Question

Sofia invested $360 in an account in the year 2005, and the value has been growing

exponentially at a constant rate. the value of the account reached $420 in the year
2009. determine the value of the account, to the nearest dollar, in the year 2015.

Answer 1

well, from 2005 to 2009 is 4 years, so, in 4 years that account went from 360 to 420 at a rate "r", let's find that rate.

`\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$420\\ P=\textit{initial amount}\dotfill &360\\ r=rate\to r\%\to (r)/(100)\\ t=years\dotfill &4\\ \end{cases} \\\\\\ 420=360(1 + (r)/(100))^(4) \implies \cfrac{420}{360}=(1 + (r)/(100))^(4)\implies \cfrac{7}{6}=\left( \cfrac{100+r}{100} \right)^4 \\\\\\ \sqrt[4]{\cfrac{7}{6}}=\cfrac{100+r}{100}\implies 100\sqrt[4]{\cfrac{7}{6}}=100+r`

`100\sqrt[4]{\cfrac{7}{6}}-100=r\implies 3.92899\approx r \\\\[-0.35em] ~\dotfill\\\\ \underset{2015~~ - ~~2005}{\stackrel{\textit{in the year 2015}}{t=10}}\hspace{5em}A=360\left(1 + (3.92899)/(100) \right)^(10)\implies A\approx 529.26`