Solving logarithmic inequalities containing a base variable. Complex logarithmic inequalities
A lesson on one inequality develops research skills, awakens students' thoughts, develops intelligence, and increases students' interest in work. It is best to conduct it when students have mastered the necessary concepts and have analyzed a number of particular solution techniques. logarithmic inequalities. In this lesson, students are active participants in finding a solution.
Lesson type
. A lesson in applying knowledge, skills, abilities in a new situation. (Lesson of systematization and generalization of the studied material).Lesson Objectives
:- educational : to develop skills and abilities to solve logarithmic inequalities of the specified type different ways; teach to independently obtain knowledge (students’ own activities to study and master the content educational material);
- developing : work on speech development; teach to analyze, highlight the main thing, prove and disprove logical conclusions;
- educational : formation of moral qualities, humane relations, accuracy, discipline, self-esteem, responsible attitude towards achieving the goal.
During the classes.
1. Organizational moment.
Oral work.
2. Checking homework.
Write down the following sentences in mathematical language: “The numbers a and b are on the same side of one,” “The numbers a and b are on opposite sides of the unit,” and prove the resulting inequalities. (One of the students prepared a solution in advance on the board).
3. Report the topic of the lesson, its goals and objectives.
Analyzing the options for entrance exams in mathematics, one can notice that from the theory of logarithms in exams one often encounters logarithmic inequalities containing a variable under the logarithm and in logarithm base.
Our lesson is lesson of one inequality, containing a variable under the logarithm and at the base of the logarithm, solved in different ways. They say that it is better to solve one inequality, but in different ways, than several inequalities in the same way. Indeed, you should be able to check your decisions. There is no better test than solving a problem in a different way and getting the same answer (you can arrive at the same systems, the same inequalities, equations in different ways). But not only this goal is pursued when solving tasks in different ways. Searching for different solutions, considering all possible cases, critically assessing them in order to highlight the most rational, beautiful, is important factor development of mathematical thinking, lead away from the template. Therefore, today we will solve only one inequality, but we will try to find several ways to solve it.
4. Creative application and acquisition of knowledge, mastering methods of activity by solving problematic problems built on the basis of previously acquired knowledge and skills in solving the inequality log x (x 2 – 2x – 3)< 0.
Here is the solution to this inequality, taken from one exam paper. Look at it carefully and try to analyze the solution. (The solution to the inequality is written down on the board in advance)
log x (x 2 – 2x – 3)< log x 1;
a) x 2 – 2x – 3 > 0; b) x 2 – 2x – 3< 1;
x 2 – 2x – 3 = 0; x 2 – 2x – 4< 0;
x 1 = - 1, x 2 = 3; x 2 – 2x – 4 = 0;
c) solution of the system
Possible student explanations:
This is not an equation, but an inequality, therefore, when moving from a logarithmic inequality to a rational one, the sign of the inequality will depend on the base of the logarithm and monotonicity logarithmic function.
With such a decision, it is possible to acquire extraneous solutions, or to lose solutions, and it is possible that with an incorrect decision, the correct answer will be obtained.
So how was it necessary to solve this inequality, in which the variable is under the sign of the logarithm and in the base of the logarithm?!
This inequality is equivalent to a combination of two systems of inequalities.
The first system of inequalities has no solutions.
The solution to the system of inequalities will be
In the proposed solution to the inequality from the exam paper, the answer was correct. Why?
Possible student answers:
Since the domain of definition of the function on the left side of the inequality consists of numbers greater than 3, therefore, the function y = log x t is increasing. Therefore, the answer turned out to be correct.
How was it possible to write down a mathematically correct solution in an exam paper?
II method.
Let's find the domain of definition of the function on the left side of the inequality, and then, taking into account the domain of definition, consider only one case
How else can this inequality be resolved? What formulas can be used?
Formula for moving to a new base a > 0, a 1
III method.
IV method.
Is it possible to apply to the inequality itself the fact that the logarithm is less than zero?
Yes. The expression under the logarithm and the base of the logarithm are on opposite sides of one, but are positive!
