All right everyone. So I thought in this video we could just do lots of examples to highlight the properties of the natural logarithms. We're going to learn by doing, hopefully by repetition. These properties become more natural and so just as a quick review of what these properties are. If I take the natural log of a product, so let's say something like xy. For logarithms, when you have a product it turns it into addition. So you have the natural log(x) remember this is of x, not times, of it's a function + the natural log(y). It's important to say of x and get used to saying the right words that you understand that this is a function. I've seen many students treat this as a product and incorrect on that. So just as a quick example with some numbers or something else. If we had the natural log of say 6, you could break that up and you can write this is the natural log of 2 x 3, so this is all happening inside the parentheses. And that becomes the natural log of 2 + the natural log of 3. And you can go ahead and check on the calculator if you want type in the log of 6 and log of 2 and log 3 and you get the same numbers when you have these. If you notice sometimes you write the parentheses, sometimes you don't just need to be very clear from context what it is, okay? So this your first rule there we're going to use all these rules, so keep these handy as you go along. If I have a quotient now log(x/y), while this becomes the log of the numerator- the log of the dinominator, so log x- log y. Just as a quick example here to show you what this is. If we did something like the natural log of, I don't know what's a good number here, 2/3 perhaps then this is the natural log of 2- the natural log of 3. All these rules follow from their corresponding properties of exponents. Here we confused about a logarithm, put it back in terms of an exponent and you'll see what I mean. And let's use last one here. Natural log of x to a power, this becomes n times the natural log of x. So x exponents we say fall down in front. All these equations are two way streets, you can also read them and write to left. So you can also have a number of front, bring it up to the exponent just as an example to show this one with some numbers. With we had ln (4) this would be natural log of 2 squared, which should be 2 x times the natural log of 2. And again, I encourage you to grab a calculator, go punch these in and show that these are actually equal. All right, so these are the rules were going to do. And remember these are not technically the rules are going to put down now but they're rate constant sinonim as well. Know that anyway, the natural log of 1 is 0. And the natural log of e is 1 because they are inverses like either 1 they cancel on. Of course, if you have ever have natural log of e to the x, well, that's just x. And if you ever have e to the ln(x), that's also just x, okay? So these are everything we need to know to go off and solve some problems. So keep these handy as we go through and let's just do lots and lots of problems. Here we go. So the first thing to do usually is to simplify. So the first thing we do is simplify. And let's do a couple examples here. What if I said without a calculator try to do this. Ln 5 + the natural log of x. So you can pause the video work on these if you want. How would you have two logarithms? How would we simplify the expression? How can we write it as a single logarithm? Will remember our rule if we have a sum of two logs this is 1 logarithm and then combine the two as a product. Nice and easy there. Let's do another one e to the 2 ln x. How would we simplify this expression? Now mistake a lot of people make here is they cancel the in the logarithm and they just say, well they are inverse is right can't we do that? And the problem is no, there's this like 2 and right there in front. So it's kind of blocking the e from hitting the natural log. So you really need to focus on this too and move it out of the way. Well, how do you do that? Remember we said if we have powers in front or numbers in front, they become powers. So you can rewrite the expression as e to the natural log of x squared. So anytime there's a little number in front blocking your way, move it as the exponent. Now you have e and the natural log are together so they would cancel and you just get back x squared. So this expression with e in the logarithm is nothing more than just x squared. It's pretty neat. How about one with three logs? Let's do natural log of x- ln of x squared + ln x to the 4th. If you like 3 logs, kind of scaring, pause video so you can simplify this expression as we go. So let's do this, so we have, let's grab two at a time. I'm going to do the first two together. So I have ln if something minus Of something. When you do that, it combines a minus becomes a quotient and you have x over x squared. There's still a natural log of x to the 4th floating around, so let's not lose that. And then we can simplify x over x squared, of course, is just the natural log of 1/x plus the natural log of x to the 4th. I have two logarithms similar to the first problem we did, I want to simplify that so that becomes the logarithm, the single logarithm of their product. This becomes x to the 4th over x, I multiply these two together. Then I get to the 4th over x, which is of course the natural log of, cancel one of them, x cubed. So this expression with three logarithms comes up quite nicely to the natural log of x cubed. And if you really wanted to simplify it, you're not wrong. If you wrote 3 times ln x, either one of these two are perfectly fine. Not one is any better than the other. They are in fact equal. Let's do another one. Let's do a scary one. Let's say e to the ln of x squared plus, and this is all in the exponent now, 2 to the natural log of y. So we have a nice multivariable function here with x's and y's. Again, then I give you variables to start with, you probably going to give back all variables upon return. So let's simplified this a little bit using laws of exponents. So e to the something plus something, just like e to the a+b. So remember when you have to a sum, it becomes the product of both these terms. So I'm going to split this up first as e to the ln of x squared