Question

$1000 were deposited into an account

with a 9% interest rate,
compounded quarterly. How many years
was it in the bank if the current amount is
$2662?
t = [?] years
A
Round to the nearest year.

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 2662\\ P=\textit{original amount deposited}\dotfill &\$1000\\ r=rate\to 9\%\to (9)/(100)\dotfill &0.09\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years\dotfill &t \end{cases}`

`2662 = 1000\left(1+(0.09)/(4)\right)^(4\cdot t) \implies \cfrac{2662}{1000}=1.0225^(4t)\implies \cfrac{1331}{500}=1.0225^(4t) \\\\\\ \log\left(\cfrac{1331}{500} \right)=\log\left( 1.0225^(4t) \right)\implies \log\left(\cfrac{1331}{500} \right)=t\log\left( 1.0225^(4) \right) \\\\\\ \cfrac{ ~~ \log\left((1331)/(500) \right) ~~ }{\log( 1.0225^(4))}\implies 11\approx t`