 Hi, I'm Zor. Welcome to your new Zor education. I would like to continue talking about logarithms. The previous lecture was kind of an introductory, so I defined what logarithm is about, and I will start from a very short summary of whatever we talked about in the previous lecture. So, first of all, logarithmic function is an inverse to an exponential function. So, you have exponential function. Now, logarithmic function is an inverse, which means that knowing the value of the function, we can find the argument. So, what is this? This is an exponent, which if I will use with a base A, I will get the argument. So, this is basically a definition of what is logarithm of x with a base A. It's the number, which if used as an exponent with a base A, will give an argument x. So, this is the definition. Now, we also know that this function the exponential function is monotonic, and it's increasing if A is greater than 1 and decreasing if A is less than 1, greater than 0. By the way, I didn't mention it before, but everywhere we assume that A is greater than 0. So, knowing the monotonicity, the monotonic character of the exponential function, we can basically state exactly the same about the inverse function. I don't remember how much attention it was given in the general theory of functions wherever I just introduced the monotonicity, but let me just very briefly make a short proof that the original function is monotonically increasing than the inverse function is also monotonically increasing. Okay, here's how to prove it in a general case. So, let's say you have a function which is monotonically increasing. Now, the function is inverse. Inverse to this one, which means if I will use g and use this and an argument, I will get this as a function variable. All right, so how can I prove that this function is also monotonically increasing? Well, very simple. Let's take two different arguments, u and v, which are related to this inequality, and I would like to prove that the corresponding variable of the functions are also in the same relationship. That's what I have to prove, right? All right, how can I do it? Here's how. Let's say this is equal to p and this is equal to q. Since g is inverse of f, if p is equal to g of u, then u is equal to f of p, right? This is inverse function to this one. If I will use, instead of an argument, I'll use the function, I will have the value of the argument. So that's basically the relationship. If p is equal to g of u, then u is equal to f of p. That's what means f and g are inverse to each other. Similarly, v is equal to, sorry, f of q, right? Now, we know this relationship. So we know this. u is less than v. So f of q is less than f of q. Now, let me ask you now. What can be the relationship knowing the f of x is monotonically increasing? What can be the relationship between p and q? Can p be equal to q? No, because then f of p will be equal to f of q. Can p be greater than q? No, because again, since f of x is monotonically increasing, if p is greater than q, then f of p will be greater than f of q, which again is not true. So the only thing which we can say is p is less than q. p which is this is less than q which is this. And that's what exactly needed to be proved. All right. So this is a general property. If the function is monotonically increasing, inverse function is also monotonically increasing. Obviously the same thing for decreasing. I don't even want to prove it. It's kind of obvious. All right. So we will use exactly this property, general property of monotonic functions when addressing the logarithms. So we know now that logarithm x is increasing whenever the corresponding exponential function is increasing and decreasing monotonically whenever the exponential function is decreasing. So these are the properties of the logarithm x. The monotonicity and direction increasing or decreasing depends on the base. All right. That's done. And now let's talk about certain properties of the logarithmic function. So that's the function. And this is inverse to this function. Now one of the things which we again can conclude from this being inverse to this, whatever is the domain in this function becomes the range of this function. And whatever is the range of this function is the main to this function. So what's the domain for a to the power of x? Well, it's defined for all real x, which means that the range of this function would be all real x, all real values. So y belongs to all real values. Now the range of this function, if you remember, it's always positive. It's either increasing or decreasing. The graph looks like this. This is for a greater than one. This is for a less than one. But it's always positive. So the range of the function where the values are is always above the x's. So that would be the domain of this function. Our plus means only positive real numbers. All right. So that we have defined. We have a domain. It's defined for any positive number. And the result of the function can be, obviously, any real number. All right. Now we still would like to mention these two restrictions for obvious reasons. Now let's have a couple of examples. Now, what is a logarithm of one with any base, which satisfies these conditions, of course, with any base a? Well, this is zero. Why? Let's again remember the definition of the logarithm. The logarithm is a number which is used as an exponent with a base a will give the argument. So let's check it out. If I will take a and put it in this particular exponent, I should get one, right? That's how we should verify it. Now let's check it out. If I will put a to the power of zero, we all know from the zero exponential function that any number in the power of zero actually is equal to one. So that's true. So that's why a loop of one with a base a is always equal to zero. Next, what is this? Again, it's supposed to be an exponent which is used with a base a will give a. So, well, obviously it's one because a to the first degree is equal to a. I'll