Question

Aiden invested $43,000 in an account paying an interest rate of 9 1/4 % compounded monthly. Hailey invested $43,000 in an account paying an interest rate of 8 7/8% compounded continuously. To the nearest dollar, how much money would Aiden have in his account when Hailey's money has doubled in value?

Answer 1

first off let's change the mixed fractions to improper fractions, and let's Hailey's account first.

`\stackrel{mixed}{9(1)/(4)}\implies \cfrac{9\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{37}{4}} ~\hfill \stackrel{mixed}{8(7)/(8)}\implies \cfrac{8\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{71}{8}} \\\\[-0.35em] ~\dotfill`

`~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill &\stackrel{43000(2)}{\$86000}\\ P=\textit{original amount deposited}\dotfill & \$43000\\ r=rate\to (71)/(8)\%\to (~~ (71)/(8)~~)/(100)\dotfill &0.08875\\ t=years \end{cases} \\\\\\ 86000=43000e^(0.08875\cdot t)\implies \cfrac{86000}{43000}=e^(0.08875t)\implies 2=e^(0.08875t)`

`\ln(2)=\ln(e^(0.08875t))\implies \log_e(2)=\log_e(e^(0.08875t))\implies \ln(2)=0.08875t \\\\\\ \cfrac{\ln(2)}{0.08875}=t\implies 7.81\approx t`

ok, now we know how long it takes for Hailey's money to double, how much money does Aiden have by then?

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$43000\\ r=rate\to (37)/(4)\%\to (~~ (37)/(4)~~)/(100)\dotfill &0.0925\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &(\ln(2))/(0.08875) \end{cases}`

`A=43000\left(1+(0.0925)/(12)\right)^{12\cdot (\ln(2))/(0.08875)}\implies A=43000\left( (4837)/(4800) \right)^{(12\ln(2))/(0.08875)}\implies \boxed{A\approx 88311}`

notice, in Hailey's amount we used the logarithmic value for "t", just to avoid any rounding issues.