I’m trying to figure out how to calculate the average percentage a year something decreases by.
For instance, say a car has went from being worth $8000 to $5000 in two years. It decreased by a total of 37.5% (3000 / 5000 x 100%).
Is there a formula to figure out exactly how much the average percentage decrease there was a year? Using trial and error, I came up with 20.94%, but it took a while. 
If x is the starting amount and y is the ending amount (less than x) after n years, the the annual % decrease is 1 - (y/x)^(1/n), i.e., 1 - n[sup]th[/sup] root of (y/x).
Does ^ mean multiply? If so, the answer using your formula is just half of the total decrease (37.5%). That’s not right.
I think nivlac neglects the factor of 100 that the “%” involves.
MadMaxine, you might look up “logarithms”, which lie at the center of your question.
You could do this with base 10 logs. These have the effect of counting the number of zeros in numbers like 10 (whose log is 1) or 1000 (whose log is 3). But they also work for numbers in between, like 320 (whose log is about 2.4). If something changes at a constant percent per year, its log is changing by a constant amount per year.
There are also base “e” logs. The number is 2.71828… and it gives base e logs a variety of special and useful properties. For this reason they are often called “natural logs”. There’s some other name for them too, but I can’t remember it. Hmmm…
nitpick: you need to multiply the answer by 100 to get a %
Edit: I see Napier was faster on the trigger finger.
What you’re looking for is a number x such that, over a period of n years, the value decreases from S[sub]0[/sub] to S[sub]0[/sub](1 + x)[sup]n[/sup]. You find this by dividing the final value by S[sub]0[/sub], setting that ratio equal to (1 + x)[sup]n[/sup], and solving for x. The solution you get will be equal to the formula that nivlac posted.
In this case, the ratio of the final value to the starting value is .675 and n = 2, so x = -0.178. In other words, the value of the car is decreasing by about 18% per year.