Question

The value of Vishal's car is depreciating exponentially.

The relationship between V, the value of his car, in dollars, and t, the elapsed time, in years, since he purchased the car is modeled by the following equation.


V=22,500⋅10^-t/12


How many years after purchase will Vishal's car be worth $10,000


Give an exact answer expressed as a base-10 logarithm.

Answer 1

-24log10(2/3)

Explanation:

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Answer 2

We have been given that the value of Vishal's car is depreciating exponentially. The relationship between V, the value of his car, in dollars, and t, the elapsed time, in years, since he purchased the car is modeled by the equation
`V=22,500\cdot 10^{-(t)/(12)}`. We are asked to find the time, when Vishal's car will be worth $10,000.

To find the time, we will equate our given equation with 10,000 as:

`10,000=22,500\cdot 10^{-(t)/(12)}`

`(10,000)/(22,500)=\frac{22,500\cdot 10^{-(t)/(12)}}{22,500}`

`(4)/(9)=10^{-(t)/(12)}`

Now we will take log on both sides.

`log_(10)\left((4)/(9)\right)=log_(10)\left(10^{-(t)/(12)}\right)`

`log_(10)\left(((2)/(3))^2\right)=-(t)/(12)`

`-(t)/(12)=log_(10)\left(((2)/(3))^2\right)`

`-(t)/(12)(-12)=(-12)log_(10)\left(((2)/(3))^2\right)`

`t=(-12)log_(10)\left(((2)/(3))^2\right)`

Using property
`log(a^b)=b\cdot log(a)`, we will get:

`t=(-12)\cdot 2log_(10)\left((2)/(3)\right)`

`t=-24log_(10)\left((2)/(3)\right)`

Therefore, the value of Vishal's car will be $10,000 after approximately
`-24log_(10)\left((2)/(3)\right)` years.