Word Problem Involving Logarithm

Question

Question 3

The price of a brand new car was $180000 on 1st January 2008. Its value depreciated in such a way

that t months after the purchase, the sale price, $P, is determined by the formula t P e 0.0186 180000 − = . Find

(a) the sale price, correct to the nearest $1000, of the car after 3 years,

(b) the number of months taken for the car to be worth less than $50000.

Answer 1

P(t) = 180,000e^-0.0186t

 

(a). P(36) = 180,000e^-0.0186(36)

 

                    = 180,000e^-0.6696

 

                    = $92,144.39 ≈ $92,000

-------------------------------------------------

(b). 50,000 = 180,000e^-0.0186t

 

       0.27777 = e^-0.0186t

 

      ln(0.27777) = ln(e^-0.0186t)

 

      ln(0.27777) = -0.0186t

 

      t = 68.86 months

 

      The car will be worth less than $50,000 if it is more than 68.86            months old. 

Answer 2

Is the formula P(t)=180000*e^-0.0186t ?