Question

A new car is purchased for $13, 000 and over time its value depreciates by one half

every 6 years. How long, to the nearest tenth of a year, would it take for the value of
the car to be $1,700?

Answer 1

since the car is depreciating by half, we can say the depreciation rate every 6 years is 50%, so

`\textit{Periodic/Cyclical Exponential Decay} \\\\ A=P(1 - r)^{(t)/(c)}\qquad \begin{cases} A=\textit{current amount}\dotfill&\$ 1700\\ P=\textit{initial amount}\dotfill &\$13000\\ r=rate\to 50\%\to (50)/(100)\dotfill &0.5\\ t=years\ \\c=period\dotfill &6 \end{cases} \\\\\\ 1700=13000(1 - 0.5)^{(t)/(6)}\implies \cfrac{1700}{13000}=0.5^{(t)/(6)}\implies \cfrac{17}{130}=0.5^{(t)/(6)}`

`\log\left( \cfrac{17}{130} \right)=\log\left( 0.5^{(t)/(6)} \right)\implies \log\left( \cfrac{17}{130} \right)=\log\left( 0.5^{(1)/(6)t} \right) \\\\\\ \log\left( \cfrac{17}{130} \right)=t\log\left( 0.5^{(1)/(6)} \right)\implies \cfrac{\log\left( (17)/(130) \right)}{\log\left( 0.5^{(1)/(6)} \right)}=t\implies 17.6\approx t`