1. **State the problem:** We have a loan of 34000 with an annual interest rate of 4.25% compounded annually. We want to find the smallest whole number of years $t$ such that the amount due is at least 63000. 2. **Formula used:** The compound interest formula is $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount after time $t$, - $P$ is the principal (initial 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 annually, $n=1$, so the formula simplifies to $$A = P(1 + r)^t$$ 3. **Plug in known values:** $$63000 \leq 34000(1 + 0.0425)^t$$ 4. **Isolate $(1 + 0.0425)^t$:** $$\frac{63000}{34000} \leq (1.0425)^t$$ $$1.852941176 \leq (1.0425)^t$$ 5. **Take natural logarithm on both sides:** $$\ln(1.852941176) \leq \ln((1.0425)^t)$$ $$\ln(1.852941176) \leq t \ln(1.0425)$$ 6. **Solve for $t$:** $$t \geq \frac{\ln(1.852941176)}{\ln(1.0425)}$$ Calculate the values: $$\ln(1.852941176) \approx 0.6162$$ $$\ln(1.0425) \approx 0.0416$$ $$t \geq \frac{0.6162}{0.0416} \approx 14.81$$ 7. **Interpretation:** Since $t$ must be a whole number and the amount must reach or exceed 63000, we round up to the next whole number: $$t = 15$$ **Final answer:** The amount due will reach 63000 or more after **15 years**.