 One of the most important things mathematicians do is they generalize concepts. So let's see where this comes from. So let's say I have a fractional expression like a to the fifth over a to the eighth. So confronted with something like this, we say no problem. We know how to handle division of exponential expressions. There's a theorem for that. So we have the equation a to the power m over a to the power n is going to be a to the power m minus n. And so this quotient is a to the power 5 minus 8. And we know how to handle subtractions like this. This is a to the power negative 3. And what? Weren't exponents supposed to tell us how many times something appeared? So how can a appear negative 3 times? To answer this question, let's rethink it. This time we'll ditch the theorem and focus only on the actual definition. So let's try and rewrite this without exponents. So a to the fifth is 5 copies of a. A to the eighth is 8 copies of a. And so when we rewrite it without exponents, we get 1 over a times a times a. Well, OK, let's write this with some exponents. This is 1 over a to the power 3. Now if we go back to the problem that we started with, we found that on the one hand, a to the power 5 over a to the power 8 looks like it should be a to the power negative 3. On the other hand, when we did this without exponents, we found that it actually was equal to 1 over a to the third. Now there's no reason why a to the power negative 3 has to mean anything. But the work that we did here suggests that we could interpret a to the power minus n as 1 over a to the power n. And that gives us an opportunity to expand what we can write in exponential notation. For example, x to the power negative 5, well that's by our definition 1 over x to the 5. Or we could take a somewhat more complicated expression, y to the power negative 3 over x to the power negative 5. So y to the power negative 3 is 1 over y to the third, and x to the negative 5 is 1 over x to the 5. Now we'll clean this up a little bit. This is a compound fraction, so we'll invert and multiply. And we'll end up with a simpler expression without negative exponents x to the 5 over x to the third. What about x to the 5 over x to the 5? Well, our theorem for the quotient of exponents says that this is going to be x to the power 5 minus 5 x to the 0. Now here's where we have to discard common sense and use only logic. In particular, x to the 0 says that x appears 0 times. So what do we actually have there? Well logic says that x to the 5 over x to the 5, anything divided by itself is equal to 1. And this suggests that a to the power 0 must be equal to 1. Well, not quite. There is one important thing we have to worry about. Because we get this result by dividing something by itself, it's important that we not try to divide by 0. And so this argument works as long as our base is not equal to 0. So we have to add the qualifier for a not equal to 0, a to the power 0 is equal to 1. Now a good student would remember this. A great student would ask the question, what happens if a is equal to 0? And for that, you'll need to take a little bit of calculus. But when you do get to calculus, you'll be able to answer the question, what happens if a is equal to 0? So we have horrible thing to power 0. The base is not equal to 0, so we can say without hesitation that the value of this horrible thing to power 0 is guaranteed to be equal to 1. How about something like square root of x? Is there some way we can write this in exponential form? Well, to begin with, because we're taking a square root, we'll want to make sure that x is greater than or equal to 0. So we'll do something that we commonly do in mathematics. We want to find something. Let's assume that we have it and see what we can say about it. Suppose square root of x is equal to x to some power a. Well, one thing we might note is that since x is greater than or equal to 0, we know that square root of x times square root of x must be equal to x itself. Over on the left-hand side, I have one factor of x, so that's x to the 1. Over on the right-hand side, if square root of x is x to the a, then I have x to the a times x to the a. But wait, my multiplication rule for exponents says that if I'm multiplying an exponential expression, I can add the exponents. So over on the right-hand side, I have x to power 2a. Well, since the expressions are supposed to be equal and the bases are equal, then it must be that the exponents are equal as well. So one must be the same as 2a, and so that tells us that a is equal to 1 half. And this generalizes. Again, for reasons that will become apparent a little bit later on, that x is greater than or equal to 0. But in general, x to power 1 over n can be defined to be the nth root of x. So let's see if we can simplify some exponential expressions, cube root of x times fifth root of x to the third. So let's see. Our definition says that we can express roots as fractional powers. So this cube root can be the power 1 third, and this fifth root is the power 1 fifth. But wait, we have x to the third to the power 1 fifth, and we do have a theorem that says what happens when we raise a power to a power, we get the product. So this x to the third to the 1 fifth is x to the 3 times 1 fifth, otherwise known as x to the power 3 fifths. And the rest of the expression remains unchanged for now. But wait, there's still more. So here I have x to the 1 third times x to the 3 fifths, and our theorem says that we can add the exponential expressions together. So that's 1 third plus 3 fifths, and here is the hardest part of the problem. 1 third plus 3 fifths is 14 fifteenths. Or let's try to rewrite this without negative or fractional exponents, x to the power negative 3 fifths. So our negative exponents, that's 1 over the same base, same power, so that gives us 1 over x to the power 3 fifths. Now to deal with this fraction 3 fifths, we'll use this product of exponents rule, and that allows us to split that fraction up into a whole number portion and a fractional part. So I need to write 3 fifths as a product of two things, so that'll be 3 times 1 fifth. And the reason that we wanted that 1 over 5 is that we have a definition for what x to the power 1 over n is. It's the nth root of x. So this x to the third to the power 1 fifth, that's the fifth root of x to the third. And so here's our final expression without negative or fractional exponents.