Ah hah, a real calculus problem :)
dV/dt = - k V
dV/V = -k dt
ln V = -k t + C
e^ln V = V = e^(-kt+C) = c e^-kt
so here at t = 0
e^0 = 1
so V = c = 70,000
and
V = 70,000 e^-kt
now when t = 10/12 years, V = 22,222
22,222 = 70,000 e^-k(.83333)
ln (22,222/70,000) = - .83333 k
-1.1474= - .83333 k
k = 1.377
so
V = 70,000 e^-1.377 t
19,999 = 70,000 e^-1.377 t
ln(19,999/70,000) = -1.377 t
t = .9098 years
By the way, are you sure it was 10 months and not ten yars?
The rate at which an item depreciates is proportional to its value at that instant. If an item is presently values at $70,000 and 10 months later it is valued at $22,222, when in years will it be valued at $19,999?
I'm not sure what formula to use...
2 answers
thank you and yes it was months
1 answer