## Algebra 2 (1st Edition)

From the exercise, the car's value after $n$ years is: $18600(1-0.155)^n$, hence our equation is: $18600(1-0.155)^n=8000\\0.845^n=\frac{8000}{18600}=\frac{40}{93}\\n=\log_{0.845}{\frac{40}{93}}\approx5.00966$ Thus, the car will have that value after approximately five years.