Answer:
Step-by-step explanation:
The depreciation of the value for a car is modeled by the equation y = 100,000(.85)^x for x years since 2000
a) for the value to be $61,412.50, then
61412.50 = 100000(.85)^x
61412.50/100000 = 0.85^x
0.614125 = 0.85^x
We would take log to base 10 of both sides. It becomes
Log 0.614125 = log 0.85^x =
x log0.85
- 0.2117 = x × - 0.07058
- 0.2117 = - 0.07058x
x = - 0.2117/- 0.07058
x = 3 years
b) 1/4 × 100000 = 25000
Therefore ,
25000 = 100000(.85)^x
25000/100000 = 0.85^x
0.25 = 0.85^x
We would take log to base 10 of both sides. It becomes
Log 0.25 = log 0.85^x =
x log0.85
- 0.602 = x × - 0.07058
- 0.602 = - 0.07058x
x = - 0.602/- 0.07058
x = 9 years
