 Well, welcome to episode 19 on exponential and logarithmic equations. We'll also talk about logarithmic scales in this episode. This is Math 1050 College Algebra, and I'm Dennis Allison. I teach mathematics here at Utah Valley State College. Let's see, in the last episode or two we've been talking about logarithms, and today we're going to look at some of the applications of logarithms, and this will extend into episode 20 as well. To look at the applications that we're going to consider today, I want to remind you of the four fundamental laws of logarithms. The first one is if we can go to the green screen, that if you take the log base B of n plus the log base B of n, you get the log base B of the product mn. What was the second law? Who remembers? Involves a quotient. The log base B of m minus the log base B of n equals the log base B of m over n. We'll be using that one today also. The third law says that if you have a constant multiple of a log base B of m, you get, who can tell me what will that become? Log base B of the quantity m to the c, exactly. I'll put that in parentheses just to set it off. And number four says if you have a, well, let's say if you have B and you raise it to the log base B of m, then m is the result. Okay, now with these four properties, we can do quite a few applications of logarithms. And let's look at the first. In fact, let's look at the list of objectives that we have here. In this episode, we want to look at solving exponential equations. We want to look at how to solve logarithmic equations. And then finally, we'll look at what we mean by logarithmic scales. So we'll finish up with that. Okay, let's go to the first example, to the first graphic of examples. And here we have three exponential equations. Now, surely these are all equations because we have two expressions that equal to each other. There's a variable involved. And so these three equations are referred to as exponential equations. Sort of like you have exponential functions, exponential equations. And some exponential equations require a little work to solve them. For example, let's look at these. In the first one, 7 to the x power is 49. I'll write that one here on the screen. 7 to the x power is 49. Well, what exponent should I put on 7 to get 49? I think x should be 2. So there's really no work involved in solving for x. But we have solved an equation for an unknown. In the second equation. Okay, now in the second example, we have 2 to the x minus 3 equals 1 half. Let's write that on the green screen. 2 to the x minus 3 equals 1 half. Now, this is another example of an exponential function because the variable is in the exponent. Excuse me, an exponential equation. And it's only slightly more complicated than that last one that we just solved. You see, this says 2 to the x minus 3 is equal to 2 to the negative 1. So if I make the bases alike, then the exponent should be alike, which means that x minus 3 equals negative 1. Now, you see, what we've done is actually more subtle than you may first realize. What we've done is we've changed the exponential equation into a linear equation. And we've been solving these sorts of equations since, say, the first year in algebra. x minus 3 is negative 1. And therefore, x is equal to, if I add 3 to both sides, x is equal to 2. So what we do is when we have a problem that looks more complicated, we change it into a different form. Rather than us going to the problem, we change the problem into something that's easier to solve. And in this case, the linear equation is very quick. If I just check that answer right below it, if I substitute in a 2 up here for x, we have 2 to the 2 minus 3 power is 2 to the negative 1 power, and that's equal to 1 half. So that does check. Okay, now, in the third example, we have 5 to the 2 plus 3x equals 5. Okay, now, if I divide both sides by 5... Excuse me, yeah, if I divide... No, I wrote that wrong, didn't I? 5 times e to the 2 plus 3x. Now, if I divide both sides by 5, I have e to the 2 plus 3x is equal to 1. And so what must this exponent be if I want this exponential expression to equal 1? Must be 0. Must be 0. So 2 plus 3x equals 0. What kind of an equation is this? This isn't an exponential equation. It's a linear equation. So you see, once again, we change this into a linear equation. So we're making the problem come to us, you see, by coming up with an equation that's easier to solve. So this says that 3x is equal to negative 2, and therefore, x is equal to negative 2 thirds. And I think if you substitute that back in here, you'll get an exponent of 0, which means 5 times 1 is equal to 5. So that will check out again. Okay, but now we ask the question, what if we're not able to get this down to a linear equation quite so easily? And let's go to the next examples and look at some problems where that would be more difficult. In the first equation here, we have 2 to the x equals 7. And I think you can see the problem is the 2 and 7. There's no common base I can convert those to, whereas before, I could somehow evaluate those fairly quickly. So let's take that first problem, 2 to the x equals 7. What I'm going to do is apply logarithms. I'm going to take a logarithm on both sides. Suppose I take log base 10, or just log on the left, and I'll take log on the right. What this does is it allows me to move the variable down onto the ground so that I can solve for it. And if I bring the x out in front, this is x times the log of 2 equals the log of 7. Now, you know, this is no longer an exponential equation. The variable is not in the exponent. What I have is a number that's a coefficient of x, the log base 2 is the coefficient of x, equals a constant on the