 Hi, I'm Zor. Welcome to a new Zor education. I would like to present you a few more general inequalities. They might seem a little bit more complex, but actually they're simple. I was just trying to mask their simplicity behind certain more complex expressions. But basically, they are based on the same principle of comparing various of functions relative to various of arguments for monotonous cases, et cetera. So you will basically see it one by one. Three examples, I have three problems. I would actually encourage you, as always, to try to solve these problems just by yourself and then listen to my lecture. Maybe you can find a faster, better solution, whatever. And maybe afterwards, after you listen to the lecture, maybe you should try it again just by yourself to make sure that you have complete control over this. Now, in any case, I would like to remind again that what's important is, for all these inequality problems, to always pay attention to allowed values of the argument, to pay attention to your non-invariant transformations if you are using them to solve the equation. Inequality, I'm sorry. So you have to be careful doing all these kind of things. All right, fine. Let's just solve problems. 5 to the power of minus x minus 25 divided by 4x squared plus 4x plus 2 greater or equal to 2. OK. I think it's quite obvious, just looking at this particular quadratic polynomial, that it's equal to this. And this is x plus 1, sorry, 2x plus 1 squared. So this is 2x plus 1 squared plus 1, right? So for those of you who used 2x plus 1 many times, and if you solved many problems, I'm sure you would get something like this many times, it would immediately strike your eyes that this is a full square of 2x plus 1. And then you add another one, which means that this expression is always strictly positive. Now this is square, which means it's non-negative. But then plus 1, it's strictly positive. So the denominator of this particular inequality of this particular fraction is always positive. Now if we want the positive result or non-negative result, then our expression in the numerator must be non-negative, right? Because for negative things divided by positive, we will get negative. Not good. For non-negative divided by positive, we will get non-negative. That's exactly what we mean. So from here, immediately without any kind of additional manipulation, we can say that 5 to the minus x minus 25 should be greater or equal to 0. That's immediate consequence. There is no restriction on any argument because in the denominator, whatever argument is, it's always positive, so we cannot divide by 0. And there is no restriction on exponent of the function 5 to some degree. You can raise 5 to any degree. Well, just as a reminder, how the monotonously increasing function y is equal to 5 to the power of x looks like. That's how it looks like. It's monotonously increasing. As x is increasing, the function is increasing. And if exponent is decreasing, the function is decreasing. Now, what does it mean? Well, it means that the greater value of the function corresponds to the greater value of the argument and vice versa. So how can we reduce this to properties of monotonous functions? Very simply, 25 is 5 squared. So I will add 5 squared to both sides. I will get 5 to the minus x greater equal to the 5 squared. So these are two values of the function. These are two arguments. So when the function is greater than another function, where this is a monotonously increasing function, well, only in case our arguments are related to the same thing. And this is obviously means x is less than minus 2. By multiplying by minus 1, since minus 1 is a negative multiplier, we have to change the sign of the equation to the opposite. Minus 1 times minus x equal to x. Minus 2 minus minus 1 times 2 would be minus 2. And this is an answer. And again, since there is no restriction on x, all x's are allowed, this is the final answer. Next. Oh, boy, it looks complicated. 1 half to the power of log x squared. Don't get scared. So first, we take log x base 2. Then the result, v squared. And then use it as an exponent for 1 half. Now, obviously, it looks like we have to use the same principle of monotonous functions. The only problem is we have different bases. This is basis 1 half, and this base is 1 8. Not good, right? But well, let's just remember that 1 8 is 1 half to the third, right? That's what we're talking about. Now, is it easier? Well, actually, it's easier except we have the double power, actually. We raised one power and then raised another power. But again, if you remember, and I'm sure you do, that if you raise a number A to the power of B and then to the power of C, then this is the same as raising number to the product of these two powers, which means that instead of this, we can write it this way. I'll just multiply. That's what it means. Now, I have clear expression on both ends. I have an exponent, exponential function, actually, with base 1 half, and exponents this and this. Now, notice that exponential function with the base 1 half is monotonously decreasing, not monotonously increasing. So you have to remember that