Welcome to negative exponents. The goal of this video is to simplify expressions involving negative exponents. When working with negative exponents, all of the basic exponent properties still apply, so we will be using all of these properties as we go through our examples. If you need to review these, you should watch the video entitled "Properties of Exponents." So here are the properties of negative exponents: a to the power of n is equal to one over a to the power of positive n. So for example, if we have three to the power of negative two, that would be equal to one over three to the positive two power, which would be one ninth. Now, for me, it's often helpful to think of three to the power of negative two as a fraction where all of this is over one. So if I move this across the fraction bar, it changes the sign of the exponent from a negative two to a positive two. It's also true that a to the n power and a to the positive n power are reciprocals. Remember, if two terms are reciprocals, their product would be one. And from what we learned in the previous video, if we're multiplying and the bases are the same, we add exponents. Well, n plus n would be zero, so a to the zero power would equal one. Or from this definition, we could rewrite a to the power of n as one over a to the power of n times a to the n. Again, this could be viewed as over one, which simplifies nicely to equal one. So whichever way you look at it, these two are reciprocals. The next property states a to the n divided by b to the m is equal to b to the positive m divided by a to the positive n. And this is kind of the idea I was alluding to above. If a factor is moved across the fraction bar, it will change the sign of the exponent. So notice how if this is moved up to the numerator, it becomes a positive m, and if this moves down to the denominator, it becomes a positive n power. Looking at a numerical example, if we had two to the power of negative two divided by five to the power of negative two, if we move this up, it changes the sign of the exponent. If we move this down, it changes the sign of the exponent. So this would equal five to the positive two power or five squared over two squared, which would equal twenty-five over four. And the last property of negative exponents: any base to an exponent is equal to the reciprocal of the base raised to the opposite exponent. So notice how if we have a over b raised to the n power, that is equal to the reciprocal b over a raised to the positive n power. So if we have four-thirds raised to the negative two power, that is the same as three over four raised to the positive two power, which would be sixteen over nine. Now in my mind, this really is not a new rule because we know the exponent on this a is one and the exponent on the b is one. So from the previous video, we know if we apply the power rule here, we would have a to the power of n over b to the power of n, which really brings us back to the previous rule where if we move these across the fraction bar, it changes the sign of the exponent. And if we wanted to, we could rewrite this as b over a raised to the n power. So all of these rules are closely related, but I would say the main rule you have to be aware of is if you move something across the fraction bar, it will change the sign of the exponent. So let's go ahead and apply these new rules with the basic properties of exponents. Here we want to simplify and write our answer without negative exponents. So here we're multiplying our bases. They are the same, so the rule is we will add our exponents. That's the product rule over here on the right. So we'd have two raised to the power of negative five plus two, which is equal to two to the power of negative three. But we cannot have negative exponents in our answer. We could make this into a fraction by putting it over one. So if we move this down to the denominator, it will equal one over two to the third power, which would equal one-eighth. For the next example, we're now dividing, so we'll apply the quotient rule where we subtract our exponents. And it's always the exponent of the numerator minus the exponent of the denominator. So this would equal x to the power of negative four. But again, to have our answer with a positive exponent, we just need to move this down to the denominator, which would equal one over x to the fourth power. On our next example, all of this is being multiplied together, so six times two, that would give us twelve. m to the negative four times m to the negative two, again we add our exponents to get m to the negative six power, and then n to the sixth times n to the first would equal n to the seventh. Again, I always find this helpful when I have a negative exponent to make it into a fraction. So in order to have a positive exponent here, we'll move the base of m down to the denominator. So we'll have twelve times n to the seventh in the numerator and m to the sixth in the denominator. Here we apply the power rule. Well, negative nine times four would be a positive thirty-six, and that would be our final simplified answer. On this next problem, there's a lot going on, and I almost take a look at this and see four problems, where we have the constants, we have the x's, we have the y's, and we have the z's, and they're all quotients. So we'll simplify each quotient separately and then combine them for our final answer. So negative fifteen divided by five would equal negative three. x to the negative one divided by x to the negative four would give us x to the third. y to the fourth divided by y to the first, four minus one would give us y to the third. And z to the negative three divided by z squared, negative three minus two would be z to the negative five. So the only issue here is we cannot have z to the negative five in our simplified answer. Remember, if this was in fraction form, we'd have a denominator of one. So we need to take this base and move it to the denominator, so our final fraction would be one times x cubed times y cubed, or just x cubed y cubed over three z to the fifth power. Okay, on this next problem, everything here is being multiplied together. So six times two would equal twelve. m to the negative four times m to the negative two, we're multiplying, so we add our exponents. This would be m to the negative six. This is n to the sixth times n to the first. That would be n to the seventh. Again, if this is in fraction form, we'd have a one in the denominator. Therefore, to make this a positive exponent, we can move it across the fraction bar or down to the denominator, which would give us twelve n to the seventh over m to the sixth. Let's go ahead and take a look at two more. Here we have powers raised to powers, so we are going to multiply our exponents. This would be x to the negative four times two, that's negative eight, and y to the negative three times two would be negative six power. So now if we move this across the fraction bar and move this up, so we'd have y to the sixth in the numerator and x to the eighth in the denominator. Okay, we have one more. There's quite a bit going on in this problem. Notice that we do have powers to powers, so we'll apply that rule first. So in our numerator, this would be a to the two times three or a to the sixth. Here we'd have a to the negative fourteenth and b to the seventh. And already on the bottom, we'd have a to the negative one times two or a to the negative two power, b to the positive two power. Now we do have an option on what we want to do next. I would probably go ahead and multiply these two together. Remember, when we multiply, we will add these exponents. We would obtain a to the eighth power, and everything else remains the same. So this next step, I'll look at this as two problems where we want to simplify the a's and the b's by applying the quotient rule, where we subtract. So here we have a to the eighth minus negative two, that would give us a to the sixth. Next, we would have b to the seventh minus two, which is b to the fifth. Again, we're not allowed to have negative exponents in our simplified answer, so we want to move this down to the denominator. So we have b to the fifth power over a to the sixth power. Okay, I hope you found this video helpful. Thank you for watching.