All right everyone. So I thought in this video we can 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. 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 x_y. For logarithms, when you have a product it turns it into addition. So you have the natural log of x, this is of x not times of, it's a function plus the natural log of 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 as the natural log of 2 times 3. So this is all happening inside the parentheses, and that becomes the natural log of 2 plus the natural log of. You can go ahead and check on a calculator if you want. Type in the log of 6, the log of 2, and the log of 3, and you'll get the same numbers when you add these up. If you notice sometimes you write the parentheses sometimes you don't. It just needs to be very clear from context what it is. So there's 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 of x divided by y, well, this becomes the log of the numerator minus the log of the denominator. So log of x minus log of y. Just as a quick example here to show you what this is. If we do something like the natural log of, I don't know, what's a good number here, two-thirds perhaps. Then this is the natural log of 2 minus the natural log of 3. All these rules follow from their corresponding properties of exponents. If you are ever confused about the logarithm, put it back in terms of an exponent and you'll see what I mean. Let's use last one here, natural log of x to a power. This becomes n times the natural log of x. So exponent's we say fall down in front. All these equations are two-way streets. You can also read them right to left. So you can also have number of front, bring it up to the exponent. Just as an example to show this one with some numbers, we had ln of 4, this would be natural log of 2 squared, which will be 2 times the natural log of 2. Again, I encourage you to grab a calculator, go punch these in and show that these are actually equal. So these are the rules we're going to do. Remember, these are not technically the rules that I'm going to put down now, but they are rate constant [inaudible] as we all know anyway. The natural log of 1 is 0 and the natural log of e is 1. It's because their inverse is like e_1, they cancel. Of course, if you have ever have natural log of e_x, well that's just x. If you ever have e_ln of x, that's also just x. 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. Here we go. So the first thing they're going to ask you to do usually is to simplify. So the first thing we do is simplify. Let's do a couple examples here. What if I said, without a calculator, try to do this, In of 5 plus the natural log of x. So you can pause the video, work on these if you want. How would you via two logarithms, how would we simplify the expression? How can we write it as a single logarithm? Well, remember our rule if we have a sum of two logs, this is one logarithm, and then combine the two as a product. It's nice and easy there. Let's do another one, e_ 2 ln of x. How would we simplify this expression? Now, mistake a lot of people make here is they cancel the e in the logarithm. They just say, "Well, their inverse is right, can we do that?" The problem is no, there is this like 2 right there in front. So it's blocking the e from hitting the natural log. So you really need to focus on this 2 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 natural log are together, so they would cancel and you just get back x squared. So this expression with an e and a 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 minus ln of x squared plus ln of x_4. If you'd like three logs, it's kind of scary, pause the video, see if you can simplify this expression as we go. So let's do this. So we have, let's see. Let's grab of two at a time. So we're going to view the first two together. So I have ln of something minus ln of something. When you do that, it combines, the minus becomes a quotient and you have x over x squared. There's still a natural log of x_4 floating around. Let's not lose that. Then we can simplify x over x squared, of course, is just the natural log of 1 over x plus the natural log of x_4. 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_4 over x, I multiply these two together, and I get to the 4_x, which is, of course, the natural log of, cancel one of them, I execute. So this expression with three logarithms is of quite nicely to the natural log of x cubed. If you really wanted to simplify it, you're not wrong. If you wrote 3 times ln of 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_ln of x squared plus, and this is all in the exponent now, 3 to the natural log of y. So you have a nice multivariable function here with xs and y'. Again, I want to give you variables to start with, you probably going to give me back all variables upon return. So let's simplify this a little bit using laws of exponents. So I have e to the something plus something. This is like e_a plus b. So remember when you have e to a sum, it becomes the product of both these terms. So I'm going to split this up first as e_ln of x squared and then times e_3 natural log of y. e and ln cancel, so you just get back x squared for your first term. Then once again, I have this three in the way. So I need to write this three as a power on the y, cancel it and get 3y is extremely important. Now you have to move the three up and over to become an exponent on the y. Now e and ln cancel and you're left with x squared y_3. So a much nicer, simpler expression. So these are some of the rules that we're going to use to solve. So let's do a couple, solve for the given equation. So step 1 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 a problem before with a couple of logarithms. Let's simplify first. It's a multi-step problems. Simplify, grab the first two, just say, ''No, I'm not dealing with three logarithms. No way.'' So you change it and say x divided by x squared. You have a difference of logarithms that becomes a quotient, plus log of 3 equals 0. You can clean that up a little bit. x over x squared is just 1 over 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 over x times 3, which is 3 over x which is 0. Now I want to solve for x. So remember log is track, 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_ln of 3 over 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 the e, we're taking e as base. e and natural log cancel. That's why you did that. That's why you introduced e. The inverse function of the natural logarithm, and you get a good old 3 over x. Now you get e_0. e_0, which is 1. You get 3 over x is 1. You can cross multiply and get x equals 3. Again, you can go check through in calculator if you want, but you can check that x equals 3 does actually solve this problem. Let's do another one. Use different colors so we can 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 you 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_1.5 and the square roots of x_1.5. This two in front, I've already seen that a number in front gets in the way. So I just have to be so moved eventually. So why don't I use my rules of logarithms to dump it on as an exponent of three? So this is 3 squared which all right is 9 in the next step. Get used to doing that move. It's going to be pretty common if the numbers in front, it really should be written as an exponent