if the base is greater than 1, then it's increasing. If the base is less than 1, like in this case, the function is monotonously decreasing. For a monotonously decreasing function, the greater the argument, the smaller the function. And vice versa, which means that the relationship between two exponents is opposite to the relationship to functions, which means that I can write the following less than 3 log 2 of 8x. All right? So far so good. Now, let's simplify this. Remember, log of AB is equal to log A plus log B. That's the main property of the logarithm. That's basically the reason why logarithms were invented. The logarithm of a product is the sum of logarithms. So instead of this, I can write log 8 plus log x. Now, what's log 8 base 2? Which power should I use to raise the 2 into to get 8? 3, obviously, right? 2 to the 3rd is 8. So instead of this, I can write this. Now, let's bring everything to one side. And we will get, well, almost at quadratic inequality. And I'll tell you why it's almost. So it's minus 3 log x minus 9, 3 times 3 is 9, less than 0. If log x is a new variable, let's call it y, y is equal to log x base 2, then our inequality is just this, which we know how to solve, right? So that's what we're going to do. Now, when is a quadratic polynomial is less than 0? Well, if it has two solutions where it's equal to 0, now the horns of this parabola is upwards directed, which means that in between the two solutions where this particular quadratic polynomial where it's equal to 0, in between them, it's below 0. That's what we need. So we have to find basically these two points, right? When is this particular polynomial equals to 0? The solutions are, I'll just use the formula for solutions of quadratic equation, when this is equal to 0, it's 3 plus minus square root of 9 plus 4 times 9, 36 plus 36. Now, this is 45. 45 is 9 times 5, square root of 9 is 3, our factor out 3. So that's my two roots. So between these two roots, my polynomial is less than 0, which means considering y is a log x, I have this. That's what the solution is. So what does it mean for x? Well, don't forget that log x is a monotonously increasing function. The graph is this, the greater the argument, the greater the function, and vice versa, which means that if I will do this, what is the value of this particular expression if I want to put it into the logarithm function? I would like everything to express as log 2. What should I put here? 2 to the power of 3 times 1 plus. To get this using log 2, I should put 2 to the power, right? Because what's the logarithm of this? It's a power which I have to raise 2 to get this. And this is the power, so basically that's what it is. Similarly, it's log 2 of 2 to the power 3, 1 minus square root of 5 over 2. And now I can use the property of log x base 2, that it's monotonously increasing function, and say that my arguments are in exactly the same relationship as my functions. And these are my arguments. Now, this is the final answer. And the only thing which I have to really check is what's the allowed values. Now, allowed value for x here and here obviously are positive x, because log doesn't exist for 0 or negative x. Now, this is positive because 2 raised to any power would always give the positive number. That's the property of the exponential function 2 to the power of x. The graph is always above level 0. So this is the final answer. x should be between these two values. 2 to the power of 3 times 1 minus square root of 5 over 2 and plus. That's it for this problem. And the last one, the last one is log base 2 third log base 3 seconds of x plus 1 over x minus 1 greater than 0. So this looks like we have to really think about what allowed values are. Well, allowed values for this one is that this is supposed to be positive. So first of all, x plus 1 over x minus 1 should be greater than 0. Now, when is this greater than 0? I don't want to multiply by x minus 1. That would be probably a mistake, because x minus 1 is something which contains unknown x. And that's not an invariant transformation. But what we can do is something which we did before. Consider points where both x plus 1 and x minus 1 equal to 0. This is 1. That's when x minus 1 is equal to 0. This is minus 1. That's x plus 1 is equal to 0. Now, these two points are dividing our numbers in three intervals greater than 1 when everything is positive. Then 1 is supposed to be excluded, because that's where the denominator is equals to 0. But then from 1 to minus 1 going that way, it would be negative, because x plus 1 would be positive and x minus 1 would be negative. And finally, it will be positive again here, since it's below the minus 1. So this area and this area, not including the point 1, are the answer to this particular inequality. So it's x greater than 1 unionized with x less than or equal to minus 1. No, equal is not right. Strictly, because we have strictly here. All right, so this is a restriction on x based on existing of the inner logarithm. Now, how about the outer logarithm? Again, this logarithm exists only if whatever is underneath under this logarithm is positive. So what we have