Question

Skylar is going to invest $69,000 and leave it in an account for 6 years. Assuming the

Answer 1

$~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \ 105000\\ P=\textit{original amount deposited}\dotfill & \ 69000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &6 \end{cases}$

$105000=69000e^{\frac{r}{100}\cdot 6}\implies \cfrac{105000}{69000}=e^{\frac{3r}{50}}\implies \cfrac{7}{46}=e^{\frac{3r}{50}} \\\\\\ \log_e\left( \cfrac{35}{23} \right)=\log_e\left( e^{\frac{3r}{50}} \right)\implies \ln\left( \cfrac{35}{23} \right)=\cfrac{3r}{50}\implies 50\cdot \ln\left( \cfrac{35}{23} \right)=3r \\\\\\ \cfrac{50\cdot \ln\left( \frac{35}{23} \right)}{3}=r\implies 6.9976\approx r\implies \stackrel{\%}{7.00}\approx r$