algebra and trigonometry

Question

A machine depreciates in value each year by 8% of its value at the beginning of the year. After how many years will its value first be less than half its original value?

Answer 1

Let V(t) be the value of the machine at time t in years.  Let V(t=0) = V₀, the initial value of the machine.

 

      V(t) = V₀*(1 - 0.08)^t = V₀*(0.92)^t         [8% = 0.08]

 

We want to know when V(t) = (1/2)V₀

 

      (1/2)V₀ = V₀*(0.92)^t

 

      1/2 = (0.92)t

 

      log(1/2) = log(0.92^t) = t*log(0.92)       [Remember that log(a^b) = b log(a)]

 

      log(1/2)/log(0.92) = t ≈ 8.3 years