Question

A principal of 1400 is invested at 5. 5, compounded annually. How many years will it take to accumulate 3000 or more

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 3000\\ P=\textit{original amount deposited}\dotfill &\$1400\\ r=rate\to 5.5\%\to (5.5)/(100)\dotfill &0.055\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &t \end{cases}`

`3000 = 1400\left(1+(0.055)/(1)\right)^(1\cdot t) \implies \cfrac{3000}{1400}=(1.055)^t\implies \cfrac{15}{7}=1.055^t \\\\\\ \log\left( \cfrac{15}{7} \right)=\log(1.055^t)\implies \log\left( \cfrac{15}{7} \right)=t\log(1.055) \\\\\\ \cfrac{ ~~ \log\left( (15)/(7) \right) ~~ }{\log(1.055)}=t\implies 14.23\approx t\qquad \textit{about 14 years, 2 and a half months }`