The rationalization method allows you to move from an inequality containing complex exponential, logarithmic, etc. expressions, to an equivalent simpler rational inequality.

Therefore, before we start talking about rationalization in inequalities, let's talk about equivalence.

equivalence

Equivalent or equivalent called equations (inequalities) whose sets of roots coincide. Equations (inequalities) that do not have roots are also considered equivalent.

Example 1 The equations and are equivalent, since they have the same roots.

Example 2 The equations and are also equivalent, since the solution to each of them is the empty set.

Example 3 The inequalities and are equivalent, since the solution to both is the set .

Example 4 and are unequal. The solution to the second equation is only 4, and the solution to the first equation is both 4 and 2.

Example 5 The inequality is equivalent to the inequality , since in both inequalities the solution is 6.

That is, in appearance, equivalent inequalities (equations) can be very far from similarity.

In fact, when we solve complex, long equations (inequalities), like this, and get the answer, after all, we have in our hands nothing more than an equation (inequality) equivalent to the original one. The look is different, but the essence is the same!

Example 6 Let's remember how we solved inequality before getting acquainted with the method of intervals. We have replaced the original inequality with a set of two systems:

That is, inequality and the last set are equivalent to each other.

Also, we could, having in hand the collection

replace it with the inequality , which can be solved in a jiffy by the interval method.

We have come close to the method of rationalization in logarithmic inequalities.

Rationalization method in logarithmic inequalities

Let's consider the inequality.

We represent 4 as a logarithm:

We are dealing with a variable base of the logarithm, therefore, depending on whether the base of the logarithm is greater than 1 or less than 1 (that is, we are dealing with an increasing or decreasing function), the inequality sign will remain or change to "". Therefore, there is a combination (combination) of two systems:

But, ATTENTION, this system should be solved taking into account the ODZ! I deliberately did not load the ODZ system so that the main idea would not be lost.

Look, now we will rewrite our system like this (we will move everything in each line of inequality to the left side):

Doesn't this remind you of anything? By analogy with example 6 we will replace this set of systems by the inequality:

Having solved this inequality on the ODZ, we will get the solution of the inequality .

Let us first find the ODZ of the original inequality:

Now let's decide

The solution of the last inequality, taking into account the ODZ:

So, here it is, this "magic" table:

Note that the table works under the condition

where are functions of ,

- function or number,

- one of the characters

Note also that the second and third rows of the table are consequences of the first. In the second line 1 is represented before as , and in the third line 0 is represented as .

And a couple more useful consequences (I hope you can easily understand where they come from):

where are functions of ,

- function or number,

- one of the characters

Rationalization method in exponential inequalities

Let's solve the inequality.

Solving the original inequality is equivalent to solving the inequality

Answer: .

Table for rationalization in exponential inequalities:

– functions of , – function or number, – one of the signs The table works under the condition . Also in the third, fourth lines - additionally -

Again, in fact, you need to remember the first and third lines of the table. The second line is a special case of the first, and the fourth line is a special case of the third.

Rationalization Method in Inequalities Containing Modulus

Working with inequalities of type , where are functions of some variable, we can be guided by the following equivalent transitions:

Let's solve the inequality ” .

A Here offer more consider a few examples on the topic “Rationalization of inequalities”.

Sections: Mathematics

Often, when solving logarithmic inequalities, there are problems with a variable base of the logarithm. So, 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 set. Even with given quadratic functions, the population solution may require a lot of time.

An alternative, less time-consuming way of solving this standard inequality can be proposed. To do this, we take into account the following theorem.

Theorem 1. Let 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 the set X, then .

Let's get back to inequality. Let's move on to the decimal logarithm (you can go to any with a constant base greater than one).

Now we can use the theorem, noticing in the numerator the increment of functions and in the denominator. So it's true

As a result, the number of calculations leading to the answer is reduced by about 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 , , .

Passing to (2) we will have:

Example 2

Comparing with (1) we find , , .

Passing to (2) we will have:

Example 3

Since the left side of the inequality is an increasing function for and , then the answer is set .

The set of examples in which Terme 1 can be applied can be easily expanded if Terme 2 is taken into account.

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 scheme: the product is less than zero when the factors are of different signs. Those. we consider a set of two systems of inequalities in which, as was 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 of O.D.Z.

The method of replacing the increment of a function with an increment of the argument, taking into account Theorem 2, turns out to be very convenient when solving typical C3 USE problems.

Example 6

Example 7

. Let's denote . Get

. Note that the replacement implies: . Returning to the equation, we get .

Example 8

In the theorems we use, there is no restriction on the classes of functions. In this article, as an example, the theorems were applied to the solution of logarithmic inequalities. The following few examples will demonstrate the promise of the method for solving other types of inequalities.

