 Welcome to episode 18 of our Math 1050 class on College Algebra. I'm Dennis Allison and I teach mathematics here at UVSC. In the last episode, we introduced logarithms near the very end of that episode. And if you remember sort of the fundamental idea of a logarithm as a logarithm as an exponent, let me just work a couple problems real quickly here at the beginning to sort of tie today's episode in with the last episode. If we go to the green screen, you remember that if we have an expression something like 4 to the second power is 16, we say this is written in exponential form. But there is another way of expressing this called logarithmic form. And what we do is to take the base over here, base 4, and write that as a subscript. And the number 16 we put inside and the exponent 2 becomes the logarithmic value over here. So the logarithm is equal to the exponent. And we call this logarithmic form. And everything that we discuss in this episode and in the next episode is going to be about logarithms and basically this notion of a logarithm as being this exponent written on the ground. Let me just ask you a couple questions here that also pertain to the last episode. Suppose I were to write the log base 6 of the square root of 6 equals x. Can anyone tell me what x should be there? Steven? One-half. It's one-half. Tell us how you knew that. Well, because you can switch that and make it 6 to the x power is equal to the square root of 6. Exactly. Okay, let me just write that up here first. So in other words, we have an expression given as a logarithmic in its logarithmic form. So what Steven is suggesting is that we convert it to its exponential form because you and I have been working with exponential forms for a long time. Logarithmic forms we just introduced in the last episode, so they're rather new to us. So 6 to the x power, you remember x becomes the exponent equals the square root of x. So then what did you do, Steven? Well, if you take a number to the one-half power, it's the same as taking the square root of it. Yeah, the square root of 6 means that 6 to the one-half power. So it looks like x should be a half. So the number over here should be one-half. Exactly. Now, let's take one more like that. What if we said the log of x is equal to 1,000? Now, I don't have a subscript given here. So what's the base that we assume when I don't put a subscript there? 10. Base 10, yeah. So this is referred to as a common log, or a common logarithm. And if we convert this to exponential form, we'd take, whoops. That's a big number. Yeah, that's a big number, isn't it? Yeah, well let's go ahead and finish this. It's bigger than I expected it to be. So we have 10 to the 1,000 is equal to x. Oh my gosh, well that's a very big number, yes. But we can still work it. Write it out first. Yeah, you know what I actually meant to do was put the 1,000 inside as you can probably imagine. But the answer is 10 to the 1,000 power. So that's okay. Okay, now let's go to the objectives for this course, for this episode. And we have three objectives. First of all, we want to establish four laws of logarithms that we'll discuss first. Then we will show that the inverses of exponential functions are logarithmic functions. So we'll establish that fact using the laws of logarithms. And then we'll finally conclude by discussing the graphs of logarithmic functions and transformations of those logarithmic graphs. Okay, well the four laws of logarithms. Let's look at the four laws right here, the four laws, right here. Four laws of logarithms. Now here's the first law. It says if you take the log base A of a number m and if you add the log base A of a number n, you get the log base A of... Now you would think that if you're adding the log of m and the log of n that you might get the log of m plus n, but it's actually m times n. And the reason for this is you see what we're really doing is adding exponents. A logarithm is an exponent, as we've just mentioned. So if I add the exponent here and the exponent here, I get the exponent on the product. Let me just give you an example. Suppose we were to take the log base two of four and add on the log base two of eight. What is the log base two of four? Two? It's two, yeah. Because remember this is an exponent. In the exponent you would put on base two to get four for an answer. So the question is what exponent would you put on base two to get four for an answer? And the exponent would be two. Two squared is four. So this number is two. And what exponent would you put on base two to get eight? Three. Three. So this logarithm, or this exponent now, is three. So we should get five. Well you see if I use law number one, the log base two of four plus the log base two of eight is the log base two of the product thirty-two. And as a matter of fact, if you take the exponent five and take two to the fifth power, you will get thirty-two. So I've added the exponent on two to get four plus the exponent on two to get eight. And I get the exponent on two that gives me thirty-two. Or in this case three plus two is five. Let me just work another example of that. What if we said the log base seven of five plus the log base seven of four? Now right off hand, I don't know what the log base seven of five is, or what the log base seven of four is, but can anyone tell me what is the sum of those two? Log base seven of twenty. It's the log base seven of twenty, and actually I don't know the value of that number either, but these two expressions are equivalent. This sum is equal to the log base seven of twenty. That's using the first law of logarithms. Okay, now for the second law of