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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.

You invest $20,000 in an account where the interest is continuously compounding. If in 10 years you have $30,000, find the interest (growth) rate and the doubling time.

Learn how to calculate the continuous compounding interest rate and the doubling time for a $20,000 investment that grows to $30,000 in 10 years. This problem uses exponential growth, logarithmic functions, and continuous compounding. Suitable for students in Grades 10-12.

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