1. **State the problem:** We have a loan of 6000 at an interest rate of 7% compounded annually. We want to find the amount owed after 1 year and after 2 years. 2. **Formula used:** The compound interest formula is: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount owed after time $t$ - $P$ is the principal (initial loan amount) - $r$ is the annual interest rate (decimal) - $n$ is the number of times interest is compounded per year - $t$ is the number of years Since interest is compounded yearly, $n=1$. 3. **Calculate amount owed after 1 year:** Given $P=6000$, $r=0.07$, $n=1$, $t=1$: $$A = 6000 \left(1 + \frac{0.07}{1}\right)^{1 \times 1} = 6000 (1 + 0.07)^1 = 6000 (1.07)$$ 4. **Intermediate step showing cancellation (if any):** No cancellation needed here since it's multiplication. 5. **Calculate the value:** $$A = 6000 \times 1.07 = 6420$$ 6. **Calculate amount owed after 2 years:** Using the same formula with $t=2$: $$A = 6000 (1.07)^2$$ 7. **Intermediate step showing expansion:** $$A = 6000 \times 1.07 \times 1.07$$ 8. **Calculate the value:** $$A = 6000 \times 1.1449 = 6869.4$$ **Final answers:** - Amount owed after 1 year: $6420$ - Amount owed after 2 years: $6869.4$