to do is we have to do this. It should be greater than 0. So only if this, which is under this logarithm, is positive, this logarithm exists. Now, let's just think about when this particular condition is true. The graph of function log base three seconds is monotonously increasing logarithm. It's asymptotically close to y-axis, and then monotonically increasing. So this is point 1, because logarithm of 1 is 0. Now, we need greater than 0, which means x plus 1 over x minus 1 should be greater than 1. Now, notice that this is a stricter solution than the one before. So whatever we had before, we can completely discard and leave only this one, because this is a solution which is much more strict. It's supposed to be greater than 1, and definitely it will be greater than 0. So this is the only restriction on x based on just allowed values. But let's find out exactly what this means in more precise terms. Let's subtract 1 from both sides. You will have this. Common denominator x plus 1 minus x minus 1 divided by x minus 1 greater than 0. And this is x minus x 1 minus minus 1, so it's 2. Now, when is it greater than 0? Obviously, if x is greater than 1, then this will be positive 2. So this condition is sufficient for our problem to be basically solvable. Now, considering this is satisfied, and we will apply this particular restriction later on, let's think about how to solve this. Well, first of all, we know that if we want logarithm of base 2 third to be greater or equal to 0, this is the graph of the function y is equal to log 2 third x. Now, because 2 third is less than 1, the logarithm goes down instead of up, right? So it's monotonously decreasing function, which means greater argument, smaller the function. Now, we need this to be greater than 0, which means whatever is under this logarithm, which is this piece, should be here. Now, this is greater or equal to 0, which means this point is included. This point is not included. So from this equation, we see that log 3 second of x plus 1 over x minus 1 should be less than or equal to, this is point 1, 1 and greater, strictly greater than 0. That's what we get. So from this, we get this. Now, let's not forget that whatever we do, this always applies. We are only considering x in this particular area. Now, how to solve this? Well, very easy. 0 is logarithm of 1. And 1 is logarithm of 3 second by base 3 second, right? Logarithm of 1 is 0 because 3 seconds to the power of 0 gives you 1. And 1 is logarithm of 3 second because 3 second to the power of 1 will give you 3 seconds. So now, I have logarithm everywhere, and this is the logarithm with a base greater than 1. The graph of this logarithm is increasing, which means that I can put exactly the same relationship between whatever is under this logarithm. So equation between the logarithms implies exactly the same equation between the arguments because the logarithm, in this case, is monotonously increasing. So that's what I have. And this is the last equation, which I really have to consider, right? OK, so it's two inequalities, less inequality Now it's two inequalities. The first one is the left part. So it's x plus 1. Now let's not forget that x is always greater than 1. That's where we are talking about looking for argument, right? So x minus 1 is positive, so I can always multiply by the positive value. So x plus 1 would be greater in this case than x minus 1. And any x, which belongs to this area, is satisfying this particular inequality because 1 is always greater than minus 1, and we add to both parts x. Now, so this is not a restriction on x at all. How about this? Well, x plus 1 greater less than or equal 3 seconds x minus 1. Again, I'm multiplying by a positive x minus 1. It's positive because I'm already dealing with this particular area, which means let's multiply by 2. So it's 2x plus 2 less than 3x minus 3. So subtract 2x would be x plus 3 would be greater than 5. And obviously, this is satisfied as well. So this is the final answer. All right, so I hope you did not get scared by some crazy formulas. Again, that was just my way of masking the simplicity behind complexity. And the complexity is very easily dissolvable by plain logic. Think about what we were using in these problems. We were using the properties of monotonous functions. So whenever two sides of the inequality are basically the values of the same monotonous functions, then the same inequality between the functions can be transferred into the corresponding inequality between the arguments. If it's monotonously increasing function, then the sign of inequality is retained. If it's monotonously decreasing function, then the sign of inequality is reversed. Well, other than that, be very, very careful with a lot of values. Well, that's it for today. This completes my exercise of general inequalities lecture number two. I will probably supply this with certain exams. And I do recommend you actually to go back to the unizor.com and try to solve these inequalities yourself just by yourself without looking at my lectures and see if you will have the same answers as I have received. That's it. Thank you very much and good luck.