Hello and welcome to another Algebra 1 lesson by EMath Instruction. My name is Kirk Weiler and today we'll be doing Unit 6, Lesson 4 on simplifying fractions that involve exponents. So very often we are tasked with writing equivalent exponential expressions.
We started doing that in the last lesson, right? And we looked at two very important exponent laws. Today we're going to start to develop the third very important exponent law which really has to do with what happens when we divide two things that have the same base but are raised to exponents. Now of course division and fractions are tied together and so all of this is sort of in the context of simplifying fractions which also involve exponents. So let's get right into it and really look at the key exponents and fractions.
And a little hand up here. We now have two exponent properties. we will develop a third that involves fractions. First, we need to understand a very basic idea involving fractions. So, let's take a look at exercise number one.
Simplify each of the following fractions. Alright, so what I'd like you to do is pause the video now, look at each one of these fractions, and see if you can figure out how each one of them simplifies. Well look, fractions and division are the same thing. So when I look at this, I literally should think about it as what is 7 divided by 7. And of course, 7 divided by 7 is 1, because 1 times 7 is 7. Right?
In letter B, I have 5 to the second divided by 5 to the second. Now if you'd like to think about that as 25 divided by 25, you go right ahead. But the plain fact is 5 to the second divided by 5 to the second is equal to 1, right? Now in both of those two cases, we're dealing with just numbers, numerical examples.
Letter C, we have something that's purely in terms of a variable, n divided by n. Right, well now you can even think of division as, well how many times does that thing go into that thing? Well how many times does one n go into one n?
Well it goes in there one time. And finally, as if it wasn't enough, right, our last one mixes numbers with variables, 10 times x to the third divided by 10 times x to the third. Well, again, it's got to be 1 because 10 times x to the third times 1 is 10 times x to the third, right? And this illustrates something that is critically important to understand about fractions, right? Any quantity, any quantity.
Except for zero, when divided by itself is equal to one. This idea is critical to simplifying fractions, even ones without exponents. Another way to put this is any fraction whose numerator and denominator are the same is equal to the number one. X divided by X, 22 divided by 22, Y over Y, X times Y over X times Y. Any fraction whose numerator and denominator are the same expression is equal to 1. So let's see how we can use that idea to actually simplify just a normal numerical fraction.
Exercise number 2. Consider the fraction 18 over 30 or 18 thirtieths. Letter A, what is the greatest common factor, also known as the greatest common divisor, of the numerator and denominator? Alright, this should be easy enough to think about. Pause the video now and just come up with the greatest common factor of 18 and 30. Alright, well it's equal to 6, and you can either think of it as what is the largest number that divides into both 18 and 30, or you could think about it this way, right?
Whoops. 18 is equal to 5 times 6. Oh. That's not true.
18 is equal to 3 times 6, and 30 is equal to 5 times 6, right? This is the largest common factor of both of those. Now let's take a look at letter B.
Write the numerator and denominator in factored form using the GCF from A. This is simple enough, right? So remember, The substitution property of equality says, look, if 18 is equal to 3 times 6, and 30 is equal to 5 times 6, then I can replace both 18 and 30 with 3 times 6 and 5 times 6 respectively.
Alright? Now, the final part of this problem, letter C. Using B, simplify 18 over 30 by writing it as the product of two fractions, one of which is equal to 1, right?
Now, this is important. Before we kind of... I want to remind you of something that's critically important that some of you may have forgotten.
And it really hinges on the rest of what we do today, which is how you multiply fractions. So just really quick, if I had something like this, 3 fourths times 7 fifths, what you should remember literally from 3rd grade on is that to multiply two fractions, you multiply their tops and you multiply their bottoms. So in this case, I'd get 21 20s. All right.
But look, equality is a two-way road. Equality is a two-way road. So if I can go in this direction, I can also go in this direction.
All right. That's incredibly important to understand, right? I can multiply two fractions by multiplying their numerators and multiplying their denominators. I can unmultiply two fractions by going this direction.
Now how does that link into what we're doing right now? Well, I know that 18 divided by 30 is the same as 3 times 6 divided by 5 times 6. But that's the same as 3 divided by 5 times 6 divided by 6, right? I can literally unmultiply these fractions, and now this is just 3 divided by 5 times 1. And that's just 3 fifths.
Now don't get me wrong, right? The vast majority of times that you've written a fraction in simplest form, you've looked at it and you've said, alright, what's the largest number that goes into both? That's 6, and I'll just divide that out and I'll get 3 fifths. But what is truly going on here, what is truly going on when you've done this all along, is this process, right? is this process.