logarithms, for number two, the log base A of m minus the log base A of n is equal to the log base A of, well now let's see, when I added the two logarithms I got the product. So if I subtract the two logarithms I should get the log of the, what would you guess? The quotient. The quotient, yeah. Good guess. m over n. I'll put that in parentheses just to set it off. Let's take an example of this. Suppose we were to take the log of a thousand minus the log of ten. Now since I'm not showing a subscript, of course these are base ten common logarithms, and according to this rule I should get the log of a thousand over ten, which is the log of a hundred. Okay, so if I divide a thousand over ten, by ten, then I can reduce that to be the log of a hundred. Now tell me what is the log of a thousand? It's three. Right. Let's see, Ginny maybe you could just sort of explain how you got that. I wrote ten to the x is equal a thousand. Okay, so she wrote ten to the x equals a thousand, where the x is the logarithm value. And I know that that's three. Right, ten to the third, ten to the third is a thousand. So she said, well if this logarithm is the exponent, the exponent is three, so she knew the answer would be three. And let's see, Susan, what's the log of ten? I have no idea. Jeff, what's the log of ten? It's one. Okay. And let's see, I'll do this one on my own. Three minus one is two. Okay, two. But guess what? The log of a hundred is two. So here's an example that sort of demonstrates the rule. But basically the purpose of the rule is to say what we're doing is subtracting two exponents. And when you subtract two exponents, you get the exponent that goes on the quotient. And this is a logarithm, so it's an exponent. And it's the exponent that goes on the quotient of the two. Okay, that is law number two. Let's go to law number three. Law number three says if you have a constant multiple of a logarithm, let me put parentheses around that to indicate this number is outside. So if I have a logarithm and I multiply it by some real number, that's the same thing as bringing the real number inside the logarithm. But you don't put it in front of the m, you put it as an exponent on m. So it's a totally different world here in logarithms. Things don't turn out the way you might first suspect. Let me demonstrate this. Suppose we were to take two times the log base three of twenty-seven. Well, actually that's pretty big. Let's say the log base three of one-third. The log base three of one-third. Now, according to this rule, how would I rewrite that if I bring the two inside? Log base three of one-ninth? Well, it'd be the log base three of one-third squared, which is one-ninth, exactly. So if you bring the two inside, you don't multiply by two, but you make it an exponent. Now let's see if this is really true. What is the log base three of one-third? That is, what is the exponent you'd put on three to give one-third for the answer? Negative one. Negative one, right. So what this really says here is two times negative one. Okay, now, do we get, let's see, that product is negative two. Do we get negative two on this side? Well, if I reduce this, let's see, if I reduce this, one-third squared is, as Stephen just said, is one-ninth. So what exponent would you put on three to give you not nine, but to give you one-ninth? I think that would be negative two, yeah. Three to the negative two is one-ninth. So negative two equals negative two. So this is not approved, but it's a demonstration of that rule. Okay, now we come to the fourth law of logarithms. And you know, if you look in your textbook, you will see other properties of logarithms listed in the section on laws of logarithms, and technically you could make this list longer, but I think the four laws that we're listing here are the ones that are most fundamental, and you can get by, you can get by with just these four. Suppose we have A raised to the log base A of M power. Now you notice I'm putting the logarithm up in the air like an exponent, because the logarithm is an exponent. A raised to the log base A of M, that has to be the same base as what I have down here. This answer is M. So if you take a number A and you raise it to a logarithm base A of M, you'll get M. I think maybe if I move that over a little bit, I can put an example right here beside it. Suppose we have two to the log base two of eight. What will we get? Well, can anyone tell me what is the log base two of eight? Three. It's three, yeah. Because this is the exponent that you put on base two to get eight for the answer. So that's a three. So that's two to the third power, and two to the third power is eight. Yeah, so that's exactly what the rule says. As long as the base here and the base in the logarithm are the same, then the answer will be whatever is inside the logarithm. In this case, it'll be eight. Okay, here's a different problem. What would be E raised to the natural log of seven? You may say, well, Dennis, we don't know what the natural log of seven is. That's okay. I think you'll know what this answer is. Because when I say natural log LN, what base am I referring to? Base E. And so if the base here, in the base in the exponential expression, is the same as the base in the logarithm, then the answer should be seven. Right, should be seven using this rule. Now, this last rule, I think we can actually give a more formal proof for. Let me just erase this. And what I want to do is to prove the rule right here. Because it's sort of a, probably the one that's the most striking, I think, of these four. So let's see. Suppose I want to figure out what is A raised to the log base A