 and try a couple more difficult problems here today. So number 6 looks very similar to number 4 that we did earlier except you'll notice inside the parentheses I see a G to the 4th in the numerator and a G in the denominator and I also see an H to the 6th in the numerator and an H to the 4th in the denominator. So you have some options here. What I am going to show you though is I'm going to simplify all the stuff that's in these parentheses first because I like things to be smaller and then I'm going to square everything with this exponent on the outside. Now you are more than welcome to square everything first but that's going to make it all bigger and grosser and not as pretty and then you can simplify things from there. So either way we should end up with the same answer but I think it's easier to simplify first. So again let's start with those coefficients so we're going to divide 12 by 4 and get 3. Then now let's look at the exponents. This is G to the 4th over G to the again there's an imaginary 1 in there so G to the 1st. The rule is when you divide you subtract your exponents so that's going to be G to the 3rd because 4 minus 1 gives me 3 top minus bottom. Then I'm going to have H to the 6 minus 4 which is 2 and then all of that is still going to be squared. And again if you don't really understand how I got those numbers out of there, those exponents out, you want to go back and look at what we did with problem number 3 back here on the first video. Okay. All right so now we're going to go ahead and distribute our exponent through to everything and that's going to give me 3 squared. Then I'm going to have G to the 3rd squared so remember that's G to the 3 times 2 and then I'm going to have H squared squared so that's H to the 2 times 2. So when I simplify all of that my answer is going to be 9 then G to the 6th and H to the 4th. I have all my exponents that are positive and I only see one of each variable so that is my final answer. Okay so I challenge you to try it the other way, distribute your 2 through and then simplify everything and see if you get the same answer. But if you're happy with this then that's fine too. Okay problem number 7 doesn't look so bad but you'll notice on our B variable we have a negative exponent. So we're going to have to use the rule that with negative exponents they get moved. So because this one's technically in the numerator because we can think of all this is over 1, that means this negative exponent is going to get moved to the denominator and become positive. If the negative exponent was in the denominator it would get moved to the numerator and again become positive. So my final answer for this one is going to be A squared over B to the 3rd. And that's my final answer because I have an A variable and a B variable. Those are different variables so I'm done. But I could not leave my answer how I started with this problem because the directions specifically say all exponents must be positive. Okay fantastic problem number 8 I see a negative X or negative coefficient here. That's perfectly fine. We just can't have negative exponents. And technically in the denominator I have a coefficient of 1. So negative 3 divided by 1, well that's just going to be negative 3. And here I see that I have some M's and I have some N's. So again if we break those up and look at those separately I'm going to end up with M to the 5th over M to the 4th. So that's going to give me M to the 5 minus 4. And then I'm going to have N to the 2 minus 8. Now you notice that 8's bigger than 2 so that means this N variable is eventually going to end up in the denominator. But we'll just stick with this for now. That's my equals here. So that's going to give me negative 3 M to the 1 and N to the negative 6. So now we're back to a problem like problem number 7. We have a negative exponent here. So I'm going to move that piece and that piece only to the denominator. My negative 3 again is a coefficient so that doesn't get moved. So that'll be negative 3 M. We can put that one up there if we want to but we don't have to. And then that's going to be N to the 6th. Because once you move that exponent it becomes positive. And that's my final answer. Okay. Again if you don't want to go through all these different steps you could just look to see that the N to the 8th is the bigger one. So that's where my N is going to have to go. But again I prefer to follow these step by step so you don't get too confused with all these different rules. Okay. Number 9 is very similar to problems that we did before with just problem number 8 actually. Except this one's got negative exponents. Here's one in the denominator. Here's one in the numerator. Here's one in the numerator and the denominator. And we have 3 variables with this one instead of just 2. So that's okay. We'll just deal with it the same way we did before. So start with our coefficients. Negative 40 over 6. Well 6 doesn't go into 40 evenly but we can reduce that down. So negative 40 over 6. Let's see a 2 will go into both of those. So that'll leave us with negative 