and then times e to the 3 natural log of y. e and ln cancel, so you just get back x squared for your first term. And then once again I have this 3 in the way, so I need to write this 3 as a power on the y. Can't cancel it and get 3y, it's extremely important. You have to move the 3 up and over to become an exponent on the y. Now, e ln cancel and you're left with x squared y cubed. So a much nicer, simpler expression. So these are just some of the rules that we're going to use to solve. So let's do a couple of, solve the given equation. So in step one here, natural log of x minus the natural log of x squared plus the natural log of 3 is 0. Solve for x. What value of x makes this true? With three logarithms, that is tricky, however, it should look familiar, we did problem before with a couple algorithms. Let's simplify first, it's a multi step problems. Simplify, grab the first two. You say no, I'm not dealing with three logarithms, no way. So you change it and say x divided by x squared. And you have a difference of logarithms, it becomes a quotient plus log of 3 equals 0. You can clean it up a little bit. X over x squared is just 1/x plus log of 3 is 0. I have two logs, better than what I started with, but I could do better. Two logs added together, become a single log of the product, 1/x times times 3, which is 3/x, which is 0. And now I want to solve for x. So remember log is tracked. Log is blocking my access to this x, and just like anything else, you want to undo the operations. So how do you undo ln? We have e to the ln of 3/x. Now remember, whatever you do to one side, you must do to the other side. So this is called exponentiation. We raise both powers to e. We take e as their base. e and the natural log cancel, and that's why you did that. That's why you introduced to e, the inverse function of the natural logarithm. And you get good 3/x, and now you get e to the 0. e to the 0 is just 1, you get 3/x is 1. You can cross multiply and get x equals. So x equals 3, and again, you can go check through a calculator if you want, but you can check that x equals 3 does actually solve this problem. All right, let's do another one. Do different colors so we could see it. Maybe I'll do it on the side of the screen. So let's do the natural log of the square root of x minus 2 natural log of 3 equals 0. Solve for x. Again, feel free to pause the video here, try it on your own before we watch me go through it. As always, when we see a square root of x, that doesn't really help me so much. I really want to think of this as the natural log of x to the 1/2. Remember square roots are x to the 1/2. And this 2 in front, I've already seen that a number in front gets in the way, it's just going to be some moved eventually. So why don't I use my rules of logarithms to dump it on as an exponent of 3? So this is like 3 squared, which I'll write as 9 in the next step. Get used to doing that move, it's going to be pretty. How many phone numbers in front it really should be written as an exponent on the 3 on the variable inside the log. Okay, so this is a little cleaned up, anytime you see square root immediate convert it to one-half, 2 in front we moved it, we're looking good. Let's do our simplification of differences, so it becomes a single logarithm of x to the one-half, divided by 3 squared, let's just write that as 9 = 0. Same thing as before, lets exponentiate both sides, so we take e to the ln a half over 9 = e to the 0. E and ln they cancel so you get x to the one-half over 9 is e to the 0 which is 1, move the 9 to the other side, you get x to the one-half or square root of x is 9, this is like square root of x here. So now what do you do? You have to get rid of a square root, what's the opposite of a square root square both sides, when you square both sides, you got x = 9 squared, better known as 81. Nice job, another one little tough, let's do another example, what if I had the natural log of x, quantity squared, minus 1 = 0? In this example, what can we do here? So I want to solve for x, should I take e to both sides first? Is that really what's happening to x first? Not really, there's the square root is sort of being applied to the whole expression. Now this minus 1 as well, so maybe I'll hold off on taking a e at the moment, let's move the 1 over, let's do that first. Natural log of x squared = 1, and then now be careful this 2 is on the outside of the x, this is extremely important, extremely important so I can't bring the two out front. Only when it's inside on the exponent on x does the rule work, when it's the the quantity squared just means ln of x times ln of x, so it doesn't work, it can't bring that down. However, it's just like something squared equals 1, so what would you do in this case to solve for the something that's being squared, we would take a square root, that's all fine as well. When you do that, ln of X is, now be careful, it's plus or minus 1, plus or minus 1, when you take a square root you have to put plus or minus. Another way to see this, if you get nervous about taking plus or minus, another way to see this, we can do this here called way number 2, they all give you the same answer, but maybe you saw this is a difference of squares. This is like a squared, minus b squared, and that's a difference of squares, which is (a + b)(a- b). There's this formula here about factoring squares, so if you call ln of x, a and you realize that 1 squared is just 1, then you have a- b, so you could factor this ln of x minus 1, and you have ln of x + 1, and that's equal to 0. And so you have two things that multiply together, they give you 0, you set each one of them equal to 0, so you have ln of x minus 1 is 0 or, ln of x plus 1 is 0, and that just gives you that ln of x is 1, or ln of x is minus 1, which is exactly the same way. So both ways lead to the same conclusion, once you have from here, we're going to exponentiate. You could treat as separately, you can treat them together, it doesn't quite matter, but if you do it separately, you get e to the ln of x is 1, or to the ln of x is minus 1. Cancel, either one or e to the minus 1, I should say, exponentiate both sides and you get x = e to the 1, which is of course just e or x equals e to the -1. E to the -1 is