Question

Suppose that $3000 is placed in a savings account at an annual rate of 7.2%, compounded monthly. Assuming that no withdrawals are made, how long will it

take for the account to grow to $4587?
Do not round any intermediate computations, and round your answer to the nearest hundredth.

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4587\\ P=\textit{original amount deposited}\dotfill &\$3000\\ r=rate\to 7.2\%\to (7.2)/(100)\dotfill &0.072\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}`

`4587=3000\left(1+(0.072)/(12)\right)^(12\cdot t) \implies \cfrac{4587}{3000}=(1.006)^(12t)\implies \cfrac{1529}{1000}=1.006^(12t) \\\\\\ \log\left( \cfrac{1529}{1000} \right)=\log(1.006^(12t))\implies \log\left( \cfrac{1529}{1000} \right)=t\log(1.006^(12)) \\\\\\ \cfrac{\log\left( (1529)/(1000) \right)}{\log(1.006^(12))}=t\implies 5.92\approx t\qquad \textit{about 5 years and 11 months}`