### Exponential and Logarithmic Functions

Finding the Initial Amount in a Word Problem on Continuous Compound Interest

Austin wants to buy a bond that will mature to \$5500 in seven years. How much should he pay for the bond now if it earns interest at a rate of 2% per year, compounded continuously?

Do not round any intermediate computations, and round your answer to the nearest cent.



Answer :

Certainly! Let's go through the problem step-by-step to determine how much Austin should pay for the bond now, given the bond will mature to [tex]$5500 in seven years with an interest rate of 2% per year, compounded continuously. ### Step-by-Step Solution 1. Identify the given information: - Future Value (\( A \)): $[/tex]5500
- Annual Interest Rate ([tex]\( r \)[/tex]): 2% or 0.02 (in decimal form)
- Time ([tex]\( t \)[/tex]): 7 years

2. Understand the formula for continuous compounding:
The formula for continuous compounding is:
[tex]\[ A = P \cdot e^{rt} \][/tex]
where:
- [tex]\( A \)[/tex] is the amount of money accumulated after n years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately equal to 2.71828).
- [tex]\( r \)[/tex] is the annual interest rate (decimal).
- [tex]\( t \)[/tex] is the time the money is invested for in years.

3. Rearrange the formula to solve for [tex]\( P \)[/tex]:
[tex]\[ P = \frac{A}{e^{rt}} \][/tex]

4. Substitute the given values into the rearranged formula:
[tex]\[ P = \frac{5500}{e^{0.02 \cdot 7}} \][/tex]

5. Evaluate the exponent:
[tex]\[ 0.02 \times 7 = 0.14 \][/tex]

6. Raise [tex]\( e \)[/tex] to the power of 0.14:
[tex]\[ e^{0.14} \approx 1.15027 \][/tex]

7. Divide [tex]$5500 by the result of \( e^{0.14} \): \[ P = \frac{5500}{1.15027} \approx 4781.47 \] ### Conclusion After evaluating all the steps, Austin should pay approximately $[/tex]4781.47 for the bond now. This ensures that the amount will grow to $5500 in seven years at an interest rate of 2% per year, compounded continuously.