other side, log 7. This is a linear equation. It just doesn't look quite as simple as the linear equations that I solved from that last graphic, but I have a multiple of x is equal to a constant. So if I solve for x, this says x is equal to the log of 7 divided by the log of 2. Now, this is what I would refer to as the exact answer. This is the exact answer. But if you want to know approximately how much it is, for example, is this number bigger than 2, is this number bigger than 3, would have to carry out a division using our calculator. So let me just move my calculator in on top of this problem. If you can zoom in on this. Okay, so I'm going to take the log of 7 and I'm going to divide it by the log of 2. And now, of course, this decimal is rounded off. When you divide these two logarithms, you actually get an infinite decimal expansion, and the calculator rounds it off to about 9 decimal places, or 10 significant digits. So we'll say this is about 2.807, about 2.807 for my approximate answer. So going back here to the green screen, x is approximately 2.807, and this is my approximate answer as opposed to the exact answer that was given above. Now on an exam, I may ask you to give me the exact answer and an approximate answer, in this case out to 3 decimal places. So you'd want to show this answer and this answer as well. Now, you notice at the beginning of this problem, I took a log on both sides, and I chose to take the common log on both sides because I had that available on my calculator, and so I was able to make this division. What other logarithm could I have taken right here so that I could have used my calculator? Could have taken a natural log. Okay, now over here on the side, let me just try reworking the same problem using natural logarithms. So we have 2 to the x equals 7. Now what if I were to take the natural log on the left and the natural log on the right? Of course natural log means log base e, and so if I move the x out in front, like I did before, x log 2, equals natural log of 7, and so x is equal to the natural log of 7 over the natural log of 2. This is the exact answer if you're using natural logarithms, and it looks different than that one, but I think when I approximate it, I think I'm going to get the same result. So to figure out what is the approximate answer, let's go back to our calculator, and if you can zoom in on this one more time, I'm going to try dividing the natural log of 7 divided by the natural log of 2. So the natural log of 7 divided by the natural log of 2 is, and I don't know if it shows up so well at home, but I have exactly the same digits, 2.807, I should say, and so we do end up with the same approximate answer, 2.807, and therefore when you give the exact answer, it could be given in either form, and when you give the approximate answer, depending on how many decimal places you want to give, these answers look the same down here. Okay, let's go to the next log, the next exponential equation where we'll have to use logarithms to solve it. This one says 15 to the 2 plus x power equals 1 third to the x power. Okay, let's solve that one right here. 15 to the 2 plus x power equals 1 third raised to the x. Now, you see, we don't have a common base that I can express both of these in, and so I'm going to need to use logarithms for this. So I'm going to try taking the common log on both sides, log of 15 to the 2 plus x power, and on the other side, log of 1 third raised to the x. Now, if I move my exponents out in front, 2 plus x times log 15 equals x log 1 third. Now, it's kind of hard to get used to this, but you see, log 15 is a constant. Just like 2 is a constant, the log of 15 is a constant. We don't see exactly its decimal representation. The variable is x. And on the other side, log of 1 third is a constant and the variable is an x. Now, what I want to do is to solve for x in this linear equation. And the way I solve for x is I need to multiply this out and get all my x's on one side. So I'm going to call this 2 times the log of 15 plus x log 15 equals... Now, what can I do with the log of 1 third? What's another way of expressing the log of 1 third? The log of 3 to the negative 1. The log of 3 to the negative 1. Okay, let's think of this as 3 to the negative 1. And if I bring my exponent out in front, that would be minus x log 3. So what we're doing is we're changing 1 third to 3 to the negative 1. The exponent comes out in front and when I put it in front, I'll just put it in front of the x and make it minus x log 3. Okay, so I have x's in this expression, x log 15, and I have x's in my expression on the right. I want to get these together. So let's get them both on the left-hand side. x log 15 plus x log 3. And the one that has no x's in it, I want to put it on the other side. So I'll subtract 2 log 15. So I get negative 2 log 15. Okay, so all my x's are on one side. I'm going to factor out the x. And I have log 15 plus log 3 in parentheses all multiplied by x equals negative 2 log 15. Okay, so I have a multiple of x equals a constant on the other side. So to solve for x, just like I would in any linear equation, I divide by the coefficient of x. It seems funny to call this the coefficient because it's number one, it's on the right of x. It's not on the left. And also, it's written with logarithms. We tend to think this is a variable, but it's just a constant. So I'm going to divide by that. And I have minus 2 log 15 divided by the log of 15 plus the log of 3. By the way, can anyone think of a way to write this denominator as a single logarithm? See, we have a question, Jenny. Yeah, that's what I was going to wonder. Could you just make that the log of 45? Exactly. See, whenever