To solve for the missing values assuming continuously compounded interest, we will use the formula for continuous compounding:
[tex]\[ A = P \cdot e^{(r \cdot t)} \][/tex]
Where:
- [tex]\( A \)[/tex] is the amount of money accumulated after time [tex]\( t \)[/tex] years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial investment).
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal).
- [tex]\( t \)[/tex] is the time the money is invested for.
- [tex]\( e \)[/tex] is the base of the natural logarithm, approximately equal to 2.71828.
Given:
- Initial investment ([tex]\( P \)[/tex]) = [tex]$11625.64
- Annual percentage rate (\( r \)) = 3.8% = 0.038 (as a decimal)
- Amount after 10 years (\( A \)) = $[/tex]17,000
### Amount After 10 Years
First, let's confirm the amount after 10 years. From the given data:
[tex]\[ A = 11625.64 \times e^{(0.038 \times 10)} \][/tex]
From calculations:
[tex]\[ A = 11625.64 \times e^{0.38} \approx 16999.99 \][/tex]
Thus, the amount after 10 years is:
[tex]\[ \$ 17,000 \][/tex]
### Time to Double
To find the time to double the initial investment, we set [tex]\( A = 2P \)[/tex]:
[tex]\[ 2P = P \times e^{(r \cdot t_{\text{double}})} \][/tex]
Simplifying, we get:
[tex]\[ 2 = e^{(r \cdot t_{\text{double}})} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ \ln(2) = r \cdot t_{\text{double}} \][/tex]
So,
[tex]\[ t_{\text{double}} = \frac{\ln(2)}{r} \][/tex]
Given [tex]\( r = 0.038 \)[/tex]:
[tex]\[ t_{\text{double}} = \frac{\ln(2)}{0.038} \approx 18.24 \][/tex]
Hence, the time to double the investment is approximately [tex]\( 18.24 \)[/tex] years.
### Summary
- Initial Investment: \[tex]$11625.64
- Annual Percentage Rate: 3.8%
- Time to Double: 18.24 years
- Amount After 10 Years: \$[/tex]17,000
