Certainly! To determine the future value of a deposit when the interest is compounded continuously, we use the formula for continuous compounding:
[tex]\[ A = P \times e^{rt} \][/tex]
where:
- [tex]\( A \)[/tex] is the future value of the investment/loan, including interest.
- [tex]\( P \)[/tex] is the principal investment amount (the initial deposit).
- [tex]\( r \)[/tex] is the annual interest rate (decimal).
- [tex]\( t \)[/tex] is the time the money is invested for, in years.
- [tex]\( e \)[/tex] is the base of the natural logarithm, approximately equal to 2.71828.
In this scenario:
- The initial deposit ([tex]\( P \)[/tex]) is \[tex]$1,000.
- The annual interest rate (\( r \)) is 1.6%, which is 0.016 as a decimal.
- The number of years (\( t \)) the money is invested is 18 years.
Plugging these values into the formula, we get:
\[ A = 1000 \times e^{(0.016 \times 18)} \]
First, we compute the exponent:
\[ 0.016 \times 18 = 0.288 \]
Next, we calculate \( e^{0.288} \):
\[ e^{0.288} \approx 1.33376 \]
Then, we multiply this result by the principal amount, \( P \):
\[ A = 1000 \times 1.33376 = 1333.76 \]
Therefore, the future value of the deposit after 18 years, rounded to the nearest cent, is \$[/tex]1333.76.