Often, when solving logarithmic inequalities, there are problems with a variable logarithm base. Thus, an inequality of the form
is a standard school inequality. As a rule, to solve it, a transition to an equivalent set of systems is used:
The disadvantage of this method is the need to solve seven inequalities, not counting two systems and one population. Already with these quadratic functions, solving the population can take a lot of time.
It is possible to propose an alternative, less time-consuming way to solve this standard inequality. To do this, we take into account the following theorem.
Theorem 1. Let there be a continuous increasing function on a set X. Then on this set the sign of the increment of the function will coincide with the sign of the increment of the argument, i.e. , Where 
.
Note: if a continuous decreasing function on a set X, then .
Let's return to inequality. Let's move on to the decimal logarithm (you can move on to any with a constant base greater than one).
Now you can use the theorem, noticing the increment of functions in the numerator 
and in the denominator. So it's true
As a result, the number of calculations leading to the answer is reduced by approximately half, which saves not only time, but also allows you to potentially make fewer arithmetic and careless errors.
Example 1.
Comparing with (1) we find 
, 
, .
Moving on to (2) we will have:
Example 2.
Comparing with (1) we find , , .
Moving on to (2) we will have:
Example 3.
Since the left side of the inequality is an increasing function as and 
, then the answer will be many.
The many examples in which Theme 1 can be applied can easily be expanded by taking into account Theme 2.
Let on the set X the functions , , , are defined, and on this set the signs and coincide, i.e. , then it will be fair.
Example 4.
Example 5.
With the standard approach, the example is solved according to the following scheme: the product is less than zero when the factors are of different signs. Those. a set of two systems of inequalities is considered, in which, as indicated at the beginning, each inequality breaks down into seven more.
If we take into account theorem 2, then each of the factors, taking into account (2), can be replaced by another function that has the same sign in this example O.D.Z.
The method of replacing the increment of a function with an increment of argument, taking into account Theorem 2, turns out to be very convenient when solving typical C3 Unified State Examination problems.
Example 6.
Example 7.
. Let's denote . We get
. Note that the replacement implies: . Returning to the equation, we get
.
Example 8.
In the theorems we use there are no restrictions on classes of functions. In this article, as an example, the theorems were applied to solving logarithmic inequalities. The following several examples will demonstrate the promise of the method for solving other types of inequalities.
Logarithmic inequalities
In previous lessons, we got acquainted with logarithmic equations and now we know what they are and how to solve them. Today's lesson will be devoted to the study of logarithmic inequalities. What are these inequalities and what is the difference between solving a logarithmic equation and an inequality?
Logarithmic inequalities are inequalities that have a variable appearing under the logarithm sign or at its base.
Or, we can also say that a logarithmic inequality is an inequality in which its unknown value, as in a logarithmic equation, will appear under the sign of the logarithm.
The simplest logarithmic inequalities have the following form:
where f(x) and g(x) are some expressions that depend on x.
Let's look at this using this example: f(x)=1+2x+x2, g(x)=3x−1.
Solving logarithmic inequalities
Before solving logarithmic inequalities, it is worth noting that when solved they are similar to exponential inequalities, namely:
First, when moving from logarithms to expressions under the logarithm sign, we also need to compare the base of the logarithm with one;
Secondly, when solving a logarithmic inequality using a change of variables, we need to solve inequalities with respect to the change until we get the simplest inequality.
But you and I have considered similar aspects of solving logarithmic inequalities. Now let’s pay attention to a rather significant difference. You and I know that the logarithmic function has a limited domain of definition, therefore, when moving from logarithms to expressions under the logarithm sign, we need to take into account the range of permissible values (ADV).
That is, it should be taken into account that when solving a logarithmic equation, you and I can first find the roots of the equation, and then check this solution. But solving a logarithmic inequality will not work this way, since moving from logarithms to expressions under the logarithm sign, it will be necessary to write down the ODZ of the inequality.
In addition, it is worth remembering that the theory of inequalities consists of real numbers, which are positive and negative numbers, as well as the number 0.
For example, when the number “a” is positive, then you need to use the following notation: a >0. In this case, both the sum and the product of these numbers will also be positive.
The main principle for solving an inequality is to replace it with a simpler inequality, but the main thing is that it is equivalent to the given one. Further, we also obtained an inequality and again replaced it with one that has a simpler form, etc.
When solving inequalities with a variable, you need to find all its solutions. If two inequalities have the same variable x, then such inequalities are equivalent, provided that their solutions coincide.
When performing tasks on solving logarithmic inequalities, you must remember that when a > 1, then the logarithmic function increases, and when 0< a < 1, то такая функция имеет свойство убывать. Эти свойства вам будут необходимы при решении логарифмических неравенств, поэтому вы их должны хорошо знать и помнить.
Methods for solving logarithmic inequalities
Now let's look at some of the methods that take place when solving logarithmic inequalities. For better understanding and assimilation, we will try to understand them using specific examples.
We all know that the simplest logarithmic inequality has the following form:
In this inequality, V – is one of the following inequality signs:<,>, ≤ or ≥.