put direct proof that this is zero. This is the check. Next, log a of one over a. Okay, what is this? What power should they use to get one over a? Well, obviously minus one. So that's why it's equal to minus one. Next, this is something which I have already proven in the previous lecture. Basically what I have proven in the previous lecture was this. But if x is equal to a, and we know that log a base a is equal to one, then this is true. Like for example, log two to the tenths degree base two is equal to ten. Just as an example. Okay, these are all trivial things which are directly followed from the definition of the libraries. That's why I don't really go into more details of this. I'm trying to very quickly put as much trivial stuff as possible. Next, log a one over x. Let's think about what is one over x? It's x to the power of minus one, right? And considering the previous rule which you were just mentioning, it would be log x to the power of minus one, which is minus. Remember this is delta. Delta goes outside of the logarithms. Minus log a. Next, log x one over a. What if I have one over a in the base? Well, quite frankly, it's the same thing. Now, why? This actually means maybe a little explanation. How can I prove that this is this? Well, basically I have to do this. I have to take the base, raise it into this power, and check if I get x, right? But now let's think about this way. This is minus. Minus is basically reversed. So this is one over a minus one log x a, right? Because subsequent raising into power is equivalent to multiplication. Minus one is multiplied by this. Now, what is one over a to the power minus one? Minus one is reversing. So one over a is converted to a. So we get a log x a. And this is x by definition of log x by the base a. OK, that is done. Next. OK, you remember from the previous lecture I have proven that log of the product is equal to sum of logs. Now, how about division? How about log u over v? Well, let's think about it. u over v can always be represented as u times v to the power of minus one, right? Now, this is a product, so I can convert it into sum. So it will be log u plus log of v to the minus one equals. And by the previously proven theorem, log of one over v, v to the power of minus one, is minus log a v. So if we multiply, we add logarithms. If we divide, we subtract logarithms. u minus, u minus logarithms. Now, this is an interesting thing. Log b for the base a times log c base b is equal to log c base a. In a way, it looks like it has nothing to do with this, but it looks like b over a times c over b is equal to c over a, right? When you can reduce it. It looks like this. I mean, it's good for memorization, but don't mix these two. Nothing can come. It's just for mnemonics. In any case, how can I prove that? Well, to prove that something, logarithm of c base b is actually this, I have to prove that if I raise a to this power, I will get c. Well, let's prove it. a to the power log b a times log cb. What is it equal to? Now, again, if you multiply two powers, it's basically a consecutive raising into power first into this one, a to the power log b a, and then to this one, log cb equals. Now, what is this? a to the power of log b base a. This is a definition of logarithms, and it's supposed to be equal to b, right? Logarithm is the power, logarithm of b by the base a is the power which if used as an exponent with a base a will get b. So that's what b is. And we retain log cb. Now, what is this? Exactly the same thing. By definition of logarithm c by the base b, this is supposed to be c. So that's why a to this power gives c. So that's why this is correct. This is just a nice consequence. Why? Well, let's just remember the previous one. Log b base a times log b of c is equal to log cb, right? That's what... No, sorry, a. That was I have just proved before. Now, what if c is equal to a? a, b, and c can be anything, right? So I put a. So on the left I have this. On the right I have this, which is equal to 1 as we have already proven before. So that's why it's true. For a greater than 1, the behavior of the logarithm function, if x is greater than 1, log x a greater than 0. If x equals to 1, then log x is equal to 0, and if x less than 1, log x is less than 0. Now, why is that? Well, it's actually a very simple result of applying the monotonicity function. Now, we know this, right? The logarithm of 1 is always equal to 0. Now, with a greater than 0, greater than 1, sorry, the function logarithm x is monotonically increasing, right? So if I increase x beyond the point 1, going to greater than 1, my value of the logarithm should increase. So it should be greater than 0. And on the opposite side, to the less than 1 area, then logarithm should also decrease because the function logarithm x by the base a greater than 1 is monotonically increasing. So if I go from some point to the right, it's increasing to the left, it's decreasing. Very similarly. If a is less than 1, I have a similar story, but now it's decreasing, right? Because the corresponding exponential function is decreasing. So inverse logarithmic is also decreasing. So if it's decreasing, then we should have this. Again, at point x equals to 1, we know that logarithm is equal to 0. Now, if I move to the right, increasing value of the x of the argument, the logarithm should decrease from 0 down, less than 0. And if I go to the left, decreasing the argument, considering that the function is monotonically decreasing, the function value should increase. So it goes up above the other. That basically concludes the main properties of logarithms. What we still have left is graph, and that's easy. That would be next lecture. And then, a certain number of problems which we will do together. Register on theunisor.com if you would like really to get to the gutsy bit, because there are exams, and exams are very important. You can take exams if you are a registered student, and I do recommend you to do this. Thank you very much.