you add two logs together, you get the log of the product. So it might look simpler to call this negative 2 log 15 over log 45. You know, as a matter of fact, you might go further and say, Dennis, couldn't we get rid of the coefficient negative 2 and put it inside as an exponent? Would that make it look simpler? Well, it might make it look a little shorter, but I'm not sure that having a negative 2 as an exponent is really anything that simplifies it. So I think I'll leave that negative 2 out in front. So I would say this is the exact answer to this problem. Negative 2 log 15 over log 45. If you left your answer in the form we had here before we simplified the denominator, I would accept that as the exact answer for this problem as well. But what is the approximate answer? Let's see, Matt, I notice you're working on your calculator there. Are you evaluating this? Yes. What are you getting for this approximate value? Let's say up to about three decimal places. Negative 1.423. Negative 1.423 is what Matt got. I'm going to divide that on my calculator up here and let's just see how he came up with that answer. So I'm going to put this right on the top of the problem. And if you'll zoom in, I'm going to take negative 2 times the log of 15, and I'm going to divide this by the log of 45. Equals. And there we have negative 1.423. And Matt rounded up to make it 423. Let me do this another way. Using this first answer that I came up with, I'm going to take negative 2 log 15. But now I'm going to make a mistake here and I want someone to tell me what the mistake is I'm going to make. If the camera can zoom out for just a moment, let's look at this answer that's right below the calculator. I'm going to enter negative 2 log 15 divided by log 15 plus log 3. So here's how I'm going to do it. If you can zoom now back in on the calculator, I'm going to say divided by log 15 plus log 3. Now, when I push enter, I'm not going to get the same answer. And now I get negative 0.954 something. What did I do wrong when I entered that? We should have put parentheses around my divisor. See, I want to divide by the sum of the log of 15. Oh, and also, I notice I put a times, not a plus there. That was just a typo in my part. There's a multiplication, not a plus. So that's also a mistake. So let's go back and do this correctly. This is negative 2 log 15. Divided by the quantity, and I'm going to open parentheses. There's a little button here that shows open parentheses. And now log 15 plus log 3. Close parentheses. And now we should get the same answer that Matt got just a moment ago and that we got here in class. And there it is, negative 1.423 when we round off. So the reason I point this out is because students, I think, sometimes forget to put in the parentheses when they're doing a division by a sum. The other mistake I made is I just hit a wrong key and I hit a multiplication up there instead of a plus, but that was my own fault. Not something you would probably be doing at home. Okay, so we have two different exact answers. Either form would be okay. And then we have one approximate answer written out to three decimal places. Okay, let's go back to the graph. You can look at the third problem that asks us to solve an exponential equation. This says e to the 2x minus e to the x minus 6 is 0. Let's write this one on our green screen. E to the... Well, let's see, for some reason my marker isn't working. Let me pick up another marker down here. Try it again. e to the 2x minus e to the x minus 6 is 0. What are some of the things that makes this problem look different from the others that we've just been working? We have two different terms with x's in it. So this is still an exponential equation, but we have the x in exponents in two different terms. We have more terms. We have three terms here instead of two terms that we had before. And you know what I'm thinking is this one looks a lot like a quadratic equation. It's an exponential equation, but it looks quite a bit like a quadratic equation. I'm thinking maybe we could factor this into e to the x and e to the x. By the way, what's the product of e to the x times e to the x? e to the 2x. Yeah, let's just write that out. e to the x times e to the x. The rule says you add the exponents and you get e to the 2x. So this product will give me e to the 2x. Now, if I factor this like a quadratic expression, this is in what's called a quadratic form, I think I should put a negative 3 here and a plus 2 over there. Let's check this and see. This product is 2 e to the x. This product is negative 3 e to the x. So when I combine them, I get negative e to the x. And by the way, the last product will give me negative 6. So I think we have been able to factor this, something we didn't see come up in any of the previous problems. And this product is 0. So either e to the x plus 2 is 0 or e to the x minus 3 is 0. Now in this first case, this says that e to the x is negative 2. This is impossible. Because if you look at the graph of e to the x, if you graph the function f of x equals e to the x, let me just remind you what it looks like. Here's the x-axis. My graph is a little small. I hope you can see this at home. If I graph e to the x, you remember the target points were, first of all, to go up one. And if I go to the right one, I should go up e, which is about 2.7. And if I go to the left one, I should go up 1 over e. And to make a long story short, the graph is going to look like this. Now my question is, when does the graph equal negative 2? Here's negative 2 down here. When does that graph come down to negative 2? It never does. So there's no solution. An exponential expression has to give me a positive answer. It