1. Stating the problem: We need to find the compound interest on a principal of 4000 borrowed at an annual interest rate of 6% compounded daily for 2 years. 2. Identify the variables: - Principal $P = 4000$ - Annual interest rate $r = 0.06$ - Time $t = 2$ years - Compounding frequency $n = 365$ times per year (daily compounding) 3. Use the compound interest formula: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where $A$ is the amount after interest. 4. Substitute the values: $$A = 4000 \left(1 + \frac{0.06}{365}\right)^{365 \times 2}$$ 5. Calculate the inside term: $$1 + \frac{0.06}{365} = 1 + 0.00016438356 = 1.00016438356$$ 6. Calculate the exponent: $$365 \times 2 = 730$$ 7. Calculate the amount: $$A = 4000 \times (1.00016438356)^{730}$$ 8. Calculate the power term: $$ (1.00016438356)^{730} \approx 1.12749$$ 9. Calculate the final amount: $$A = 4000 \times 1.12749 = 4509.96$$ 10. Calculate the compound interest: $$\text{Interest} = A - P = 4509.96 - 4000 = 509.96$$ Final answer: The compound interest for 2 years compounded daily is approximately $509.96$.