Answer :
To determine how long it will take to double your initial investment of [tex]$7500 at an annual interest rate of 4.25%, compounded semi-annually, we need to use the compound interest formula:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after \( t \) years, including interest.
- \( P \) is the principal amount (the initial sum of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times the interest is compounded per year.
- \( t \) is the time the money is invested for in years.
Here are the known values:
- \( P = 7500 \)
- \( r = 4.25 / 100 = 0.0425 \)
- \( n = 2 \) (since the interest is compounded semi-annually)
- \( A = 2P = 2 \times 7500 = 15000 \) (since we want to double the investment)
We need to solve for \( t \).
Rearranging the compound interest formula to solve for \( t \):
\[ 2P = P \left(1 + \frac{r}{n}\right)^{nt} \]
Dividing both sides by \( P \):
\[ 2 = \left(1 + \frac{r}{n}\right)^{nt} \]
Taking the natural logarithm (ln) of both sides:
\[ \ln(2) = \ln\left[\left(1 + \frac{r}{n}\right)^{nt}\right] \]
Using the logarithm power rule:
\[ \ln(2) = nt \cdot \ln\left(1 + \frac{r}{n}\right) \]
Solving for \( t \):
\[ t = \frac{\ln(2)}{n \cdot \ln\left(1 + \frac{r}{n}\right)} \]
Substitute the known values:
\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{0.0425}{2}\right)} \]
Evaluating this expression yields:
\[ t \approx 16.482024930376596 \]
To round this to the nearest half year:
\[ t \approx 16.5 \]
Thus, it will take approximately 16.5 years to double your initial investment of $[/tex]7500 at an annual interest rate of 4.25%, compounded semi-annually.