(Describer) Titles: Welcome to Algebra 1 - Negative Exponents

Hey, guys. Welcome to Algebra 1. Today's lesson focuses on negative exponents. Your knowledge on the laws of exponents and quotients and products will get you through this lesson. Ready? Let's go.

(Describer) On a screen, x-cubed over x to the fifth.

Okay. Think about this problem in two separate parts. First thing I'll do for the right side of it is expand it out.

(Describer) She writes.

X cubed divided by X to the 5th, I know for my numerator that means that I have X times X times X. When raising X to the 3rd power, I'm just multiplying by three Xs. For the denominator, if I have X to the 5th, then that means X times X times X times X times X. Let's make that fraction bar longer. For the denominator, with X to the 5th power, that means I'm multiplying five Xs together, okay? Now, knowing what you do about simplifying fractions, if you have the same thing in the numerator and the denominator, then cancel things out. I can cancel out one X up top with one X in the bottom, then another on top with another on the bottom. I'll cancel out another on the top and bottom. Now here's a little trick. We don't have a value of 0 on top, but you know there's always that invisible factor of 1? I'm going to throw up here that this is like X times X, times X, times 1. Just throw that out there. I have canceled out these three Xs up top with these three X's on the bottom, and what I'm left with is a 1 up top and I'm left with X times X on the bottom. Right? Now, I could represent that differently. I could say, well, 1 over X times X, that's really the same thing as 1 over X squared. Right? Okay. That's how the right side looks. I'll box that in to focus on that for the right. Now go back to the left side, how it started. We had X cubed divided by X to the 5th. One of the laws of exponents is, with a setup like this, when we're dividing powers, we can subtract the exponents. I could represent this as X to the 3 minus 5, because I need to subtract those exponents to simplify them. What is 3 minus 5? Let's come off to the side. 3 minus 5, that's negative 2. Knowing what you do about integer operations-- or just put that in the calculator if want. X to the 3 minus 5, that's the same thing as X to the negative 2. What that means is X to the negative 2nd power is really the same thing as 1 over X to the 2nd power. Sounds a little crazy at first. Let's do a couple more to make it make sense. Let's take this one. Going to apply that same process. So X to the 4th, I know that's the product of four Xs: X times X times X times X. Okay? Then on my denominator, X to the 7th, that's the product of seven Xs. So X times X times X times X times X times X times X. Let's count that. One, two, three, four, five, six, seven--yes. I'm going to expand, draw that fraction line longer. On the last one, I added that factor of 1, because multiplying by 1 doesn't change anything. That's how I'll start. So X times X times X times X, I'll multiply times 1-- throw that in there, okay? Didn't change the problem at all. Now, let's start canceling things out. I cancel out one X in the numerator with one in the denominator; cross them out. I cancel out another X in the numerator with one in the denominator; one in the numerator with one in the denominator; and another one in both places. And what I'm left with is a 1 in the numerator and then the product of three Xs in my denominator. So I'm left with 1 over X times X times X. I'm going to represent that differently, and say, well, this is the same thing as 1 over X to the 3rd power, right? So we expanded it out and looked at it that way. Let's look at how this problem started, this X to the 4th over X to the 7th. Knowing the laws of exponents, when you have to find the quotient of powers, subtract your exponents. I have X to the 4th, divided by X to the 7th, so that's X to the 4 minus 7, right? So 4 minus 7-- come off to the side-- that's negative 3. I could represent that as X to the negative 3. So X to the negative 3rd power means the same thing as 1 over X to the positive 3rd power. Are you starting to notice the pattern? Let's try another one. X cubed over X to the 8th. I know I can represent X cubed as X times X times X, and I'll throw in that times 1, because it didn't change anything. I have the product of eight Xs on the bottom, so X times X times X times X times X times X times X times X. Let's double check: three, four, five, six, seven, eight-- we do have eight Xs up there. I'll start canceling things out, let's start simplifying. I cancel out one X each in the numerator and the denominator; another one in the numerator with one in the denominator; and another in the numerator with one in the denominator. When you do that, you're left with that 1 in the numerator, and you have one, two, three, four, five Xs in your denominator. You can simplify that, write that a little differently. I'll leave my 1 in the numerator. And since I'm multiplying X by itself five times in my denominator, then I can represent that as X to the 5th, okay? We handled the right side; now let's handle the left side. X to the 3rd divided by X to the 8th, apply that law of exponents. For the quotient of powers, you know you subtract your exponents, so I need to simplify that as X to the 3 minus 8. Go ahead and clean that up, come off to the side. What is 3 minus 8? That's negative 5. So that means that X to the 3 minus 8 simplifies to X to the negative 5th power. So X to the negative 5th means the same thing as 1 over X to the 5th. Did you see that pattern? Each and every time, if you notice, when you raise a value to a negative power, it's the same thing as 1 over that value to the positive power. That's how you handle negative exponents. Let's look at this rule. Any value raised to the negative power-- any negative power-- it's the same thing as 1 over raising that value to a positive power, except for one value: 0.

