Question

A car depreciates in value at an average rate of 12% per year. If the value of the car was originally $45,000, in how many years will it be worth $8,000? (Round to the nearest year.)

Answer 1

`\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\dotfill & \$ 8000\\ P=\textit{initial amount}\dotfill &45000\\ r=rate\to 12\%\to (12)/(100)\dotfill &0.12\\ t=years \end{cases}`

`8000 = 45000(1 - 0.12)^(t) \implies \cfrac{8000}{45000}=(1 - 0.12)^(t)\implies \cfrac{8}{45}=0.88^t \\\\\\ \log\left( \cfrac{8}{45} \right)=\log(0.88^t)\implies \log\left( \cfrac{8}{45} \right)=t\log(0.88) \\\\\\ \cfrac{\log\left( (8)/(45) \right)}{\log(0.88)}=t\implies \boxed{14\approx t}`