can't even give me 0 for the answer. So it's got to give me a positive answer. Certainly not negative 2. So my conclusion from all this is there's no solution to this portion of the equation. But the other side says e to the x could equal 3. So I'm looking for an exponent that I can put on base e to give me 3. I don't see any way to make these two bases look just like, not any obvious way right now. So I'm going to take a logarithm on both sides, and I want to be able to use my calculator. So which logarithm do you suspect we should use here? Natural log, because there's a base e involved in the exponential equation. So I'm going to take the natural log of e to the x, and I'm going to take the natural log of 3. You see, if these two expressions are equal, they should have the same natural logarithm. And so this says x natural log of e is the natural log of 3. And what is the natural log of e? It's 1, so this is just x. So we get x equals ln of 3. This is my only exact answer. This is the only exact answer we get because we didn't get any answer over here on the other side. And if I approximate it, let's say we approximate it to 3 decimal places, I'm going to put my calculator in here if you can zoom in on this. And we want to approximate the natural log of 3, so we'll just enter natural log of 3. And we get about 1.098. And if I round that off, we'll call it 1.099. So that's my approximate answer. We'll call that 1.099 right there. So back to the green board, 1.099 is the approximate answer given to 3 decimal places. Okay, so once again, we have an exact answer and we have an approximate answer. The exact answer, because there's nothing better than the exact answer, but when we ask how big is that number, for example, is that more than one? Is it more than one and a fourth? Is it more than one and a half? It's hard to say until we approximate it. So both answers are important that we be able to calculate. Okay, I'm going to take one more exponential equation and I want someone to tell me how to graph this. I'm going to keep the camera on the green screen. What about this problem? Suppose I have 2 times 9 to the x plus 3 equals 18. Now do I need logarithms to solve this problem? I don't think so. It may be difficult to tell in the beginning, but I think there's a way that I can make these have the same bases, and I don't have to use logarithms to solve it. What would you do first to solve this problem, Matt? Divide by 2, yeah, and so now we have 9 to the x plus 3 equals 9. Yeah, so now what? Well, then I know that 9 raised to the 1 equals 9. So I would find something that would make the exponent 1. So what would the exponent have to be? Negative 2. Okay, actually Matt's a little bit ahead of us. If there's a first power over here, x plus 3 is going to be 1. So now here's our linear equation, and then when I solve for x, x is negative 2. You were exactly right. So in this case I don't have to use logarithms, but if I erase this and change one number, what if I change this number not to 18 but to 20? Now here's a problem where I have to use logarithms to solve it. So you see it's very easy to go from the relatively simple problems to the ones that are a bit more complex. In this case, we'd still divide by 2. 9 to the x plus 3 power is equal to 10, but the problem is I no longer have a common base between these two. So at this point we have to shift gears, go to logarithms, and just to be different, what if I take a natural log on both sides? Even though there's a 10 in the problem, let's take a natural log. Natural log of 9 to the x plus 3 equals the natural log of 10. I'll bring the x plus 3 out in front times the natural log of 9 equals the natural log of 10. Then I'll multiply this out so that I get x times the natural log of 9 plus 3 times the natural log of 9 equals the natural log of 10. And there's only one term with an x in it, so I want to isolate that. So what I'll do is subtract 3 natural log of 9 from both sides, and I have x natural log 9 equals the natural log of 10 minus 3 natural log of 9. Therefore x is equal to, now this is the exact answer, so I can still put an equal sign, the natural log of 10 minus 3 times the natural log, oops, natural log of 9, over the natural log of 9. This is a perfectly good answer, although it is possible to maybe change this a bit, make it look a little bit more compact. Jenny? Why couldn't you divide when you have x plus 3 in parentheses and then you have the natural log I think you had of 9? Why can't you just divide from there that natural log of 9? Actually you could, that would be an alternative. What Jenny is suggesting is we could have isolated the x perhaps faster if rather than multiplying this out if we had just divided by the natural log of 9, call this ln of 10 over ln of 9 equals x plus 3. I think that probably would have been a little quicker. And that would be equivalent to this. You see, what you would get is if you divide this out you'd have the natural log of 10 over the natural log of 9 minus 3. And that's the answer that you would arrive at up there. I think Jenny is exactly right, that would have been a little bit shorter. To get the approximate answer, I don't think I'll do this in the interest of time, but what you'd want to do is think of parentheses around all this. When you enter this in your calculator, open parentheses natural log 10 minus 3 natural log 9, close parentheses, and then divide by the natural log of 9. Okay, that would give you your approximation. Okay, well, I think we've covered fairly thoroughly now exponential equations. Let's look at logarithmic equations. And we have another graphic now with several problems of a different variety than what