That is, we again obtain the same set of two systems of inequalities:
All considered methods lead to a combination of two systems of inequalities. In all cases the same answer is obtained. All methods are theoretically justified.
Question to students: why do you think a question was asked in the homework that was not related to the material studied in grade 11?
Knowing the properties of the logarithm that log a b< 0 , If a And b on opposite sides of 1,
log a b > 0 if a And b on one side of 1, you can get a very interesting and unexpected way solutions to inequalities. This method is written about in the article “Some useful logarithmic relationships” in the magazine “Quantum” No. 10 for 1990.
log g(x) f(x) > 0 if
log g(x) f(x)< 0, если
(Why condition g(x) 1 is not necessary to write?)
Solution to inequality log x (x 2 – 2x – 3)< 0 looks like that:
a) x 2 – 2x – 3 > 0; b) (x – 1)(x 2 – 2x – 4)< 0;
c) solution to the system of inequality
VI method.
Interval method. (“Solving logarithmic inequalities using the interval method” is the topic of the next lesson).
5. The result of the work done.
1. In what ways was inequality resolved? How many ways to solve this
Did we find any inequalities?
2. Which one is the most rational? Beautiful?
3. What was the solution to the inequality based on in each case?
4. Why is this inequality interesting?
Qualitative characteristics of the teacher’s work in the classroom.
6. Generalization of the studied material.
Is it possible to consider this inequality as a special case of a more general problem?
Inequality of the form log g(x) f(x)<(>) log g(x) h(x) can be reduced to inequality log g(x) p(x)<(>) 0 using the properties of logarithms and the properties of inequalities.
Solve inequality
log x (x 2 + 3x – 3) > 1
by any of the methods considered.
7. Homework, instructions on its implementation
.1. Solve the inequalities (from the options for entrance exams in mathematics):
2. In the next lesson we will consider logarithmic inequalities that are solved by the interval method. Repeat the algorithm for solving inequalities using the interval method.
3. Arrange the numbers in ascending order (explain why this arrangement):
log 0.3 5; ; ; log 0.5 3 (repeat for next lesson).
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.
Collection and use of personal information
Personal information refers to data that can be used to identify certain person or connection with him.
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:
- When you submit an application on the site, we may collect various information, including your name, telephone number, address Email etc.
How we use your personal information:
- Collected by us personal information allows us to contact you and inform you about unique offers, promotions and other events and upcoming events.
- From time to time, we may use your personal information to send important notices and communications.
- We may also use personal information for internal purposes such as auditing, data analysis and various studies in order to improve the services we provide and provide you with recommendations regarding our services.
- If you participate in a prize draw, contest or similar promotion, we may use the information you provide to administer such programs.
Disclosure of information to third parties
We do not disclose the information received from you to third parties.
Exceptions:
- If necessary, in accordance with the law, judicial procedure, in legal proceedings, and/or on the basis of public requests or requests from government agencies in the Russian Federation - to disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public importance purposes.
- In the event of a reorganization, merger, or sale, we may transfer the personal information we collect to the applicable successor third party.
Protection of personal information
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.
Respecting your privacy at the company level
To ensure that your personal information is secure, we communicate privacy and security standards to our employees and strictly enforce privacy practices.
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. The presentation presents solutions to tasks C3 of the Unified State Exam - 2014 in mathematics.
Download:
Preview:
To use presentation previews, create a Google account and log in to it: https://accounts.google.com
Slide captions:
Solving logarithmic inequalities containing a variable in the base of the logarithm: methods, techniques, equivalent transitions, mathematics teacher, Secondary School No. 143 Knyazkina T.V.
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. Don't forget the ODZ of the 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 is found, all that remains is to intersect it with the solution rational inequality- and the answer is ready.
