Certainly! To find the amount [tex]\( P \)[/tex] that must be invested at an interest rate of [tex]\( 8.5\% \)[/tex] compounded continuously to obtain a balance of [tex]\( A = \$160,000 \)[/tex] in [tex]\( t \)[/tex] years, we can use the formula for continuous compounding:
[tex]\[ P = \frac{A}{e^{rt}} \][/tex]
where:
- [tex]\( A \)[/tex] is the final amount
- [tex]\( r \)[/tex] is the interest rate as a decimal
- [tex]\( t \)[/tex] is the time in years
- [tex]\( e \)[/tex] is the base of the natural logarithm
Given:
- [tex]\( A = 160000 \)[/tex]
- [tex]\( r = 0.085 \)[/tex] (which is [tex]\( 8.5\% \)[/tex])
- [tex]\( t \)[/tex] is provided for different years (3, 7, and 10 years)
We'll calculate [tex]\( P \)[/tex] for each of those time periods.
### When [tex]\( t = 3 \)[/tex] years:
[tex]\[ P = \frac{160000}{e^{0.085 \times 3}} \][/tex]
After calculating, the amount that must be invested is:
[tex]\[ P \approx 123986.64 \][/tex]
### When [tex]\( t = 7 \)[/tex] years:
[tex]\[ P = \frac{160000}{e^{0.085 \times 7}} \][/tex]
After calculating, the amount that must be invested is:
[tex]\[ P \approx 88250.01 \][/tex]
### When [tex]\( t = 10 \)[/tex] years:
[tex]\[ P = \frac{160000}{e^{0.085 \times 10}} \][/tex]
After calculating, the amount that must be invested is:
[tex]\[ P \approx 68386.39 \][/tex]
### Summary Table
| Years ([tex]\( t \)[/tex]) | Amount to Invest ([tex]\( P \)[/tex]) |
|----------------------|-------------------------------|
| 3 | \[tex]$123,986.64 |
| 7 | \$[/tex]88,250.01 |
| 10 | \[tex]$68,386.39 |
So, for the given years, the amounts that must be invested at an interest rate of \( 8.5\% \) compounded continuously to reach a balance of \$[/tex]160,000 are:
- For 3 years: \[tex]$123,986.64
- For 7 years: \$[/tex]88,250.01
- For 10 years: \$68,386.39
