No need for confusion:
This is an exponential function
value= 23000(.88)^t , where t is the number of years
2. when t=2
value = 23000(.88)^2 = $17811.20
3.
2300 = 23000(.88)^t
.1 = .88^t
take log of both sides
log .1 = log .88^t
log .1 = t log .88
t = log .1/log .88 = 18.01 or appr 18 years
The Question is:
The value of a new car depreciates at a rate of 12% per year.
1)Write an equation to represent the approximate value of a car purchased for $23 000.
2)Determine the value of the car two years after it is purchased.
3)Approximately how many years will it take until the car is worth $2300?
I have received answers that differ for this question. I request feedback for further understanding what must be done.
The following is an example of how the answers I've received are different:
ex.1
a) V= C - rtC,
Eq: V = C(1 - rt).
V = value.
C = cost.
r = rate expresed as a decimal.
t = time in years.
b) V = 23000(1 - 0.12*2),
V = 23000 * 0.76 = 17480.
c) V = 23000(1 - 0.12t) = 2300,
23000(1 - 0.12t) = 2300,
Divide both sides by 23000:
1 - 0.12t = 0.1,
-0.12t = 0.1 - 1 = -0.90,
t = -0.90 / -0.12 = 7.5 yrs.
OR
a) A=23000(1-0.12)^1
A= 20240
b) A=23000(1-0.12)^2
A=17 811.20
c) 2300 = 23000(.88)^n
0.1 = (.88)^n
n = (log 0.1) / (log 0.88)
=18.01 years
2 answers
Great this helps alot, was worried I screwed up something
2 answers