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This unit begins with a fundamental treatment of exponent rules and the development of negative and zero exponents. We then develop the concepts of exponential growth and decay from a fraction perspective. Finally, percent work allows us to develop growth models based on constant percent rates of change. Geometric sequences are tied to exponential growth in the last lesson.
Get Access to Additional eMath Resources Register and become a verified teacher for greater access. Already have an account? Log in Math Hello and welcome to another common core algebra one lesson by E math instruction. My name is Kirk weiler, and today we're going to be doing unit 6 lesson number one, simplifying expressions involving exponents. Before we begin this very important topic, let me just remind you that you can find the worksheet and a homework assignment that go with this video by clicking on the video's description or my visiting our website at WWW dot E math instruction dot com. As well, don't forget about the QR codes at the top of every worksheet. If you scan those with your smartphone or a tablet, it will bring you right to the video associated with the worksheet. All right, let's begin. Now previously in this course, we had talked about exponents and how they behaved and how to think about them. So the first exercise is really just a review of a lot of different basic skills associated with exponents. What I'd like you to do is actually treat this as a warmup problem. Pause the video, take whatever time you need, maybe up to ten minutes and work through each one of these 6 problems, and then we'll go through each one of them in turn. Okay? Pause the video now and see what you can remember about exponents. All right, let's work through them. Letter ray says represent 6 to the third as an extended product don't evaluate the product. All right, no problem, right? 6 times 6 times 6. Thank you for telling me to not evaluate the product. So just at its most basic, what does 6 to the third mean? That's it. All right. Letter B, we have a little bit of function notation going on. If F of X equals two times X cubed plus 7, what's F of negative one? Well, remember how we interpret function notation wherever there's an X we put in a negative one, keep in mind your order of operations, we have to do negative one cubed first, right? That's negative one times negative one times negative one. That all ends up being a grand total of negative one. So that will be two times negative one plus 7, now the multiplication, that gives me negative two plus 7, and that gives me positive 5. All right, just a little bit of function notation. All right, letter C, if X to the third times X to the 5th is written in the form of X to the N, what is the value of N? Write extended products if you don't remember the exponent rule. Well, let me just remind for you the exponent rule. We had an excellent role that said that X to the a times X to the B was X to the a plus B so that means X to the third times X to the 5th is X to the 8th, so N is 8. Remember to always answer whatever question is being asked. Don't just leave it as X to the 8th. They wanted the value of the exponent, and that's 8. Letter D is even trickier. Something like this could easily show up, let's say on an SAT problem. If the expression 5 X to the third squared is written in the form a times X to the B, what's the value of a plus B? Well, a lot of people get confused when it looks like this, but don't forget, you know, at its essence, what that is is 5 X to the third times 5 X to the third, right? That's what it means to square something. Now I can use the associative property of multiplication to rearrange this a little bit. And rewrite it like this. 5 times 5 times X to the third times X to the third. But that will be 25 times X to the 6th using that property from over here. Now what that means is that a is equal to 25 and B is equal to 6, but a plus B, which is what they ask for. Is equal to 31. All right, letter E, little bit of what's known as scientific notation. If the length of a rectangle is three times ten to the 5th meters. And its width is two times ten to the fourth meters, what is its area written in scientific notation? Well, as we should know, the area of any rectangle is its length times its width. So in this case, that will be three times ten to the 5th times two times ten to the fourth, but again, we can use that associative property, whoops of multiplication to actually multiply the three and the two, and then the ten to the 5th times ten to the fourth, three times two is obviously 6. I'm going to use the time symbol there. Here we can use that exponent property again. It doesn't matter if the basis a number like ten or a variable like X but 5 plus four is 9, so we'll have 6 times ten to the 9th and that'll be square meters. Don't forget your units. All right, finally, letter F, rewrite the product three X squared squared times two X to the 5th, cubed as an equivalent expression and simplest exponential form. Look, if you don't know what else to do, all we start with the basics. So three X squared squared is this. Two X to the 5th cubed. Is this. Right? Now we can use the associative property of multiplication to group all the numbers together. IE three three two two two. And all the variables X squared X squared X to the 5th. X to the 5th, X to the 5th, all of this three times three is 9 times two is 18 times two, actually. Maybe I'll just do 9 times 8. That'll be easier, right? Three times three is 9, two times two times two is 8. So that'll be 72 X and now we can add two two 5 5 5. Those 5s all come out to 15. Plus four is 19. 