SOLUTION: Carl bought a car for $5,000. The value of the car is decreasing by 9% each year. Janet bought a car for $3,000. The value of her car is decreasing by 3% per year. How long does it

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Question 1204541: Carl bought a car for $5,000. The value of the car is decreasing by 9% each year. Janet bought a car for $3,000. The value of her car is decreasing by 3% per year. How long does it take for the value of the cars to be approximately the same?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
9% per year decrease means keeping 91% each year.
3% per year decrease means keeping 97% each year.

5000%2A0.91%5Ex=3000%2A0.97%5Ex
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log%28%285000%29%29%2Bx%2Alog%28%280.91%29%29=log%28%283000%29%29%2Bx%2Alog%28%280.97%29%29
log%28%285000%29%29-log%28%283000%29%29=x%2Alog%28%280.97%29%29-x%2Alog%28%280.91%29%29
3%2Alog%28%285%29%29-3%2Alog%28%283%29%29=x%28log%28%280.97%29%29-log%28%280.91%29%29%29
x=3%28log%28%285%29%29-log%28%283%29%29%29%2F%28log%28%280.97%29%29-log%28%280.91%29%29%29
x=3%280.69897-0.47712%29%2F%28-0.013228%2B0.0409586%29
cross%28x=24%29

---

Retry -------
log%28%285000%29%29%2Bx%2Alog%28%280.91%29%29=log%28%283000%29%29%2Bx%2Alog%28%280.97%29%29
log%28%285%2F3%29%29=x%2Alog%28%280.97%29%29-x%2Alog%28%280.91%29%29
log%28%285%2F3%29%29=x%28log%28%280.97%2F0.91%29%29%29
x=%28log%28%285%2F3%29%29%29%2F%28log%28%2897%2F91%29%29%29
x=0.221849%2F0.02773
highlight%28x=8.00%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Carl bought a car for $5,000. The value of the car is decreasing by 9% each year. Janet bought a car for $3,000. The value of her car is decreasing by 3% per year. How long does it take for the value of the cars to be approximately the same?

He makes this problem so complex and still got the wrong answer. It's NOT 24 (highlight%28x=24%29) years 
as he claims.

Let the number of years it takes for both cars to depreciate to the same amount, be y
Then the yearly depreciation for the $5,000 car is, 5,000(1 - .09)y  
And, the yearly depreciation for the $3,000 car is, 3,000(1 - .03)y
We then set one equal to the other to get:  
                                               
                                                     matrix%281%2C3%2C+y%2C+%22=%22%2C+log+%28%28.91%2F.97%29%2C+3%2F5%29%29 ---- Convert to logarithmic form
Number of years it'll take for the value of the cars to be approximately the same, or y  = 8