# Exponential Depreciation and linear equations help

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

Question goes as so:

A 2004 Mercedes costs $50,000 and the car depreciates a total of 37% in the first 5 years. Find the formula for the exponential equation and linear equation. I know I need to use y = ab^x but I'm wondering how I find the correct percent of depreciation (1-.37)^5 doesn't seem to be correct as it depreciates 37 percent over 5 years. Do I just divide 37 by 5, or is it different. - If$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

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?)