1. The problem asks for the principal amount deposited in a savings account with 4.5% annual interest compounded monthly to reach 10,000 after 8 years. 2. Use the compound interest formula: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where - $A$ is the amount after time $t$ (10,000) - $P$ is the principal (unknown) - $r$ is the annual interest rate (4.5% or 0.045) - $n$ is the number of compounding periods per year (12 for monthly) - $t$ is the time in years (8) 3. Substitute values: $$10000 = P \left(1 + \frac{0.045}{12}\right)^{12 \times 8}$$ 4. Simplify the base inside parentheses: $$1 + \frac{0.045}{12} = 1 + 0.00375 = 1.00375$$ 5. Calculate the exponent: $$12 \times 8 = 96$$ 6. Calculate the growth factor: $$1.00375^{96} \approx 1.432364$$ 7. Solve for $P$: $$P = \frac{10000}{1.432364} \approx 6985.89$$ Thus, you need to deposit approximately $6985.89$ as the principal.