Question

Camden is going to invest $600 and leave it in an account for 12 years. Assuming the

interest is compounded continuously, what interest rate, to the nearest hundredth of
a percent, would be required in order for Camden to end up with $920?

Answer 1

`~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$920\\ P=\textit{original amount deposited}\dotfill & \$600\\ r=rate\to r\%\to (r)/(100)\\ t=years\dotfill &12 \end{cases}`

`920=600e^{(r)/(100)\cdot 12}\implies \cfrac{920}{600}=e^{(3r)/(25)}\implies \cfrac{23}{15}=e^{(3r)/(25)}\implies \log_e\left( \cfrac{23}{15} \right)=\log_e\left( e^{(3r)/(25)} \right) \\\\\\ \ln\left( \cfrac{23}{15} \right)=\cfrac{3r}{25}\implies 25\cdot \ln\left( \cfrac{23}{15} \right)=3r\implies \cfrac{25\cdot \ln\left( (23)/(15) \right)}{3}=r \\\\\\ 3.5620\approx r\implies \stackrel{\%}{3.56}\approx r`