1. **State the problem:** Benjamin and Zachary each invest 99,000 in accounts with different interest rates and compounding methods. We want to find how much longer it takes for Benjamin's money to triple compared to Zachary's. 2. **Formulas and rules:** - For continuous compounding, the amount after time $t$ is given by $$A = P e^{rt}$$ where $P$ is the principal, $r$ is the interest rate (as a decimal), and $t$ is time in years. - For quarterly compounding, the amount after time $t$ is $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where $n=4$ (quarters per year). - We want to find $t$ such that $A = 3P$ (tripling the investment). 3. **Benjamin's investment (continuous compounding):** - Interest rate: 7 1/2 % = 7.5% = 0.075 - Equation: $$3P = P e^{0.075 t_B}$$ - Divide both sides by $P$: $$3 = e^{0.075 t_B}$$ - Take natural logarithm: $$\ln(3) = 0.075 t_B$$ - Solve for $t_B$: $$t_B = \frac{\ln(3)}{0.075}$$ 4. **Zachary's investment (quarterly compounding):** - Interest rate: 7 7/8 % = 7.875% = 0.07875 - Quarterly compounding: $n=4$ - Equation: $$3P = P \left(1 + \frac{0.07875}{4}\right)^{4 t_Z}$$ - Divide both sides by $P$: $$3 = \left(1 + 0.0196875\right)^{4 t_Z} = (1.0196875)^{4 t_Z}$$ - Take natural logarithm: $$\ln(3) = 4 t_Z \ln(1.0196875)$$ - Solve for $t_Z$: $$t_Z = \frac{\ln(3)}{4 \ln(1.0196875)}$$ 5. **Calculate values:** - $\ln(3) \approx 1.098612289$ - $\ln(1.0196875) \approx 0.019494$ 6. **Compute times:** - $$t_B = \frac{1.098612289}{0.075} \approx 14.648$$ years - $$t_Z = \frac{1.098612289}{4 \times 0.019494} = \frac{1.098612289}{0.077976} \approx 14.086$$ years 7. **Find difference:** - $$t_B - t_Z = 14.648 - 14.086 = 0.562$$ years 8. **Round to nearest hundredth:** - Difference is approximately **0.56 years**. **Final answer:** It takes Benjamin approximately 0.56 years longer to triple his money than Zachary.