Question

If $2000 becomes $8000 in 3 years
(compounded monthly), what is the interest rate?

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 8000\\ P=\textit{original amount deposited}\dotfill &\$2000\\ r=rate\to r\%\to (r)/(100)\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &3 \end{cases}`

`8000 = 2000\left(1+( ~~ (r)/(100) ~~ )/(12)\right)^(12\cdot 3)\implies \cfrac{8000}{2000}=\left( 1+\cfrac{r}{1200} \right)^(36) \\\\\\ 4=\left( \cfrac{1200+r}{1200} \right)^(36)\implies \sqrt[36]{4}=\cfrac{1200+r}{1200} \\\\\\ 1200\sqrt[36]{4}=1200+r\implies 1200\sqrt[36]{4}-1200=r\implies \stackrel{ \% }{47.11}\approx r`