Question

A new car is purchased for $19,000 and over time its value depreciates by

one half every 5 years. How long, to the nearest tenth of a year, would it take
for the value of the car to be $7,000?
i

Answer 1

so, the car depreciates by half every 5 years, or namely it depreciates by 50% every 5 years, so

`\textit{Periodic/Cyclical Exponential Decay} \\\\ A=P(1 - r)^{(t)/(c)}\qquad \begin{cases} A=\textit{current amount}\dotfill &\$7000\\ P=\textit{initial amount}\dotfill &\$19000\\ r=rate\to 50\%\to (50)/(100)\dotfill &0.5\\ t=\textit{elapsed time}\\ c=period\dotfill &5 \end{cases}`

`7000=19000(1-0.5)^{(t)/(5)}\implies \cfrac{7000}{19000}=0.5^{(t)/(5)}\implies \cfrac{7}{19}=0.5^{(t)/(5)} \\\\\\ \log\left( \cfrac{7}{19} \right)=\log\left( 0.5^{(t)/(5)} \right)\implies \log\left( \cfrac{7}{19} \right)=t\log\left( 0.5^{(1)/(5)} \right) \\\\\\ \log\left( \cfrac{7}{19} \right)=t\log( \sqrt[5]{0.5})\implies \cfrac{\log\left( (7)/(19) \right)}{\log( \sqrt[5]{0.5})}=t\implies 7.2\approx t`