1. **Problem statement:** We have three compound interest problems to solve. 2. **Formula for compound interest:** $$A = P(1 + r)^t$$ where $A$ is the amount after $t$ years, $P$ is the principal, $r$ is the annual interest rate (as a decimal), and $t$ is the time in years. --- ### (a) Find the interest rate $r$ given $P=7500$, $A=8195.45$, $t=3$. 3. Substitute known values: $$8195.45 = 7500(1 + r)^3$$ 4. Divide both sides by 7500: $$\frac{8195.45}{7500} = (1 + r)^3$$ 5. Simplify fraction: $$\frac{\cancel{8195.45}}{\cancel{7500}} = 1.0927267 = (1 + r)^3$$ 6. Take the cube root of both sides: $$1 + r = \sqrt[3]{1.0927267}$$ 7. Calculate cube root: $$1 + r \approx 1.0300$$ 8. Solve for $r$: $$r = 1.0300 - 1 = 0.0300 = 3.00\%$$ --- ### (b) Find principal $P$ given $A=5000$, $r=0.04$, $t=5$. 9. Use formula: $$5000 = P(1 + 0.04)^5$$ 10. Calculate $(1.04)^5$: $$1.04^5 = 1.2166529$$ 11. Divide both sides by $1.2166529$: $$P = \frac{5000}{1.2166529}$$ 12. Simplify fraction: $$P \approx 4109.14$$ --- ### (c) Find time $t$ given $P=10000$, $A=30000$, $r=0.01$. 13. Use formula: $$30000 = 10000(1 + 0.01)^t$$ 14. Divide both sides by 10000: $$3 = (1.01)^t$$ 15. Take natural logarithm of both sides: $$\ln(3) = t \ln(1.01)$$ 16. Solve for $t$: $$t = \frac{\ln(3)}{\ln(1.01)}$$ 17. Calculate values: $$\ln(3) \approx 1.0986, \quad \ln(1.01) \approx 0.00995$$ 18. Compute $t$: $$t \approx \frac{1.0986}{0.00995} = 110.4$$ years --- **Final answers:** - (a) Interest rate $r = 3.00\%$ - (b) Principal $P \approx 4109.14$ - (c) Time $t \approx 110.4$ years