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To solve this problem, we’ll use the continuous compound interest formula:

\[
A = P e^{rt}
\]

where:
- $ A $ is the amount of money accumulated after time $ t $,
- $ P $ is the initial principal,
- $ r $ is the interest rate,
- $ t $ is the time in years, and
- $ e $ is the base of the natural logarithm, approximately equal to 2.718.

We are given:
- $ A = 22141 $,
- $ P = 13800 $,
- $ t = 60 $ months (which we’ll convert to years: $ t = 60/12 = 5 $ years),
- $ r $ is unknown.

### Steps to Solve for $ r $

1. Substitute the known values into the formula:
   \[
   22141 = 13800 \cdot e^{5r}
   \]

2. Divide both sides by 13800:
   \[
   \frac{22141}{13800} = e^{5r}
   \]

3. Take the natural logarithm (ln) of both sides to solve for $ r $:
   \[
   \ln\left(\frac{22141}{13800}\right) = 5r
   \]

4. Solve for $ r $ by dividing both sides by 5:
   \[
   r = \frac{\ln\left(\frac{22141}{13800}\right)}{5}
   \]

Let’s calculate the value.The interest rate $ r $ is approximately $ 0.0946 $, or 9.46% per year when compounded continuously.

Would you like further details on this solution, or have any additional questions?

Here are five related questions you might consider:
1. How would the solution change if the interest were compounded monthly instead of continuously?
2. What is the formula for compound interest if it’s compounded $ n $ times per year?
3. How does continuous compounding differ from simple compounding in terms of growth rate?
4. If the interest rate $ r $ were doubled, what would be the new accumulated amount after 5 years?
5. How can we use logarithmic properties to solve for time $ t $ if $ r $ and $ A $ are known?

**Tip:** Continuous compounding is ideal for scenarios where investments are growing constantly, like certain financial or biological growth models.

Use the continuous compound interest formula to find the indicated value.
A = $22,141; P = $13,800; t = 60 months; r = ?

Find the interest rate for a continuous compounding investment where A = $22,141, P = $13,800, and t = 60 months. This problem covers key algebraic concepts, exponential functions, and continuous compounding suitable for Grades 11-12.

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