perfectly fine, although you often tend to see it written as 1 over e, just a matter of personal preference, if you want negative exponents. So two answers on this one, two answers, it's kind of nice, and perhaps not surprising, given that we started off with something being squared, we got two answers however, you want to do it, take a square root. Just remember when you take a square root, you've got a plus or minus, or see it as factored, and then you factor there as well. Let's do another one, so let's do more square roots because they can get confusing, so we have ln of the squared of x is the square root of the e ln of x. This one scary, we have natural log two of them, we have square roots, gross. So once we square roots, let's go right to our we know what to do, we convert this to a half, and be careful here, the x is being taken the square root of in the first expression on the left, and the second expression it's the whole thing, it's ln of x to the one-half. So this one-half the first one natural log of x to the one-half, since on the x, sure I can bring that down, that's perfectly fine. The second one I cannot, I can't, so I have ln of x to the one-half. All right, so what should we do here? I have an x that's trapped inside a natural log, I'll probably need a e at some point, but first I see this one-half the square going on here, that's a little tricky, so why don't we get rid of that first? Why don't we square both sides? So I'll do one-half, natural log of x, the whole thing squared, = the natural log of x to the one-half the square root, sort of squared and you just get back itself. So if I square both sizes, square root goes away. Did it make life any better, so let's see the two hits both pieces, so we have one-half, squared which is one-fourth, then be careful I have natural log of x times the natural log of x, so natural log of x squared is the natural log of x. Okay, I think we can work with this, let's move things around, so I have one-fourth natural log of x squared, let's subtract the ln x over again expression equal to 0. Normally that's a pretty common thing you want to do some things equal to 0, we have a lot of rules for that, so whenever you have x is on both sides and equation, it's good practice to move everything with an x to one side so you can have it equal to 0. Okay so let's see now they have stuff in common, so they each have an ln of x, so I would get one-fourth natural log of x, left over and then minus 1, equals 0. Convince yourself this is true, right? If you bring this back in, so both pieces, then you get the original expression, you start off with, and then once you have this, you have two things that multiply together it could be 0, so what do we do? We break it off and we say ln of x is 0 or one-fourth. Ln of x minus 1 is 0, so you sort of split, you say, well, they're going to be 0, the whole product has to be 0 then, well, one of them has to equal 0. You don't know which one could be both, it could be 1, so you solve. Take e to both sides again, you get e to the ln x = e to 0, so you get x equals e the 0, which is x equals 1. Or, you get one-fourth ln of x = 0, so let's let's see here, if we do, we can move the one over so ln of x is 1, there's one-fourth still floating around. Now we multiply the whole thing by 4, I'll come up here so I have some room, ln of x is 4, let's take e to both sides, and you get x = e to the fourth, so this is interesting, so are two solutions. Two solutions here, are x = 1 and x = e to the 4th, and you can just go back and check and make sure everything looks good if you plug it in, you can get these to work, so it's kind of nice. All right, let's do maybe like one more, just getting stuff, let's see what we can do here. Let's see one more how about, 3 ln of x,- ln 3 of x = 0. Okay, so this is like 3 of x and parentheses and of course x is by itself there. So I have last things going on here, this was like medium difficulty, see if you can solve this, if you'd like, pause the video, let's see what we can do. I have a 3 in front as well as kind of in my way, so it's probably best to perhaps rewrite this right off the bat as ln (3x). You could break this off and do ln of 3 + ln(x) all with the minus sign out front, but I don't think you need to. In this case here, what should we do? I see log of a difference, let's combine that, let's simplify that, that becomes a quotient, the first term is the numerator, the second term is the denominator, and at least something cancels. It was assigned are on the right track instead of three x's, you only have two, and you get a 3 on the bottom, and then last but not least, if we have exponential, let's take e to both side, so we have e to the Ln, and e to the 0, so we've done that a couple times. So you have x squared over 3, e to the 0 is 1, over 3 over to get x squared is 3. Now you're about to take a square root here, careful, careful, careful, so x = plus or minus the square root of 3. So are we done, is that our answer? Be careful here, be careful, be careful, be careful, if you got plus or minus three if you got all the way down here, that's really good, but you gotta be careful. X is not equal to negative root 3, why? Why am I throwing this one away is called extraneous solution, go back to the original functions, well you always want to check your work? You always want to check your work, if you plug in x as negative or 3, now root 3 is some number, who cares, but it's a negative, so we throw in a negative, that's not in the domain of Logarithm, the logarithm graph, remember what it is, it only takes in, its only allowed to take in positive numbers. It looks something like this, sure it outputs negative number sure, but only takes in negative numbers. So ln of x, so it's domain is from zero to infinity with parentheses. You cannot plug in a negative number, so while algebraically you get back to solutions, we have an extraneous solution. So the true answer is X equals route three positive route three only, so we throw this one away from one solution. Always go back and check your work and make sure the answers you find actually do work. Be careful of extraneous solutions, they can be a little tricky. Alright, great job on this one, see you next time.
