A new car is purchased for 23600 dollars. The value of the car depreciates at 8.75% per year. To the nearest tenth of a year, how long will it be until the value of the car is 7100 dollars?

Answer 1

Hi Scy,

7100 = 23600(1-0.0875)^t A = P(1 + r)^t

7100 = 23600(0.9125)^t

(7100/23600) = 0.9125^t Divide by 23600

log (7100/23600) = log (0.9125)^t Log of both sides

log (7100/23600) = t log (0.9125) Properties of Logs

log (7100/23600)/log (0.9125) = t ≈ 13.1 Divide by log (0.9125)

about 13.1 years

Enjoy