of m power. So let's just look at this exponent for a moment. The log base A of m. Now, of course, we don't know what A is or what m is for the moment. But this has some value. Let's say we'll call it x for the moment. Now, how would I rewrite this in its exponential form? A to the x equals m. These two things mean the same thing. Now, what if I replace x with what it's equal to? x is equal to the log base A of m. So A raised to the x, that's the log base A of m power. So all I've done is replace x with what it's equal to. I should still get the same answer, m, and there's our rule. That sounds like a little bit of sleight of hand. But if we kind of summarize it in a different way, it says what we're doing is we're putting a logarithm in the exponential position because the logarithm is an exponent. So if you put it where it naturally goes as an exponent on base A, then you will get the number m for the answer. Okay, we have these four laws spelled out on the next graphic. So let's go to that and just show them collectively here. These are the four laws. I think the only four laws that we need to know is to work the rest of the problems in this episode and in the next episode. First of all, number one says if you add two logarithms together and they're the same base, the log base A of m plus the log base A of n, you get the log base A of the product mn. Number two is sort of the corollary to that that says if you subtract two logarithms with the same base, log base A of m minus the log base A of n, you get the log base A of the quotient m over n. Number three says that if you multiply a logarithm by a constant, you can bring the constant inside by making it an exponent. C times the log base A of m is the log base A of m to the c power. And number four, this last one that we just derived, says if you take A and raise it to the log base A of a number m, then that expression is equal to m. Okay. Now, let's go to the next graphic and we'll see some examples where we can apply these rules. Okay, here we have an example where we want to rewrite several expressions in simpler logarithmic terms and we're going to be using those laws of logarithms that we just saw. Now, let's take, for example, the log base five of 25x cubed. Now, if we change that to a simpler logarithm expression, the first thing I notice is this is a product of two factors, 25 and x cubed. So if I use the very first law of logarithms, the log of a product is the sum of two separate logs. So this will be the sum of the log base five of 25 and the log base five of x cubed. Now, we can reduce each of these. Well, what is the log base five of 25? Two. It's two. Yeah, because you remember, what were you going to say, Jenny? The wrong answer. Oh, the wrong answer, okay. Well, this is an exponent, right? A logarithm's an exponent. It's the exponent you'd put on this base to get 25, so I think we want a square so that'll be a two. And what about this expression? We have the log base five of x cubed. Now, I don't think we can actually give a value to this because we don't know what x is, but there is another way I can write it that makes the logarithm look simpler using one of our four laws of logarithms. David, it looks like you have an idea. What are you thinking? I wasn't thinking. Oh, he wasn't thinking. Okay, okay. I just thought he was thinking. Well, let's see. Now, look, we have an exponent inside, and if there's an exponent inside, I could bring it out as a coefficient. So I'll put that in front and say three log base five of x. Now, you might say, well, Dennis, you're actually using this backwards because when we stated the law, it stated that if you have a coefficient in front, it comes in as an exponent. But you see, it's really a two-way street. If you have an exponent inside, you can bring it out. And if you have a coefficient outside, you can bring it in as an exponent. So here is an expression that's equal to the log base five of 25x cubed, and this has a simpler logarithm expression in it. Now, when you compare these two, the one on the left and the one on the right, you may ask, well, which one technically is simpler? Because this one looks pretty simple. It's just a single logarithm. This one, in a sense, looks more complex. And I guess it's really debatable which one is simpler. This certainly has the simpler logarithm in it. But what's significant here is that we can change logarithms to other forms. We can change what's on the left to the right. And if I go backwards, I can change what's on the right back to what's on the left. Okay, let's do this next one. This is a natural logarithm, so the base is what? Base e. So whenever you see ln, that means it's the natural exponential function, base e. E is about 2.718, more or less. Okay, which law of logarithms would you use to begin to change this? The second law we covered. The second law, yeah. Because the log of a quotient is the... David, what would you write here? Let's see, log... See, the log of a quotient. Pardon me? I'm sorry. What were you saying? Go ahead. Well, we have the log, natural log of a quotient. And by the second law of logarithms, I can write this as a difference. What would the difference be? You could put... Would it be log e? Would you start off log e? No, I'd put natural log of 6. Natural log of 6. Natural log of the numerator. Natural log of e. Minus the natural log of e squared. That was in the denominator. Okay. Now, if you were to write log base e, that's the same thing as ln, which means a little bit more compact. In other words, if you were to write log base e of 6 minus the log base e of e squared, that would be fine. It's