SOLUTION: Jasmine's car was worth $11,000 at the beginning of 2000 and the value of the car decreased exponentially, decreasing by 20% each year. a. Write a function f that determines th

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Question 1124886: Jasmine's car was worth $11,000 at the beginning of 2000 and the value of the car decreased exponentially, decreasing by 20% each year.
a. Write a function f that determines the value of Jasmine's car (in dollars) in terms of the number of years t since the beginning of 2000.
b. How many years after the beginning of 2000 was Jasmine's car worth $6,000?
c. How many years after the beginning of 2000 was Jasmine's car worth $3,400?
Found 2 solutions by MathLover1, josmiceli:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Jasmine's car was worth $11000 at the beginning of 2000 and the value of the car decreased exponentially, decreasing by 20% each year.


use depreciation formula:
v+=+c%281-r%29%5Et
where:
v = future value
c = current value
r = depreciation rate per period
t+= number of periods
a. Write a function f that determines the value of Jasmine's car (in dollars) in terms of the number of years t since the beginning of 2000.
since the beginning of 2000 to this year, there is 18
v+=+c%281-r%29%5E18
if car was worth $11000 at the beginning of 2000 , and the value decreasing by 20%=0.20 each year
this year is worth:
v+=+11000%281-0.20%29%5E18
v+=+11000%280.80%29%5E18
v+=+198.16
b. How many years after the beginning of 2000 was Jasmine's car worth $6000?
6000+=+11000%281-0.20%29%5Et
6000+%2F11000=%280.80%29%5Et
6+%2F11=%280.80%29%5Et
log%286+%2F11%29=log%28%280.80%29%5Et%29
log%280.5454545454545455%29=t%2Alog%280.80%29
t=log%280.5454545454545455%29%2Flog%280.80%29
t=2.7 ->little over 2 and half years; so, it was in 2003

c. How many years after the beginning of 2000 was Jasmine's car worth $3400?
3400+=+11000%281-0.20%29%5Et
3400+%2F11000=%280.80%29%5Et
34+%2F110=%280.80%29%5Et
log%2817+%2F55%29=log%28%280.80%29%5Et%29
log%280.3090909090909091%29=t%2Alog%280.80%29
t=log%280.3090909090909091%29%2Flog%280.80%29
t=5.3 ->little over 5 years and three months; so, it was in at beginning of 2006

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(a)
+V+=+11000%2A%28+1+-+.2+%29%5Et+
(b)
+6000+=+11000%2A.8%5Et+
+.54545+=+.8%5Et+
Take the log base 10 of both sides
+-.2632+=+t%2A%28-.0969+%29+
+t+=+2.7162+ yrs
——————————-
(c)
+3400+=+11000%2A.8%5Et+
+.3091+=+.8%5Et+
Take log of both sides
+-.5099+=+t%2A%28-.0969+%29+
+t+=+5.2623+ yrs
———————————-
Get 2nd opinion if needed