To solve this, we can use the formula for continuous compounding interest:

$A = P e^{rt}$

where:

• $A$ is the amount of money in the account after $t$ years,
• $P$ is the initial amount invested,
• $r$ is the annual interest rate (as a decimal),
• $t$ is the time in years,
• $e$ is the base of the natural logarithm, approximately equal to 2.71828.

Given values:

• $P = 3100$
• $r = 0.08$
• $t = 5$

Let's substitute these values into the formula:

$A = 3100 \times e^{(0.08 \times 5)}$

Step 1: Calculate the exponent

$0.08 \times 5 = 0.4$

Step 2: Compute $e^{0.4}$

Using the approximate value of $e$: $e^{0.4} \approx 1.4918$

Step 3: Multiply by the principal

$A = 3100 \times 1.4918 \approx 4624.58$

Therefore, the amount in the account after five years, rounded to the nearest cent, is $4624.58.

Would you like more details, or have any other questions?

Here are some related questions:

1. What would the balance be if the investment period was 10 years instead of 5?
2. How would the balance change if the interest rate were 10% instead of 8%?
3. How does continuous compounding compare to annual compounding for the same investment?
4. Can we determine the doubling time for this investment using continuous compounding?
5. How much interest is earned in the five years?

Tip: Continuous compounding uses the natural exponential base $e$, which grows investments slightly faster than standard compounding intervals.