1. **State the problem:** Jaxon and London each invest 77000 in accounts with different interest rates and compounding frequencies. We want to find how much longer it takes for London's money to double compared to Jaxon's, rounded to the nearest hundredth of a year. 2. **Formula for compound interest doubling time:** The amount after time $t$ years with principal $P$, rate $r$ (as decimal), compounded $n$ times per year is: $$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$ To find doubling time $t$ when $A = 2P$: $$ 2 = \left(1 + \frac{r}{n}\right)^{nt} $$ Taking natural logarithm: $$ \ln(2) = nt \ln\left(1 + \frac{r}{n}\right) $$ Solving for $t$: $$ t = \frac{\ln(2)}{n \ln\left(1 + \frac{r}{n}\right)} $$ 3. **Convert interest rates to decimals:** - Jaxon's rate: 6 1/2 % = 6.5% = 0.065 - London's rate: 6 1/8 % = 6.125% = 0.06125 4. **Calculate doubling time for Jaxon:** - $r = 0.065$ - $n = 4$ (quarterly) $$ t_J = \frac{\ln(2)}{4 \ln\left(1 + \frac{0.065}{4}\right)} $$ Calculate inside logarithm: $$ 1 + \frac{0.065}{4} = 1 + 0.01625 = 1.01625 $$ Calculate $t_J$: $$ t_J = \frac{0.693147}{4 \times \ln(1.01625)} = \frac{0.693147}{4 \times 0.016121} = \frac{0.693147}{0.064484} \approx 10.75 \text{ years} $$ 5. **Calculate doubling time for London:** - $r = 0.06125$ - $n = 12$ (monthly) $$ t_L = \frac{\ln(2)}{12 \ln\left(1 + \frac{0.06125}{12}\right)} $$ Calculate inside logarithm: $$ 1 + \frac{0.06125}{12} = 1 + 0.005104167 = 1.005104167 $$ Calculate $t_L$: $$ t_L = \frac{0.693147}{12 \times \ln(1.005104167)} = \frac{0.693147}{12 \times 0.005091} = \frac{0.693147}{0.061092} \approx 11.35 \text{ years} $$ 6. **Find how much longer it takes for London:** $$ t_L - t_J = 11.35 - 10.75 = 0.60 \text{ years} $$ **Final answer:** It takes approximately **0.60 years** longer for London's money to double than Jaxon's.