Municipal Autonomous Educational Institution "Yarkovskaya Secondary School"

Educational project

Solving logarithmic inequalities by the rationalization method

MAOU "Yarkovskaya secondary school"

Shanskikh Daria

Leader: math teacher

MAOU "Yarkovskaya secondary school"

Yarkovo 2013

1) Introduction………………………………………………………….2

2) Main part………………………………………………..3

3) Conclusion……………………………………………………..9

4) List of used literature…………….10

5) Applications……………………………………………………………11-12

1. Introduction

Often, when solving USE tasks from part “C”, and especially in tasks C3, there are inequalities containing logarithmic expressions with an unknown at the base of the logarithm. Here is an example of the standard inequality:

As a rule, to solve such tasks, the classical method is used, that is, the transition to an equivalent set of systems is applied.

With the standard approach, the example is solved according to the scheme: the product is less than zero when the factors are of different signs. That is, a set of two systems of inequalities is considered, in which each inequality breaks down into seven more. Therefore, a less time-consuming method for solving this standard inequality can be proposed. This is a rationalization method known in the mathematical literature as decomposition.

During the implementation of the project, I set the following goals: :

1) Master this decision technique

2) Practice solving skills on tasks C3 from training and diagnostic work in 2013.

Project objectiveis the study of the theoretical justification of the method of rationalization.

Relevancework lies in the fact that this method allows you to successfully solve the logarithmic inequalities of part C3 of the Unified State Examination in mathematics.

2. Main part

Consider a logarithmic inequality of the form

font-size:14.0pt; line-height:150%">, (1)

where font-size:14.0pt;line-height:150%"> The standard method for solving such an inequality involves parsing the two cases into areas of acceptable inequality values.

In the first case when the bases of the logarithms satisfy the condition

font-size:14.0pt; line-height:150%">, the inequality sign is reversed: font-size:14.0pt;line-height:150%"> In the second case when the base satisfies the condition, the inequality sign is preserved: .

At first glance, everything is logical, let's consider two cases and then combine the answers. True, when considering the second case, a certain discomfort arises - you have to repeat the calculations from the first case by 90 percent (transform, find the roots of auxiliary equations, determine the intervals of the monotonicity of the sign). A natural question arises - is it possible to combine all this somehow?

The answer to this question is contained in the following theorem.

Theorem 1. logarithmic inequality

font-size:14.0pt;line-height:150%">is equivalent to the following system of inequalities :

font-size:14.0pt; line-height:150%"> (2)

Proof.

1. Let's start with the fact that the first four inequalities of system (2) define the set of admissible values ​​of the original logarithmic inequality. Let us now turn our attention to the fifth inequality. If font-size:14.0pt; line-height:150%">, then the first factor of this inequality will be negative. When reducing by it, you will have to change the sign of inequality to the opposite, then you get the inequality .

If , That the first factor of the fifth inequality is positive, we reduce it without changing the inequality sign, we get the inequality font-size:14.0pt;line-height: 150%">. Thus, the fifth inequality of the system includes both cases of the previous method.

The term has been proven.

The main provisions of the theory of rationalization method.

The rationalization method consists in replacing the complex expression F(x ) to a simpler expression G(x ) under which the inequality G(x )EN-US" style="font-size:14.0pt;line-height:150%;font-family:Calibri">F(x )0 in the expression domain F(x).

Let's single out some expressions F and their corresponding rationalizing expressions G , where u , v , , p , q - expressions with two variables ( u > 0; u ≠ 1; v > 0, > 0), a - fixed number (a > 0, a ≠ 1).

Proof

1. Let logav - logaφ > 0, that is logav > logaφ, and a > 0, a ≠ 1, v > 0,

φ > 0.

If 0< a < 1, то по свойству убывающей логарифмической функции имеем v < φ . Hence, the system of inequalities holds

a -1<0

v – φ < 0

Whence follows the inequality (a – 1)( v – φ ) > 0 true on the domain of the expressionF = logav - logaφ.

If a > 1, That v > φ . Therefore, we have the inequality ( a – 1)( v – φ )> 0. Conversely, if the inequality ( a – 1)( v – φ )> 0 on the range of acceptable values ( a > 0, a ≠ 1, v> 0, φ > 0),then on this domain it is equivalent to the combination of two systems.

a – 1<0 a – 1 > 0

v – φ < 0 v – φ > 0

Each system implies the inequalitylogav > logaφ, that is logav - logaφ > 0.

Similarly, we consider the inequalities F< 0, F ≤ 0, F ≥ 0.

2. Let some number A> 0 and A≠ 1, then we have

logo v- loguφ = EN-US" style="font-size:14.0pt;line-height:150%">v - 1)( u- 1)(φ -u).

