 Alright, today we're going to be working on some properties of exponents that you guys should have worked on as well, just to review some of those, and you can always look in your book too if you want a list of those handy, that might be a good thing to do. Alright, problem number one is a problem that just says h to the fourth times h to the seventh times h to the third, and make sure also we read the directions so it says the final answer should show each different variable written only once with this resulting exponent, and all exponents should be positive. So what that means is right now I see three h's, I have to write that so it's only one h, and because we're multiplying these h's that means we're going to go ahead and add those exponents, so we're going to end up with h to the fourth plus seven plus three, and when you go ahead and add those three numbers together that gives us h to the fourteenth. I have my one variable written once with an exponent that is positive, so I know that is my final answer. Okay, problem number two, I see two different variables on that one as well as some coefficients, so what we're going to do is just multiply our coefficients together, since all these parentheses mean we're going to be multiplying, so we're going to multiply five by six, and that's going to give us thirty, then I need to multiply my a's together, and then my b's together, because remember we can only have each variable written once, and again parentheses mean multiplying, so multiplying means we have to add those exponents, so we'd have a to the four plus two, and we'd have b to the three plus eight, and you don't have to write this stuff out every single time, but I think it's good for the first few times for you to get the hang of these problems, so that'll give us thirty, and then we'd have a to the sixth, and b to the eleventh. Again, I have my variables written only once, and all my exponents are positive, so I'm done, it's my final answer. Okay, number three also has two variables to it, but you'll notice this one is a dividing problem, so that means we're going to need to use the rule that when you divide, you subtract your exponents, and we can kind of break these up so you can see them separately, so those are m's there, and these are our n's over here, and of course I kind of wrote over each other, so hopefully you guys have problems printed out, you can see those, so that's going to give us m to the four minus three, because those are both of my exponents on the m's, and then we'd have an n to the eight minus two, since those are my n exponents, and you notice I did the top minus the bottom, that's the order you always have to go in, I know the four is bigger than the three, but if they were reversed, then well we'll deal with one of those later, but you always do the top minus the bottom, so that's going to give us m to the first, and n to the sixth when you subtract those, and you can leave it like that, or if you don't want to put that one exponent, you can write it as m n to the sixth, so either of those two answers would be good, again I'm only seeing one variable, each variable once with a positive exponent on it, okay problem number four is similar, but I don't see any common variables in here, I see an a, I see a b, and I see a c, so all those variables are different, but what I do notice on this one is there is a power on the outside, so now we've got to go to our next rule, and this one actually combines two of them together, where you have to distribute this power through, so that means it's going to go to the three, the a to the fourth, the b to the sixth, the five, and the c squared, and I also have to then use the rule that when I have a power to power I multiply those numbers, okay got a lot to remember here, and don't forget your coefficients, because those are very easy to forget in front there, so I'm going to end up with, let's write all this out, come on, all right, three to the third, then we'd have a to the four times three, since we multiply those exponents powers to powers, then we'd have b to the six times three, again powers to powers, but then you notice again in our denominator I've got a coefficient here, so this is going to be five to the third, that's one thing that gets tricky about these, is you have to remember your coefficients act different than your variables do, so here I'm doing a power, where here I'm multiplying, okay, so you've got to keep those straight, and then we'd have c to the two times three, because again I've got a power of two here and a power of three here, so I have to multiply those, okay when I simplify all this that gives me three squared I believe is 27, and then I'd have a to the four times three is 12, b to the six times three is 18, divided by five to the third is 125, and we'd have c to the sixth power, so I have three variables, but they're all different, so I'm good there, all the exponents are positive, I'm good there, and my coefficients here 27 over 125, I don't think that can be reduced at all, so that's going to be our final answer, if those two numbers did have a number in common that you could factor out you can reduce it down, but that's good for this problem, okay number five, so I'm now going to combine what we just did in problem four with our distributing of exponents along with what we did in problem number two, which is the multiplying things, so these problems get more and more complicated as you go, and that's okay, just take it piece by piece, so the first piece I see here is I need to distribute that exponent through, we'll write that out in a second, then the other piece I see is I need to distribute that power through, so we're going to do those two pieces separately, and then we're going to combine our answers together in the end, so I'm going to distribute my third power through, so that's going to give me two to the third, and I've got m to the fourth to the third, but again power to power you multiply, so that'll be m to the four times three, then I'd have n to the seventh to the third power, so that'll be seven times three in there, then times, because remember that's what all these parentheses mean, distribute the two power through or the square through, so that'll be five squared times m to the, now this one doesn't have an exponent on it, so remember there's always an imaginary one there, so that's going to be one times two, and then times n to the, that'll give us a three times two, okay, now we have to simplify this and then we'll combine those answers together, so that's going to give us two to the third, which is eight, m four times three is twelve, n to the seven times three is twenty-one, and then times five squared is twenty-five, m to the one times two is two, and n to the three times two is six, okay, so now we're to a problem just like number two, so we're going to multiply our coefficients together first, so eight times twenty-five is, I believe, two hundred, then we have our m's together that we can combine, so that'll be m to the twelve plus two is fourteen, and then n to the twenty-one plus six, and I'm adding again because we're multiplying those, so that'll be n to the twenty-seven, and that right there is our final answer to that part of the problem. We'll do the remaining worksheet on another video just in case you want to review this one again and review your properties again, then you can come back and finish that one up.