To determine how much should be invested now at an interest rate of 6.5% per year, compounded continuously, to have [tex]$3000 in two years, we use the formula for continuous compounding:
\[ A = P e^{rt} \]
Where:
- \( A \) is the final amount (in this case, $[/tex]3000),
- [tex]\( P \)[/tex] is the initial amount (the amount we need to find),
- [tex]\( r \)[/tex] is the annual interest rate (0.065),
- [tex]\( t \)[/tex] is the time the money is invested (2 years),
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately equal to 2.71828).
Rearranging the formula to solve for [tex]\( P \)[/tex]:
[tex]\[ P = \frac{A}{e^{rt}} \][/tex]
1. Substitute the known values into the formula:
[tex]\[ P = \frac{3000}{e^{0.065 \times 2}} \][/tex]
2. Calculate the exponent:
[tex]\[ 0.065 \times 2 = 0.13 \][/tex]
3. Raise [tex]\( e \)[/tex] to the power of 0.13:
[tex]\[ e^{0.13} \approx 1.13883 \][/tex]
4. Divide [tex]\( 3000 \)[/tex] by the result from step 3:
[tex]\[ P = \frac{3000}{1.13883} \][/tex]
5. Complete the division:
[tex]\[ P \approx \frac{3000}{1.13883} \approx 2634.29 \][/tex]
Therefore, the amount that should be invested now to have [tex]$3000 in two years at an interest rate of 6.5% per year, compounded continuously, is approximately $[/tex]2634.29 when rounded to the nearest cent.
