4. An initial amount of \( \$ 2300 \) is deposited in an account. In 5 years, the investment grows to \( \$ 4237.61 \). a. Find a formula for the value of the investment as a function of time, assuming exponential growth. Report any calculated values correct to two decimal places. Your formula should begin with " \( y= \) ". b. What annual interest rate does the account pay?

To solve this problem, we will follow these steps: ### Part a: Find the formula for the value of the investment 1. **Identify the known values:** - Initial amount (principal) \( P = 2300 \) - Amount after 5 years \( A = 4237.61 \) - Time \( t = 5 \) years 2. **Use the formula for exponential growth:** The formula for exponential growth is given by: \[ A = P e^{rt} \] where: - \( A \) is the amount of money accumulated after time \( t \), - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate (as a decimal), - \( t \) is the time in years, - \( e \) is the base of the natural logarithm. 3. **Substituting the known values into the formula:** \[ 4237.61 = 2300 e^{5r} \] 4. **Solve for \( e^{5r} \):** \[ e^{5r} = \frac{4237.61}{2300} \] 5. **Calculate \( \frac{4237.61}{2300} \):** Let's calculate this value. Now, I will perform the calculation for \( \frac{4237.61}{2300} \). Calculate the value by following steps: - step0: Calculate: \(\frac{4237.61}{2300}\) - step1: Convert the expressions: \(\frac{\frac{423761}{100}}{2300}\) - step2: Multiply by the reciprocal: \(\frac{423761}{100}\times \frac{1}{2300}\) - step3: Multiply the fractions: \(\frac{423761}{100\times 2300}\) - step4: Multiply: \(\frac{423761}{230000}\) The calculation gives us: \[ e^{5r} \approx 1.842439 \] 6. **Take the natural logarithm of both sides:** \[ 5r = \ln(1.842439) \] 7. **Calculate \( \ln(1.842439) \):** Now, I will calculate the natural logarithm of \( 1.842439 \). Calculate the value by following steps: - step0: Calculate: \(\ln{\left(1.842439\right)}\) - step1: Convert the expressions: \(\ln{\left(\frac{1842439}{1000000}\right)}\) - step2: Rewrite the expression: \(\ln{\left(1842439\right)}-6\ln{\left(10\right)}\) The calculation gives us: \[ 5r = \ln(1.842439) \approx 0.620576 \] 8. **Solve for \( r \):** \[ r = \frac{0.620576}{5} \approx 0.1241152 \] 9. **Convert \( r \) to a percentage:** \[ r \approx 0.1241152 \times 100 \approx 12.41\% \] 10. **Write the formula for the value of the investment:** Now we can express the value of the investment as a function of time \( t \): \[ y = 2300 e^{0.1241152t} \] ### Part b: What annual interest rate does the account pay? The annual interest rate is approximately \( 12.41\% \). ### Final Answers: a. The formula for the value of the investment is: \[ y = 2300 e^{0.1241152t} \] b. The annual interest rate is approximately \( 12.41\% \).