Answer: approximately 8.89 years
Explanation: Let
[tex]v_n = \text{value of the car after year }n[/tex]
Since the car loses 15% of the value after one year,
New value of car after one year = (previous value) - (15% of previous value)
New value of car after one year = (previous value) - 0.15 × (previous value)
New value of car after one year = 0.85 × (previous value) (1)
Based on our representation, if [tex]v_n [/tex] represents the new value, then [tex]v_{n-1}[/tex] represents the previous value. Using equation (1),
[tex]v_n = 0.85v_{n - 1}[/tex] (2)
Moreover, note that [tex]v_0[/tex] represents the initial value of the car. So, using equation (2),
[tex]v_1 = 0.85v_0[/tex]
[tex]v_2 = 0.85v_1 = 0.85(0.85v_0) = (0.85)^2 v_0[/tex]
[tex]v_3 = 0.85v_2 = 0.85((0.85)^2 v_0) = (0.85)^3 v_0[/tex]
.
.
.
So, doing this for n times, we have
[tex]v_n = (0.85)^n v_0[/tex] (3)
In the problem, the initial value of the car is $28,000 and we need to find the value of n such that after n years the value of the car is $6,600.
So, [tex]v_0 = 28,000, v_n = 6,600[/tex]. Using equation (3),
[tex]v_n = (0.85)^n v_0
\\ 6,600 = (0.85)^n (28,000)
\\ (0.85)^n (28,000) = 6,600
\\
\\ (0.85)^n = \frac {6,600}{28,000}
\\
\\ \ln ((0.85)^n) = \ln(\frac {6,600}{28,000})
\\ n(\ln (0.85)) = \ln(\frac {6,600}{28,000})
\\
\\ \boxed{n = \frac{\ln(\frac {6,600}{28,000})}{\ln (0.85)} \approx 8.89 \text{ years} }[/tex]