(Describer) Negative Exponents: a to the negative r power equals 1 over a to the r power.

Let's get a pen. When you raise 0 to a negative power, this gets a little heavier than Algebra 1, so you won't really see this too much in this course. You can just focus on this rule, knowing that A will never be 0. Any value, except 0 raised to a negative power, means the same thing as 1 over that value to a positive power, that same power but positive. Let's try a couple problems. We have 5X and we raise that to the negative 2nd power. First, since I know my rule that anything raised to a negative power means the same as 1 over raising that same value to a positive power, I'm going to rewrite this as 1 over 5X to the positive 2nd power, because those values mean exactly the same thing. I'll work with that positive exponent to help me make my problem simpler. I'm going to keep simplifying this. In my denominator, I have a product and I'm raising it to a power, so I apply that rule of exponents. On the side, I'll do a little work. I raise that 5 to the 2nd power and that X to the 2nd power. We'll put our work here. We'll raise our 5 to the 2nd power and that X to the 2nd power. Let's keep chugging along. 5 squared, I know that's 25, and X squared, it kind of took care of itself; you got X squared. Once you're there, then you're all done. You can't take it any further. So 5X raised to the negative 2nd power means the same thing as 1 over 25X squared. Let's apply this again, give you a little practice here. We have X to the 5th, Y to the 8th over X squared, Y to the 10th. This one's a little different, but remember, don't memorize my steps, memorize my process-- that's what you want to learn. I have a quotient and I have a product up top. I'm dividing it by a product on the bottom. I have some of the same bases: an X and Y on top, X and Y on the bottom. That's my cue to group the like things together. I'm going to handle the Xs first. I need X to the 5th over X squared, and then I'll multiply that by Y to the 8th over Y to the 10th. I broke it up to help me deal with it. Knowing what I do about the laws of exponents, with a quotient of powers, I must subtract the exponents to make it simple. Here, I have a 5 and a 2, so it's 5 minus 2-- come off to the side for that--5 minus 2, that's 3. So as far as my Xs go, I have X to the 3rd power. For my Ys, I have an 8 and a 10; I need to subtract those. So 8 minus 10, that's negative 2. Our negative exponent popped up in this problem. So Y to the negative 2nd power. I'm going to keep going across with this one. X cubed, I don't have a negative exponent, I don't need to do anything. I'll come underneath. For the negative exponent, for Y to the negative 2nd power, I know I can rewrite that as 1 over Y to the positive 2nd power. Those values mean exactly the same thing. In algebra, you don't want negative exponents as your answer. When you've taken it as far as you can go, your exponents should be all positive. That's why we kept going. You could stop here and you wouldn't be wrong necessarily, but you'll see it written like this. Go ahead and multiply those two terms together, and you'll see it written as X cubed over Y squared. That's how you'll see that answer written-- we multiplied our numerators together. X cubed times 1, that's just X cubed. And you remember fractions, you can always make something whole a fraction by putting it over 1. 1 times Y squared, that's Y squared. That's how the problem went from this to this. But either one is really acceptable. Okay? Let's keep going. We have C to the 8th, D to the negative 5th over 3 and we're raising that entire quotient to the 3rd power. We apply a couple of laws of exponents here. I'll start by breaking this up, separating my numerator from my denominator. I'm going to write this as C to the 8th, D to the negative 5th, and I'm raising that to the 3rd power, over 3 to the 3rd. I just distributed that exponent throughout those pieces in my quotient. Now that I see the numerator, it's a product, I need to keep on carrying that exponent throughout. I need to raise C to the 8th to the 3rd power, so C to the eighth to the 3rd. And I need to raise D to the negative 5th to the 3rd power, so D to the negative 5th to the 3rd power. For my denominator, I have 3 to the 3rd. I can take care of that: 3 to the 3rd, that's 27. So the denominator is done; we've got to clean up the numerator here. So you know about raising a power to a power, right? Here you need to multiply your exponents together. So 8 times 3, that's 24. So that first part became C to the 24th, and negative 5 times 3, that's negative 15, so D to the negative 15th. And then we'd already taken care of that 27. Not quite done yet because we have that negative exponent so we've got to keep cleaning this up. Let me shift a little work around here, get some more space. Since we've taken this problem from this step to this step, I'll get rid of this up top to focus on where we are right now. I think I'll also get rid of that. Now, oop! I do want to see my work here. Let's bring this work up top. Let's get rid of that slash-through, because we didn't want to slash through that. Let's keep going. Get the pen back, here we go. All right. We need to handle that negative exponent, that negative 15. Now, here's when you can use a shortcut. The rule tells us, if I rewrite the rule-- if we have X to the negative A power--I'll change it up-- that's the same as 1 over X to the positive A power. As a shortcut, you can think about this is, if your negative exponent starts out on top in a quotient, drop it to the bottom and make it positive. Had the negative exponent been at the bottom, we'd need to bring it up top and make it positive. It's like you're shifting its position, like a shortcut. What that means is I can rewrite this as C to the 24th. Because this part of the term has a negative exponent, I'll bring it to the bottom. I still have that 27, so I'll drop that D to the bottom, changing that exponent to a positive 15. You've applied that same rule for negative exponents; you just took a shortcut to get there.

