A new car is purchased for 23900 dollars. The value of the car depreciates at 9.25% per year. To the nearest tenth of a year, how long will it be until the value of the car is 6000 dollars?

9514 1404 393

  14.2 years

Step-by-step explanation:

The value is described by the exponential function ...

  y = (initial value)·(decay factor)^x

where y is the value after x years.

The decay factor is (1 - annual depreciation) = 1 - 0.0925 = 0.9075, so we want to find x when ...

  6000 = 23900·0.9075^x

  6000/23900 = 0.9075^x . . . . . . . . . . . . . . divide by 23900

  log(6000/23900) = x·log(0.9075) . . . . . . . take logarithms

  x = log(6000/23900)/log(0.9075) ≈ 14.2396

It will be about 14.2 years until the value of the car is $6000.

step-by-step explanation:

just plug in each number for x and see which ones work and don't

y=mx+