Solve the inequality: Solution First, let's write out the OD of the logarithm. The first two inequalities are satisfied automatically, but the last one will have to be written down. Since the square of a number is equal to zero if and only if the number itself is equal to zero, we have: x 2 + 1 ≠ 1; x2 ≠ 0; x ≠ 0. It turns out that the ODZ of a logarithm is all numbers except zero: x ∈ (−∞0)∪(0 ;+ ∞). Now we solve the main inequality: We make the transition from the logarithmic inequality to the 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)
Transforming Logarithmic Inequalities Often the original inequality is different from the one above. This is easy to fix by standard rules working with logarithms. Namely: Any number can be represented as a logarithm with a given base; The sum and difference of logarithms with the same bases can be replaced by one logarithm. 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, general scheme solutions to logarithmic inequalities are as follows: Find the ODZ of each logarithm included in the inequality; Reduce the inequality to a standard one using the formulas for adding and subtracting logarithms; Solve the resulting inequality using the scheme given above.
Solve the inequality: Solution Let's find the domain of definition (DO) of the first logarithm: Solve by the method of intervals. Find the zeros of the numerator: 3 x − 2 = 0; x = 2/3. Then - the zeros of the denominator: x − 1 = 0; x = 1. Mark zeros and signs on the coordinate line:
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 let's transform the second logarithm so that there is a two at the base: As you can see, the threes at the base and in front of the logarithm have been canceled. We got two logarithms with the same basis. Add them up: log 2 (x − 1) 2
(f (x) − g (x)) (k (x) − 1)
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. Answer: x ∈ (−1; 2/3)∪(1; 3)
Solving USE-2014 tasks type C3
Solve the system of inequalities. Solution. ODZ: 1) 2)
Solve the system of inequalities 3) -7 -3 - 5 x -1 + + + − − (continued)
Solve the system of inequalities 4) Common decision: and -7 -3 - 5 x -1 -8 7 log 2 129 (continued)
Solve the inequality (continued) -3 3 -1 + − + − x 17 + -3 3 -1 x 17 -4
Solve the inequality Solution. ODZ:
Solve the inequality (continued)
Solve the inequality Solution. ODZ: -2 1 -1 + − + − x + 2 -2 1 -1 x 2
LOGARITHMIC INEQUALITIES IN THE USE
Sechin Mikhail Alexandrovich
Small Academy of Sciences for Students of the Republic of Kazakhstan “Iskatel”
MBOU "Sovetskaya Secondary School No. 1", 11th grade, town. Sovetsky Sovetsky district
Gunko Lyudmila Dmitrievna, MBOU teacher"Soviet Secondary School No. 1"
Sovetsky district
Goal of the work: study of the mechanism for solving logarithmic inequalities C3 using non-standard methods, identifying interesting facts logarithm
Subject of study:
3) Learn to solve specific logarithmic inequalities C3 using non-standard methods.
Results:
Content
Introduction………………………………………………………………………………….4
Chapter 1. History of the issue……………………………………………………...5
Chapter 2. Collection of logarithmic inequalities ………………………… 7
2.1. Equivalent transitions and the generalized method of intervals…………… 7
2.2. Rationalization method……………………………………………………………… 15
2.3. Non-standard substitution……………….................................................... ..... 22
2.4. Tasks with traps……………………………………………………27
Conclusion……………………………………………………………………………… 30
Literature……………………………………………………………………. 31
Introduction
I am in 11th grade and plan to enter a university where the core subject is mathematics. That’s why I work a lot with problems in part C. In task C3, I need to solve a non-standard inequality or system of inequalities, usually related to logarithms. When preparing for the exam, I was faced with the problem of a shortage of methods and techniques for solving exam logarithmic inequalities offered in C3. Methods that are studied in school curriculum on this topic, do not provide a basis for solving C3 tasks. The math teacher suggested that I work on C3 assignments independently under her guidance. In addition, I was interested in the question: do we encounter logarithms in our lives?
With this in mind, the topic was chosen:
“Logarithmic inequalities in the Unified State Exam”
Goal of the work: study of the mechanism for solving C3 problems using non-standard methods, identifying interesting facts about the logarithm.
Subject of study:
1) Find the necessary information about non-standard methods solutions to logarithmic inequalities.
2) Find additional information about logarithms.
