000 {\left(1 + \frac{r}{4}\right)}^{32}$
Divide both sides by $3 , 000$ to get $1.52381 = {\left(1 + \frac{r}{4}\right)}^{32}$

Now, there are a couple of ways to solve this, with either roots or logarithms. Since you're in pre-calculus, I'll assume your teacher wants you to use logs.

Take the $\log$ of both sides: $\log 1.52381 = \log {\left(1 + \frac{r}{4}\right)}^{32}$
Remember the property of logarithms that says $\log {a}^{b} = b \log a$.
Bring the $32$ in front of the expression on the right:
$\log 1.52381 = 32 \log \left(1 + \frac{r}{4}\right)$

Then divide both sides by $32$: $\frac{\log 1.52381}{32} = \log \left(1 + \frac{r}{4}\right)$

Simplify: $.0057166 = \log \left(1 + \frac{r}{4}\right)$

Using the inverse property of logarithms, change the expression to read: ${10}^{.0057166} = 1 + \frac{r}{4}$

Simplify the left: $1.02325 = 1 + \frac{r}{4}$

Solve for $r$ by subtracting $1$ and multiplying by $4$.

$r = .053000$, or in percent notation, r=5.3%.