just that standard notation is to use this ln rather than log base e. Okay, now, can I reduce either one of these? Jenny, which one or both can we reduce? The log of e squared to just 2. Exactly. Very good. Tell us how you arrived at that. Because ln is base e, and if you're raising it, it's by that fourth law... Right. ...that you wrote down that said that if you have it to raise a power to the same power... Yeah, now, you remember this natural log is an exponent. It's the exponent you'd put on this base. I'll just pencil it in right below it. It's the exponent you'd put on this base to get e squared. Well, I think you'd want to put a 2 as the exponent on e to get e squared. So this expression is equal to 2. Now, the first one I can't reduce. That'll be the natural log of 6, but then minus 2. Here's another way of thinking of it. Let me just rewrite this expression right below it. ln e squared. Now, someone may be thinking, now, if you have an exponent inside a logarithm, can't you bring that out in front? Well, yes, we can, and we would write it as 2 ln of e. And then at this point would say, what is ln of e? That's the exponent you put on base e to get e. That's a first power. So this is ln of 6 minus 2 times 1, and that's ln of 6 minus 2. Ginny found a faster way of getting there because she reduced this using the definition of a logarithm. Here, I used a property of a logarithm, but then eventually I had to go to the definition of a logarithm to get rid of the natural log there. But we get the same answer either way. Okay, now, let's go to the last example. And let's see, let's go back to that graph. You can look at example C. Yeah, we have the log of 10 squared to 10. Now, you know, actually, I should probably put parentheses around this so that it's not misinterpreted because without parentheses, this could be misconstrued as being the log of 10 multiplied times the square root of 10. But what I really mean to do is take the logarithm of this product right here. Okay, well, without understanding, let's continue. Can anyone think of a way of evaluating that using properties of logs? The first law. The first law? Okay, so how would you write this, Stephen? Log of 10 plus the log of root 10. The log of 10 plus the log of the square root of 10. Okay, and what is the log of 10? One. Is one. And what's the log of the square root of 10? One half. One half. And so our answer is three halves. We get rid of all the logarithms that way. Can anyone think of another way of getting the same answer? Let's just go down below. What I'm thinking is, what if we actually combine these? What is 10 times the square root of 10? It's 10 to the what power? Three halves. 10 to the three halves power, yeah. You see, you're multiplying 10 to the first times 10 to the one half, and the rule says when you multiply like basis, you add the exponent. So one plus a half is three halves. And then the log of 10 to the three halves is three halves, and there we have the same answer. Now I'm not saying either procedure is the best way to do it. It's just these are two different ways of applying various laws of logarithms to arrive at this result. Okay, so on an exam, if you took one extra step or one less step than I did, that's fine as long as you're using the rules correctly and you arrive at the correct answer. Okay, let's go to the next graphic, and we'll see another, a little bit different example, but still uses the same laws of logarithms. And let's see, I'd like to be able to write on that graphic if I can. Here we go. Okay, this example says suppose that we have some base, I'll call it B, and the log base B of two is 0.35. And suppose the log base B of nine is 1.12. And this is for some base B, but we don't know exactly what the base is. So the question is to compute each of the following. What would be the log base B of 18? The log base B of 18. Well, now we know the log base B of two and we know the log base B of nine, and I'm thinking that this is the log base B of nine plus the log base B of two, because nine times two is 18. I'm thinking of factoring 18 and then writing the log of each factor separately. And the log base B of 18, I think I can write that in just below here, is 1.12. And the log base B of two is 0.35. And if you add those together, you get 1.47. Now, here's the purpose of this activity. If you know a few logarithms, like in this case I know two logarithms, I can compute other logarithms based upon those. So by knowing just one or two, you can begin to generate a whole list of logarithmic values. Here are some others that we can calculate. I'll put that answer over here, 1.47. By the way, you'll see all these worked out on the website. So if I say something too fast or if you miss writing something down, you can always look on the website and get the details there. The next question is, what's the log base B of 0.25? Well, now let's see. We don't have any fractions or decimals given. We're given the log base B of two and the log base B of nine. Do you see any connection with 0.25 and either two or nine? I'm thinking, here's what's going through my mind. I'm thinking this is the log base B of one-fourth. Because 0.25 is a fourth. And one-fourth is the log base B of two to the negative two power. And that's going to be negative two log base B of two. Okay, here I'm using a property of logarithms. And this number we know, it's 0.35. So this is negative two times 0.35. And that tells me this is negative 0.70. That's the logarithm of log base B of 0.25. Okay, so given as a decimal, we don't recognize it as being related to two or nine, but when we convert it to a fraction and