Question

when the principle "P" earns an annual interest rate "r" compounded "n" times per year, the amount "A" after "t" years is A=P(1+r/n)^nt. Rounding the answer to the nearest tenth, how long will it take $1200 to grow to $6000 at 4% annual interest compounded monthly?

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 6000\\ P=\textit{original amount deposited}\dotfill &\$1200\\ r=rate\to 4\%\to (4)/(100)\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &t \end{cases}`

`6000 = 1200\left(1+(0.04)/(12)\right)^(12\cdot t) \implies \cfrac{6000}{1200}=\left( \cfrac{301}{300} \right)^(12t)\implies 5=\left( \cfrac{301}{300} \right)^(12t) \\\\\\ \log(5)=\log\left[ \left( \cfrac{301}{300} \right)^(12t) \right]\implies \log(5)=t\log\left[ \left( \cfrac{301}{300} \right)^(12) \right] \\\\\\ \cfrac{\log(5)}{ ~~ \log\left[ \left( (301)/(300) \right)^(12) \right] ~~ }=t\implies 40.3\approx t`