3) Learn to solve specific C3 problems using non-standard methods.
Results:
The practical significance lies in the expansion of the apparatus for solving C3 problems. This material can be used in some lessons, for clubs, and elective classes in mathematics.
The project product will be the collection “C3 Logarithmic Inequalities with Solutions.”
Chapter 1. Background
Throughout the 16th century, the number of approximate calculations increased rapidly, primarily in astronomy. Improving instruments, studying planetary movements and other work required colossal, sometimes multi-year, calculations. Astronomy was in real danger of drowning in unfulfilled calculations. Difficulties arose in other areas, for example, in the insurance business, compound interest tables were needed to different meanings percent. The main difficulty was multiplication, division multi-digit numbers, especially trigonometric quantities.
The discovery of logarithms was based on the properties of progressions that were well known by the end of the 16th century. Archimedes spoke about the connection between the terms of the geometric progression q, q2, q3, ... and the arithmetic progression of their exponents 1, 2, 3,... in the Psalm. Another prerequisite was the extension of the concept of degree to negative and fractional exponents. Many authors have pointed out that multiplication, division, exponentiation and root extraction in geometric progression correspond in arithmetic - in the same order - addition, subtraction, multiplication and division.
Here was the idea of the logarithm as an exponent.
In the history of the development of the doctrine of logarithms, several stages have passed.
Stage 1
Logarithms were invented no later than 1594 independently by the Scottish Baron Napier (1550-1617) and ten years later by the Swiss mechanic Bürgi (1552-1632). Both wanted to provide a new, convenient means of arithmetic calculations, although they approached this problem in different ways. Napier kinematically expressed the logarithmic function and thereby entered into new area theory of function. Bürgi remained on the basis of considering discrete progressions. However, the definition of the logarithm for both is not similar to the modern one. The term "logarithm" (logarithmus) belongs to Napier. It arose from a combination of Greek words: logos - “relation” and ariqmo - “number”, which meant “number of relations”. Initially, Napier used a different term: numeri artificiales - “artificial numbers”, as opposed to numeri naturalts - “natural numbers”.
In 1615, in a conversation with Henry Briggs (1561-1631), a professor of mathematics at Gresh College in London, Napier proposed taking zero as the logarithm of one, and 100 as the logarithm of ten, or, what amounts to the same thing, simply 1. This is how they appeared decimal logarithms and the first logarithmic tables were printed. Later, Briggs' tables were supplemented by the Dutch bookseller and mathematics enthusiast Adrian Flaccus (1600-1667). Napier and Briggs, although they came to logarithms earlier than everyone else, published their tables later than the others - in 1620. The signs log and Log were introduced in 1624 by I. Kepler. The term “natural logarithm” was introduced by Mengoli in 1659 and followed by N. Mercator in 1668, and the London teacher John Speidel published tables of natural logarithms of numbers from 1 to 1000 under the name “New Logarithms”.
The first logarithmic tables were published in Russian in 1703. But in all logarithmic tables there were calculation errors. The first error-free tables were published in 1857 in Berlin, processed by the German mathematician K. Bremiker (1804-1877).
Stage 2
Further development of the theory of logarithms is associated with more widespread use analytical geometry and infinitesimal calculus. By that time, the connection between the squaring of an equilateral hyperbola and natural logarithm. The theory of logarithms of this period is associated with the names of a number of mathematicians.
German mathematician, astronomer and engineer Nikolaus Mercator in an essay
"Logarithmotechnics" (1668) gives a series giving the expansion of ln(x+1) in
powers of x:
This expression exactly corresponds to his train of thought, although, of course, he did not use the signs d, ..., but more cumbersome symbolism. With the discovery of the logarithmic series, the technique for calculating logarithms changed: they began to be determined using infinite series. In his lectures "Elementary Mathematics with highest point vision", read in 1907-1908, F. Klein proposed using the formula as the starting point for constructing the theory of logarithms.