on the three on the variable inside the log. So this is a little cleaned up. Anytime you see square root immediate convert it to 1.5, two in front. We moved it. We're looking good. Let's do our simplification of differences. So it becomes a single logarithm of x_1.5 divided by 3 squared. Let's just write that as 9 equals 0. Same thing as before. Let's exponentiate both sides. So we take e_ln a half over 9 equals e_0, e and ln, they cancel. So you get x_one-half over 9 is e_0, which is 1. Move the nine to the other side, you get x_one-half, the 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 the square root. What's the opposite of the square root? Square both sides. When you square both sides, you get x equals 9 squared, better known as 81. [inaudible] little tough. Let's do another example. What if I had the natural log of x quantity squared minus 1 equals 0? In this example, what could we do here? So I want to solve for x. Should I take either both sides first? Is that really what's happening to x first? No, not really. There's the square is being applied to the whole expression. I have this minus one as well. So maybe I'll hold off on taking the e at the moment. Let's move the one over, let's do that first. Natural log of x squared equals 1. Then now be careful, this two is on the outside of the axis. It's 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 quantity squared, just means ln of x times ln of x, so that it doesn't work, it can't bring that down. However, it's just 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 one. 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 could do with this here called way number 2. They all give you the same answer, but maybe you saw this. This is a difference of squares. This is like a squared minus b squared and that's a difference of squares, which is a plus b, a minus 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 minus b. So you could factor this as ln of x minus 1, and then you have ln of x plus 1, and that's equal to 0. So you have two things that multiply together, they give you zero. 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 it from here we're going to exponentiate. You can treat it separately, you can treat them together. It doesn't quite matter. But if you do it separately, you get e_ln of x is 1, or e_ln of x is minus 1. Cancel e_1 or e_minus 1. I should say exponentiate both sides, and you get x equals e_1, which is a cautious e, or x equals e_minus 1. Either minus one is perfectly fine, although you'd 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 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 got to put plus or minus or see it as factored, and then you factor there as well. Now, let's do another one. So let's do more square roots because they can get confusing. So we have ln of the square root of x is the square root of the ln of x. This one's scary. We have natural log, two of them. We have square roots. So once we see square roots, let's go right to our, we know what to do. We convert this to a half. Be careful here, the x is being squared, taking the square root in the first expression on the left, and the second expression, it's the whole thing, it's ln of x_one-half. So this one-half, the first one natural log of x_one-half is it's 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_one-half. 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 equals the natural log of x_one-half the square root squared, and you just get back itself. So if I square both sides, the square root goes away. Did I 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. 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 of x over and get an expression equal to 0. Normally, that's a pretty common thing you want to do, you set things equal to 0, we have a lot of rules for that. So whenever you have xs on both sides of an equation, it's good practice to move everything with an x to one side, so you can have it equal to 0. So let's see. Now they have stuff in common, so they each have an ln of x. So I will get one-fourth natural log of x leftover and then minus 1 equals 0. Convince yourself this is true. If you bring this back in to both pieces, then you get the original expression you started off with. Then once you have this, you have two things that multiply together to give you zero. So what did 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 split it and you say, well they're going to be zero, the whole product has to be zero. Then well, one of them has to be equal to zero. You don't know which one, it could be both, it could be one. So you solve, take e to both sides again and you get e_lnx is equal to e_0. So you get x equals e_0, which is x equals 1, or you get one-fourth ln of x equals 0. So let's see here. What can we do? We can move the one over. So ln of x is 1, there's a 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 equals e_4. So this is interesting. So two solutions here are x equals 1 and x equals e_4. 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 that's nice. Let's do maybe one more. This is getting tough. Let's see what we can do here. Let's see one more. How about 3ln of x minus ln 3 of x equals 0. So this is like 3 of x in parentheses and, of course, x is by itself there. So I have lots of things going on here. This one's like medium difficulty. See if you can solve this. If you'd like, pause the video and see if we can do. I have the three in front as well. It's in my way. So it's probably best to perhaps rewrite this right off the back as ln 3 of x. You could break this off and do ln of the 3 plus ln of x, all with the minus sign at front. But I don't think you need to. In this case here, what should we do? I see a log of a difference. Let's combine that, lets simplify that. That becomes a quotient. The first term is the numerator, second term is the denominator, and at least something cancels. So it was assigned on the right track. Instead of 3xs you now only have two and you get a three on the bottom. Then last but not least, if we have the exponential, let's take e to both sides, so we have e_ln and e_0. So we've done that a couple times. So we have x squared over 3, e_0 is 1, move the three over to get x squared is 3. Now, you're about to take a square root here. Careful, careful, careful. So x equals 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've got all the way down here, that's really good. But you got to be careful, x is not equal to negative root 3. Why am I throwing this one away? It's called extraneous solution. Go back to the original function. This is why I always want to check your work. If you plug in x as negative root three, now root three is some number, who cares? But it's a negative. So if we throw in a negative, that's not in the domain of logarithm. The logarithm graph, remember what it is? It's only allowed to take in positive numbers. It looks something like this. It outputs negative number, sure, but only takes in negative numbers. So ln of x. So its domain is from zero to infinity with parentheses, you cannot plug in a negative number. So while algebraically you get back two solutions, we have an extraneous solution. So the true answer is x equals positive root 3 only. So we throw this one away for 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. Great job on this one. See you next.