18 is 3 times 6, 30 is 5 times 6, so that's the same as 3 fifths times 6 sixths, that's the same as 3 fifths times 1, which is 3 fifths. And that is really critical to understand, because in a moment, we're going to start bringing variables into it. And that moment is now.
We can use the same idea in the last problem to simplify fractions involving variables and exponents. Let's take a look at exercise number 3. Whoops! Let's come back to it. Here we go.
Wish I could do that on command. Exercise number three. Consider the fraction 2x to the 5th divided by 6x to the 3rd. Letter A. What is the greatest common factor of 2x to the 5th and 6x to the 3rd?
If needed, write both in expanded form. All right, well, let's actually do this together. So, 2x to the 5th.
is 2 times x times x times x times x times x. That's what it is, literally. And 6x to the third is 3 times 2 times x times x times x. Alright, I'm just writing these things in their factored form. So what is common between these two?
Well, the 2 is common. These x's are all common, right? So what is that?
The greatest common factor is 2x to the third. Right? That's my GCF. Okay, now one way I can think about my GCF is I can think about it as what is the greatest common factor of 2 and 6, and that's 2, right? And then I can say, well, I've got an x to the 5th and I've got an x to the 3rd.
There is an x to the 3rd that sits inside of an x to the 5th, right? It's x to the 3rd times x to the 2nd, so my x to the 3rd is what's common. Now let's take a look at letter B. Simplify the fraction using the technique from exercise 2. Okay, so here's how I'm going to think about it, right? 2x to the 5th divided by 6x to the 3rd is 2x to the 3rd.
times x to the second, all over two x to the third, times three. Right? That's all, right?
I've just taken my GCF and I've taken it and I've made two x to the fifth and to 2x to the third times x squared, and 6x to the third is 2x to the third times 3. But now this is equal to 2x to the third divided by 2x to the third times x to the second divided by 3. And that is now 1 times x to the second divided by 3, and that's just x to the second divided by 3, right? It's exactly the same process, right? We are finding the greatest common factor of the numerator and divider and literally what a lot of math teachers will say is we are canceling them out. You'll even get some math teachers that will kind of do this and just say well there's our final answer.
I would encourage you though instead of doing this to really go through this process, really understand what's going on here. We're really just using exponent laws. Maybe we're kind of running them backwards a little bit, right?
We're making x to the fifth into x to the third times x to the second, right? Then we're turning that six into a two times three. Then we're finding out what's common between those two, rewriting it as the product of two fractions where one of them is equal to one, and then one times that other fraction just gives us that other fraction.
No different than simplifying 18 thirtieths. Well, maybe a little bit more complicated. All right.
It is important to be able to be fluent with this type of simplification. And exercise four gives us some nice practice. Although there are some tricky things in here. So let's take a look. Exercise number four.
Simplify each of the following fractions by rewriting as the product of two fractions, one of which is equal to one. Use the technique of identifying the greatest common factor of the numerator and denominator. Alright, well the first one, x to the second divided by x to the sixth.
Let me write that out. Now you've got to watch out on something like this. It'd be super duper easy to be like, oh, well that's x to the third because six divided by two is three.
And yet, really, what we have in this situation is x to the second times x to the fourth. all over, I'm going to say x to the second times 1. If you really feel the need, you can always put a times 1 in there anytime you want, right? But definitely x to the second times x to the fourth is x to the sixth.
That's just that exponent property that says when I multiply two things with the same base, I add their exponents together. But now I can rewrite this as x to the second divided by x to the second times x to the fourth divided by 1. And that's 1 times x to the fourth divided by 1, which is x to the fourth divided by 1, which is just x to the fourth. Keep in mind, dividing by 1 doesn't do anything, right?
8 divided by 1 is 8. x divided by 1 is x. x to the fourth divided by 1 is x to the fourth. All right, why don't you try letter b?
See how that one goes for you. Play around with it a little bit. All right, well we have 10m to the 9th divided by 2m to the 4th. Let me write, whoops, that was weird. Okay, or not.
Here we go. 10... m to the ninth divided by two m to the fourth.
Okay, well, I can write that as two m to the fourth times five m to the fifth, all divided by two m to the fourth times one. Hello, times one. That red thing is still there, times one.
Okay, all right, and again, let's talk about this, right? Definitely the same thing, right? I got two times five.
5 is 10, 10, 10. m to the 4th times m to the 5th is m to the 9th. And then I've got that 2m to the 4th down here as well, right? So that's just going to be 2m to the 4th divided by 2m to the 4th all times 5m to the 5th over 1, right?