4. From inequality UV- uφ > 0 should UV > uφ. Let the number a > 1, thenloga UV > logauφ or

( u – φ) loga u > 0.

Hence, taking into account change 1b and the conditiona > 1 we get

( v – φ)( a – 1)( u – 1) > 0, ( v – φ)( u – 1) > 0. Similarly, we prove the inequalities F< 0,

F ≤ 0, F ≥ 0.

5. The proof is similar to Proof 4.

6. The proof of substitution 6 follows from the equivalence of the inequalities | p | > | q | and p 2 > q 2

(|p|< | q | и p 2 < q 2 ).

Let us compare the volume of solving inequalities containing a variable at the base of the logarithm by the classical method and the rationalization method

3. Conclusion

I believe that the tasks that I set for myself in the performance of work have been achieved. The project is of practical importance, since the method proposed in the work makes it possible to significantly simplify the solution of logarithmic inequalities. As a result, the number of calculations leading to the answer is reduced by about half, which saves not only time, but also allows you to potentially make fewer arithmetic and careless errors. Now, when solving C3 problems, I use this method.

4. List of used literature

1. , – Methods for solving inequalities with one variable. – 2011.

2. - Mathematics guide. - 1972.

3. - Mathematics for the applicant. Moscow: MTSNMO, 2008.

Ezhova Elena Sergeevna
Job title: mathematic teacher
Educational institution: MOU "School №77"
Locality: Saratov
Material name: methodical development
Subject: Rationalization method in solving inequalities in preparation for the exam "
Publication date: 16.05.2018
Chapter: complete education

Obviously, the same inequality can be solved in several ways. Luckily

in a chosen way or, as we used to say, in a rational way, any

inequality will be solved quickly and easily, its solution will be beautiful and interesting.

I would like to consider in more detail the so-called rationalization method when

solving logarithmic and exponential inequalities, as well as inequalities containing

variable under the module sign.

The main idea of ​​the method.

The method of changing factors is used to solve inequalities reduced to the form

Where is the symbol

» denotes one of four possible inequality signs:

When solving inequality (1), we are only interested in the sign of any factor in the numerator

or denominator, and not its absolute value. Therefore, if for some reason we

it is inconvenient to work with this multiplier, we can replace it with another

coinciding with it in the region of definition of inequality and having in this region

the same roots.

This determines the main idea of ​​the multiplier replacement method. It is important to fix that

the fact that the replacement of factors is carried out only under the condition that the inequality is reduced

to the form (1), that is, when it is required to compare the product with zero.

The main part of the replacement is due to the following two equivalent statements.

Statement 1. The function f(x) is strictly increasing if and only if for

any values ​​of t

) coincides with

sign with the difference (f(t

)), that is, f<=>(t

(↔ means sign match)

Statement 2. The function f(x) is strictly decreasing if and only if for

any values ​​of t

from the domain of the function difference (t

) coincides with

sign with the difference (f(t

)), that is, f ↓<=>(t

The justification of these assertions follows directly from the definition of strictly

monotonic function. According to these statements, it can be established that

The difference of degrees in the same base always coincides in sign with

the product of the difference between the indicators of these degrees and the deviation of the base from unity,

The difference of logarithms in the same base always coincides in sign with

the product of the difference between the numbers of these logarithms and the deviation of the base from unity, then

The fact that the difference of non-negative quantities has the same sign as the difference

squares of these values, allows the following substitutions:

Solve the inequality

Solution.

Let's move on to an equivalent system:

From the first inequality we get

The second inequality holds for all

From the third inequality we get

Thus, the set of solutions to the original inequality:

Solve the inequality

Solution.

Let's solve the inequality:

Answer: (−4; −3)

Solve the inequality

Let us bring the inequality to a form in which the difference between the values ​​of the logarithmic

Let's replace the difference in the values ​​of the logarithmic function with the difference in the values ​​of the argument. IN

the numerator is an increasing function, and the denominator is decreasing, so the inequality sign

will change to the opposite. It is important not to forget to take into account the scope

logarithmic function, so this inequality is equivalent to a system of inequalities.

Numerator roots: 8; 8;

Denominator root: 1

Solve the inequality

Let us replace in the numerator the difference between the modules of two functions by the difference between their squares, and in

the denominator is the difference between the values ​​of the logarithmic function and the difference between the arguments.

In the denominator, the function is decreasing, which means that the inequality sign will change to

opposite.

In this case, it is necessary to take into account the domain of definition of the logarithmic

We solve the first inequality by the interval method.

Numerator roots:

Denominator roots:

Solve the inequality

Let us replace in the numerator and denominator the difference between the values ​​of monotone functions by the difference

values ​​of arguments, taking into account the domain of definition of functions and the nature of monotonicity.

