Question

A car loses 15% of its value per year. After how many years will a $28,000 car be worth $6,600? (Please show work)

Answer 1

Losing 15% value per year is that same as retaining 85% of value. We can create a table recursively multiplying by the 85% for retained value. The car hits a value of $6,600 between year 8 and 9, specifically during the 11th month of year 9.
Value Year
28000 0
23800 1
20230 2
17195.5 3
14616.18 4
12423.75 5
10560.19 6
8976.16 7
7629.73 8
6485.27 9

Answer 2

Answer: approximately 8.89 years

Step-by-step explanation: Let


`v_n = \text{value of the car after year }n`

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
`v_n` represents the new value, then
`v_(n-1)` represents the previous value. Using equation (1),


`v_n = 0.85v_(n - 1)` (2)

Moreover, note that
`v_0` represents the initial value of the car. So, using equation (2),


`v_1 = 0.85v_0`

`v_2 = 0.85v_1 = 0.85(0.85v_0) = (0.85)^2 v_0`

`v_3 = 0.85v_2 = 0.85((0.85)^2 v_0) = (0.85)^3 v_0`
.
.
.

So, doing this for n times, we have


`v_n = (0.85)^n v_0` (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,
`v_0 = 28,000, v_n = 6,600`. Using equation (3),


`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} }`