then we convert it to an exponential notation, then I can use a lot of logarithms to allow me to evaluate that. Oh, yeah. In fact, we can do another one in just a minute here. Let's go ahead and do this last one and then I'll make up one or two more. Okay, the log base B of the square root of two-thirds, let's see, let's write that in right here. We have the log base B of two. We have the log base B of nine. How are we going to do this? Well, first of all, how are we going to get rid of the square root? Right. Put a one-half power. In fact, why don't we just go ahead and move the power out in front. I think people know what we're doing here. We're writing this as two-thirds of the one-half power. The one-half power comes out in front. So this is the log base B of two-thirds. Okay, where can we go with that? Well, you can divide it by the quotient rule. Okay, so what will we write? You'd have log base two, log base B is two, minus log base B of one-third. Well, let's see now, if you guys say minus. Or excuse me, three. Yeah, we'll have to put a three there. Okay, now we know the log base B of two, but do we know the log base B of three? It would be to the one-half power, log base B of nine to the one-half. Nine to the one-half power, yeah. So let's write that in right here. One-half times, now this log we know, this is 0.35 minus, and this is nine to the one-half power. That's what three is. So if you put the one-half out in front, then you'll be taking the log of nine, and the log of nine is 1.12. So I'm taking one-half of the log of nine to get the log of three. So this will be one-half times 0.35 minus 0.56. I think I'm just about out of room there, but that's going to be one-half of negative 0.21. Let me just go up here and finish that up. One-half of negative 0.21, and that's going to be negative 0.105. So that's the answer for this problem. Okay, Ginny, you were just saying, would we have problems like this on the test? Yes, we would. And you know, one of the purposes of this is, we are computing logarithms of values that we couldn't find on a calculator. Let me just make up a new set of values. Suppose I have the log base B of, let's say we call this base C. And the log base C of four is, I'll just make up something, 0.14. And suppose the log base C of five is 0.118. So what would be the log base C of ten if you're given those two? How could you get the log base C of ten? Five plus five is ten. Can you do that? Five plus five is ten. Also five times two is ten. So if we think of this as being five times two, see if you have a product inside, you can break it up as a sum of logarithms. So I could write this as the log base C of five plus the log base C of two. Now how can I get the log base C of two? The square root of four. Well take the square root of the four. Yeah. So I could write this as the log base C of five plus one-half the log base C of four because that's actually C of four to the one-half power and I brought the one-half power out in front. Now we know all these numbers. This is 0.18 and this is one-half of 0.14. So we get 0.18 plus 0.07 and that's 0.25. That's the log of ten, log base C of ten. Okay, one more like this. Given those same values, how could I find the log of 25 times the square root of 20 base C? The log of 25 times the square root of 20. Well this is all made up of fives and fours. So I'm thinking we could write this as the log base C of 25 plus one-half the log base C of 20. And this is two times the log of five, log base C of five, and this is one-half of the log base C of four plus the log base C of five. Now we have it broken down into all logs of fours and fives. So this will be two times 0.18 plus one-half of 0.14 plus 0.18 and that's 0.09 plus one-half of 0.32. What is half of 0.32? 0.16. Now what's 0.16 plus 0.09? 0.25. 0.25. You know that's the same answer we got in an earlier problem, but that was for a different base as well. Yes. Two times 0.18, 0.09. Two times, oops, back up here. Yeah, you're right. I made a mistake there. I took half of it and I should have doubled it. Let's back up. That's okay. This should be, thank you, 0.36 plus, now over here we said that was 0.32. We took half of it as 0.16. Oh, you're absolutely right. Different answer, 0.52. Yeah, thank you. Okay, so that's the logarithm of 25 times the square root of 20 using these two given values. Okay, next idea. You know, on your calculator, you only have two logarithm buttons. You have a log base 10 or the common log button and you have the log base E or the natural log button. So you may wonder, Dennis, what if I wanted to calculate on my calculator like the log base three of a number or the log base seven of a number, we don't have buttons on our calculator for that. Well, number one, there would be an infinite number of buttons required to do that because there's log base two, log base three, log base four, log base five. I mean, there's no end to it. But it turns out you don't need any others because you can calculate any logarithm base using the two bases on your calculator, using a formula that we call the change of base formula. Let me just show you how the formula goes and then I'll show you where it comes from. If you have the log base B of M, that's equal to the log base A of M divided by the log base A of the old base B. Now, imagine that base A is a base that's available on your calculator, either base 10 or base E. Then you could convert any other base to a common log or a natural log and you could divide it out. Now, let me first of all show you where this comes from. We'll just write a little proof right below it here. Suppose I'm given the log base