Numerator roots:

Denominator roots:

The most commonly used substitutions (excluding O D 3).

a) Change of sign-constant multipliers.

b) Replacing non-constant factors with the modulus.

c) Replacing non-constant factors with exponential and logarithmic

expressions.

Solution. ODZ:

Replacing multipliers:

We have a system:

In this inequality, the factors

be considered as differences of non-negative values, since the expressions 1

ODZ can take both positive and negative values.

We have a system:

Replacing multipliers:

We have a system:

Replacing multipliers:

We have a system:

Replacing multipliers:

We have a system:

As a result, we have: x

rationalization method(decomposition method, multiplier replacement method, replacement method

functions, sign rule) consists in replacing the complex expression F(x) with a more

a simple expression G(x) for which the inequality G(x)

0 is equivalent to the inequality F (x

0 in the domain of expression F(x).

Sections: Mathematics

The practice of checking examination papers shows that the greatest difficulty for schoolchildren is the solution of transcendental inequalities, especially logarithmic inequalities with a variable base. Therefore, the lesson summary presented to your attention is a presentation of the rationalization method (other names are the decomposition method (Modenov V.P.), the method of replacing factors (Golubev V.I.)), which allows you to reduce complex logarithmic, exponential, combined inequalities to a system of simpler rational inequalities. As a rule, the method of intervals as applied to rational inequalities by the time the topic “Solution of logarithmic inequalities” was studied was well mastered and worked out. Therefore, students with great interest and enthusiasm perceive those methods that allow them to simplify the solution, make it shorter and, ultimately, save time on the exam for solving other tasks.

Lesson Objectives:

• educational: actualization of basic knowledge when solving logarithmic inequalities; introduction of a new way of solving inequalities; improvement of decision skills
• Educational: development of mathematical horizons, mathematical speech, analytical thinking
• Educational: education of accuracy and self-control.

DURING THE CLASSES

1. Organizational moment. Greetings. Setting lesson goals.

2. Preparatory stage:

Solve inequalities:

3. Checking homework(No. 11.81*а)

When solving the inequality

You had to use the following scheme for solving logarithmic inequalities with a variable base:

Those. There are 2 cases to consider: the base is greater than 1 or the base is less than 1.

4. Explanation of new material

If you look at these formulas carefully, you will notice that the sign of the difference g(x) – h(x) coincides with the sign of the difference log f(x) g(x) - log f(x) h(x) in the case of an increasing function ( f(x) > 1, i.e. f(x) – 1 > 0) and is opposite to the sign of the difference log f(x) g(x) - log f(x) h(x) in the case of a decreasing function (0< f(x) < 1, т.е. f(x) – 1 < 0)

Therefore, this set can be reduced to a system of rational inequalities:

This is the essence of the rationalization method - to replace the more complex expression A with a simpler expression B, which is rational. In this case, the inequality В V 0 will be equivalent to the inequality А V 0 on the domain of the expression А.

Example 1 Let us rewrite the inequality as an equivalent system of rational inequalities.

I note that conditions (1)–(4) are the conditions for the domain of definition of the inequality, which I recommend finding at the beginning of the solution.

Example 2 Solve the inequality by rationalization method:

The domain of definition of the inequality is given by the conditions:

We get:

It remains to write the inequality (5)

Subject to domain

Answer: (3; 5)

5. Consolidation of the studied material

I. Write down the inequality as a system of rational inequalities:

II. Express the right side of the inequality in the form of a logarithm in the desired base and go to the equivalent system:

The teacher calls to the board the students who wrote down the systems from groups I and II, and invites one of the strongest students to solve the home inequality (No. 11.81 * a) using the rationalization method.

6. Verification work

Option 1

Option 2

1. Write down a system of rational inequalities for solving inequalities:

2. Solve the inequality by the rationalization method

Grading Criteria:

3-4 points - "satisfactory";
5-6 points - "good";
7 points - "excellent".

7. Reflection

Answer the question: which of the known methods for solving logarithmic inequalities with a variable base will allow you to make better use of time on the exam?

8. Homework: No. 11.80 * (a, b), 11.81 * (a, b), 11.84 * (a, b) solve by the rationalization method.

Bibliography:

1. Algebra and the beginning of analysis: Proc. For 11 cells. general education Institutions /[S.M. Nikolsky, M.K. Potapov, N.N. Reshetnikov, A.V. Shevkin] - 5th ed. - M .: Education, JSC "Moscow textbooks", 2006.
2. A.G. Koryanov, A.A. Prokofiev. Materials of the course "Preparing good students and excellent students for the exam": lectures 1-4. - M .: Pedagogical University "First of September", 2012.