 Hi, I'm Zor. I would like to continue talking about logarithms, and primarily, right now, this is about graphs. So, graph of the logarithmic function. The key to this point is the fact that logarithmic function, rather, is an inverse function to exponential function. Now, we all know how exponential function looks like. So, if this is exponential function, its graph is for a greater than one, it goes like this, and for a less than one, it goes like this. So, if necessary, just refresh your memory and go to the lectures about exponential function. Fine. Next, we have to understand that logarithmic function is an inverse by definition of the exponential function. And as I was explaining in lectures about monotonic functions and inverse functions, the graph of the inverse function is symmetrical to the graph of the original function relative to the angle bisector between x and y axis. So, this is angle bisector. Symmetrical relative to this line. Well, let me try. Symmetrical to this one would be something like, symmetrical to this point would be this point, also one. Symmetrical to this would be this, and symmetrical to this would be this. So, this is y is equal to log xA, where A is greater than one. Now, symmetrical to this line relative to this axis would be this. So, these two lines symmetrically are these two lines. So, this is y is equal to log x, where A is less than one, greater than zero. So, basically, I finished. I mean, these are graphs of the logarithmic function. In as much as exponential function asymptotically approaching the x-axis, the logarithmic function will asymptotically approach the vertical y-axis here, with A greater than one. With A less than one, it's the other way around, same as the exponential function is asymptotically approaching the x-axis. It would be here this. Okay, now, if this is the qualitative, so to speak, picture of how the graph of the logarithmic function looks like, then there are certain quantitative issues. And the most important quantitative issue is the following. What happens with the graph of the logarithmic function if A is changing? Well, let's go again to the graph of the exponential function. Now, for greater than one A, the exponential function is increasing. Now, if A is greater, then the exponential function would be growing faster or diminishing faster, because the base is greater. So, let me just use this. This is greater A. Now, similarly, so it looks like the whole graph actually is turning this way when A is growing. Now, the corresponding logarithmic will also be turning, but actually the opposite, because it's symmetrical, right? If this goes this way, symmetrical goes this way. So, most likely, I will have to have something like this as a graph. So, as A is growing, exponential function turns this way, symmetrical logarithmic function will go this way. And actually, I would like to algebraically prove it. What I mean is, if I am in this area of the argument x greater than one, then the greater base will correspond to a smaller logarithm. That's what I would like to prove. Purely algebraically. And here is how we do it. Consider, first of all, that the base is greater than one, and I have two bases. I have this base, which is A, and this base, which is B. And B is greater than A, and both are greater than one. So, we are talking about the logarithmic function, which is increasing. Also, I would like to consider, first, this particular interval, x greater than one. And I would like to prove that under these conditions, the logarithm of the x base A is greater than logarithm x base B. How can I prove that? Here is how. One of the properties of the logarithmic function can allow me to wipe out this, because we don't really need this graph anymore. We have already drawn it. Now we are algebraically trying to research. One of the properties of the logarithms, which we were talking about, was the following. Logarithm B A times logarithm x B is equal to logarithm x A. Remember, mnemonics, which I was talking about, B over A, x over B. B is reduced, so we have x over A. This is just mnemonic. You don't have to think about any kind of meaningful relationship between this and this. So anyway, in this particular case, let's think about this. This is logarithm A. This is logarithm B. And it looks like I'm multiplying logarithm by the base B by something to get this. Now, how about this? Well, obviously, log B A is greater than 1 since A is less than B. Now, why is that? This is a bigger number. This is a smaller number. Obviously, I have to raise it to a power greater than 1. But we can actually do it more strictly if you wish. Now, the way how to do it strictly is that if you have B greater than A, then you know that A to the power log B A is equal to B. Because that's the definition of the logarithm. So what we can say is A to the first degree, first power is less than B. And B I will represent as this one. Now, we know that exponential function is monotonically increasing for A greater than 1. Now, since it's monotonically increasing function, we know that increase of the argument is always coming together with increase of the function. And increase of the