SOLUTION: please solve help me solve this word problem: Lucy bought a car for $25,000. The value of the car depreciates by about 13% each year. when will the value of the car be $10,000 ?

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: please solve help me solve this word problem: Lucy bought a car for $25,000. The value of the car depreciates by about 13% each year. when will the value of the car be $10,000 ?       Log On


   

Question 570809: please solve help me solve this word problem: Lucy bought a car for $25,000. The value of the car depreciates by about 13% each year. when will the value of the car be $10,000 ?
please help me. and please show me the work to get the answer because i have no idea how to solve this problem.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose for now the car starts out being worth x
Let the yearly depreciation = +a+ ( .13, actually )
At end of 1st yr it is worth
+x+-+a%2Ax+
+x%2A%28+1+-+a+%29+
At end of 2nd yr
+x+-+a%2Ax+-+a%2A%28+x+-+a%2Ax+%29+
+%28+x+-+a%2Ax+%29%2A%281+-+a+%29+
+x%2A%28+1+-+a+%29%5E2+
At end of 3rd yr
+x%2A%28+1+-+a+%29%5E3+
------------------------
You can see the progression here
In your problem, +x+=+25000+ and
+a+=+.13+
I can say
+25000%2A%28+1+-+.13+%29%5En+=+10000+
where +n+ is the number of yrs car is owned
+%28+1+-+.13+%29%5En+=+10000+%2F+25000+
+.87%5En+=+.4+
+n%2Alog%28.87%29+=+log%28.4%29+
+n+=+log%28.4%29%2Flog%28.87%29+
+n+=+%28+-.39794+%29+%2F+%28+-.06048+%29+
+n+=+6.579+
and +.579%2A12+=+6.96+
So, after 6 yrs ( 6 yrs and 7 months ),
the car is worth $10,000
---------------------------
I can check this
+25000%2A%28+1+-+.13+%29%5E6.579+=+10000+
+.87%5E6.579+=+.4+
+.4000342+=+.4+
close enough
The key to understanding this problem is to see
that each years depreciation is on last year's
DEPRECIATED value. Hope this helps.