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 

​$15,924.4515,924.45.

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

Part 4

The balance after 5 years is 

​$19,007.9219,007.92.

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

Part 5

The balance after 10 years is 

​$25,530.04

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

Part 6

​c) The doubling time is 

enter your response here 

years.

​(Simplify your answers. Round to one decimal place as​ needed.)

Let's break down the problem step-by-step:

### Part (a)
The general formula for continuous compounding is:

\[
P(t) = P_0 e^{rt}
\]

where:
- $ P(t) $ is the amount after time $ t $ years,
- $ P_0 $ is the initial principal,
- $ r $ is the annual interest rate (as a decimal), and
- $ t $ is the time in years.

Given:
- $ P_0 = 14,152 $
- $ r = 5.95\% = 0.059 $

The exponential function describing the amount in the account after time $ t $ is:

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

### Part (b)
We need to calculate the balance after 1, 2, 5, and 10 years using the formula $ P(t) = 14,152 e^{0.059t} $.

1. **After 1 year**:
   \[
   P(1) = 14,152 e^{0.059 \times 1} = 14,152 \times e^{0.059}
   \]

2. **After 2 years**:
   \[
   P(2) = 14,152 e^{0.059 \times 2} = 14,152 \times e^{0.118}
   \]

3. **After 5 years**:
   \[
   P(5) = 14,152 e^{0.059 \times 5} = 14,152 \times e^{0.295}
   \]

4. **After 10 years**:
   \[
   P(10) = 14,152 e^{0.059 \times 10} = 14,152 \times e^{0.59}
   \]

Let's calculate these values.

### Part (c)
The doubling time $ T $ for continuous compounding can be found using the formula:

\[
T = \frac{\ln(2)}{r}
\]

where:
- $ \ln(2) \approx 0.693 $
- $ r = 0.059 $

Now, I will compute all these values.### Results

**Part (b):** The balance after:
1. **1 year**: $15,012.09
2. **2 years**: $15,924.45
3. **5 years**: $19,007.92
4. **10 years**: $25,530.04

**Part (c):** The doubling time is approximately **11.7 years**.

If you have any questions or need further details, feel free to ask!

### Related Questions:
1. How would the formula change if the interest rate were compounded quarterly instead of continuously?
2. What is the formula to calculate the amount after $ n $ years if the interest is compounded monthly?
3. How would you find the time required to triple the investment with the same interest rate?
4. What is the effective annual rate corresponding to a continuous compounding rate of 5.9%?
5. How does the doubling time change if the interest rate is increased to 6.5%?

**Tip:** For continuous compounding, the formula $ P(t) = P_0 e^{rt} $ can be used to easily find the amount at any time $ t $ without needing complex periodic calculations.

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?

Discover how to calculate the exponential growth of a $14,152 investment at a 5.9% continuous interest rate. This guide provides detailed calculations for balances after 1, 2, 5, and 10 years, as well as the formula to determine the doubling time. Suitable for high school students learning about continuous compounding and exponential growth.

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