When the base of a given logarithm is greater than one (a>1), making the transition from logarithms to expressions under the logarithm sign, then in this version the inequality sign is preserved, and the inequality will have the following form:
which is equivalent to this system:
In the case when the base of the logarithm is greater than zero and less than one (0 This is equivalent to this system: Let's look at more examples of solving the simplest logarithmic inequalities shown in the picture below: Exercise. Let's try to solve this inequality: Solving the range of acceptable values. Now let's try to multiply its right side by: Let's see what we can come up with: Now, let's move on to converting sublogarithmic expressions. Due to the fact that the base of the logarithm is 0< 1/4 <1, то от сюда следует, что знак неравенства изменится на противоположный: 3x - 8 > 16; And from this it follows that the interval that we obtained entirely belongs to the ODZ and is a solution to such an inequality. This is the answer we got: Now let's try to analyze what we need to successfully solve logarithmic inequalities? First, concentrate all your attention and try not to make mistakes when performing the transformations that are given in this inequality. Also, it should be remembered that when solving such inequalities, it is necessary to avoid expansions and contractions of the ODZ of inequality, which can lead to the loss or acquisition of extraneous solutions. Secondly, when solving logarithmic inequalities, you need to learn to think logically and understand the difference between concepts such as a system of inequalities and a set of inequalities, so that you can easily select solutions to the inequality, while being guided by its DL. Thirdly, to successfully solve such inequalities, each of you must perfectly know all the properties of elementary functions and clearly understand their meaning. Such functions include not only logarithmic, but also rational, power, trigonometric, etc., in a word, all those that you studied during school algebra. As you can see, having studied the topic of logarithmic inequalities, there is nothing difficult in solving these inequalities, provided that you are careful and persistent in achieving your goals. To avoid any problems in solving inequalities, you need to practice as much as possible, solving various tasks and at the same time remember the basic methods of solving such inequalities and their systems. If you fail to solve logarithmic inequalities, you should carefully analyze your mistakes so as not to return to them again in the future. To better understand the topic and consolidate the material covered, solve the following inequalities: Maintaining your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please review our privacy practices and let us know if you have any questions. Personal information refers to data that can be used to identify or contact a specific person. You may be asked to provide your personal information at any time when you contact us. Below are some examples of the types of personal information we may collect and how we may use such information. What personal information do we collect: How we use your personal information: We do not disclose the information received from you to third parties. Exceptions: We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as unauthorized access, disclosure, alteration and destruction. To ensure that your personal information is secure, we communicate privacy and security standards to our employees and strictly enforce privacy practices. Lesson objectives: Didactic: Educational: develop memory, attention, logical thinking, comparison skills, be able to generalize and draw conclusions Educational: cultivate accuracy, responsibility for the task being performed, and mutual assistance. Teaching methods:
verbal ,
visual ,
practical ,
partial-search ,
self-government ,
control. Forms of organization of students’ cognitive activity:
frontal ,
individual ,
work in pairs. Equipment:
a set of test tasks, reference notes, blank sheets for solutions. Lesson type: learning new material. During the classes 1. Organizational moment. The topic and goals of the lesson, the lesson plan are announced: each student is given an assessment sheet, which the student fills out during the lesson; for each pair of students - printed materials with tasks; tasks must be completed in pairs; blank solution sheets; support sheets: definition of logarithm; graph of a logarithmic function, its properties; properties of logarithms; algorithm for solving logarithmic inequalities. All decisions after self-assessment are submitted to the teacher. Student's score sheet 2. Updating knowledge. Teacher's instructions. Recall the definition of a logarithm, the graph of a logarithmic function, and its properties. To do this, read the text on pp. 88–90, 98–101 of the textbook “Algebra and the beginnings of analysis 10–11” edited by Sh.A Alimov, Yu.M Kolyagin and others. Students are given sheets on which are written: the definition of a logarithm; shows a graph of a logarithmic function and its properties; properties of logarithms; algorithm for solving logarithmic inequalities, an example of solving a logarithmic inequality that reduces to a quadratic one. 3. Studying new material. Solving logarithmic inequalities is based on the monotonicity of the logarithmic function. Algorithm for solving logarithmic inequalities: A) Find the domain of definition of the inequality (the sublogarithmic expression is greater than zero). Learning element #1. Goal: consolidate the solution to the simplest logarithmic inequalities Form of organization of students' cognitive activity: individual work. Tasks for independent work for 10 minutes. For each inequality there are several possible answers; you need to choose the correct one and check it using the key. Learning element #2. Goal: consolidate the solution of logarithmic inequalities using the properties of logarithms. Teacher's instructions. Remember the basic properties of logarithms. To do this, read the text of