Question

For accounting purposes, the value of assets (land, buildings, equipment) in a business depreciates at a set rate per year. The value, V(t), of $408,000 worth of assets after t years, which depreciate at 18% per year, is given by the formula V(t) = V0(b)t. What is the value of V0 and b, and when rounded to the nearest cent, what is the value of the assets after 8 years?

Answer 1

V(t) = $408,000 - ( $408,000 x 18%)
       = $334,560 - ($334,560 x 18%)
       = $274,339.20 - ($274,339.20 x 18%)
       = $224,958.14 - ($224,958.14 x 18%)
       = $184,465.68 - ($184,465.68 x 18%)
       = $151,261.86 - ($151,261.86 x 18%)
       = $124,034.73 - ($124,034.73 x 18%)
       = $101,708.48 - ($101,708.48 x 18 %)
        = $83,400.95

The new carrying value of the asset on the current year is deducted with the depreciation rate to get the carrying value of the next year

Answer 2

For this case we have an equation of the form:
 
$V (t) = V0 * (b) ^ t$ 
 Where,
 v0: initial value in assets
 b: depreciation rate
 t: time in years.
 Substituting values we have:
$V (t) = 408000 * (0.82) ^ t$
 For year 8 we have:
$V (t) = 408000 * (0.82) ^ 8 V (t) = 83400.94703$
 Rounding off we have:
 V (t) = 83401
 Answer:
 the value of V0 and b are:
 V0 = $ 408,000
 b = 0.82
 the value of the assets after 8 years is:V (t) = 83401 $