we've just seen. Now these three problems are equations because we have expressions that are set equal to each other and there's an unknown that we'd like to figure out. But rather than having exponentials, we have logarithms, the variables are inside logarithms. So we call these logarithmic equations and we solve these using the same rules, the same four basic laws of logarithms that we've been applying already. And later in this episode and in the next episode we'll see applications where equations like this come up. So we're just preparing for those by finding out how to solve them now. In the first equation we have the log base 3 of x plus 1 equals 4. Let's solve that one right here. The log base 3 of x plus 1 equals 4. Now I would call this a logarithmic equation because the variable is inside of a logarithm. Anyone have a suggestion on how we could solve this? Form an exponential equation from this. Yeah, change this to exponential 4 because we're used to looking at exponential statements much more than we are looking at logarithmic statements. Let's see, Matt, what would be the exponential version of this? 3 raised to the 4 equals x plus 1. x plus 1, yes, because all of that is inside the logarithm. This is sometimes referred to as the argument. So x plus 1 is the argument, so x plus 1 is equal to the exponential quantity. So this says that 81 is equal to x plus 1. By the way, what sort of equations are we solving here? Linear. These are linear equations, yes, exactly. So once again, we've made the problem come to us by changing it to a form that we like better, linear equation and linear equation. Now it's fairly simple to get out of this. x is just equal to 80. So that's our solution. And by the way, this is the exact solution and there's no approximation, of course, involved in this. So that's the only answer we'd have to give. Okay, now in the other two examples, the problems get a little bit more complicated. Let's take part B. The log of x minus the log of x plus 2 is equal to 2. I'll write that one on the green screen. The log of x minus the log of the quantity x plus 2 is equal to 2. Well, this one's a bit more complicated than the last problem because it's a logarithmic equation, but I have two different logarithms involved in both of them. How could I somehow get a single logarithmic expression here? David, what would you do? How could you get a single logarithm there? Can we pull the log x out anyway? Well, we can't factor the log out, but we do have a difference of logs. Now one of our properties tells us a way to reduce that. Jeff, what are you thinking of? We could change that to the log of x over x plus 2. Right, yes. So that was our second property of logarithms that says if you have the log of m minus the log of n, it's the log of m over n. So now I have one logarithm, and I have a more complicated expression inside, but it's only one logarithm. Now, you might ask the question, Dennis, which one of these is simpler? Well, in one sense, the first one is simpler because I had simpler arguments, simpler logarithmic expressions. This one's simpler because there's only one logarithm. So it's sort of in the eyes of the beholder, which one is simpler. But for our purposes, I think we like this one better because the next thing I'm going to do is change this to exponential form. Let's see, Ginny, what is the exponential version of this logarithmic equation? It would be 10 raised to the 2 is equal to x over x plus 2. Equals x over x plus 2. Okay, now this is no longer a logarithmic equation because we have no logarithms. If anything would call this a rational equation, because I have a ratio of polynomials, I have a rational expression. So this is a rational equation. We haven't seen one of those in this episode, whereas up above, I had a logarithmic equation there and there. So once again, I've changed it to a form that I find easier to solve. You might say, gee, is that easier to solve? Yes, I think it is because look, x over x plus 2 is equal to x over x plus 2. And how do you go about solving equations like this back in say intermediate algebra? What do you do? Both sides by x plus 2? Yeah, I'm going to multiply both sides by x plus 2. So this is 100 times x plus 2 and I'll multiply on the right by x plus 2 and of course they cancel there, so I have x. Now, what kind of an equation is this? This is an irrational equation. This is a linear equation. So the problem just keeps changing. This is a linear equation now, but this is the one that we'll actually choose to solve. So we had a logarithmic equation, a rational equation, a linear equation, and now I think we're ready to resolve all this. We have 100x plus 200 is equal to x. So that says that 99x is equal to negative 200. And that says that x, I'll put it over here on the side, x must be negative 200 over 99. This is my exact answer. Now, if on your calculator you were to divide this out, this would be an infinite repeating decimal and your calculator wouldn't be able to display an infinite decimal because it's not infinitely wide. So this is the exact answer. You wouldn't want to give this as an in an approximate form unless I ask you to give it as an approximate. So our final answer is negative 200 over 99. I don't think there's anything that'll reduce there, so that's the final result. Okay, once again I took a logarithmic equation. At this time I made it a rational equation and then we got a linear equation out of that. Okay, in the last example for logarithmic equations, we have the natural log of x plus the natural log of x plus 6 equals 2 times the natural log of x plus 4. I'll write that one here. We have the natural log of x plus the natural log of the