Common Core Algebra II.Unit 4.Lesson 5.Method of Common Bases

Remember, a lot of students like to ignore terminology, but this is the base. So right now literally we have the equation to the X equals 16 to the first, if you will. But they don't have common basis. One of them has a base of two. One of them has a base of 16. But you see 16 can be thought of as two to the fourth. So if we rewrite this as two to the X equals two to the fourth, then we can say X must be equal to four. Right? The idea is very simple here. If the bases are the same, if the bases are the same, then the exponents must be the same. Okay, this is due to the fact that exponential expressions are or exponential functions are one to one. Meaning that if the exponent, if what the exponent is gives you the same output, then the exponent has to be the same. I don't know how well that really turned out. But let's keep going. These first few are quite simple. Three to the X, we want to common base, 27, is three to the third. 

Therefore, X equals three. But you might just say to yourself, look, I know three to the third is 27. So X is three. That's great. Let's look at one that's more challenging. Here, we have 5 to the X here, we have one 25th. Well, one thing I know for certain is that one 25th is the same as one over 5 squared. So that means we have 5 to the negative two. And therefore, X is negative two. Oh. Now, up to this point, each time we looked at the left hand side, that was the base we used. On the other hand, here we might look at the 16 and the four. And the first thing that might occur to us is that 16 is the same as four squared. So I have four squared to the X is equal to four. Now I can use that property of exponents that says, hey, let's multiply these two together. And I get four to the two X is equal to four to the first. That means I can now set my two X equal to one and get X equals one half. All right. Now, some of you that really picked up on that whole fractional exponents thing could have also said, well, look, I mean, I can use 16. Because I know that four is the same as 16 to the one half. 

So you could have easily used 16 as a base. If you said, okay, I got 16 to the X equals 16 to the one half. Then X equals one half. All right. But if not, you go with the more common base, which is four. All right, take a look at what we've done. Pretty easy, maybe so easy that you don't really see the point. And then we'll go on to some that are a little bit more challenging. All right, I'm going to clear this out. All right. Exercise number two says simplify each of the following exponential expressions. Notice none of these are equations. I'm not asking you to solve any equation. I'm just asking you to simplify these. So pause the video for a minute and ask yourself how can you simplify these exponential expressions? All right. 

Well, in each one of these cases, we're using the exponential rule or exponent rule or property that says when we have something raised to an exponent, raised to another exponent, we simply multiply the two exponents so we have two to the three X here, we'd have three to the adex to times four X here, we'd have 5 to the negative three X plus 7. Because remember, we're going to have to multiply these two exponents together. So they're going to distribute. And likewise, here we'd have four to the negative three. Plus three X squared. Nothing very special about this particular problem. You might even wonder what it has to do with what we're doing in this lesson today. But we'll get to that in just a moment. This exponent law is going to be critical for what we're doing next. 

So let's take a look at some method of common based problems that are a little more challenging. I'm going to clear this out and pause the video if you need to. All right, here we go. All right, solve each of the following equations by finding a common base for each side. All right. So I take a look at this equation and I look at the 8 and I look at the 32, and I try to come up with a common base. Now, for me, what I know is that 8 is equal to two to the third, and I know that 32 is equal to two to the 5th. So I'm going to use two as my common base. So I'm going to write 8 as two to the third, but then that's being raised to the X and then I'm going to write 32 as two to the 5th. 

Now I'm going to use that great property of exponents. That we were just reviewing. And I'll get two to the three X equal to two to the 5th. That means at this point I can say that three X is equal to 5 and X must be 5 thirds. Now, I want you to keep in mind what we're not doing. We're not canceling the twos or anything like that. We're not doing that. We're just recognizing the fact that if I take two and erase it to a number, I take two and I raise it to another number, if those two things are equal, then the two exponents have to be equal as well. We're not canceling the two. All right. Let's take a look at letter B what's going to be the common base here. Think about it for a moment. All right, well hopefully you figured out it would be three. You see 9 is equal to three squared. 27 is equal to three cubed. So we can write this left hand side as three squared to the two X plus one. And we can write this side as three cubed. 

Again, I can use that great property. Whoops. Apparently I can use it incorrectly. You can use that great property that allows me to now multiply these don't forget to distribute that too. So it's now going to be three to the four X plus two. Equals three to the third, once I have those common bases, I can, if you will, drop them, not cancel them, just drop them. And I'll get four X plus two is equal to three. Four X is equal to one. And X is equal to one fourth. Isn't that great? Okay, letter C is more of a challenge, but why don't you pause the video and see if you can pull it off yourself? All right. Well, hopefully you figured out a good common base of 5. You could probably do other common bases, but a common base of 5 works out well. It works out well because a 125 is 5 to the third. And one 25th is 5 to the negative two. So we're going to say 5 to the third. All raised to the X is equal to 5 to the negative two. 

All raised to the four minus X again, just using that property of exponents multiplying them. Gives me this. Now that we've got that common base, I can get rid of it. And I'm sure you can finish it out. Isn't that cool? Now, let me be very clear about something. Really clear about something. All right? If I had something like this, let's say 5 to the X equals 8. This method would be fairly useless. Although not impossible, it is extremely difficult to write 5 and 8 with a common base. Pretty difficult. It can be done. I don't want to claim it can't be done. But it is very, very difficult. So generally speaking, this method only works if there is that convenient and relatively obvious common base. So look for bases like two, three. Not so much forks, then you probably would have a base of two. But two, three, 5, 6, 7, et cetera. All right, pause the video now. And then I'll clear the text. All right, here we go. Let's take a look at a multiple choice problem. 

