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
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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
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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: \[ A = 2000 \left( 1 + 0.015 \right)^{20} \] \[ A = 2000 (1.015)^{20} \]
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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
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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
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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: \[ A = 3000 \left( 1 + 0.0041667 \right)^{96} \] \[ A = 3000 (1.0041667)^{96} \]
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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.