A new imaginary number? $x^c = -x$ Being young, I don't have much experience with imaginary numbers outside of the basic usages of $i$. As I was sitting in my high school math class doing logs, I had an idea of something that would allow solving for logs with negative bases or with negative arguments. Taking a similar idea to i, this number, when used as an exponent, would result in a negative number. In other words, as the title says, $x^c = -x$.

For starters, I just quickly mentally ran through simple examples of how it could be used in these cases, (ex. $\log(-100) = 2c$) but I soon began to wonder if it could have other applications and or just if it could be anything else.  

I'm also not 100% sure what to tag it, so I kinda guessed.
Robert's answer tells you correctly how to find, given an $x$, a complex $c$ such that $x^c=-x$.

However, it is worth pointing out that you cannot find a single $c$ such $x^c=-x$ will hold for every $x$. There's no existing number that achieves this, and if you try to extend the number system by simply postulating that $c$ exists (which I think was the point of the question), you run into this:
$$ 0 = \frac08 = \frac{-4+4}{8} = \frac{4^c+4}8 = \frac{2^c2^c+4}8 = \frac{(-2)(-2)+4}8 = \frac{4+4}8 = 1 $$
So if you want your $c$ to exist, but don't want to collapse everything into $0=1$, then you have to discard the exponentiation rule $(ab)^c=a^cb^c$. And without that rule, the result isn't really worthy of being called exponentiation in the first place.

I think you have been fooled by a common introduction to complex numbers, which goes like this: "There is no real number whose square is $-1$, but assume we had such a number anyway and call it $i$. Then we can calculate with the assumption $i^2=-1$, and, ta-dah! we have invented a new kind of numbers". This presentation, however, sweeps an important point under the carpet, namely that we had better be sure that assuming $i^2=-1$ doesn't allow us to prove falsehoods such as $0=1$ about the real numbers that we already know and love -- because if it does that means that we haven't invented a new kind of numbers but simply deceived ourselves.

For $i^2=-1$ it turns out that it's not too difficult to prove that it doesn't lead to nonsense. But even when this proof is shown, something is still swept under the carpet here, namely that if we assume $i^2=-1$ and the familiar rule that every number is either 0 or positive or negative (and the usual rules for positive and negative numbers), then we can still derive nonsense. So what's really true is that we can get a $\sqrt{-1}$ if and only if we're willing to drop what we know about positive and negative numbers (or "less than"/"greater than") when we're working with complex numbers.

Extending the concept of number is almost always connected to a loss of some rules that used to hold before the extension. How far we can push the extension depends on how many rules we're willing to let go of. So you can have your $c$, but the cost you must pay is the rules for how exponentiation works. And if you pay that cost, having $c$ becomes pretty useless anyway, so the whole expedition turns out to be pointless.

(But of course we couldn't have known that before we went and checked. The idea was good enough; it just didn't work. Perhaps your next one will.)
$$\log(-100)=\log(100i^2)$$
$$\log(-100)=2+2\log(i)$$