Chapter 8.7, Problem 40PPS

(a)

To determine

To find: The car be worth in $18$ months.

Answer to Problem 40PPS

The car is worth in months $v_1=13995.99$ .

Explanation of Solution

Given information:

The given function is $v\left(t\right)=18500e^{-0.186t}$ .

Calculation:

The value of the car after $18$ months of purchase is $v_1$ .

  $t=\frac{18}{12}=1.5v\left(t\right)=v_1$

  $v\left(t\right)=18500e^{-0.186t}$

Use $TI83$ to calculate to computer the value.

  

Therefore the value of car in worth is $v_1=13995.99$

(b)

To determine

To find: The car be worth half of its original value.

Answer to Problem 40PPS

The car be worth half of its original value $t_1=3.7$ .

Explanation of Solution

Given information:

The given function is $v\left(t\right)=18500e^{-0.186t}$ .

Calculation:

Consider the value of $t_1$ year from the day of purchase the value of the car will be half of its original value.

Substitute the value of $v\left(0\right)=18500$ is

  $\begin{array}{c}\ln e^{-0.186t_1}=\ln \frac{1}{2}\\ -0.186t_1\ln e=\ln \frac{1}{2}\\ t_1=-\frac{1}{0.186}\ln \frac{1}{2}\end{array}$

Therefore $t=t_{1}v\left(t\right)=\frac{18500}{2}$

The natural logarithm of both sides

  $t_1=3.7$

Use $TI83$ to calculate to computer value

  

Therefore, the car is worth half of its original value $t_1=3.7$ .

(c)

To determine

To find: The car be worth less than $\$1000$ .

Answer to Problem 40PPS

The after $t_2>11.5$ year from the day of purchase the value of the care will be less then $\$1000$

Explanation of Solution

Given information:

The given function is $v\left(t\right)=18500e^{-0.186t}$ .

Calculation:

Calculate the equation,

  $\begin{array}{c}\ln e^{-0.186t_2}<\ln \frac{10}{85}\\ -.0186t_2\ln e^{}<\ln \frac{10}{85}\\ t_2>-\frac{1}{.0186}\ln \frac{10}{85}\end{array}$

The natural logarithm of both sides

  $t_2>11.5$

Use $TI83$ to calculate to computer value

  

  $t_2>11.5$

Therefore after $t_2>11.5$ year from the day of purchase the value of the care will be less then $\$1000$