Question

Question 2 of 15, Step 1CorrectThe value of a machine, V, at the end of years is given by V = C(1 - 1), where is the original cost of the machine and r is the rate of depreciation. A machine thatoriginally cost $19,600 is now valued at $15,528. How old is the machine if r = 0.12? Round your answer to two decimal places.

Answer 1

$V=C(1-r)^t$

If C = $19600, V = $ 15528 and r = 0.12, we have:

$\begin{gathered} 15528=19600(1-0.12)^t \\ 15528=19600\cdot0.88^t \\ \frac{15528}{19600}=0.88^t \\ \frac{1941}{2450}=0.88^t \\ log(\frac{1,941}{2,450})=log(0.88^t) \\ log(\frac{1,941}{2,450})=t\cdot log(0.88^) \\ t=\frac{log(\frac{1,941}{2,450})}{log(0.88^)} \\ t=\frac{-0.101}{-0.056} \\ t\approx1.82\text{ years} \end{gathered}$