72 X to the 19th. So how'd you do? How'd you do on these? Hopefully well, all of them really required just two things. A basic knowledge of what positive exponents represent. And that basic exponent law that says add two exponents when you multiply quantities that have the same base. Okay. But we're going to go on now and start to simplify exponential expressions that involve division. So I'm going to clear out the screen, pause the video now and write down anything you need to. Okay, here it goes. All right, let's move on. So today, what we're going to be doing is we're going to be playing around with expressions that look like this. We're going to try to simplify things that look like two X to the 6th divided by four X squared. And letter a says write this expression as the product of two fractions, one of which is equal to one. Well, before we even do this, I want to make sure that we have a little primer, okay? And that little primer is how we multiply fractions. All right, so if I gave you this problem, two thirds times 7 fifths, right? Would you know how to do it? And hopefully the answer is yes. Because you've learned this back in middle school, right? To multiply two fractions is almost the easiest thing on earth because we simply multiply the tops. And we multiply the bottoms. Okay? Multiply the numerators, multiply the denominators, using the correct terminology tops and bottoms. But the idea here is that we can also un multiply. If we can go this way, we can go that way because equality is a two way road. It's like ah, you can go this way. You can go that way. So let's take a look at this top fraction. Okay? We've got two X to the 6th divided by four X squared. Now, think about this for a minute, right? That's two times X to the 6th. But on the bottom, it's two times two X squared. All right? And in fact, we can write one of these fractions equal to one as long as the thing in the top, the thing in the numerator is the same as the thing in the denominator. Well, what's common to both things in the numerator is that factor of two. So this is actually a GCF problem. And what's common to both the numerator and the denominator in terms of X is X squared. So think about this first second. Let me write it down and then let's talk about it. This and this product. Are equivalent. All right, how do I know that they're equivalent? Because if I multiply them back out, I get this again. So this entire fraction is equal to one. Anything divided by itself, except for zero. Is equal to one. So it's beautiful because now I can simplify the expression by saying that that first fraction is equal to one, the second fraction is equal to X to the fourth divided by two, and so overall the fraction is equal to X to the fourth divided by two. Now that probably seems a little bit complicated and I agree it does seem a little bit complicated. But it's the way that we reduce fractions, right? It's even the way that you reduce just normal fractions. Like if I had something like four, 6. And I know what you've learned how to do is to divide both top and bottom by two. But we can really relook at four 6 as two two two divided by two times two divided by three, right? Where two divided by two is one and two divided by three is what it is. Let's do divided by three. And one times two divided by three is two divided by three, or two thirds. The real question becomes well, how do you how do you identify all of this? How do you know what gives you that fraction? And that's what we're going to work on in the coming problems. So a lot that's important on this screen. Pause the video now, think about it. All right. And then we're going to clear out the text. Okay, here we go. Let's keep working on this. Exercise three says rewrite each expression as the product of two fractions, one of which is equal to one. Then write it as an equivalent but simpler expression. All right, let's take a look at letter a. This is kind of cool. I have 5 to the third down here in 5 to the 7th up here. That means that there's actually a 5 to the third hanging out in both of them, right? The question is, what else do we have? Well, in the numerator, what would we multiply 5 to the third by to get 5 to the 7th? Thinking about that exponent law would have to be 5 to the fourth. Right? 5 to the third times 5 to the fourth would be 5 to the 7th. 5 to the third times one will be 5 to the third. But now all of this is equal to one. And of course we don't really need this thing down here. So we just get 5 to the fourth. Same deal up here. We have X to the fourth in the numerator X to the tenth and the denominator. Now if you think about GCF, because that's what we're really doing here. We're finding the greatest common factor of the numerator and denominator. And that's what's going in our numerator and denominator in this first fraction. What would we have here? We'd have an X to the 6th, right? X to the fourth times one is X to the fourth. X to the fourth times X to the 6th, this X to the tenth. This now is equal to one times one divided by X to the 6th, and that gets gives us one divided by X to the 6th. Watch out. It would be tempting to say that's X to the 6th, but it's not, right? One divided by X to the 6th is very different than X to the 6th divided by one. Oh, letter C looks horrible. All right, let's take them a bit at a time. Here I have X to the fourth here I have X to the first, so the best we're going to get in that fraction where they have the common the GCF is an X to the first. Then I have Y to the 8th and Y to the tenth. The best that we're going to get there, which is quite a bit better, is Y to the 8th. So really, these things are GCS. They're the greatest common factor of the numerator and denominator. Now, what do I have left over? Well, if I look at the numerator, I took care of the Y to the 8th, but I still have an X to the third. And in the denominator I took care of the X to the first, but I still have a wider second. This entire fraction is now equal to one. That's X cubed divided by Y squared. And so it simplifies into X cubed divided by Y squared. All right. And that's it. Okay. The key though, when you're separating these single fractions into two fractions, one of which is equal to one, is to find that GCF to find the largest thing the largest sort of term that sits inside of both the numerator and nominator. And we'll get more work on that and even more complicated problems. So pause the video now and write down whatever you need to. All right, let's keep solving problems. Next page. Holy cow, rewrite each of the following equivalent exponential expressions in simplified exponential form. Oh. Wow, this is crazy. But let's do it. Let's jump right in. We'll do the easier one together, letter a and then I'll have you have you really kind of pound on letter B by yourself, and then we'll go through it. Now, what do we do on this? Well, my piece of advice is always, when you don't know what to do, do anything you can do, just make sure it's not wrong. So one thing that I certainly know is not wrong is if I rewrite that numerator as three X squared times three X squared times three X squared. That's what it means to cube something. Now the denominator, well, that's just 9 X to the fourth. Remember, it's not 9 X times 9 X times 9 X times 9 X times 9 X because only the X is being raised to the fourth. Now I want to simplify this a little bit. Three times three times three is 27. X squared times X squared times X squared is X to the 6th. And now we're really where we were in the last prong, right? I want to look at the numerator. First look at the numbers. 