Stage 3
Definition of a logarithmic function as an inverse function
exponential, logarithm as an exponent of a given base
was not formulated immediately. Essay by Leonhard Euler (1707-1783)
"An Introduction to the Analysis of Infinitesimals" (1748) served to further
development of the theory of logarithmic functions. Thus,
134 years have passed since logarithms were first introduced
(counting from 1614), before mathematicians came to the definition
the concept of logarithm, which is now the basis of the school course.
Chapter 2. Collection of logarithmic inequalities
2.1. Equivalent transitions and the generalized method of intervals.
Equivalent transitions
, if a > 1
, if 0 <
а <
1
Generalized method intervals
This method most universal for solving inequalities of almost any type. The solution diagram looks like this:
1. Bring the inequality to a form where the function on the left side is 
, and on the right 0.
2. Find the domain of the function .
3. Find the zeros of the function , that is, solve the equation 
(and solving an equation is usually easier than solving an inequality).
4. Draw the domain of definition and zeros of the function on the number line.
5. Determine the signs of the function on the obtained intervals.
6. Select intervals where the function takes the required values and write down the answer.
Example 1.
Solution:
Let's apply the interval method
where
For these values, all expressions under the logarithmic signs are positive.
Answer:
Example 2.
Solution:
1st way . ADL is determined by inequality x> 3. Taking logarithms for such x in base 10, we get
The last inequality could be solved by applying expansion rules, i.e. comparing factors to zero. However, in in this case easy to determine intervals of constant sign of a function
therefore, the interval method can be applied.
Function f(x) = 2x(x- 3.5)lgǀ x- 3ǀ is continuous at x> 3 and vanishes at points x 1 = 0, x 2 = 3,5, x 3 = 2, x 4 = 4. Thus, we determine the intervals of constant sign of the function f(x):
Answer:
2nd method . Let us directly apply the ideas of the interval method to the original inequality.
To do this, recall that the expressions a b- a c and ( a - 1)(b- 1) have one sign. Then our inequality at x> 3 is equivalent to inequality
or
The last inequality is solved using the interval method
Answer:
Example 3.
Solution:
Let's apply the interval method
Answer:
Example 4.
Solution:
Since 2 x 2 - 3x+ 3 > 0 for all real x, That
To solve the second inequality we use the interval method
In the first inequality we make the replacement
then we come to the inequality 2y 2 - y - 1 < 0 и, применив метод интервалов, получаем, что решениями будут те y, which satisfy the inequality -0.5< y < 1.
From where, because
we get the inequality
which is carried out when x, for which 2 x 2 - 3x - 5 < 0. Вновь применим метод интервалов
Now, taking into account the solution to the second inequality of the system, we finally obtain
Answer:
Example 5.
Solution:
Inequality is equivalent to a collection of systems
or
Let's use the interval method or
Answer:
Example 6.
Solution:
Inequality equals system
Let
Then y > 0,
and the first inequality
system takes the form
or, unfolding
quadratic trinomial factored,
Applying the interval method to the last inequality,
we see that its solutions satisfying the condition y> 0 will be all y > 4.
Thus, the original inequality is equivalent to the system:
So, the solutions to the inequality are all
2.2. Rationalization method.
Previously method rationalization of inequality was not solved, it was not known. This is the "new modern" effective method solutions to exponential and logarithmic inequalities" (quote from the book by S.I. Kolesnikova)
And even if the teacher knew him, there was a fear - does the Unified State Exam expert know him, and why don’t they give him at school? There were situations when the teacher said to the student: “Where did you get it? Sit down - 2.”
Now the method is being promoted everywhere. And for experts there is guidelines, associated with this method, and in the "Most Complete Editions of Model Options..." solution C3 uses this method.
WONDERFUL METHOD!
"Magic Table"
In other sources
If a >1 and b >1, then log a b >0 and (a -1)(b -1)>0;
If a >1 and 0 if 0<a<1 и b
>1, then log a b<0 и (a
-1)(b
-1)<0;
if 0<a<1 и 00 and (a -1)(b -1)>0. The reasoning carried out is simple, but significantly simplifies the solution of logarithmic inequalities. Example 4.
log x (x 2 -3)<0
Solution:
Example 5.
log 2 x (2x 2 -4x +6)≤log 2 x (x 2 +x ) Solution: Example 6.