Which of the following represents the solution set to two to the X squared minus three equals 64? Well, play around with this one a little bit. See what you can get. All right. Well, when the number on this side is prime, which two is. Pretty much that has to be the common base. And it is. 64 is two to the 6th, if you think about it. Now that we've got our common bases, we're going to get X squared minus three is equal to 6. Add three to both sides. We're going to get X squared equals 9, take the square root, and remember we're going to get both plus and a minus root when we do that. X equals plus or minus three. And both are completely valid. And of course, you can check these equations. 

Easily check them like every other equation. If I took like, let's say the X equals negative three root and I checked to see whether or not this was true. I could take that negative three square it. Subtract three. That would be negative three squared, which would be 9. Minus three is 64. And then I get two to the 6 equals 64. And that checks. All right. So the method is pretty straightforward on the back side of the sheet. There's always backs out of the sheet. We're going to be using it to solve some other problems, maybe some higher order problems, but essentially that's the technique. Find a common base for both sides of the equation. Manipulate and set exponents equal. Pause the video now and write down anything you need to. Okay. Here we go. All right. 

Two exponential curves. Y equals four to the X plus 5 halves, and Y equals one half to the two X plus one, are shown below. They intersect at point a a rectangle has one vertex of the origin, and the other at a, as shown. So that's not a square. It's just garden variety rectangle. I suppose it could be a square. But I don't think it is. We want to find its area. So we're going to try to figure out the area of this rectangle. What are aces fundamentally, what do we need to know about a rectangle to find its area? This is easy enough. Pause the video for now and write down what you need to know to find a rectangle's area. All right, well, I don't care how you phrase it. You got to know its length. And width, or its base, and its height. I almost felt the word hyper off. I'd be embarrassing for a math teacher. All right. I'll let her be. How would knowing the coordinates of point a help us find this area or the area? Well, think about that for a minute. I knew the coordinates of point a, how would it help me find the area? I wonder. Well, let's talk about it. 

Now, the Y coordinate, the Y coordinate would simply be the height. And the X coordinate, well, it would be negative, obviously, from the picture, so maybe it's absolute value, would be the base. All right, this is find the area of the rectangle algebraically using the method of common bases show your work carefully. Um. Oh, I get it. You see, I need to find that intersection point. So I need to solve this equation. Four to the X plus 5 halves is equal to one half to the two X plus one. Well, what I'd like you to do is pause the video now and try to solve that to find the value of X then think about finding the value of Y as well. All right, let's do it. Shuffle my notes a little bit so that I can make sure I see the proper answer at the end. Well, the common base is a base of two. Four is two squared. So I'll write that. One half is two to the negative one.

 All right, doing a little simplifying. We'd get two to the two X plus 5 is equal to two to the negative two X minus one. Wow, I almost forgot to distribute that. Now I can set those two exponents equal. And I'll get four X is equal to negative 6. And X is equal to negative three halves. All right, fair enough. Let's figure out what Y is equal to. Um, I think I'll go with this one. Y is equal to four to the negative three halves plus 5 halves, be careful. Those are all in the exponent, which would be four to the two halves, which would be four to the first, which is four. So we now know the coordinates of that intersection are negative three halves. And four, but that means it's area it's going to have a base length of three halves, a height of four, and we're going to have an area of 6. All right. 

This is totally the kind of problem that I could see being maybe even upwards of a 6 point problem on a common core assessment like the New York State regions. Granted, I doubt there would be part a in part B but there could be. Most of the time though you'd have to just simply know, well, if I find that intersection point, it's going to help me find the area. Pause the video now and write down anything you need to. All right, great. It's clear it. And let's move on. Okay, one more problem, a little multiple choice here. At what X coordinate will the graph of Y equals 25 to the X minus a intersect the graph of Y equals one over 125 to the three X plus one. Show the work that leads to your choice. All right, so notice all four choices have X solved for in terms of a that makes some sense. All right? So why don't you pause the video and see if you can figure out which one of these is the correct choice? All right. 

One of the things I like about this problem is although not impossible to use your calculator. If we didn't have that X minus a there, if we had X minus three, well then you could really put the two things in your calculator, graph, use your intersect command, et cetera to find the intersection. Here, though, to find the intersection, we're going to set the two equations equal. And we're going to use the method of common basis. Now that common base is a base of 5, we've seen both of these a couple times today. 25 is obviously 5 squared. So the X minus a one 25 is 5 to the minus three. All right, so we got our common base of 5. When we distribute those inner powers, we get 5 to the two X minus two a is equal to 5 to the negative 9 X minus three. We've now got our common basis. 

So we're going to set two X minus two a equal to negative 9 X minus three. No matter what we're trying to do here, never lose sight of what you're trying to do. That can happen in math often, right? We're trying to solve for X so I'm going to bring this negative 9 X to the side. It's going to give me 11 X minus two a is equal to negative three. And then I'm going to add a two a to both sides here. And that's going to give me 11 X equals two a minus three. But now, of course, it's easy enough to just defy and get X equals two a minus three divided by 11. Choice two. All right. And that's it. All right, so the method of common basis, not too bad. Pause the video now, write down anything you have to. And then we'll wrap up. Okay, here we go. So as I mentioned at the outset, there are very few algebraic techniques that really allow us to work with exponential equations. 

The method of common base is a very restrictive method, meaning that it doesn't work very often. But when it does work, it's really neat. And it works when you can look at two sides of an exponential equation and write both sides with the same numerical base. If you can do that, then the method of common basis works great to help you solve exponential equations. For now, I'd like to thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk Weiler, and until next time, keep thinking and keep solving problems.