Question

A new car is purchased for 15100 dollars. The value of the car depreciates at 12. 25% per year. To the nearest year, how long will it be until the value of the car is 2300 dollars?​

Answer 1

`\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\dotfill &\$2300\\ P=\textit{initial amount}\dotfill &\$15100\\ r=rate\to 12.25\%\to (12.25)/(100)\dotfill &0.1225\\ t=years \end{cases} \\\\\\ 2300=15100(1 - 0.1225)^(t)\implies \cfrac{2300}{15100}=(1 - 0.1225)^(t)\implies \cfrac{23}{151}=0.8775^t`

`\log\left( \cfrac{23}{151} \right)=\log(0.8775^t)\implies \log\left( \cfrac{23}{151} \right)=t\log(0.8775) \\\\\\ \cfrac{~~\log\left( (23)/(151) \right)~~}{\log(0.8775)}=t\implies 14.4\approx t\qquad \textit{about 14 years and 146 days}`