Question

The dollar value v(t) of a certain car model that is t years old is given by the following exponential function. v(t)=27500(0.88)^t Find the initial value of the car and the value after 13 years. Round your answers to the nearest dollar as necessary.

Answer 1

Answer: Initial value of car = $ 27500

And, the value of car after 13 years is $5219.242

Step-by-step explanation:

Here, the given function that models the price of car after t years,

$v(t) = 27500(0.88)^t$

Since, initially,  t = 0

Thus, the initial value of the car,

$v(0) = 27500(0.88)^0=27500$

Now, after 13 years, t = 13

Thus, the value of car after 13 years,

$v(13)=27500(0.88)^{13}$

= $27500\times 0.189790617123$

= $5219.24197088\approx 5219.242$

Answer 2

Answer:

Current value is $27500 and the value after 13 years will be $20417.

Step-by-step explanation:

The dollar value v(t) of a certain car model in t years is given by the exponential function $v(t)=27500\times(0.88)^{t}$

Now we have to find the initial value and the value after 13 years.

Therefore to calculate the initial value of the car v(0)=27500\times(.88)^{0}

= 27500×1 (since $x^{0}=1$

So the current value of the car is $27500.

Now we will calculate the value of car after 13 years.

v(13) = $27500(0.88)^{13}$

Now we take the log on both the sides of the equation

$logv(13)=log\left \{ 27500\times (.88)^{13} \right \}$

           $=log 27500+13log(.88)$

                  = 4.44 + 13log(88÷100)

                  = 4.44 + 13( log88 - log100)

                  = 4.44+ 13(1.94-2)

     log v(13)= 4.44 - 13(.056)

                  = 4.44- 0.72

                  = 3.72

⇒ v(13) = $10^{3.72}$ = 20417.38 ≈ $20417