Question

The value V of a Honda Civic DX that is

t years old can be modeled by$V\left(t\right)=16,775{(0.905)}^t$

(a) According to the model, when will the car be worth \(15,000?

(b) According to the model, when will the car be worth \)8000?

(c) According to the model, when will the car be worth $4000?

Short answer

a)The price of car becomes $15000 in 0.9yrs

b)The price of car becomes $8000 in 7.41yrs

c)The price of car becomes $4000 in 14 yrs

Step-by-step solution

Given information

We are given a function which represents the price of car in t years$V\left(t\right)=16,775{(0.9052)}^t$

Part a) Step 1: Substitute V(t)=15000 and solve

We get,

$$\begin{gathered} V\left(t\right)=16,775{(0.905)}^t \\ 15000=16755{(0.905)}^t \\ 0.9090={(0.905)}^t \\ \ln (0.9090)=t(\ln 0.905) \\ t=\frac{\ln (0.9090)}{\ln (0.905)} \\ t=0.955yrs \end{gathered}$$

The price of car becomes $15000 in 0.9yrs

Part b) Step 1: Substitute V(t)=8000 and solve

We get,

$$\begin{gathered} V\left(t\right)=16500{(0.905)}^t \\ 8000=16775{(0.905)}^t \\ 0.4769={(0.905)}^t \\ \ln (0.4769)=t(\ln 0.905) \\ t=\frac{\ln (0.4769)}{\ln (0.905)} \\ t=7.41yrs \end{gathered}$$

The price of car becomes $8000 in 7.41 yrs

Part c) Step 1: Substitute V(t)=4000 and solve

We get,

$$\begin{gathered} V\left(t\right)=16775{(0.905)}^t \\ 4000=16775{(0.905)}^t \\ 0.2384={(0.905)}^t \\ \ln (0.2384)=t(\ln 0.905) \\ t=\frac{\ln (0.2384)}{\ln (0.905)} \\ t=14.36 \end{gathered}$$

The price of car becomes $4000 in 14 yrs