B of M that I'd like to evaluate and I don't know how much that's equal to, let's call this X for the moment. Now, what is the equivalent expression for this in exponential form? B to the X is equal to M? B to the X is equal to M. But let's say base B is not one of the bases available on my calculator. So what I'm going to do is I'm going to take the logarithm on both sides of this equation and I'm going to use a base that I know. Let's say I'm going to use, let's say base A is one of the bases available on my calculator and I'm going to take log base A on the other side as well. So on the left I took log base A of B to the X and on the right I took log base A of M. Now, if I bring the X out in front this says X log base A of B equals log base A of M. And if I solve for X X is equal to the log base A of M divided by the log base A of B. But X was equal to the log base B of M. So this is the log base B of M because we were given X was that in the very beginning and now we've calculated X to be this ratio. So the log base B of M is the ratio that we have up here called our change of base formula. Let's go to the graphic up here on the screen the next graphic to come up and we'll see this expressed. The change of base formula as we've just arrived at says the log base B of M is the ratio of the log base A of M over the log base A of the old base B. Now, we can use this to calculate some other logarithmic values and the problem we have there is to find the log base 4 of 15. Now if we come back to the green screen I'm going to do this on my calculator if you can zoom in on this calculator. Okay, so the problem that we had on the graphic was to find the log base 4 of 15 but you notice we don't have a log base 4 button anywhere on our calculator. So what I'm going to do is change this to a logarithm base that I'm familiar with. So let's take the log base 10 or just log of 15 but then I have to divide it by the log of the old base 4. So I'll just take this ratio and evaluate it in this form. So I'll take the log of 15 divided by the log of 4 and the answer is 1.9534 I don't know if you can see that very well on the screen or not but it's about 1.95. Now, just below it let me write this another way. We have the log base 4 of 15 and this time I'm going to change it to the natural log of 15 over the natural log of 4 because I do have a natural log button on my calculator. So I'm going to try taking the natural log of 15 and dividing it by the natural log of 4 and you see we get the very same answer. 1.953445298 So if I had yet some other base like a log base 3 button I could have taken this ratio using a change of base formula. So you know what this tells me is rather than having not enough buttons on my calculator I really have too many. I don't really need both of those. I really only need one logarithm button and I can find any other logarithm base. The reason these two are given is because common logs and natural logs come up probably 95% of the time in applications so they've been kind enough to give them both on our calculators. If we go back to that graphic on the change of base formula let's look at the answers that were computed or is written on the screen. The log base 4 of 15 if you take log 15 over log 4 you get this is rounded off 1.95344 and if you take the log base 4 of 15 the natural logarithm you get approximately the same number again. So this is a formula that allows us to compute logarithms in any other base. Okay now yet another application of these laws of logarithms is the fact that if you have an inverse function excuse me if you have an exponential function like 2 to the x its inverse function will be a logarithm function log base 2 of x. Now it's been several episodes now since we talked about graphs of exponential functions and we have yet to graph a logarithmic function but the fact is that these two functions are inverses and if you think back to some material we covered quite a bit earlier in this course one way of finding out if two functions are inverses is to take their composition and if I take the composition of these two functions on x what I should get for an answer is x if they're inverse functions that is these two functions should undo what each other does and I should get x. Now if I take this composition I think that's exactly what I get. Here's how. The composition of f and g means to take f of g of x. That's just what the composition means. And then to evaluate this I'm going to substitute on the inside for g of x and I'll put in the log base 2 of x. And then what does f do to anything? Well whatever number I put in here the answer will be 2 raised to that power so what does f do to this? It'll be 2 raised to this power the log base 2 of x. Now by our properties of logarithms what is 2 raised to the log base 2 of x equal to? It's equal to x. So it looks like these two functions are inverses because when I took their composition on x I got x. Let's try it the other way around. Suppose I take the composition g of f of x. So that's g parentheses f of x and to evaluate this I'll substitute for f of x f of x is 2 to the x so this is g of 2 to the x and what does g do to that? Well g takes any number and it takes the log base 2 of it so g will take the log base 2 of 2 to the x. Now using another property of logarithms how can I reduce this? If we bring the x out in front this is x log base 2 of 2 and what is the log base 2 of 2? 