# Step-by-Step Solution We are to find the monthly payment necessary to amortize a $184,600 loan at 4.8% annual interest over 30 years. --- ## 1. **Identify the Amortization Formula** The monthly payment \( P \) for a loan with: - Principal \( L \) - Monthly interest rate \( r \) - Number of payments \( n \) is: \[ P = \frac{L \cdot r}{1 - (1 + r)^{-n}} \] --- ## 2. **List the Known Values** - Loan amount (\( L \)): $184,600 - Annual interest rate: 4.8% - Number of years: 30 ### Convert to Monthly Terms: - Monthly interest rate (\( r \)): \( \frac{4.8\%}{12} = \frac{.048}{12} = .004 \) - Number of monthly payments (\( n \)): \( 30 \times 12 = 360 \) --- ## 3. **Plug in the Values** \[ P = \frac{184{,}600 \times .004}{1 - (1 + .004)^{-360}} \] --- ## 4. **Calculate Denominator** \[ (1 + .004)^{-360} = (1.004)^{-360} \] First, calculate \( 1.004^{360} \): - \( \ln(1.004) \approx .003992 \) - \( 360 \times .003992 = 1.43712 \) - \( e^{1.43712} \approx 4.208 \) - Therefore, \( (1.004)^{360} \approx 4.208 \) - So, \( (1.004)^{-360} \approx \frac{1}{4.208} \approx .2376 \) Now, denominator: \[ 1 - .2376 = .7624 \] --- ## 5. **Calculate Numerator** \[ 184,600 \times .004 = 738.40 \] --- ## 6. **Compute Monthly Payment** \[ P = \frac{738.40}{.7624} \approx 968.48 \] --- ## 7. **Final Answer** **The monthly house payment is:** \[ \boxed{\$968.48} \] --- ## **Summary Table** | Description | Value | |--------------------------|------------| | Principal | $184,600 | | Annual Interest Rate | 4.8% | | Monthly Interest Rate | .004 | | Number of Payments | 360 | | Monthly Payment | $968.48 | --- **Final Answer:** **$968.48 per month** (rounded to the nearest cent)