Question

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

interest is compounded continuously, what interest rate, to the nearest hundredth of
a percent, would be required in order for Skylar to end up with $105,000?

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 (r)/(100)\\ t=years\dotfill &6 \end{cases}`

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