27 and 9. Their GCF is just 9. Then I look at X to the 6th and X to the fourth, their GCF is X to the fourth. Right? Now the question is, what do I have left over? Well, I already took care of the 9, but I still need a three to get me up to 27. That's an X to the 6th. I need an X squared. Now here, I already have the 9 X to the fourth, so I just need a one. All of this fraction is equal to one. And that's just equal to three X squared. So that is my final answer. Three X squared. That's it? All right, let her be looks like a beast, and you do have to both manipulate the numerator and the denominator before you can even start simplifying. So pause the video now and take up to 5 minutes to work on letter B, okay? All right, let's go through it. So letter B okay, what we're going to do is we're going to first rewrite that out as 5 X squared, Y cubed times 5 X squared, Y cubed, and that's just what it means to square something. And likewise, the denominator ten XY times ten X, Y all right. Let's manipulate a little bit. Without formally writing down the associative and commutative properties, we're going to do 5 times 5 and get 25. X squared times X squared will give us X to the fourth. Y cubed times Y cubed will give us Y to the 6th. And then the denominator ten times ten will be 100. X times X will be X squared, and Y times Y will be Y squared. In order to simplify this now, we're going to multiply two fractions together, one of which is one. Look at the numbers. We've got 25 and a hundred. 25 goes into both of them. We have an X to the fourth and an X squared, the biggest thing that's common to both is X squared. Y to the 6th Y of the second, again, biggest thing is Y to the second. Now what do we have left? Well, in the numerator, we've taken care of the 25, and I'll need to have an X squared to bump that up to an X to the fourth, and I'll need a Y to the fourth to get that up to a Y to the 6th. Almost looks like Y and the denominator I have a 25, but I'll need to multiply that by four to get a hundred. But I've already got the X squared and the Y squared taken care of, if you will, I'd need to multiply by one, but I don't really need to do that. So all of this fraction is equal to one, all that fraction is equal to X squared Y to the fourth divided by four. So that's my final answer. X squared Y to the fourth. Divided by four. Talk about some exponent work. This is good stuff though. It's especially good to help you eventually prepare for taking algebra two. Okay? So I'm going to clear out the screen. You pause the video and write down as much as you need to. Okay, here we go. All right, let's go through one more problem, where we really try to understand the reasoning property. The reasoning behind all of this. Now, in this problem, it shows diagram on how to simplify one of these sort of more complicated fractions. And for each transition, it says give a reason a rule of property, et cetera, just a general description even that justifies the manipulation. The point here isn't to get the technical language exactly, correct? It's to explain, well, what's happening? What are we doing? Right? So, for instance, how do we explain going from this step to this step? And for me, that is simply the definition. Of an exponent. That's just what an exponent means. All right? Now what did we do when we went from this step to this step? We grouped those two twos together and we grouped those access together. We grouped that four and those two X's together. And then those two fours and that X together. But what does that mean when we group multiplication in a different order than when it was originally brute? That does have a technical name. That's the associative. Property. Of multiplication. So being able to rearrange multiplication in any order we want is the associative property of multiplication. Then we just ungrouped this fraction we put that here. And we put actually not just that. Sorry, that was bad. Let me expand my circle a bit. All right. We put all of this here, and then we left the rest of it, if you will, this part of it here. Now, what would we what would we say? How do we justify that? And I will justify that by simply saying, multiplication, a fractions. In other words, that just is the way fractions multiply. Then, of course, eventually that's equal to one, and we get one divided by 16 X so it's important. A lot of times students kind of walk through mathematics, which is manipulating symbols, right? Not really having good reasoning on why things are working the way that they're working. Here, though, we see all right, we use the properties of exponents to expand, if you will, then we use the associative property of multiplication to rearrange in whatever order we feel the need to. And then we use the property of multiplying fractions to literally, if you will un multiply a fraction. To get it to the product of two fractions, one of which is equal to one. All right. So I'm going to clear this out, write down what you need to. Okay, here we go. Let's finish up the lesson. Exponents are extremely important. They represent repeated multiplication, and that's something that we're going to see so much in unit 6 that in fact we call the unit exponents exponents and more exponents. So obviously they're kind of important. Simplifying exponential expressions, fractions that involve exponents in the numerator and the denominator is an important skill. Not so, so much for common core algebra one, although we will need it for the next few lessons. But it's a critical skill in algebra two. So it's good to have exposure for the first time at this level. For now, let me thank you for joining me for another common core algebra one lesson by email instruction. My name is script weiler and until next time, keep thinking. I keep solving problems. Remove Ads |
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