So that's the fraction 1 times 5m to the 5th over 1. And that's just 5m to the 5th. I know that that's a lot of work to get down there, but again, it's really the process we're doing as opposed to just going like this. Right?
Ah, they just cancel out. They just go away. Why do they go away?
They go away because 2m to the 4th divided by 2m to the 4th is 1, and then 1 times 5m to the 5th is 5m to the 5th. Now, we've got some trickier ones coming up here, especially letter D. What I'd like you to do is pause the video now and see if you can do letter C on your own, and then we'll do letter D together, because there's something in there that can really trick people up.
Pause the video now. All right, well, let's take a look at letter C. We've got 6y to the third divided by 4y to the seventh. In terms of the numbers, right, we've got 2y to the third, 2y to the third, all right, that's our GCF, times 3 in the top, and times 2y to the fourth. in the bottom.
All right, let me just step back here and think about this. I've got 2y to the third times 3. 3 times 2 is 6. Then I've got that y to the third. I've got 2 times 2, which is 4. I've got y to the third times y to the fourth. That's y to the seventh.
It all looks good. In which case, I get 2y to the third divided by 2y to the third times 3 over 2y to the fourth. So that ends up being 1 times 3 over 2y to the fourth, and that's just 3 over 2y to the fourth. All right. Now the final one is a bit tricky, and let's take a look at why.
We have 4x cubed. All over 12x to the eighth. So when I think about that largest number that goes into both 4 and 12, it's 4. All right.
Then when I think about the largest power, right, that's x cubed and that's x cubed. And then the denominator, we would have 3 times x to the fifth. Now let's take a look at that, right? The top hasn't changed at all. It's 4x cubed.
The bottom I have 4x cubed times 3x to the fifth. So it's kind of difficult to sort of unmultiply these fractions when I don't have anything else in the numerator, right? That is where it's extremely important to think of it like this.
4x cubed times 1. Let me move my hypercube out of the way. I want to think of that as 4x cubed times 1 so that I can then rewrite this as 4x cubed divided by 4x cubed times 1 divided by 3x to the 5th. That's 1 times...
1 over 3x to the fifth and that's 1 over 3x to the fifth. Alright. At no point in time should something in the denominator in the bottom of the fraction suddenly be looking like it's a whole number.
The last thing you would ever want to do at this point, right, let's say that you did put this one there. The last thing you'd ever want to do is cross those out and be like the answer is 3x to the fifth. Because 3x to the fifth is by default in the numerator. 3x to the fifth would be 3x to the fifth over one.
1, right? But our answer is actually 1 over 3x to the 5th. So you really got to watch that, right? When everything cancels in the numerator, there's really still a 1 there.
Now, it's very important, right? In that exercise, we had numbers, we had variables, we had all sorts of stuff. It's really important to understand how to simplify simple cases to develop important patterns, right?
We want to be able to do this a little bit faster than what we've been doing. So let's take a look at exercise number five. Simplify each of the following.
Now before we go ahead and do that, I want you to take a look every one of these situations right. We've just got the same variable raised to a power over the same variable raised to a power. We've got no other numerical numbers in there.
We don't have 5x to the third, etc. So when I look at letter A and I see x to the 10th, divided by x to the third, right? That's going to be x to the third times x to the seventh all over x to the third times 1. And that's going to be x to the seventh over 1 or just x to the seventh, right? And I can really think of it as I've got x to the third here dividing an x to the third out here and leaving just an x to the seventh in the numerator, all right?
Why don't you go ahead and do letter B? Alright, we should see a very similar result. I have n to the 12th divided by n to the 8th, which would be n to the 8th times n to the 4th, all over n to the 8th times 1. Those two will cancel and be equal to 1, and that's just going to be n to the 4th over 1. or n to the fourth.
Again, it's as if we've got n to the eighth dividing an n to the eighth in the numerator and leaving us with only an n to the fourth in the numerator, right? It's really n to the fourth divided by one, but you know, it's like we wouldn't write five divided by one, we'd just write five. Now, in both of those cases, the power of the numerator was larger than the power of the denominator. Let's look at two cases where it's the opposite. Power in the denominator is higher than the power in the numerator.
So in letter C we've got y to the third divided by y to the ninth. So that could be thought of as y to the third times one over y to the third times y to the sixth. Those two will divide out to be one and we'll be left with one divided by y to the sixth.
Again, it's like y to the third in the numerator is dividing out y to the third in the denominator, but we end up with a 1 in the numerator. Okay? Very, very important.
