Question

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"
How many years after purchase will Vishal's car be worth $10,000?
Give an exact answer expressed as a base-10 logarithm.
years

Answer 1

The number of years after purchase at which Vishal's car will be worth $10,000 is
`Log_(10)\left ((9)/(4) \right )^(12)` years

Explanation:

The relationship is given as follows

Value, V of the car = 22500×
`10^(-t/12)`

Therefore, when the car is $10,000 we will have

$10,000 = 22,500×
`10^(-t/12)`

Which will give;

`(10000)/(22500) =(4)/(9) = 10^{-(t)/(12)}`

Hence;

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

Therefore;

`-(t)/(12)* Log_(10) 10 = Log(4)/(9)`

`\because Log_(10) 10 = 1, \ we \ have; \ -(t)/(12) = Log_(10)(4)/(9)`

Which gives;

`t = -12 * Log_(10)(4)/(9) \ or \ t = Log_(10)\left ((4)/(9) \right )^(-12)`

`\therefore t = Log_(10)\left ((9)/(4) \right )^(12)` years

Evaluated, the above equation becomes t = 4.226 years

Therefore, the number of years after purchase at which Vishal's car will be worth $10,000 =
`Log_(10)\left ((9)/(4) \right )^(12)` years.

Answer 2

t = -24*log(2/3)

Explanation:

The expression is:

V = 22,500*10^(-t/12)

Replacing with V = 10,000 and isolating t, we get:

10,000 = 22,500*10^(-t/12)

10,000/22,500 = 10^(-t/12)

4/9 = 10^(-t/12)

(2/3)² = 10^(-t/12)

2*log(2/3) = -t/12

-12*2*log(2/3) = t

t = -24*log(2/3)