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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.)2011-09-23
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    @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!2011-09-23
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    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?2011-09-23
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    50,000 / 31500 = b^5 ?2011-09-23
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    Yup, that works for the second part. Now solve for $b$.2011-09-23
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    @anon I believe I wrote that down backwards. It should have been `31,500 / 50000 = b^5`. `b = .9117` (rounded to 4 decimal places). So the final equation would be: `y = 50,000(.9117)^t` if I did it correctly. Then I can move onto the linear equation which should be easier. If you post an answer, I can mark it correct :) - thanks for the help!2011-09-23
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    Ah, whoops. I'll type something up.2011-09-23

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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 :)2011-09-23