Question

Dale has $2,000 to invest. He has a goal to have $5,200 in this investment in 10 years. At what annual rate compounded continuously will Dale reach his goal?​

Answer 1

`~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 5200\\ P=\textit{original amount deposited}\dotfill & \$2000\\ r=rate\to r\%\to (r)/(100)\\ t=years\dotfill &10 \end{cases}`

`5200 = 2000e^{(r)/(100)\cdot 10} \implies \cfrac{5200}{2000}=e^{(r)/(10)}\implies \cfrac{13}{5}=e^{(r)/(10)} \\\\\\ \log_e\left( \cfrac{13}{5} \right)=\log_e\left( e^{(r)/(10)} \right)\implies \log_e\left( \cfrac{13}{5} \right)=\cfrac{r}{10} \\\\\\ \ln\left( \cfrac{13}{5} \right)=\cfrac{r}{10}\implies 10\ln\left( \cfrac{13}{5} \right)=r\implies \stackrel{ \% }{9.56}\approx r`