Question
In Exercises 4 and 5, identify a, b, and x. Then write the exponential function that models the situation. Finally, use the function to answer the question.
4. You invest $2000 in an investment that earns 6% interest, compounded quarterly. How much will the investment be worth after 5 years?
5. You invest $3000 in an investment that earns 5% interest, compounded monthly. How much will the investment be worth after 8 years?
6. The formula that financial managers and accountants use to determine the value of investments that are subject to compounding interest is A = P(1 + r/n)^nt where A is the final balance, P is the initial deposit, r is the annual interest rate, n is the number of times the interest is compounded per year and t is the number of years. Redo Exercises 4 and 5 using this formula.
4. You invest $2000 in an investment that earns 6% interest, compounded quarterly. How much will the investment be worth after 5 years?
5. You invest $3000 in an investment that earns 5% interest, compounded monthly. How much will the investment be worth after 8 years?
6. The formula that financial managers and accountants use to determine the value of investments that are subject to compounding interest is A = P(1 + r/n)^nt where A is the final balance, P is the initial deposit, r is the annual interest rate, n is the number of times the interest is compounded per year and t is the number of years. Redo Exercises 4 and 5 using this formula.
Answers
GPT-4o mini
To address your request, we'll follow the steps outlined in Exercises 4 and 5 by identifying \( a \), \( b \), and \( x \) for both situations, writing the exponential functions, and then using those functions to calculate the final balances.
### Exercise 4
1. **Identifying values for Exercise 4:**
- \( P = 2000 \): Initial investment
- \( r = 0.06 \): Annual interest rate (6%)
- \( n = 4 \): Number of times interest is compounded per year (quarterly)
- \( t = 5 \): Number of years
2. **Exponential function:**
The formula for compound interest is:
\[
A = P \left( 1 + \frac{r}{n} \right)^{nt}
\]
Plugging in the identified values:
\[
A = 2000 \left( 1 + \frac{0.06}{4} \right)^{4 \times 5}
\]
Simplifying:
\[
A = 2000 \left( 1 + 0.015 \right)^{20}
\]
\[
A = 2000 (1.015)^{20}
\]
3. **Calculating the final amount:**
\[
A \approx 2000 \times 1.346855007 = 2693.71
\]
**Final amount:** After 5 years, the investment will be worth approximately **$2693.71**.
---
### Exercise 5
1. **Identifying values for Exercise 5:**
- \( P = 3000 \): Initial investment
- \( r = 0.05 \): Annual interest rate (5%)
- \( n = 12 \): Number of times interest is compounded per year (monthly)
- \( t = 8 \): Number of years
2. **Exponential function:**
\[
A = P \left( 1 + \frac{r}{n} \right)^{nt}
\]
Plugging in the identified values:
\[
A = 3000 \left( 1 + \frac{0.05}{12} \right)^{12 \times 8}
\]
Simplifying:
\[
A = 3000 \left( 1 + 0.0041667 \right)^{96}
\]
\[
A = 3000 (1.0041667)^{96}
\]
3. **Calculating the final amount:**
\[
A \approx 3000 \times 1.432364654 = 4297.09
\]
**Final amount:** After 8 years, the investment will be worth approximately **$4297.09**.
---
### Summary of Results
- In Exercise 4, the final investment value after 5 years is **$2693.71**.
- In Exercise 5, the final investment value after 8 years is **$4297.09**.
### Exercise 4
1. **Identifying values for Exercise 4:**
- \( P = 2000 \): Initial investment
- \( r = 0.06 \): Annual interest rate (6%)
- \( n = 4 \): Number of times interest is compounded per year (quarterly)
- \( t = 5 \): Number of years
2. **Exponential function:**
The formula for compound interest is:
\[
A = P \left( 1 + \frac{r}{n} \right)^{nt}
\]
Plugging in the identified values:
\[
A = 2000 \left( 1 + \frac{0.06}{4} \right)^{4 \times 5}
\]
Simplifying:
\[
A = 2000 \left( 1 + 0.015 \right)^{20}
\]
\[
A = 2000 (1.015)^{20}
\]
3. **Calculating the final amount:**
\[
A \approx 2000 \times 1.346855007 = 2693.71
\]
**Final amount:** After 5 years, the investment will be worth approximately **$2693.71**.
---
### Exercise 5
1. **Identifying values for Exercise 5:**
- \( P = 3000 \): Initial investment
- \( r = 0.05 \): Annual interest rate (5%)
- \( n = 12 \): Number of times interest is compounded per year (monthly)
- \( t = 8 \): Number of years
2. **Exponential function:**
\[
A = P \left( 1 + \frac{r}{n} \right)^{nt}
\]
Plugging in the identified values:
\[
A = 3000 \left( 1 + \frac{0.05}{12} \right)^{12 \times 8}
\]
Simplifying:
\[
A = 3000 \left( 1 + 0.0041667 \right)^{96}
\]
\[
A = 3000 (1.0041667)^{96}
\]
3. **Calculating the final amount:**
\[
A \approx 3000 \times 1.432364654 = 4297.09
\]
**Final amount:** After 8 years, the investment will be worth approximately **$4297.09**.
---
### Summary of Results
- In Exercise 4, the final investment value after 5 years is **$2693.71**.
- In Exercise 5, the final investment value after 8 years is **$4297.09**.