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I'm currently working on a problem regarding depreciation of car value.

Question goes as so:

A 2004 Mercedes costs $50,000 and the car depreciates a total of 37% in the first 5 years. Find the formula for the exponential equation and linear equation.

I know I need to use y = ab^x but I'm wondering how I find the correct percent of depreciation (1-.37)^5 doesn't seem to be correct as it depreciates 37 percent over 5 years. Do I just divide 37 by 5, or is it different.

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If $y=ab^x$ is the formula, $y(0)=50,000$ is the initial value, and $y(5)=(1-0.37)y(0)$ is the value after five years, can you solve for $a$ and $b$? (That's a rhetorical question: yes, you can.) –  anon Sep 23 '11 at 5:11
    
@anon So I should just make it into two points and go from there? (0, 50000) & (5, 31500) then work it like I would if I had points? I'll give it a go, thanks! –  jwf Sep 23 '11 at 5:16
    
You don't actually have to plot any points. All you need to do is a bit of algebra. What happens when you plug $x=0$ in the formula and set it equal to $50,000$? What happens when you take $50,000$ and substitute it for$y(0)$ and then plug in $x=5$ into the formula for the second equation I gave? –  anon Sep 23 '11 at 5:18
    
50,000 / 31500 = b^5 ? –  jwf Sep 23 '11 at 5:21
    
Yup, that works for the second part. Now solve for $b$. –  anon Sep 23 '11 at 5:24
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1 Answer

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Plug known values into the formula and see what it says about the unknowns: $$y(0)=ab^0=50,000\quad\implies a=50,000;$$ $$y(5)=ab^5=(1-0.37)y(0)\quad\implies b=\sqrt[5]{0.63}=0.9117339... $$ (Also, what does the question mean by "and linear equation"? Linear equation for what exactly?)

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Well we found out the exponential depreciation, I was also required to find the linear depreciation (which was much easier). To do so, all I did was find the slope from the points and then use point-slope formula to figure out the rest :) –  jwf Sep 23 '11 at 5:42
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