1. **Problem Statement:** A loan of 10000 is amortized by equal annual payments for 30 years at an effective annual interest rate of 9%. We need to find the year in which the interest portion of the payment is most nearly equal to one-half of the payment. 2. **Formula for the annual payment (A):** $$A = P \times \frac{i(1+i)^n}{(1+i)^n - 1}$$ where: - $P = 10000$ (principal) - $i = 0.09$ (interest rate) - $n = 30$ (number of years) 3. **Calculate the annual payment:** $$A = 10000 \times \frac{0.09(1.09)^{30}}{(1.09)^{30} - 1}$$ Calculate $(1.09)^{30}$: $$ (1.09)^{30} \approx 13.2687 $$ Then: $$A = 10000 \times \frac{0.09 \times 13.2687}{13.2687 - 1} = 10000 \times \frac{1.1942}{12.2687} \approx 10000 \times 0.0973 = 973.0$$ 4. **Interest portion in year $k$:** The outstanding balance after $k-1$ payments is: $$B_{k-1} = P(1+i)^{k-1} - A \times \frac{(1+i)^{k-1} - 1}{i}$$ Interest portion in year $k$ is: $$I_k = i \times B_{k-1}$$ 5. **Condition for interest portion to be half the payment:** $$I_k = \frac{A}{2}$$ Substitute $I_k$: $$i \times B_{k-1} = \frac{A}{2}$$ $$B_{k-1} = \frac{A}{2i}$$ 6. **Set $B_{k-1}$ equal to $\frac{A}{2i}$ and solve for $k$:** $$P(1+i)^{k-1} - A \times \frac{(1+i)^{k-1} - 1}{i} = \frac{A}{2i}$$ Rearranged: $$P(1+i)^{k-1} - \frac{A}{i}(1+i)^{k-1} + \frac{A}{i} = \frac{A}{2i}$$ $$\left(P - \frac{A}{i}\right)(1+i)^{k-1} = \frac{A}{2i} - \frac{A}{i} = -\frac{A}{2i}$$ 7. **Calculate constants:** $$P - \frac{A}{i} = 10000 - \frac{973}{0.09} = 10000 - 10811.11 = -811.11$$ 8. **Solve for $(1+i)^{k-1}$:** $$-811.11 \times (1.09)^{k-1} = -\frac{973}{2 \times 0.09} = -\frac{973}{0.18} = -5405.56$$ $$ (1.09)^{k-1} = \frac{5405.56}{811.11} = 6.67$$ 9. **Take natural logarithm:** $$ (k-1) \ln(1.09) = \ln(6.67)$$ $$ k-1 = \frac{\ln(6.67)}{\ln(1.09)} = \frac{1.897}{0.0862} = 22.0$$ 10. **Final answer:** $$k = 23$$ **The interest portion of the payment is most nearly equal to one-half of the payment in year 23.**