quantity x plus 6 equals 2 times the natural log of the quantity x plus 4. Now what I've tried to do in these examples is to find various problems that illustrate different ideas and how to go about solving these sorts of problems. This one is a little different from either of the previous two logarithmic equations. Now, I'm going to begin on the same note. I'm going to combine these two logarithms into a single log. And if I add two logarithms, I get the natural log of the product. So this is the natural log of the quantity x times x plus 6. Equals, and over here I have a coefficient. Where's another place we could hide the coefficient? As an exponent to the x plus 4. Bring it in as an exponent. So this is the natural log of x plus 4 squared. Okay, now what makes this problem different is this time, I'm going to say, well, I've got the natural log of this number equals the natural log of this number. Well, if I have the same natural logarithm values, the two numbers inside must be the same. So I'm going to set this product equal to x plus 4 squared. So I have x times x plus 6 equals x plus 4 squared. So I'm thinking if the two natural logs are the same, the two arguments must have been the same from the beginning. So this time I just eliminate the natural logs. Of course, when I multiply out the left-hand side, I'm going to get an x squared. And when I square the right-hand side, I'm going to get an x squared. So it looks like this should be a quadratic equation. It's an equation, and I have some x squared terms involved. So let's multiply this out. We have x squared plus 6x. And let's see, what is x plus 4 squared? Plus 8x plus 16. Plus 8x plus 16. Now, I said it looked like it's going to be a quadratic equation, but actually the x-squares cancel off. So we're down to 6x equals 8x plus 16. Now, this is actually now a linear equation. So here we are, changing from one form to another until we get something that we would like to solve. If I collect all my x's on the right, I have 2x. And if I subtract off the 16, minus 16, and so I get x equals negative 8. x equals negative 8. Unfortunately, you know what? It's sort of sad. I'm going to have to throw that answer out. There's no solution for this problem. Does anyone know why? Because it'll leave you with a negative natural log, and it never is. Yeah, exactly. You see, if I take this number and substitute it back in up here, I'm going to have the natural log of a negative 8. The natural log of negative 8. Over here, I'll have the natural log of negative 2. And here I'll have the natural log of negative 4. Well, you can't take natural logs, or any logs, of negative numbers. Let me show you why. I'll just move that answer up here to the top. And I want to make myself some space. So we thought the answer was negative 8, but then we had to throw it out. Here's the reason. If I graph the function f of x equals ln of x. I'm choosing ln because these are all natural logarithms. And if I graph that function, you remember this is the inverse function for e to the x. So when I graph the natural log function, I'm going to steal the target points from the function e to the x. But I'm going to move them sideways. So starting at the origin, I go over 1. And if I go up 1, I go over e, about 2.7. And if I go down 1, I go over 1 over e, or about a third, roughly. And so my graph looks like this. Now, looking at this graph, what would you say is the natural log at 1? What's the natural log at 1? 0. Is 0. Yeah, the function of that is 0. This is e right here. What's the natural log of e? 1. Is 1. If I were to go over here to 3, just to take a wild guess, could anyone just, based on this graph, what would you say is the natural log of 3? 2, possibly? Yeah, it's actually a little bit less than 2. But since my graph isn't just perfect, it's kind of hard to tell. It's actually a little bit less than 2. You could find it on your calculator. But let me ask you this. What's the natural log of negative 8? Here's negative 8 out here. Well, let's see, there's no graph out here. So there's nothing for me to locate it, to locate a function value. As a matter of fact, negative 8 is not in the domain. If I were to take this graph and press it onto the x-axis, the domain would only be the positive numbers, not negative numbers. That's why I threw out negative 8. Although in the algebra I came out with a negative 8, it's what's referred to as an extraneous root in mathematics. That is, it's an extra root, but it doesn't really solve the problem. And that's because I can't put in negative 8 there. I can't put in negative 8 there. And by gosh, I can't even put in negative 8 over there. Can't take the natural log of negative 4. So it's a number that has to be thrown out. There was no solution for the third part, part C of that example. Okay, so we have to be aware that sometimes extraneous roots come up. And you might ask, you might wonder, how is it that negative 8 could have been introduced? If negative 8 wasn't a solution in the beginning, how did we ever get to negative 8 later on? Well, in the beginning when I wrote that natural log equation, there was a subtle assumption in there that the x's couldn't be negative or even 0. But when I changed it to a quadratic equation and then a linear equation, that assumption was dropped out and we were looking for every possible solution. So negative 8 may be a solution to the quadratic and to the linear equation, but it's not a solution to the natural log equation because of the restriction on the domain there. Okay, so now you know a little bit about solving exponential and logarithmic equations, but really the only way you learn how to do these accurately is to do the