To determine the interest rate for a deposit that grows from [tex]$4500 to $[/tex]5994.79 over a period of 4 years with monthly compounding, we will use the compound interest formula. The detailed steps are as follows:
1. Identify the variables in the compound interest formula:
[tex]\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\][/tex]
where:
- \( A \) is the future value of the investment, which is $5994.79.
- \( P \) is the principal investment amount, which is $4500.
- \( r \) is the annual interest rate (as a decimal).
- \( n \) is the number of times the interest is compounded per year, which is 12 (monthly compounding).
- \( t \) is the number of years the money is invested, which is 4 years.
2. Rearrange the formula to solve for \( r \):
First, isolate the term involving \( r \).
[tex]\[
\left(1 + \frac{r}{n}\right)^{nt} = \frac{A}{P}
\][/tex]
Then:
[tex]\[
1 + \frac{r}{n} = \left(\frac{A}{P}\right)^{\frac{1}{nt}}
\][/tex]
3. Calculate the specific value:
Substitute the known values into the equation:
[tex]\[
1 + \frac{r}{n} = \left(\frac{5994.79}{4500}\right)^{\frac{1}{12 \times 4}}
\][/tex]
4. Solve for \( \frac{r}{n} \):
Compute the expression on the right-hand side first:
[tex]\[
1 + \frac{r}{12} = \left(\frac{5994.79}{4500}\right)^{\frac{1}{48}}
\][/tex]
By inserting the given values and solving this:
[tex]\[
1 + \frac{r}{12} = 1.09995339 \ (\text{approx})
\][/tex]
5. Isolate \( r \):
[tex]\[
\frac{r}{12} = 1.09995339 - 1
\][/tex]
[tex]\[
\frac{r}{12} = 0.005993165958726454
\][/tex]
Multiply both sides by 12 to find \( r \):
[tex]\[
r = 0.07191799150471745 \ (\text{approx})
\][/tex]
6. Convert \( r \) to a percentage:
To express \( r \) as a percentage, multiply by 100:
[tex]\[
r \approx 7.19\%
\][/tex]
Therefore, the interest rate is approximately 7.19%.
