Multiplying & dividing powers (integer exponents) | Mathematics I | High School Math | Khan Academy

Let's get some practice with our exponent properties, especially when we have integer exponents. So let's think about what ( 4^{-3} \times 4^{5} ) is going to be equal to. I encourage you to pause the video and think about it on your own.

Well, there's a couple of ways to do this. One, you say, "Look, I'm multiplying two things that have the same base." So this is going to be that base, 4, and then I add the exponents: ( 4^{-3 + 5} ), which is equal to ( 4^{2} ). And that's just a straightforward exponent property.

But you can also think about why that actually makes sense. ( 4^{-3} ) power; that is ( \frac{1}{4^{3}} ), or you could view that as ( \frac{1}{4 \times 4 \times 4} ). And then ( 4^{5} ), that's ( 4 ) multiplied together ( 5 ) times, so it's ( 4 \times 4 \times 4 \times 4 \times 4 ).

So notice, when you multiply this out, you're going to have five ( 4 )s in the numerator and three ( 4 )s in the denominator. Three of these in the denominator are going to cancel out with three of these in the numerator. So you're going to be left with ( 5 - 3 ) or ( -3 + 5 ) ( 4 )s.

So this ( 4 \times 4 ) is the same thing as ( 4^{2} ). Now let's do one with variables. So let's say that you have ( a^{-4} \times a^{2} ). What is that going to be?

Well, once again, you have the same base; in this case, it's ( a ). And since I'm multiplying them, you can just add the exponents. So it's going to be ( a^{-4 + 2} ), which is equal to ( a^{-2} ). And once again, it should make sense.

This right over here, that is ( \frac{1}{a \times a \times a \times a} ) and then this is ( \times a \times a ). So that cancels with that; that cancels with that, and you're still left with ( \frac{1}{a \times a} ), which is the same thing as ( a^{-2} ).

Now let's do it with some quotients. So what if I were to ask you, what is ( 12^{-7} / 12^{-5} )? Well, when you're dividing, you subtract exponents if you have the same base. So this is going to be equal to ( 12^{-7 - (-5)} ). You're subtracting the bottom exponent, and so this is going to be equal to ( 12^{-7 + 5} ), well that’s ( 12^{-2} ).

And once again, we just have to think about why this actually makes sense. Well, you can actually rewrite this ( \frac{12^{-7}}{12^{-5}} ); that's the same thing as ( 12^{-7} \times 12^{5} ). If we take the reciprocal of this right over here, you would make the exponent positive, and then you get exactly what we were doing in those previous examples with products.

So let's just do one more with variables for good measure. Let's say I have ( \frac{x^{20}}{x^{5}} ). Well, once again, we have the same base and we're taking a quotient. So this is going to be ( x^{20 - 5} ) because we have this ( 5 ) in the denominator.

So this is going to be equal to ( x^{15} ). And once again, you could view our original expression as ( x^{20} ) and having ( x^{5} ) in the denominator. Dividing by ( x^{5} ) is the same thing as multiplying by ( x^{-5} ), and so here you just add the exponents. Once again, you would get ( x^{15} ).