homework problems, try solving these on your own. And you may make a few mistakes, but you learn from your mistakes and I think you'll get good at it. Okay, let's change topics. And I want to show you another article that I pulled out of a Salt Lake Tribune just recently. I think I showed you several articles that represented exponential problems. Here's an article that I think could use some logarithmic application. So if you can show the green screen, I'll just show you this came from the Daybreak section back on May the 1st, 2002. So I've had this for quite a while. It's from the Food section and the title was called Mile to Wild. And now if you go further down, go further down here, the article is actually down here at the very bottom of the page and the article has to deal with measuring the heat in various peppers and how hot some peppers can get. Is there a way to actually classify them mathematically how hot a pepper can be? So they discuss in the article something referred to as the Schofield heat scale right here and I won't go into all the details of scientifically how you measure the heat index of a pepper, but it's the graph that I'd like to show you. Now this may not show up very well on the camera, but the scale goes from zero at the bottom up to 600,000, 600,000 what they call Schofield units. Schofield is the name of the person who invented this technique for measuring the heat of a chili pepper. Well they start off at zero and at zero we have a yellow bell pepper. Now frankly I don't remember anyone who's jumped up from the table and run out of the room looking for a glass of water because they bit into a yellow pepper. I mean yellow peppers just don't have any heat to them. I don't know about green peppers and those red peppers, those bell peppers, but I think they're all classified as being rather low. Now if you go up a little bit higher you may recognize this anaheim pepper and its heat index varies from 500 units to 2,500 units and there's a little line that goes down here very close to zero. So what they're saying is there's not that much heat in an anaheim pepper and so on the scale you can hardly tell it from zero. But the next one above it, have to follow the line up to it, is the serrano pepper and the serrano pepper has 8,000 to 22,000 scofield units. Why do you think it varies from 8,000 to 22,000? Stages of its growth. Stages of its growth. Yeah, what could be other factors that would affect the heat index of the pepper? Personal taste? Well, I don't know if personal taste enters into this but it could be just the environment it was grown in whether it got more water, less water, what sort of nutrients it got. So probably you could find one serrano pepper a little hotter than another but they say generally they vary from 8,000 to 22,000 units. So if we put that on the graph, look, it's way down here near the bottom too because that first marker is at 50,000 units. So these peppers are clumped together. Now the next one to come to is the Thai pepper. Now I've had experience with Thai peppers and to me they are hot. Now for some people they may say, no, Dennis, you're such a wimp but I find Thai peppers to be really hot and Thai peppers vary from 50,000 to 100,000 Schofield units and here on the scale, well they haven't marked off at 100,000, at 100,000 but look, the scale goes on up to 600,000. That's because we haven't gotten to the really biggie here the red savina habanero pepper and its Schofield unit measure is 350,000 to 577,000. Okay, Stephen, oh, yeah, there so you can see that. Stephen's nodding his head, yes, he's bit into those. Stephen, how did you find them? Tasty or? Well, it stays in your mouth for about three hours afterwards and it's very hot. And you're probably still flush in the face the next morning when you get up from that pepper. I don't think I've ever had one of those but here's the problem that I posed to you. When I draw these on the scale or when the newspaper drew them on the scale this one is way up above all the others and we have three of them clumped together near the bottom. So what would be another way that I could represent this so that they weren't so spread out? So let me suggest an alternative way of representing them. So if you take a look at this graphic that I've prepared I have just picked a number in the range for each pepper. For the Anaheim pepper I set about 2,000. For the Serrano pepper about 15,000. You get Serrano peppers in Mexican food quite a bit. Thai peppers, of course in Thai food you might bite into a Thai pepper and for me it was an experience I haven't forgotten. My eyes watered everything. But then here's the red Savina Habanero pepper and I called it 400,000. The range was a little bit broader than that obviously but here's how the newspaper has shown these. The Anaheim pepper was at 2,000. So I've marked off a horizontal scale 0, 100,000, 200,000, 300,000 on up and so the Anaheim pepper is going to be right here. I'll just put N-A-N for Anaheim. By the way there was the yellow bell pepper right at 0. It would be right on top of that. Then the Serrano pepper at 15,000, okay if that's 100,000, 50,000 15,000 would be right about here. I'll put S-E-R for the Serrano pepper and then there's the Thai pepper at 75,000. Well 75,000 is right about here and there's the Thai pepper and then there's the red Savina habanero pepper at 400,000. Now you see it's distorted way up above. I'll just say red there for that. So these things are not clustered together at all. So here's an alternative. Suppose I were to label my scale 0, 1, 2, 3. I'll try to write it a little bigger so you can see it there. 