20 over 3 for a coefficient. If you prefer to write that as a decimal obviously go ahead but this is ugly as enough as it is. Now we have to do our top variable minus our bottom variable here with those exponents. So I'm going to have x to the 2 minus, you know, I'm going to slide a parenthesis in here and then put a negative 4 because remember that if we have minus a minus that's going to turn into a positive. Then I'm going to have y to the, again start with the top, negative 3 minus 4, minus 4. And then with my last one I'm going to have z to the minus 5 or negative 5 minus a negative 2. So again the double negative here is going to become a positive but we'll deal with that one in just a second. Okay so that gives me negative 23rds, the coefficient gets carried down. I'm going to have x to the, that ends up being a 6 because again like I talked about before, 2 minus a minus 4 is 2 plus 4 which is 6. I'm going to have y to the negative 3 minus 4 is negative 7. And then I'm going to have z to the negative 5 minus a negative 2. So negative 5 plus 2 gives me a negative 3. Okay so I've now got my variables written once but you'll notice that these last 2 have negative exponents on them so that means we have to write them in the denominator. Now since our coefficient was a fraction I'm going to go ahead and stick the x to the 6 in the numerator with my negative 20 and then I'm going to stick the y to the 7th and the z to the 3rd in my denominator with my 3. If you prefer to leave them separate again that's fine. That's just a matter of how you want to write them. So negative 20x to the 6 since that was a positive exponent and 3y to the 7th z to the 3rd since those were negative exponents they get kicked to the denominator and that's our final answer. All right the last one, whoo it's the ugliest one yet, so we have negative coefficients as well as some negative exponents as well as a negative exponent on the outside. Okay so where do we want to start with this one? Again I think it's smart to simplify everything inside the parentheses first. So let's take a look at those coefficients, a negative 30 over a negative 12. Well 12 doesn't go into 30 very evenly and you notice the negative negative that's going to be a positive but I do notice that 3 will go into both of them. So if I divide negative 30 by, well negative 3 technically, that's going to end me up with a 10 and then a 4 and the denominator, oh I guess the 6 is going to go into both of them too. That's okay we can reduce it in the next step. Then we're going to have j to the second, j to the negative 5. So that'll be j to the second minus a negative 5 and then we're going to have a k to the negative 2, let's see, minus 4 and all of this junk is still to the negative 3 power, we'll get to that in a second. All right so let's simplify that coefficient again since I didn't pick the biggest one to simplify it first. So that'll give me 5 halves, if I do 2 minus a negative 5, that gives me j to the 7th and then I'm going to have k to the negative 2 minus 4 is a negative 6 and again, all of that is still to the negative 3rd. Okay so simplifying again what's inside these parentheses first, I'm going to go ahead and put this k to the 6 in the denominator since it's a negative exponent and then the j to the 7th in the numerator. And you'll see in a second my method to my madness, so 5j to the 7th over 2k to the 6th. All of that is still to the negative 3rd power. Okay now because this negative 3rd power goes to everything, it's basically just going to flip flop this whole fraction on the inside. So everything that's in the numerator goes to the denominator, everything that's in the denominator goes to the numerator. So what this is going to do then, I'm just going to flip this whole fraction on the inside which is why I wanted to write it as one big fraction. 2k to the 6th over 5j to the 5th, oops sorry that should be a 7, my fault. Erase that out of there. j to the 7, then all of that is to the 3rd power. Okay now we have something that we can do like we did before so we want to distribute this power to everything and I'm just going to do this in one last step since I'm running out of room here. So 2 to the 3rd is 8, k to the 6th to the 3rd, so power to power we multiply. That'll be k to the 18 divided by 5 to the 3rd is 125 and j to the 7th to the 3rd, again power to power you multiply will give me j to the 21 and finally we are done with this problem but this one really makes sure that you understand what's going on with all these different rules and you may have had a better way to do this than the way I did it and we may end up with the same answer, at least we shouldn't end up with the same answer but just remember, you know take it stop by stop whichever way you're going to do it is fine. I think this way is the easiest again simplifying what's inside the parentheses first and then going from there but if you find a better way then that's fantastic too. All right, thank you for listening to this about your exponents and we'll deal with some fractional exponents next.