The answer here is not y to the sixth. It's 1 over y to the sixth. Why don't you go ahead and do letter D?
Alright, very very similar. I've got t to the second divided by t to the fifth. Alright, and I can think of that as t to the second times one, all over t to the second times t to the third.
Those two divide out to be equal to one, and I just have one divided by t to the third. Alright, so big difference between whether the higher power is in the numerator and the higher power is in the denominator. So, let's kind of generalize that now in exercise number six. Fill in the following.
A. When the power of the numerator is greater than the power of the denominator, then what do we get? Well, look back at letter A and B and see if you can fill in this law for what we get when the power of the numerator is greater than the power of the denominator. Pause the video now.
Alright, well in this case, right, what do we have? We had x to the 10th over x to the 3rd, we got x to the 7th. 10 minus 3 is 7. 12 minus 8 is 4. So x to the a, divided by x to the b.
is going to be x to the a minus b. All right? That's when the number up here, right, the exponent in the top is bigger than the exponent in the bottom. On the other hand, right, When we had the other situation over here, right, we had y to the third over y to the ninth. Now we had nine minus three, which gave us six, and we had one over y to the sixth.
Here we had t to the second by t to the fifth and we got 5 minus 2 which is t to the third but it's in the denominator so when we've got the power of the denominator larger than the power of the numerator We end up getting 1 over x to the b minus a, right? You just got more in the denominator so the result stays in the denominator. Here you have more in the numerator, the result stays in the numerator.
Alright, let's use these quick patterns to simplify these a little bit faster. Let's take a look at exercise 7. Simplify each of the following fractions using the patterns developed in exercise 6. Alright, well what I'd like you to do is do letters A and B on your own, and then we'll revisit on C and D. Pause the video now. All right, let's take a look.
The first problem, let me write it a little bit bigger, x to the eighth divided by x to the fifth. Power in the numerator is larger than the power in the denominator, so that's just gonna be x to the third. Now in letter B, we have y to the fourth divided by y to the sixth. It's like four y's here cancel four y's there.
And I have one over y to the second, right? So six minus four is two, one over y to the second. By the way, notice, right? Notice how similar this is to the first exponent law, right? The first exponent law said when we multiply two things with the same base, we add their exponents.
Here, it's looking like when we divide. two things with the same base, we subtract their exponents. Hmm.
Coincidence? I don't think so. All right, now let's take a look at a couple where there's fractions involved that include numbers. So in letter C, we have 8 twelfths, and I'm going to undo sort of this multiplication of fractions. In other words, it's always okay if you have something like 8t to the seventh divided by 12t to the fourth.
To rewrite it as 8 twelfths... times t to the seventh divided by t to the fourth. Now I can simplify 8 twelfths the old-fashioned way.
I can say well alright 4 divides into 8 two times, divides into 12 three times, so this is going to be 2 thirds. On the other hand, t to the seventh divided by t to the fourth is going to be t to the third over 3. So that's 2t to the third over 3. That's it, right? I can separate out the sort of purely numerical fraction from the fraction that involves my variables. Why don't you go ahead and play around with letter d? All right, let's take a look.
So in letter D, same situation. I'm going to rewrite this as 15 divided by 3 times r. Maybe I'll write it as r to the first divided by r to the fifth.
Well, 15 divided by 3 is just 5. And r to the first divided by r to the fifth, because the power is larger in the denominator, I get r to the fourth. Now, I could easily see somebody looking at this and thinking, well, I don't know. Do I multiply the one by five, or do I multiply the r to the fourth by five? Well, keep in mind, right, that five is the same as five over one, or five firsts. Therefore, I just multiply the numerators, multiply the denominators.
And we have 5 over r to the fourth, not 1 over 5r to the fourth. All right, let's summarize. So today, we used properties of exponents along with what I would claim is the fundamental property of multiplying fractions to simplify fractions that involve exponents. And one of the most important things we kind of saw here was that we could break apart fractions so that we took greatest common divisors or greatest common factors, however you want to look at it, right, of both the numerator and denominator, separate them out into a fraction that's equal to one, and then the fraction that remains is our simplified version.
From that we develop these great patterns that allow us to subtract exponents and leave the result either in the numerator, if the power in the numerator is greater, or in the denominator if the power in the denominator is greater. So you're going to play around with that a little bit on homework, then we're going to develop this a little bit more into the third exponent law next time I see you. But for now, I just want to thank you for joining me for another Algebra 1 lesson by eMath Instruction.
My name is Kirk Weiler, and until I see you again, keep thinking and keep solving problems.