function comes together with increase of the argument. So if I know this, it means that my exponential parts are also in the same relationship. So 1 is less than logarithm. Logarithm is greater than 1. So proof that, what do we see now? We see that this logarithm X at the base B should be multiplied by something greater than 1 to get this. Which means, obviously, that this is greater than this. Log X. A is greater than log X. Since I have to multiply this by a factor greater than 1 to get this, it's supposed to be less. So, let's just think again what we have proven. That this graph of the logarithmic function, if I take a different base B, if A is less than B, then the logarithm goes above. So it's logarithm, logarithm at this base is greater. Now, on this particular segment from 0 to 1, situation is reversed. So the logarithm of the bigger base should be higher, which means greater than the logarithm on the smaller base. Now, how can we prove that? Actually, we will just reduce one problem to another, as mathematicians usually do. So, again, we have this, and now we have X in this interval. And we would like to prove that logarithm of the bigger base is greater than logarithm of the smaller base. Now, how can we do this? Well, very simply, don't forget that X is less than 1, and if you remember, logarithms of any number which is less than 1 are negative, right? So, instead of doing that, I will use another variable, let's say T, which is equal to 1 over X. Now, if X is in this particular interval, then T is obviously greater than 1, right? It's inverse. Now, but I know that if I consider this and this, whatever I have just proved before was this. Now, these are positive numbers, because T is greater than 1, and this is, I can just prove in this, right? But now, let's think about, back to the X terminology, what is T? T is 1 over X, right? So, let's replace it with this, and let's replace it with this. And now, let's remember that logarithms of 1 over X is minus logarithm of X, right? That was one of the properties in the previous lecture. So, minus log XA is greater than minus log XB. Now, we don't need these minuses, right? And if you remember, if we multiply the inequality by minus 1, then the side of the inequalities should be changed to the opposite, right? So, log XA is greater, change to opposite, less than log XB. And that's exactly what needs to be proved. So, what we have actually done right now, we not only built the general form of the graph, but we also quantitatively evaluated it. So, let me again draw the graph. This is 1, and this is general form of Y is equal to log X, where A is greater than 1. So, it's increased. Now, I don't even want to talk about the graph when A is less than 1 because it's completely symmetrical and all relationships are obviously symmetrical as well. So, not only we have drawn this qualitative picture, so to speak, but we also have the quantitative evaluation. If A is growing, then the graph is turning this way. So, this goes to this, and this goes closer to this. So, that actually kind of gives you a very nice picture of how the graph behaves. There is one more very interesting detail about this. Some time ago, when I was talking about exponential functions, I was talking about graph and the tangent at the point X equals 0. This tangent for greater base, this is A to the power of X. The greater base A, the graph is steeper. And now, what I also pointed out, that somewhere between the values of 2 and 3 for the base, this particular angle is exactly 45 degree. Now, the value when it is equal to 45 degree is actually irrational number E, which is approximately 2.721. So, the graph of the function E to the power of X has this tangent exactly at 45 degrees. Now, let's consider the symmetrical graph, symmetrical relative to the bisector of the angle between the axis. Now, this is also 45 degree, and this is 45 degree. So, this line should be parallel. So, the tangential principle is supposed to be preserved between this line and the tangent. So, if I will have a tangent here for logarithm, which is based the same number E, and if you remember, we called it natural logarithm, and there is a special notation for this, natural logarithm is logarithm base E, where E is this irrational number. It should also have this tangential line, this tangent at 45 degrees. So, whenever your base of the logarithm is greater than E, E is 2.71, etc., so let's say it's 3 or 4, etc., it's increasing, then the graph should go in such a way that this tangent goes a little bit, it's tilted down, like in this particular case. And whenever you have smaller base than E, let's say, logarithm by the base 2, for instance, yes, it goes actually above, tangential line will be greater than 45 degrees. So, this is a complete picture of the graph, and that basically is all I wanted to talk about as a theoretical material. I mean, I also had a couple of problems, but these are easy. So, next lecture, probably more than one, I will devote to different problems related to logarithms. So, good luck, register and take all the exams which are available on the website, Unisor.com, and until the next time, thank you very much.