Question 3 John takes out a loan of \( \$ 11000 \) that charges \( 9 \% \) interest compounded monthly. If John makes \( \$ 100 \) monthly payments, determine how long it will take him to pay off the loan. Round your answer up. John will pay off the \( \$ 11000 \) loan after Time Value of Money Solver Enter the given values. \( \mathrm{N}:=\quad 0 \)

We start with the formula for the payment on an amortized loan: \[ \text{Payment} = \frac{P i}{1 - \left(1+i\right)^{-n}}, \] where \( P \) is the principal loan amount, \( i \) is the monthly interest rate, and \( n \) is the number of monthly payments. **Step 1. Identify the given values.** \[ P = 11000, \quad \text{Annual interest rate} = 9\% \quad \Longrightarrow \quad i = \frac{0.09}{12} = 0.0075, \] \[ \text{Monthly Payment} = 100. \] **Step 2. Write the payment formula with the values.** \[ 100 = \frac{11000 \times 0.0075}{1 - \left(1.0075\right)^{-n}}. \] **Step 3. Compute the numerator.** \[ 11000 \times 0.0075 = 82.5. \] So the equation becomes: \[ 100 = \frac{82.5}{1 - \left(1.0075\right)^{-n}}. \] **Step 4. Solve for \(1 - (1.0075)^{-n}\).** Multiply both sides by the denominator: \[ 100 \left[ 1 - \left(1.0075\right)^{-n}\right] = 82.5. \] Divide both sides by 100: \[ 1 - \left(1.0075\right)^{-n} = \frac{82.5}{100} = 0.825. \] Then, isolate the exponential term: \[ \left(1.0075\right)^{-n} = 1 - 0.825 = 0.175. \] **Step 5. Solve for \(n\) using logarithms.** Take the natural logarithm of both sides: \[ \ln\left( \left(1.0075\right)^{-n} \right) = \ln(0.175). \] Using the logarithm rule \(\ln(a^b) = b \ln(a)\), we have: \[ -n \ln(1.0075) = \ln(0.175). \] Now solve for \(n\): \[ n = -\frac{\ln(0.175)}{\ln(1.0075)}. \] **Step 6. Compute the value of \(n\).** Calculate the logarithms (using a calculator): \[ \ln(0.175) \approx -1.743668, \] \[ \ln(1.0075) \approx 0.007471. \] Thus, \[ n \approx -\frac{-1.743668}{0.007471} \approx \frac{1.743668}{0.007471} \approx 233.36. \] Since \(n\) must be a whole number of months and we round up to ensure the loan is fully repaid, \[ n = 234 \text{ months}. \] John will pay off the \(\$11000\) loan after \(234\) months.