To determine how long it will take for [tex]$27,000 to grow to $[/tex]31,340.37[tex]$ when the interest rate is 5% compounded quarterly, we can use the compound interest formula:
\[ A = P \left(1 + \frac{r}{k}\right)^{kt} \]
Here:
- \( A \) is the final amount ($[/tex]31,340.37).
- [tex]\( P \)[/tex] is the principal amount ([tex]$27,000).
- \( r \) is the annual interest rate (0.05).
- \( k \) is the number of times interest is compounded per year (4, quarterly).
- \( t \) is the number of years, which we need to find.
First, we plug the known values into the formula:
\[ 31,340.37 = 27,000 \left(1 + \frac{0.05}{4}\right)^{4t} \]
Next, simplify the term inside the parenthesis:
\[ 1 + \frac{0.05}{4} = 1 + 0.0125 = 1.0125 \]
Now our equation is:
\[ 31,340.37 = 27,000 \cdot 1.0125^{4t} \]
To solve for \( t \), we first isolate the exponential part by dividing both sides by 27,000:
\[ \frac{31,340.37}{27,000} = 1.0125^{4t} \]
\[ 1.1614962963 = 1.0125^{4t} \]
Take the natural logarithm (ln) on both sides to bring down the exponent:
\[ \ln(1.1614962963) = \ln(1.0125^{4t}) \]
Using the property of logarithms \( \ln(a^b) = b \cdot \ln(a) \):
\[ \ln(1.1614962963) = 4t \cdot \ln(1.0125) \]
Next, solve for \( t \):
\[ t = \frac{\ln(1.1614962963)}{4 \cdot \ln(1.0125)} \]
From the given answers, we know the following values:
\[ \ln(1.1614962963) \approx 0.14907017685257912 \]
\[ \ln(1.0125) \approx 0.01242251999855711 \]
Thus,
\[ t = \frac{0.14907017685257912}{4 \cdot 0.01242251999855711} \]
\[ t = \frac{0.14907017685257912}{0.04969007999422844} \]
\[ t \approx 2.9999987295229453 \]
Therefore, approximately 3 years will be needed for the $[/tex]27,000 to grow to [tex]$31,340.37$[/tex] at an annual interest rate of 5% compounded quarterly.
