Question

1. A person invests $300 in an account that earns 1.88% annual interest compounded quarterly. Find when the value of the investment reaches $800.

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 800\\ P=\textit{original amount deposited}\dotfill &\$300\\ r=rate\to 1.88\%\to (1.88)/(100)\dotfill &0.0188\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years \end{cases}`

`800 = 300\left(1+(0.0188)/(4)\right)^(4\cdot t) \implies 800=300(1.0047)^(4t)\implies \cfrac{800}{300}=1.0047^(4t) \\\\\\ \cfrac{8}{3}=1.0047^(4t)\implies \log\left( \cfrac{8}{3} \right)=\log(1.0047^(4t))\implies \log\left( \cfrac{8}{3} \right)=t\log(1.0047^(4)) \\\\\\ \cfrac{ ~~ \log\left( (8)/(3) \right) ~~ }{\log(1.0047^(4))}=t\implies 52.29\approx t`