Answer:
a) The car will be worth $8000 after 2.9 years.
b) The car will be worth $6000 after 4.2 years.
c) The car will be worth $1000 after 12.7 years.
Step-by-step explanation:
The value of the car after t years is given by:
[tex]V(t) = 14651(0.81)^{t}[/tex]
According to the model, when will the car be worth V(t)?
We have to find t for the given value of V(t). So
[tex]V(t) = 14651(0.81)^{t}[/tex]
[tex](0.81)^t = \frac{V(t)}{14651}[/tex]
[tex]\log{(0.81)^{t}} = \log{(\frac{V(t)}{14651})}[/tex]
[tex]t\log{(0.81)} = \log{(\frac{V(t)}{14651})}[/tex]
[tex]t = \frac{\log{(\frac{V(t)}{14651})}}{\log{0.81}}[/tex]
(a) $8000
V(t) = 8000
[tex]t = \frac{\log{(\frac{8000}{14651})}}{\log{0.81}} = 2.9[/tex]
The car will be worth $8000 after 2.9 years.
(b) $6000
V(t) = 6000
[tex]t = \frac{\log{(\frac{6000}{14651})}}{\log{0.81}} = 4.2[/tex]
The car will be worth $6000 after 4.2 years.
(c) $1000
V(t) = 1000
[tex]t = \frac{\log{(\frac{1000}{14651})}}{\log{0.81}} = 12.7[/tex]
The car will be worth $1000 after 12.7 years.
