To solve for the correct interest rate with continuous compounding, we can use the formula:

$A = P e^{rt}$

where:

• $A$ is the amount after time $t$,
• $P$ is the initial principal,
• $r$ is the interest rate (growth rate),
• $t$ is the time in years,
• $e$ is the base of the natural logarithm.

Given Data

• Initial amount, $P = 20,000$
• Amount after 10 years, $A = 30,000$
• Time, $t = 10$ years

1. Set up the equation:
   $30,000 = 20,000 \cdot e^{10r}$

2. Solve for $r$: Divide both sides by 20,000: $1.5 = e^{10r}$

   Take the natural logarithm of both sides: $\ln(1.5) = 10r$

   Solve for $r$: $r = \frac{\ln(1.5)}{10}$

3. Calculate $r$ to two decimal places.

Let me calculate this.The correct interest rate, rounded to two decimal places, is 4.05%.

Let me know if you'd like further details or if you have any questions.

Here are some related questions to explore:

1. How does continuous compounding differ from annual compounding?
2. How can the formula be modified if compounding occurs quarterly instead of continuously?
3. What is the doubling time for a 4.05% continuously compounded rate?
4. How would a change in the time period affect the final amount?
5. What is the effect of increasing the initial investment on the final amount?

Tip: When working with continuously compounded interest, remember that small changes in the rate or time can significantly impact the final amount.