a car was bought new for 15 000. each year the value of the car depreciates by 30% of its value of the previous year. in how many years will the car be worth only $500.?

Question

amistre64 · Answer

A = P(1-r)^t

amistre64 · Answer

A/P = (1-r)^t

log[1-r](A/P)  = t

ln(A/P)
----- = t
ln(1-r)

if i did it right

chaise · Answer

Is this not a geometric progressions?

amistre64 · Answer

to compact it for google:

(ln(500/15000))/ln(1-.3) = abt 9.54

chaise · Answer

$$a _{n}= an ^{n-1}$$

amistre64 · Answer

more like regression maybe :)

amistre64 · Answer

to dbl check

500 = 15000(1-.3)^(9.54)

500 = 499.26  is pretty good

chaise · Answer

You answer seems pretty nice. I don't know how you got it - Although I'm 90% sure this problem can be solved using geometric progression. I just can't remember how it is done.

amistre64 · Answer

the A= P(1-r)^t  is the derived from the progression, but that usually aint covered till later on in the maths

amistre64 · Answer

i just algebrad it into submission :)

anonymous · Answer

it doesn't have natural logs or progression because i haVen't learned that yet

amistre64 · Answer

a0 = 15000
a1 = 15000(1-.3)
a2 = 15000(1-.3)(1-.3)
a3 = 15000(.7)^3
a4 = 15000(.7)^4

an = 15000(.7)^n

500 = 15000(.7)^n , and solve for n

amistre64 · Answer

you have to know logs in order to solve for an exponential variable

amistre64 · Answer

the only other option is to brute math it till it succombs :)

anonymous · Answer

the answer is 9.5 years

amistre64 · Answer

yay!! .. i was close :)

actually thats an approximation, but yes

anonymous · Answer

yes