1. **State the problem:** Jeriel and Amelia each invest 72000 in accounts with different interest rates and compounding methods. We want to find how much longer it takes for Amelia's money to double compared to Jeriel's. 2. **Formulas and rules:** - For continuous compounding, the amount after time $t$ is given by $$A = P e^{rt}$$ where $P$ is principal, $r$ is the interest rate as a decimal, and $t$ is time in years. - For daily compounding, the amount after time $t$ is $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where $n=365$ (days per year). - To find doubling time, set $A = 2P$ and solve for $t$. 3. **Calculate Jeriel's doubling time:** - Interest rate: $7\frac{1}{4}\% = 7.25\% = 0.0725$ - Equation: $$2P = P e^{0.0725 t} \implies 2 = e^{0.0725 t}$$ - Taking natural log: $$\ln(2) = 0.0725 t \implies t = \frac{\ln(2)}{0.0725}$$ - Calculate: $$t = \frac{0.693147}{0.0725} \approx 9.56 \text{ years}$$ 4. **Calculate Amelia's doubling time:** - Interest rate: $6\frac{3}{4}\% = 6.75\% = 0.0675$ - Daily compounding: $$2P = P \left(1 + \frac{0.0675}{365}\right)^{365 t} \implies 2 = \left(1 + \frac{0.0675}{365}\right)^{365 t}$$ - Take natural log: $$\ln(2) = 365 t \ln\left(1 + \frac{0.0675}{365}\right)$$ - Solve for $t$: $$t = \frac{\ln(2)}{365 \ln\left(1 + \frac{0.0675}{365}\right)}$$ - Approximate inside log: $$1 + \frac{0.0675}{365} \approx 1.00018493$$ - Calculate log: $$\ln(1.00018493) \approx 0.00018493$$ - Calculate denominator: $$365 \times 0.00018493 \approx 0.0675$$ - Calculate $t$: $$t = \frac{0.693147}{0.0675} \approx 10.27 \text{ years}$$ 5. **Find difference:** $$10.27 - 9.56 = 0.71 \text{ years}$$ **Answer:** It takes Amelia approximately **0.71 years longer** to double her money than Jeriel.