the textbook on pp. 92, 103–104. Tasks for independent work for 10 minutes. KEY: 2113, maximum number of points – 8 points. Learning element #3. Purpose: to study the solution of logarithmic inequalities by the method of reduction to quadratic. Teacher's instructions: the method of reducing an inequality to a quadratic is to transform the inequality to such a form that a certain logarithmic function is denoted by a new variable, thereby obtaining a quadratic inequality with respect to this variable. Let's use the interval method. You have passed the first level of mastering the material. Now you will have to independently choose a method for solving logarithmic equations, using all your knowledge and capabilities. Learning element #4. Goal: consolidate the solution to logarithmic inequalities by independently choosing a rational solution method. Tasks for independent work for 10 minutes Learning element #5. Teacher's instructions. Well done! You have mastered solving equations of the second level of complexity. The goal of your further work is to apply your knowledge and skills in more complex and non-standard situations. Tasks for independent solution: Teacher's instructions. It's great if you completed the whole task. Well done! The grade for the entire lesson depends on the number of points scored for all educational elements: Submit the assessment papers to the teacher. 5. Homework: if you scored no more than 15 points, work on your mistakes (solutions can be obtained from the teacher), if you scored more than 15 points, complete a creative task on the topic “Logarithmic inequalities.” Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely taught in school: log k (x) f (x) ∨ log k (x) g (x) ⇒ (f (x) − g (x)) (k (x) − 1) ∨ 0 Instead of the “∨” checkbox, you can put any inequality sign: more or less. The main thing is that in both inequalities the signs are the same. This way we get rid of logarithms and reduce the problem to a rational inequality. The latter is much easier to solve, but when discarding logarithms, extra roots may appear. To cut them off, it is enough to find the range of acceptable values. If you have forgotten the ODZ of a logarithm, I strongly recommend repeating it - see “What is a logarithm”. Everything related to the range of acceptable values must be written out and solved separately: f(x) > 0; g(x) > 0; k(x) > 0; k(x) ≠ 1. These four inequalities constitute a system and must be satisfied simultaneously. When the range of acceptable values has been found, all that remains is to intersect it with the solution of the rational inequality - and the answer is ready. Task. Solve the inequality: First, let’s write out the logarithm’s ODZ: The first two inequalities are satisfied automatically, but the last one will have to be written out. Since the square of a number is zero if and only if the number itself is zero, we have: x 2 + 1 ≠ 1; It turns out that the ODZ of the logarithm is all numbers except zero: x ∈ (−∞ 0)∪(0; +∞). Now we solve the main inequality: We make the transition from logarithmic inequality to rational one. The original inequality has a “less than” sign, which means the resulting inequality must also have a “less than” sign. We have: (10 − (x 2 + 1)) · (x 2 + 1 − 1)< 0; The zeros of this expression are: x = 3; x = −3; x = 0. Moreover, x = 0 is a root of the second multiplicity, which means that when passing through it, the sign of the function does not change. We have: We get x ∈ (−∞ −3)∪(3; +∞). This set is completely contained in the ODZ of the logarithm, which means this is the answer. Often the original inequality is different from the one above. This can be easily corrected using the standard rules for working with logarithms - see “Basic properties of logarithms”. Namely: Separately, I would like to remind you about the range of acceptable values. Since there may be several logarithms in the original inequality, it is required to find the VA of each of them. Thus, the general scheme for solving logarithmic inequalities is as follows: Task. Solve the inequality: Let's find the domain of definition (DO) of the first logarithm: We solve using the interval method. Finding the zeros of the numerator: 3x − 2 = 0; Then - the zeros of the denominator: x − 1 = 0; We mark zeros and signs on the coordinate arrow: We get x ∈ (−∞ 2/3)∪(1; +∞). The second logarithm will have the same VA. If you don't believe it, you can check it. Now we transform the second logarithm so that the base is two: As you can see, the threes at the base and in front of the logarithm have been reduced. We got two logarithms with the same base. Let's add them up: log 2 (x − 1) 2< 2; We obtained the standard logarithmic inequality. We get rid of logarithms using the formula. Since the original inequality contains a “less than” sign, the resulting rational expression must also be less than zero. We have: (f (x) − g (x)) (k (x) − 1)< 0; We got two sets: It remains to intersect these sets - we get the real answer: We are interested in the intersection of sets, so we select intervals that are shaded on both arrows. We get x ∈ (−1; 2/3)∪(1; 3) - all points are punctured.
Solving Examples
3x > 24;
x > 8. 
What is needed to solve logarithmic inequalities?
Homework
Collection and use of personal information
Disclosure of information to third parties
Protection of personal information
Respecting your privacy at the company level
B) Represent (if possible) the left and right sides of the inequality as logarithms to the same base.
C) Determine whether the logarithmic function is increasing or decreasing: if t>1, then increasing; if 0
D) Go to a simpler inequality (sublogarithmic expressions), taking into account that the sign of the inequality will remain the same if the function increases and will change if it decreases.
KEY: 13321, maximum number of points – 6 points.
x2 ≠ 0;
x ≠ 0.
(9 − x 2) x 2< 0;
(3 − x ) (3 + x ) x 2< 0.
Converting logarithmic inequalities
x = 2/3.
x = 1.
log 2 (x − 1) 2< log 2 2 2 .
((x − 1) 2 − 2 2)(2 − 1)< 0;
x 2 − 2x + 1 − 4< 0;
x 2 − 2x − 3< 0;
(x − 3)(x + 1)< 0;
x ∈ (−1; 3).
