Exponential Decay and Depreciation

Question

The moment a car is purchased its value depreciates according to the law of exponential

decay. 1 year after it was purchased the value of the car was $14,400 and 4 years after it was

purchased the value of the car was $9,216.

a) What was the cars purchase price?

b) What was the car’s value 6 years after it was purchased?

c) How many years after it was purchased will the car be worth $1?

Answer 1

If you have a model of exponential function f(t) = a*b^t where b is a positive real number, we can say that it is a decay when 0 < b < 1 or t < 0 but not both. But in this problem time t can't be negative.

Let a = initial value of the car

t= time in years

1-b= annual percentage rate of decay

For t=1

f(1) = 14,400 = a*b¹

For t=4

f(4) = 9,216 = a*b⁴

Based on the first equation:

a= 14400/b

Based on the second equation:

a= 9,216/b⁴

Therefore:

14,400/b = 9,216/b⁴

b⁴/b = 9,216/14,400

b³ = 0.64

b = (0.64)^1/3 ≈ 0.86177

a) What was the cars purchase price?

14,400 = a*(0.64)^1/3

a= 14,400/(0.64)^1/3

a ≈ $16,709.72

b) What was the car’s value 6 years after it was purchased?

f(6) = 16,709.72*((0.64)^1/3)⁶

f(6) = 16,709.72*(0.64)²

f(6) ≈ $6844.30

c) How many years after it was purchased will the car be worth $1?

1= 16,709.72*(0.64)^1/3)^t

1= 16,709.72*(0.64)^t/3

1/16,709.72 = (0.64)^t/3

Get the ln (natural log) of both sides of the equation

ln(1/16,709.72) = ln(0.64)^t/3

ln(1/16,709.72) = (t/3) ln(0.64)

ln(1/16,709.72)/ln(0.64) = t/3

3*ln(1/16,709.72)/ln(0.64) = t

t ≈ 65.36 years