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The problem from the image involves calculating the time to pay off a loan, given the following details:

- Principal: $3,000
- Annual interest rate: 13%, compounded monthly
- Monthly payment: $101
- Round the final answer to two decimal places.

We will use the formula for the time to pay off a loan under monthly compounding:

\[
M = P \frac{r(1 + r)^n}{(1 + r)^n - 1}
\]

Where:
- $ M $ is the monthly payment ($101),
- $ P $ is the loan amount ($3,000),
- $ r $ is the monthly interest rate, which is $ \frac{13\%}{12} = \frac{0.13}{12} \approx 0.010833 $,
- $ n $ is the number of months (the unknown we want to solve for).

Rearranging the formula to solve for $ n $ involves taking the natural logarithm to isolate $ n $. I will perform this calculation to determine how long it will take to pay off the loan.

Let me calculate that.It will take approximately **36.04 months** to pay off the loan.

Would you like me to calculate the total interest paid over the life of the loan or explain any part of the calculation?

Here are 5 related questions you might find useful:
1. What is the formula for calculating monthly loan payments?
2. How does the interest rate affect the loan payoff time?
3. How is monthly compound interest different from simple interest?
4. What happens if you increase your monthly payments by $20?
5. How can you calculate the interest portion of each monthly payment?

**Tip**: Always round your loan payments to two decimal places to avoid under- or overpaying over time.

You've just bought a new sectional couch for $3,000, and the store you bought it from offers to let you finance the entire purchase at an annual rate of 13% compounded monthly. If you take the financing and make monthly payments of $101, how long will it take to pay the loan off? How much will you pay in interest over the life of the loan?

Learn how to calculate the time to pay off a $3,000 loan with monthly payments of $101 at a 13% annual interest rate. Understand key financial concepts such as compound interest and loan amortization. Ideal for students in Grades 11-12.

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