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We are given the following problem:

An initial investment of $18,254 is made at an interest rate of 6.2% per year, compounded continuously. The tasks are as follows:

### a) Exponential Function
To describe the amount in the account after $ t $ years, we use the formula for continuous compounding:

\[
A(t) = P e^{rt}
\]

Where:
- $ A(t) $ is the amount after $ t $ years.
- $ P = 18,254 $ is the principal (initial amount).
- $ r = 6.2\% = 0.062 $ is the annual interest rate.
- $ t $ is the time in years.
- $ e $ is Euler's number (approximately 2.718).

Thus, the function describing the amount after $ t $ years is:

\[
A(t) = 18,254 \cdot e^{0.062t}
\]

---

### b) Balance after specific years
We need to compute the balance for $ t = 1, 2, 5, 10 $ years. Using the formula:

\[
A(t) = 18,254 \cdot e^{0.062t}
\]

We can substitute the values of $ t $.

1. For $ t = 1 $:
\[
A(1) = 18,254 \cdot e^{0.062 \times 1} \approx 18,254 \cdot 1.06396 = 19,416.57
\]

2. For $ t = 2 $:
\[
A(2) = 18,254 \cdot e^{0.062 \times 2} \approx 18,254 \cdot 1.13101 = 20,642.47
\]

3. For $ t = 5 $:
\[
A(5) = 18,254 \cdot e^{0.062 \times 5} \approx 18,254 \cdot 1.34986 = 24,646.58
\]

4. For $ t = 10 $:
\[
A(10) = 18,254 \cdot e^{0.062 \times 10} \approx 18,254 \cdot 1.82178 = 33,258.87
\]

---

### c) Doubling Time
To find the time it takes for the investment to double, we can set $ A(t) = 2P $. This means we solve the equation:

\[
2P = P e^{rt}
\]

Simplifying:

\[
2 = e^{rt}
\]

Taking the natural logarithm of both sides:

\[
\ln(2) = rt
\]

So, the doubling time $ t $ is:

\[
t = \frac{\ln(2)}{r} = \frac{\ln(2)}{0.062} \approx \frac{0.6931}{0.062} \approx 11.18 \, \text{years}
\]

---

Let me know if you'd like any further explanations!

---

Here are 5 related questions for further practice:
1. How would the function change if the interest rate was 7.5% instead of 6.2%?
2. What happens to the doubling time if the interest rate decreases to 4%?
3. Calculate the amount after 15 years with continuous compounding.
4. What would be the total interest earned after 10 years?
5. How would the formula change for monthly compounding instead of continuous?

**Tip:** When dealing with continuously compounded interest, the exponential function $ e^{rt} $ plays a crucial role, with $ r $ being the key to understanding how fast your money grows!

Suppose that $18,254 is invested at an interest rate of 6.2% per year, compounded continuously. 
 a) Find the exponential function that describes the amount in the account after time t, in years. 
 b) What is the balance after 1 year? 2 years? 5 years? 10 years? 
 c) What is the doubling time?

Learn how to calculate the exponential growth of an investment of $18,254 at an annual interest rate of 6.2%, compounded continuously. Find the balance after 1, 2, 5, and 10 years, and determine the doubling time.

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