To find the balance in a continuously compounded account, we use the formula for continuous compounding:

$A = Pe^{rt}$

where:

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

Given values:

• $P = 3000$
• $r = 5.3\% = 0.053$
• $t = 3$ years

Step-by-Step Calculation

1. Substitute the given values into the formula:

   $A = 3000 \times e^{(0.053 \times 3)}$

2. Calculate the exponent part first:

   $0.053 \times 3 = 0.159$

3. Now, calculate $e^{0.159}$:

   $e^{0.159} \approx 1.1725$

4. Multiply by the principal:

   $A \approx 3000 \times 1.1725 = 3517.50$

So, the balance after 3 years is approximately $3517.50.

Would you like more details on each step, or do you have any questions?

Here are some related questions to explore further:

1. What happens if the interest rate is compounded quarterly instead of continuously?
2. How does changing the time period affect the balance in a continuously compounded account?
3. What if the initial principal was different, say $4000 instead of $3000?
4. How does continuous compounding compare to simple interest over the same period?
5. What is the impact of increasing the interest rate on the final amount?

Tip: Continuous compounding grows faster than any other type of compounding (like monthly or annually) because the interest is added at every possible moment.