Question

Jonathan purchased a new car in 2008 for $25,400. The value of the car has been

depreciating exponentially at a constant rate. If the value of the car was $7,500 in


the year 2015, then what would be the predicted value of the car in the year 2017, to


the nearest dollar?



HELP

Answer 1

well, first off let's find the rate its depreciating by, so hmm in 2008 it was $25,400 and then 7 years later it went down to $7,500, so

`\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\dotfill & \$ 7500\\ P=\textit{initial amount}\dotfill &25400\\ r=rate\to r\%\to (r)/(100)\\ t=\textit{years}\dotfill &7\\ \end{cases} \\\\\\ 7500 = 25400(1 - (r)/(100))^(7) \implies \cfrac{7500}{25400}=\left( \cfrac{100-r}{100} \right)^7`

`\cfrac{75}{254}=\left( \cfrac{100-r}{100} \right)^7\implies \sqrt[7]{\cfrac{75}{254}}=\cfrac{100-r}{100}\implies 100\sqrt[7]{\cfrac{75}{254}}=100-r \\\\\\ 100\sqrt[7]{\cfrac{75}{254}}-100=-r\implies 100-100\sqrt[7]{\cfrac{75}{254}}=r\implies \stackrel{ \% }{15.99}\approx r`

so hmmm how about in 2017? well, 2017 from 2008 would be 9 years later, so using that rate

`\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\\ P=\textit{initial amount}\dotfill &25400\\ r=rate\to 15.99\%\to (15.99)/(100)\dotfill &0.1599\\ t=\textit{years}\dotfill &9\\ \end{cases} \\\\\\ A \approx 25400(1 - 0.1599)^(9) \implies A \approx 25400( 0.8401 )^(9)\implies \boxed{A \approx 5294}`