1. So this is x times 1 r x. So you see in either order my composition of functions on x gives me x so these two functions are inverses of each other. Now let's go to the next graphic. The inverse of a function f of x equals a to the x is the logarithmic function g of x equals the log base a of x that is the inverse of an exponential function is the logarithmic function with the same base and vice versa the inverse of a logarithmic function is the exponential function with the same base. Now this allows us to graph logarithmic functions and here are a couple examples. We want to graph first of all the function g of x equals log base 2 of x and then g of x equals log base 6 of x. Okay let's go to the green screen and look at those. First of all I want to graph equals the log base 2 of x. Now we've never graphed a so-called logarithmic function before but now that we know that this is the inverse of the function f of x equals 2 to the x what I'm going to do is graph that function and then if I flip it across the diagonal line y equals x I should see the graph of the inverse function. So let's do that right here. I'm going to graph the exponential function first of all because we're quite familiar with these from earlier episodes. So we'll get our coordinate plane set up here. Here's the x-axis, here's the y-axis. Now this function, this exponential function has three target points. Can someone remind me what the target points are? Zero one. Zero one, okay we go up one then what else? One two. Okay if you go to the right one you go up the base two and if you go to the left one what happens? Up one half. You go up the reciprocal of the base one half. Okay now we're not graphing the logarithm function we're graphing the exponential function and so this function has a graph that looks like this. Does it have an asymptote? Yes. What is its asymptote? y equals zero. Right y equals zero are the x-axis. Now if I draw in a 45 degree line y equals x right along here if I flip this graph over to the other side if I flip it over to the other side I should see the inverse function appear and the inverse function will be the log base two function. One of the things we pointed out several episodes ago is that exponential functions with bases other than a one not one to the x but any other positive base these are one to one functions because they pass the horizontal line test and therefore they have inverses and it's only today that we're finding out what the inverses are. Now when I flip this over the point one two will become the point two one over on this side should be exactly the same distance either side but my diagonal line's a little bit off and the point zero one when I flip it over will become what? One zero. One zero right here and the point negative one a half let me write that one down negative one one half when I flip it over it'll become one half negative one and even more than that the horizontal asymptote when I flip it over will become a vertical asymptote it'll be the y-axis and my graph will look like this but what I'm graphing here is f inverse of x and f inverse of x is the logarithm function so what I'm graphing is g of x equals the log base two of x now you know it really takes quite a bit of time if we're going to draw the exponential function every time and then draw the logarithm function so here's what I'm going to do to speed this up rather than graphing the exponential function I'm going to keep the target points in mind and I'm going to do everything sideways so you see here where we said we're going to go up to zero one I'm going to go to the right one zero I'm going to go to the right and instead of going over one and up two I'm going to go up one and over two and instead of going back one and up a half I'm going to go down one and over a half the reciprocal of the base and I'm going to plot these three points I'm going to keep in mind that I have a vertical asymptote at the y-axis and I'm going to draw that graph and I'm never going to see this one now if on your homework you draw that graph in that's okay but just be aware that's not part of the logarithm graph that's merely an a to help you graph these so let's do the other example that was on the screen there the other problem was to find the log base six of x or to graph it so part b says to graph the function g of x equals the log base six of x okay here we go this time I'm going to graph it in a faster manner without drawing that exponential function okay so to graph this logarithm function I'm keeping in mind that it's inverse I'll write it over here it's the inverse of six to the x now for six to the x you'd normally go up one so I'm going to go to the right one because I flipped everything across that diagonal line and normally I would go to the right one and up six so now I'm going to go up one and over six one two three four five six right here this is six and how would I get to the third point? go down one and over one six and over one six I'll have to kind of squeeze it in right there and you see this function is coming in not horizontally but it looks a little bit more horizontal than the last one does and it approaches the y-axis very quickly now as the base of the logarithm gets bigger you tend to get flatter looking curves because they turn out so fast and the reason is because the corresponding exponential function turns up so fast let me just graph one more like this and then we'll look at transformations of these graphs suppose I wanted to graph this function capital F of x equals the log of x so I want to graph the so-called common log function well the base is base ten so here's one two three let's say ten is out about there here's one two three four five one two negative five okay x-axis y-axis okay now to graph this quickly what I'll do is just go over one and if I go up one I'll go over ten I will