To solve this inequality, instead of the denominator, we write (x-1-1)(x-1), and instead of the numerator, we write the product (x-1)(x-3-9 + x). Example 7.
Example 8.
2.3. Non-standard substitution. Example 1.
Example 2.
Example 3.
Example 4.
Example 5.
Example 6.
Example 7.
log 4 (3 x -1)log 0.25 Let's make the replacement y=3 x -1; then this inequality will take the form Log 4 log 0.25 Because log 0.25 Let us make the replacement t =log 4 y and obtain the inequality t 2 -2t +≥0, the solution of which is the intervals - Thus, to find the values of y we have a set of two simple inequalities Therefore, the original inequality is equivalent to the set of two exponential inequalities, The solution to the first inequality of this set is the interval 0<х≤1, решением второго – промежуток 2≤х<+ Example 8.
Solution:
Inequality equals system The solution to the second inequality defining the ODZ will be the set of those x,
for which x > 0.
To solve the first inequality we make the substitution Then we get the inequality or The set of solutions to the last inequality is found by the method intervals: -1< t < 2. Откуда, возвращаясь к переменной x, we get or Lots of those x, which satisfy the last inequality belongs to ODZ ( x> 0), therefore, is a solution to the system, and hence the original inequality. Answer: 2.4. Tasks with traps. Example 1.
Solution. The ODZ of the inequality is all x satisfying the condition 0 Example 2.
log 2 (2 x +1-x 2)>log 2 (2 x-1 +1-x)+1.
Answer. (0; 0.5)U.
Answer :
(3;6)
.
= -log 4 = -(log 4 y -log 4 16)=2-log 4 y , then we rewrite the last inequality as 2log 4 y -log 4 2 y ≤.
The solution to this set is the intervals 0<у≤2 и 8≤у<+.
that is, aggregates 
. Thus, the original inequality is satisfied for all values of x from the intervals 0<х≤1 и 2≤х<+.
.
. Therefore, all x are from the interval 0
Conclusion
It was not easy to find specific methods for solving C3 problems from a large abundance of different educational sources. In the course of the work done, I was able to study non-standard methods for solving complex logarithmic inequalities. These are: equivalent transitions and the generalized method of intervals, the method of rationalization , non-standard substitution , tasks with traps on ODZ. These methods are not included in the school curriculum.
Using different methods, I solved 27 inequalities proposed on the Unified State Exam in part C, namely C3. These inequalities with solutions by methods formed the basis of the collection “C3 Logarithmic Inequalities with Solutions,” which became a project product of my activity. The hypothesis I posed at the beginning of the project was confirmed: C3 problems can be effectively solved if you know these methods.
In addition, I discovered interesting facts about logarithms. It was interesting for me to do this. My project products will be useful for both students and teachers.
Conclusions:
Thus, the project goal has been achieved and the problem has been solved. And I received the most complete and varied experience of project activities at all stages of work. While working on the project, my main developmental impact was on mental competence, activities related to logical mental operations, the development of creative competence, personal initiative, responsibility, perseverance, and activity.
A guarantee of success when creating a research project for I gained: significant school experience, the ability to obtain information from various sources, check its reliability, and rank it by importance.
In addition to direct subject knowledge in mathematics, I expanded my practical skills in the field of computer science, gained new knowledge and experience in the field of psychology, established contacts with classmates, and learned to cooperate with adults. During the project activities, organizational, intellectual and communicative general educational skills were developed.
Literature
1. Koryanov A. G., Prokofiev A. A. Systems of inequalities with one variable (standard tasks C3).
2. Malkova A. G. Preparation for the Unified State Exam in Mathematics.
3. Samarova S. S. Solving logarithmic inequalities.
4. Mathematics. Collection of training works edited by A.L. Semenov and I.V. Yashchenko. -M.: MTsNMO, 2009. - 72 p.-
