Suppose that 

​$14 comma 15214,152 

is invested at an interest rate of 

5.95.9​% 

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?

Question content area bottom

Part 1

​a) The exponential growth function is 

​P(t)equals=14,152 e Superscript 0.059 t 14,152e 0.059t  .

​(Type exponential notation with positive exponents. Do not simplify. Use integers or decimals for any numbers in the​ equation.)

Part 2

​b) The balance after 1 year is 

​$15,012.0915,012.09.

​(Simplify your answers. Round to two decimal places as​ needed.)

Part 3

The balance after 2 years is 

​$15,924.45

​(Simplify your answers. Round to two decimal places as​ needed.)

Part 4

The balance after 5 years is 

​$enter your response here.

​(Simplify your answers. Round to two decimal places as​ needed.)

Let's break down the problem step by step:

### Part a) Exponential growth function

We are given:
- Principal amount $ P_0 = 14,152 $
- Interest rate $ r = 5.9\% = 0.059 $
- The formula for continuous compound interest is given by:
  \[
  P(t) = P_0 \cdot e^{rt}
  \]
Substituting the values into this formula:

\[
P(t) = 14,152 \cdot e^{0.059t}
\]

This is the exponential function that describes the amount in the account after $ t $ years.

### Part b) Balances after 1, 2, 5, and 10 years

We will now calculate the balance for different values of $ t $ by plugging them into the formula $ P(t) = 14,152 \cdot e^{0.059t} $.

1. **Balance after 1 year ($ t = 1 $):**
   \[
   P(1) = 14,152 \cdot e^{0.059 \cdot 1}
   \]
   Let's calculate that.

2. **Balance after 2 years ($ t = 2 $):**
   \[
   P(2) = 14,152 \cdot e^{0.059 \cdot 2}
   \]
   Similarly, let's calculate the result.

3. **Balance after 5 years ($ t = 5 $):**
   \[
   P(5) = 14,152 \cdot e^{0.059 \cdot 5}
   \]
   Now, we'll calculate this.

4. **Balance after 10 years ($ t = 10 $):**
   \[
   P(10) = 14,152 \cdot e^{0.059 \cdot 10}
   \]
   Let's compute this as well.

### Part c) Doubling time

The formula for the doubling time for continuous compounding is:

\[
T_{\text{double}} = \frac{\ln(2)}{r}
\]
Substitute $ r = 0.059 $ into this formula and compute the doubling time.

I will now calculate the required balances and doubling time.### Part b) Balances after 1, 2, 5, and 10 years:

1. **Balance after 1 year:**  
   \[
   P(1) = 15,012.09 \, \text{(rounded to two decimal places)}
   \]

2. **Balance after 2 years:**  


Suppose that $14,152 is invested at an interest rate of 5.9% 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?

Explore how to model $14,152 invested at 5.9% interest using exponential growth with continuous compounding. Learn how to find the exponential function and calculate the balance after 1, 2, 5, and 10 years, as well as determine the doubling time.

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