Mathematics, 05.10.2019 03:50, lelen2021
Find the purchase price of a car if after 3 years it was worth $15000 and after 7 years it was worth $8000
Answers: 1
![\bf \qquad \textit{Amount for Exponential Decay, \boxed{\textit{3 years}}} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&15000\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &3\\ \end{cases} \\\\\\ 15000=P(1-r)^3 \\\\[-0.35em] ~\dotfill](/tpl/images/0288/0774/64e47.png)
![\bf \qquad \textit{Amount for Exponential Decay, \boxed{\textit{7 years}}} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&8000\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &7\\ \end{cases} \\\\\\ 8000=P(1-r)^7 \\\\[-0.35em] \rule{34em}{0.25pt}](/tpl/images/0288/0774/53be6.png)
![\bf \sqrt[4]{\cfrac{8}{15}}=1-r\implies r=1-\sqrt[4]{\cfrac{8}{15}}\implies r\approx 0.15 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 15000=P(1-r)^3\implies 15000=P(1-0.15)^3\implies 15000=P(0.85)^3 \\\\\\ \cfrac{15000}{0.85^3}=P\implies 24424.995\approx P](/tpl/images/0288/0774/9c7de.png)
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