If [tex]$740 is invested at an interest rate of 11% per year and is compounded continuously, how much will the investment be worth in 7 years?

Use the continuous compound interest formula \( A = Pe^{rt} \).

A. $[/tex]742
B. [tex]$1,100
C. $[/tex]1,548
D. $1,598



Answer :

Certainly! Let's solve this step-by-step using the continuous compound interest formula:

The continuous compound interest formula is given by:
[tex]\[ A = P e^{rt} \][/tex]
where:
- [tex]\( A \)[/tex] is the amount of money accumulated after n years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( r \)[/tex] is the annual interest rate (in decimal form).
- [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.

Given values:
- [tex]\( P = 740 \)[/tex] dollars
- [tex]\( r = 0.11 \)[/tex] (since 11% = 0.11)
- [tex]\( t = 7 \)[/tex] years

Let's calculate:

1. Identify the values:
- Principal (P) = [tex]$740$[/tex]
- Annual interest rate (r) = [tex]$0.11$[/tex]
- Time (t) = [tex]$7$[/tex] years

2. Substitute the values into the formula:
[tex]\[ A = 740 \times e^{0.11 \times 7} \][/tex]

3. Calculate the exponent part:
[tex]\[ 0.11 \times 7 = 0.77 \][/tex]

4. Using the value of e (approximately 2.71828), calculate:
[tex]\[ e^{0.77} \approx 2.157 \][/tex]

5. Now multiply the principal amount by this value:
[tex]\[ A = 740 \times 2.157 \][/tex]
[tex]\[ A \approx 1598.23 \][/tex]

So, after 7 years, the investment will be worth approximately \[tex]$1598.23. Thus, the correct choice based on the given options would be the value closest to $[/tex]\[tex]$ 1598$[/tex], which matches [tex]$\$[/tex] 1,598$.

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