4, 5, 6 and what I'm going to do is take the logarithm of each of these values up here. So let me just slide this down so you can see what I'm going to do. I'm going to take the log of 2,000. I'm going to take the log of 15,000 and I'm going to take the log of 75,000. And maybe you could help me out here. Those students in the classroom, what is the log of 2,000? 3.30 3.30 approximately. Now how would you have known that was going to be a little bit larger than 3? 3 zeros in the number. 3 zeros in the number. You see the log of 1,000 is 3. So the log of 2,000 should be a little bit more than that. Now for the log of 15,000 this should also be 3. something. It's 4.17. Oh, 4.17. Oh, you're right. 4. 4.17. 4.17. And the reason it's a little bit bigger than 4 is because the log of 10,000 is 4 and this is a little bit more than 10,000. The log of 75,000. Well, this is not up to 100,000 yet so it's not 5. It's going to be 4.875. Okay, and now let's go to the red Savina Habanero. The log of 400,000. Let's see, that's one, two, three, four, five. The log of 100,000 is five. So this is 5. what? 6.02. 5.602. Now on this scale, I'm going to locate them by their logarithms. And so the Anaheim would be at 3.3. Okay, here's Anaheim. The Serrano pepper is 4.17. So that would be right about here, the Serrano pepper. The Thai pepper is about 4.875. That would be right about here. There's the Thai pepper. And the red Savina Habanero is 5.6. 5.6 is right about here. Now my point is this. Look at the difference. These peppers are now rather equally spaced and they're drawn together. Up above they were spread way out. And sometimes when you have data that is widely dispersed, in order to look at it more conveniently, you take logarithms. And this was referred to as a logarithmic scale. And now you can see them closer together. But these spaces, although they appear to be about even, represent ever-increasing spaces back on the real scale. See, we had a small change, a bigger change, and then we have a huge change. Here, they're all about the same. Now let me show you how this is applied in science. Can we go to the next graphic? Okay, here are three examples of how logarithmic scales are used in science. First of all, in chemistry, when you measure the pH of a solution, the pH is calculated by taking the negative log of the hydrogen ion concentration that's abbreviated by a square brackets H+. Now, I'm not a chemist and I won't try to explain how you measure the hydrogen ion concentration, but that H+, tells you generally how many free hydrogen ions are floating around in the solution per mole. And if this number turns out to be less than seven, it's said to be acetic. And if this number is greater than seven, it's said to be a base. Now, the hydrogen ion concentration can actually go from a number extremely small, like 0.000, let's say about 9 zeros and then a one, to something rather large, like 0.01, relatively speaking. Well, those numbers are so diverse that what you do is you take the logarithm of those and then you put a negative on there to make it positive. Let me just give you an example. Suppose the hydrogen ion concentration, H+, could vary between, let's say, 10 to the negative nine up to 10 to the negative two. I'm not sure that those are actually accurate numbers, but it's something like that. You see, this is 0.0000001. I've got nine decimal places there. The hydrogen ion concentration, let's say, is generally bigger than that, but less than or equal to, what if this number is a decimal? 0.01. 0.01, yeah, exactly. Well, you see, those numbers are so long that it's actually nicer to take the logarithm of 10 to the negative nine and the logarithm of 10 to the negative two and then inside here we have the logarithm of the hydrogen ion concentration. But what is the log base 10 of 10 to the negative nine? Negative nine, yeah. Remember, what's the exponent you put on base 10 to get 10 to the negative nine? It's negative nine. And what is the logarithm base 10 of 10 to the negative two? Negative two. So this logarithm would vary from negative nine to negative two. Now, what they do in chemistry to make these numbers positive is they just multiply through by a negative and this is referred to as the pH. That's called the pH. So it's done in an artificial manner. We take a logarithm so that we make these messy decimals into rather nice numbers and we put a negative in front of the logarithm to make the negative numbers into positive numbers and we say if the pH, which is what this is defined to be, if the pH is less than seven, we say that the solution is an acid. And if the pH is greater than seven, we say that it's a base. And if I'm not mistaken, the pH of ordinary water is seven. So if the pH varies from what typical water would be, it's either an acid or a base because it's more or less than seven. Well, you know, I think we have time for about one example before this episode ends. So let's take whole milk and the pH for whole milk is generally about 6.5. Now, that number is less than seven so that means milk is slightly acetic. So it's a lactic acid. So it's only slightly acetic. And what if we wanted to find the hydrogen ion concentration in milk? Well, if I substitute 6.5 into my equation, 6.5 then is negative log hydrogen ion concentration. Before I can solve for the hydrogen ion concentration, I better move the negative over. So negative 6.5 is the log of the hydrogen ion concentration. So changing this to exponential form based in to the negative 6.5 power is the hydrogen ion concentration. And if you multiply that out, you get approximately 0.0000003. So that's roughly the hydrogen ion concentration. We'll continue talking about logarithmic scales in the next episode. Hey, we got it right.