Simplifying quotient of powers (rational exponents) | Algebra I | High School Math | Khan Academy

So we have an interesting equation here, and let's see if we can solve for K. We're going to assume that m is greater than zero, like always. Pause the video, try it out on your own, and then I will do it with you.

All right, let's work on this a little bit. You could imagine that the key to this is to simplify it using our knowledge of exponent properties. There's a couple of ways to think about it. First, we can look at this rational expression here: m to the 7/9 power divided by m to the 1/3 power.

And the key realization here is that if I have x to the a over x to the b, that this is going to be equal to x to the a minus b power. It actually comes straight out of the notion that x to the a over x to the b is the same thing as x to the a times 1/x to the b, which is the same thing as x to the a times 1/x to the b.

That's the same thing as x to the b, which is going to be the same thing as if I have a base to one exponent times the same base to another exponent. That's the same thing as that base to the sum of the exponents a plus b, which is just going to be a minus b. So we got to the same place.

So we can rewrite this as... So we can rewrite this part as being equal to m to the 7/9 power minus 1/3 power is equal to m to the K/9. And I think you see where this is going. What is 7/9 minus 1/3? Well, 1/3 is the same thing, if we want to have a common denominator, as 3/9.

So I can rewrite this as 3/9. So 7/9 minus 3/9 is going to be 4/9. So this is the same thing as m to the 4/9 power is going to be equal to m to the K/9.

So 4/9 must be the same thing as K/9. So we can say 4/9 is equal to K/9, which tells us that K must be equal to 4, and we're all done.