Question

Ryan invested \$4,800$4,800 in an account in the year 1990, and the value has been growing exponentially at a constant rate. The value of the account reached \$6,300$6,300 in the year 1998. Determine the value of the account, to the nearest dollar, in the year 2007.

Answer 1

well, from 1990 to 1998 is 8 years, and we know the amount went from $4800 to $6300, let's check for the rate of growth.

`\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$6300\\ P=\textit{initial amount}\dotfill &\$4800\\ r=rate\to r\%\to (r)/(100)\\ t=\textit{years}\dotfill &8\\ \end{cases} \\\\\\ 6300=4800(1 + (r)/(100))^(8) \implies \cfrac{6300}{4800}=(1 + (r)/(100))^8\implies \cfrac{21}{16}=(1 + (r)/(100))^8`

`\sqrt[8]{\cfrac{21}{16}}=1 + \cfrac{r}{100}\implies \sqrt[8]{\cfrac{21}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[8]{\cfrac{21}{16}}=100+r\implies 100\sqrt[8]{\cfrac{21}{16}}-100=r\implies \stackrel{\%}{3.46}\approx r`

now, with an initial amount of $4800, up to 2007, namely 17 years later, how much will that be with a 3.46% rate?

`\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &4800\\ r=rate\to 3.46\%\to (3.46)/(100)\dotfill &0.0346\\ t=years\dotfill &17\\ \end{cases} \\\\\\ A=4800(1 + 0.0346)^(17) \implies A=4800(1.0346)^(17)\implies A \approx 8558.02`