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. $$ \mathrm{N}:=\quad 0 $$

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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:

$$
n = 234 \text{ months.}
$$
John will pay off the \$11,000 loan after 234 months.

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