Nathaniel purchased a new car in 1997 for $20,300. The value of the car has been depreciating exponentially at a constant rate. If the value of the car was $2,700 in the year 2004, then what would be the predicted value of the car in the year 2007, to the nearest dollar?

Step-by-step explanation:

We would apply the formula for exponential decay which is expressed as

A = P(1 - r)^t

Where

A represents the value of the car after t years.

t represents the number of years.

P represents the initial value of the car.

r represents rate of decay.

From the information given,

A = $2700

P = $20300

n = 2004 - 1997 = 7 years

Therefore,

20300 = 2700(1 - r)^7

20300/2700 = (1 - r)^7

7.519 = (1 - r)^7

Taking log of both sides, it becomes

Log 7.519 = 7 log(1 - r)

0.876 = 7 log(1 - r)

Log (1 - r) = 0.876/7 = 0.125

Taking inverse log of both sides, it becomes

10^log1 - r = 10^0.125

1 - r = 1.33

r = 1.33 - 1 = 0.33

The expression would be

A = 20300(1 - 0.33)^t

A = 20300(0.67)^t

Therefore, in 2007,

t = 2007 - 1997 = 10 years

The value would be

A = 20300(0.67)^10

A = $370

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!