say ten is out there and if I go down one I go over one-tenth now that's even closer to the y-axis than one-sixth was a minute ago and this graph comes in very flat it makes a very sharp turn and it comes down the y-axis and this is the graph of the common log function F of x equals log x you could graph this on your graphing calculator and this is exactly what you would see by the way speaking of calculators on the next exam there will be a portion of the exam where you will need a calculator to compute logarithmic values and so forth but for the graphing aspect you'll need to know how to draw the graphs without a graphing calculator okay can we go to the last graphic here and look at transformations of these functions in this example it says sketch these transformations of fundamental logarithmic functions and these are going to be translations up and down vertical transformations horizontal transformation stretches we're going to be flipping them around so let's take these one at a time first of all we have F of x equals two plus log base four of x what is the two going to do to the graph would it shift it? oh okay can we show the green screen what is this two going to do to the graph here David? it would shift it vertically it's going to shift it vertically because you see this is actually the same thing as writing the log base four of x plus two so this is a vertical shift so I'm going to be graphing one of my fundamental functions log base four of x but I'm going to shift it up to two units so if I mark off my scale okay so when I shift it up to two this is my new origin now from this position I'm going to graph the log base four of x I should go over one unit to the right if I go up one I should go over four units to the right let's see one two three four right about there let's take out those dots this is four and if I go down one I should go over one fourth so those are my three target points and when I draw my graph it still approaches the y-axis it just approaches it from a higher level okay next function let's graph log of the quantity x plus two so back on the green screen we have g of x equals the log of x plus two the log of x plus two well you notice this is different than before we had added a two in front now we're adding a two directly on the x what does that do to the graph Jenny? it's going to shift it to the left two okay so I'm going to be graphing the log base ten but I'm going to be shifting it to the left two so here's how we do it let's see those aren't quite even there okay so if I shift it to the left two this is my new origin right here and that says my vertical that says that my vertical asymptote which normally went through the origin is now going to be going right through negative two so this is the line x equals negative two now from this position I'm going to go over one and if I go up one I'm going to go over eight one, two, three, four, five, six, seven, eight, nine, ten so that's actually eight on the x-axis but it's ten from the new origin or from the asymptote and from the new origin if I go down when I go over one-tenth you notice this time I had to bring my vertical asymptote over with me so my graph looks like this okay we have one more function on that graphic that we want to graph this one is f of t equals two ln negative t now this one's a little fancy here not just because we're calling the variable t that's not really that significant but we have a natural logarithm function what is the two going to do to the graph? vertical stretch it's going to be a vertical stretch and what does the negative do when you put a negative directly on the variable it's a flip but David what kind of flip is that? it's a flip over the asymptote it's going to be a flip around the y-axis so we're going to flip it around the y-axis so you see we're going to have to plug in negative numbers for t so that negative t will be positive so that I can take the natural logarithm so my graph is actually going to be going off to the left-hand side okay let me take that arrow out because I think we're going to need a little bit more space here is there a restriction on t? that t has to be negative? in this case, see on those other graphs t was positive and so when I flip it over now the t's are going to be negative so what I'm going to do is I'm going to go over one unit to the left and normally if I go up one I go over e now I'm going to go up two and over e one, two, three so e is about 2.7 I'll put e right about there because I have to stretch my graph vertically so I went up two and if I go down two I go over one over e which is about a third that's a negative one third and therefore my function is going to look like this now let me just ask you how can I look at this graph and determine what is the domain what is the domain of this graph? remember if we just press the graph onto the t-axis I get the domain there is all numbers from minus infinity up to zero not including zero and what is the range? well if I press the graph onto the y-axis I get the range and it looks like it completely covers the y-axis so the range would be all real numbers okay well I think we're just about out of time here's what we've done today we looked at four fundamental laws of logarithms now those four laws are what make logarithms significant the rest of this would none of this would have happened I've been using those four laws of logarithms through all of this discussion and in the next episode you'll see us use those same four laws of logarithms again as we look at applications of logarithms using logarithmic scales and some other applications so I will see you for episode 19