Question

A new car purchased for $20.000 decreases by 10% each year. Write an exponential decay

model for the car. Then determine how long it will take the value of the car to decrease to $7000.

Answer 1

`\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\\ P=\textit{initial amount}\dotfill &20000\\ r=rate\to 10\%\to (10)/(100)\dotfill &0.1\\ t=years\\ \end{cases} \\\\\\ A=20000(1 - 0.1)^(t) \implies A=20000(0.9)^t`

how long till it's $7000?

`\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\dotfill &7000\\ P=\textit{initial amount}\dotfill &20000\\ r=rate\to 10\%\to (10)/(100)\dotfill &0.1\\ t=\textit{elapsed time} \end{cases} \\\\\\ 7000=20000(1 - 0.1)^(t) \implies \cfrac{7000}{20000}=0.9^t\implies \cfrac{7}{20}=0.9^t`

`\log\left( \cfrac{7}{20} \right)=\log(0.9^t)\implies \log\left( \cfrac{7}{20} \right)=t\log(0.9) \\\\\\ \cfrac{\log\left( (7)/(20) \right)}{\log(0.9)}=t\implies 9.96\approx t \qquad \textit{about 9 years and 11 months}`