(Describer) C to the 24th over 27 d to the 15th.

That's the trick with algebra: once you see patterns, algebra is really a bunch of shortcuts that people figured out over time by noticing patterns. Use your knowledge of patterns to really help you get through the course. It is your turn to try. Look at these three problems, pause the tape, take a few minutes and work through them. When you're ready to compare answers, press play.

(Describer) Title: Simplify. Number One: 6y to the negative second power. Number Two: a to the seventh b to the sixth, over a to 11th b. Number Three: s to the negative third, t to the fourth, over 9, all to the second power.

(female narrator) Simplify. Number 1: 6Y to the negative 2nd power. Number 2: A to the 7th, B to the 6th over A to the 11th, B. Number 3: S to the negative 3rd, T to the 4th, over nine, all to the 2nd power. You ready to get going? Let's compare our answers. For the first problem, you had 6Y and you were raising it to the negative 2nd power. You should have got 1 over 36Y squared for that one. Number 2, you had A to the 7th, B to the 6th, over A to the 11th, B. That simplified to B to the 5th, A to the 4th-- over A to the 4th. Number 3, you had S to the negative 3rd, T to the 4th, over 9, and you were squaring that quotient. That answer was T to the 8th, 81S to the 6th. To see how I did these, stay with me. Let's look at that first one. I have 6Y; I'm raising that product to the negative 2nd power. I'll apply my rule about negative exponents which tells me that I can represent this as 1 over 6Y to the positive 2nd power. Now I just need to simplify this denominator. I've got my 1 on top and that's good. I'll raise these individual pieces of this product to the 2nd power. I have 6 to the 2nd power times Y to the 2nd power. Then just simplify this a little further. The 1 is good to go up top; 6 squared, that's 36. And Y to the 2nd power took care of itself-- it's already simple. That's how I got that first answer, that 1 over 36Y squared. Need to see the second one, here we go. I had A to the 7th, B to the 6th, over A to the 11th, B. First, I broke this up. I grouped the like things together. I looked at it as A to the 7th over A to the 11th, and I was multiplying that by B to the 6th over B. Then I used the laws of exponents, to know that if I have a quotient of powers, then I need to subtract the exponents. So for the As, you have 7 and 11. Come off to the side here-- 7 minus 11, that's negative 4. The first part of that was A to the negative 4th. For the Bs, I had B to the 6th, then B. When you don't see an exponent, there's an invisible exponent of 1. Always remember, if you don't see an exponent there's a 1 there. Just write it in. For this one, B to the 6th over B to the 1st. Subtract your exponents here, so 6 minus 1, that's 5. I know this was times B to the 5th and knew I wasn't done because I had a negative exponent. Use what we know about patterns and take a shortcut here. That A to the 4th is part of the numerator now. I need to drop it to the bottom and make it positive. That is how I got that second answer. If you need to see the third one, here we go. We had S to the negative 3rd, T to the 4th, over 9. We were squaring that whole quotient. I started out by breaking it up. I took my numerator, S to the negative 3, T to the 4th squared, over 9 squared. I used what I knew about exponents to apply the rule of raising a product to a power, so I raised S to the negative 3rd power, to the 2nd power. I needed to take these pieces separately. Now I need to raise T to the 4th to the 2nd power, so T to the 4th to the 2nd power, so I handled the numerator. Now, for the denominator, 9 squared, that's 9 times 9, so that's 81. My denominator's all taken care of, but I'm not done yet. I've got to handle that numerator. So raising a power to a power, negative 3 times 2, that's negative 6. So S to the negative 6. 4 times 2 here, that's 8, so T to the 8th. Then that was all over 81. You have a negative exponent, so you're not done yet. I may be able to fit that over here. We use what we know about negative exponents for a shortcut. T to the 8th isn't going anywhere, it's fine. That S to the negative 6-- I'll take care of that. Apply the rule about negative exponents. 81 is good. I'll drop that S to the bottom, making that exponent positive. Because remember, the rule tells us if I'm raising something to a negative power, it means the same thing as 1 over that to a positive power. You can follow that pattern and know you need to change its position in the fraction. We were all done there. I hope you're feeling more comfortable with negative exponents, that you saw how products and quotients and factors-- all that knowledge helped you get through these problems. See you soon. Bye!

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