Respuesta :
first off let's change the mixed fractions to improper fractions, and let's Hailey's account first.
[tex]\stackrel{mixed}{9\frac{1}{4}}\implies \cfrac{9\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{37}{4}} ~\hfill \stackrel{mixed}{8\frac{7}{8}}\implies \cfrac{8\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{71}{8}} \\\\[-0.35em] ~\dotfill[/tex]
[tex]~~~~~~ \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 \frac{71}{8}\%\to \frac{~~ \frac{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}[/tex]
[tex]\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[/tex]
ok, now we know how long it takes for Hailey's money to double, how much money does Aiden have by then?
[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$43000\\ r=rate\to \frac{37}{4}\%\to \frac{~~ \frac{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 &\frac{\ln(2)}{0.08875} \end{cases}[/tex]
[tex]A=43000\left(1+\frac{0.0925}{12}\right)^{12\cdot \frac{\ln(2)}{0.08875}}\implies A=43000\left( \frac{4837}{4800} \right)^{\frac{12\ln(2)}{0.08875}}\implies \boxed{A\approx 88311}[/tex]
notice, in Hailey's amount we used the logarithmic value for "t", just to avoid any rounding issues.
