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 present value of an annuity formula for a loan: \[ \text{PV} = \text{PMT} \cdot \frac{1 - (1+i)^{-n}}{i} \] Here, - \(\text{PV} = \$11000\), - \(\text{PMT} = \$100\), - The monthly interest rate is \( i = \frac{0.09}{12} = 0.0075 \), - and \( n \) is the number of months. Substitute the values in: \[ 11000 = 100 \cdot \frac{1 - (1.0075)^{-n}}{0.0075} \] Multiply both sides by \(0.0075\): \[ 11000 \times 0.0075 = 100 \left( 1 - (1.0075)^{-n} \right) \] \[ 82.5 = 100 \left( 1 - (1.0075)^{-n} \right) \] Divide both sides by \(100\): \[ 0.825 = 1 - (1.0075)^{-n} \] Solve for \((1.0075)^{-n}\): \[ (1.0075)^{-n} = 1 - 0.825 = 0.175 \] Take the natural logarithm of both sides: \[ \ln \left((1.0075)^{-n}\right) = \ln (0.175) \] Utilize the power rule for logarithms: \[ -n \ln(1.0075) = \ln(0.175) \] Now, solve for \( n \): \[ n = -\frac{\ln(0.175)}{\ln(1.0075)} \] Calculating the logarithms (using approximate values): \[ \ln(0.175) \approx -1.743 \quad \text{and} \quad \ln(1.0075) \approx 0.00747 \] Thus, \[ n \approx -\frac{-1.743}{0.00747} \approx \frac{1.743}{0.00747} \approx 233.3 \] Since the number of months must be a whole number and we round up, John will pay off the loan in: \[ n = 234 \text{ months.} \]