Today we’re going to talk about terms with different bases when we multiply, and we’re going to discuss pitfalls to avoid when you’re doing some of these math problems.

We’re going to start with a problem that has four expressions. We’re going to start with 5 squared and we’re going to multiply that by 3 to the 4th power times 5 to the 6th power times 3. If you take a look at these expressions within our big expression or our big problem here, we have two buckets, if you will, of numbers. I have a 5 to the 2nd power and a 5 to the 6th power; those are what we call ‘like terms’. And we have 3 to the 4th and then a 3. So you want to combine the like terms and put them together. It may help, if you’re someone who has some difficulty remembering the rules, it might be a good habit to get into, to just combine them right away. If you are not someone who has that problem, then no worries. You can just multiply them; group them in a different way, a more simple way.

Here I’m going to actually put them in parentheses. I know when I work with students parentheses are a nice visual because it helps separate numbers and group things together. So that’s why I’m putting them in parentheses here. So I have my 5 to the 2nd times 5 to the 6th and I am multiplying here and when I multiply two expressions with the same base my rule is to add my exponents. Very common mistake would be to multiply and get 5 to the 12th power, but we’re not doing that here. We are adding so we end up with 5 to the 8th. Because remember, if I were to expand this out, 5 to the 2nd is two 5s, 5 to the 6th is six 5s. So when I add up all my fives I have eight 5s. Over here when I grouped my 3 to the 4th times 3 notice I have nothing here next to my three, but I have a 4 exponent. A common pitfall that I’ve seen students do is to accidentally add or mistakenly add the 4 plus nothing because there’s nothing there and get a 3 to the 4th. But again, if you think about expanded notation, 3 to the 4th means I have four 3s, and then I have another one which means I have five 3s. So, just because there’s nothing there doesn’t mean the exponent is a zero. It’s actually a 1. So my exponents, when I add them, will be 4 plus 1. So I have 3 to the 5th power. Good pitfall to avoid. So my answer, to simplify this expression, would be 5 to the 8th power times 3 to the 5th power. And when you’re asked to do a simplification problem they’re not asking you to give the actual number. This would be a gazillion or something. I don’t even know what it would be, but the simplified version of this multiplication problem is simply 5 to the 8th times 3 to the 5th.

Okay. The next example is going to be another four-term example. I have 6 to the 5th times 6 to the 3rd times 6 to the 2nd times 4 to the 6th. Go ahead and pause the play button and see if you can work this out on your own and when you’re ready hit the play button again and we’ll review.

So when I start this problem I take a look at which bases are the same and I have three 6s and then I have a 4. So then I’m going to combine my first three together and in order to do that, I simply add my exponents and then I’m going to bring down my 4 to the 6th. A pitfall to avoid here is to not bring down that 4 to the 6th. Very easy to get started on this; jump to our answer because we can quickly add 5, 3 and 2 to get 10 and then forget to bring down that 4 to the 6th, but the correct, simplified version of multiplying all four of those original terms together is 6 to the 10th times 4 to the 6th. We definitely don’t want to confuse them. I’ve seen students accidentally take 6 and multiply it by 4 and put 24, then 10 plus 6 would be 16. That is not the same; you do not want to do that. A definite, definite pitfall to avoid there; you do not want to do that for these problems. Your bases again are simply the term that you’re multiplying the number of times your exponent tells you do so. So 6 to the 10th is 6 multiplied 10 times. So you want to be careful not to make that mistake.

Okay. Another one that we’re going to see frequently, especially when you do standardized tests, you might get a problem that looks like this: 8 to the 3rd times 5 to the 10th power, and they’ll ask you to simplify it. And, if I look at this problem, I have an 8 over here as a base and a 5 as a base over here. It’s actually already in simplified form so you can’t do anything to simplify this. You’re done when you get a problem like that.

Okay. The next one I want to show you is a quotient of 4 to the 10th over 4 to the 3rd power, and if you think back to the rules or take a look at your graphic organizer or your notebook, when you divide numbers that have exponents, if the bases are the same, the rule is to take your exponents and subtract them. So my simplified version of this would be 4 to the 10 minus 3 which is 4 to the 7th power, and that’s it!

The next one I’d like to show you is 5 to the 14th power over 7 to the 3rd power. And, again, you’ll be asked to simplify this and when I take a look at this my base on the numerator is 5, the base on the denominator is 7 so I cannot simplify this any further. I can’t subtract my exponents because my bases are different. So it is already in simplified format.

Okay. The next type of problem you may see is 1 over 9 to the 6th power. When I see 1 over 9 to the 6th power, this is really not in simplified format. But I can make it in simplified form, if I want to write it as a base with an exponent. So if you’re looking for one base with an exponent, I definitely don’t have that right now. However, I can rewrite my 1 as a 9 to the 0 power. We learned how to use the zero property and that’s how we would write it, and then my denominator would be 9 to the 6th power. My rules tell me I can take my exponents and subtract them and now I have 9 to the 0 minus 6 which is simply 9 to the -6th power. So now I have a base with an exponent.

Okay. Let’s try another one. A couple of tricky type problems you might run into include when you combine your multiplication and your division. So here I have 5 to the 10th times 5 to the 8th and I’m going to put that over top of or divide by 5 to the 14th power. The best way to do this would be to do the numerator first, and then once you’re finished you’re going to combine that with the denominator. So I’m going to start with my numerator, and I have the same base so I can add my exponents, and my denominators going to just stay the same. So my numerator is 5 to the 18th and my denominator...oops...messed that up there...my denominator is 5 to the 14th. The rules we have learned so far tell me that I can simply subtract my exponents since the bases are the same and my answer will be 5 to the 4th power. Okay. This last one is a nice problem and you will see many problems like this. We’ll have some more practice on this type of thing later; but I am going to have an expression that multiplies four terms. I have a 7 as my coefficient, I have a to the 10th, b to the 2nd, and c to the 4th power. That is going to be my numerator. My denominator is 14 a to the 9th power, b and c to the 5th power. Okay. Go ahead and pause your video and see if you can try this on your own, and then when you’re finished working it out hit the play button and you can continue to watch me continue through solving this.

Okay. Now this one doesn’t have anything on the numerator that you could combine, and it doesn’t have anything across the denominator that you could combine. So we’re going to have to go right away to dividing our tops and bottoms. So I’m going to start with my numbers, my coefficient. If I had a fraction without everything over here on the right (without my a, b and c), and I had just a 7/14ths to divide by 7 on the top and the bottom would give me a 1 on the top and a 2 on the bottom. So I’m going to treat this entire problem like that, like individual fractions and see how that works. So I’m going to combine my ‘a’s, I’m going to combine my ‘b’s, and I’m going to combine my ‘c’s. Okay. So, we already did the coefficient of 1 and 2, and now I have a to the 10th over a to the 9th. That is going to be a to the 10 minus 9 is 1, and b to the 2nd over b. Remember earlier we looked at this example and with no exponent there, it really means there is an exponent of 1. So I have b to the 2 minus 1 which is b to the 1 and over here I have c to the 4th divided by c to the 5th which means I have c to the 4 minus 5 which is c to the negative 1. And then over here, 7/14th reduced to one half so I could write that as 1⁄2 in front or I could write it as (I’m going to leave my ones off; I don’t really need to have that right now)...I could technically write it like this. You’re going to learn more about negative exponents and how to deal with them, shortly. So I think for this problem the preferred method of writing it would be 1⁄2 a, b, c to the -1, in this example. You’ll learn later that we generally don’t leave negative exponents, but that’s a little way down the road.