Sections: Maths

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 summary of the lesson 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 of studying the topic "Solving logarithmic inequalities" is well mastered and worked out. Therefore, students with great interest and enthusiasm perceive the 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: updating basic knowledge when solving logarithmic inequalities; introduction of a new way of solving inequalities; improving solution skills
• Developing: development of mathematical horizons, mathematical speech, analytical thinking
• Educational: education of accuracy and self-control.

DURING THE CLASSES

1. Organizational moment. Greetings. Setting the goals of the lesson.

2. Preparatory stage:

Solve inequalities:

3. Checking homework(No. 11.81 * a)

When solving the inequality

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

Those. 2 cases must be considered: the base is greater than 1 or the base is less than 1.

4. Explanation of the new material

If you look at these formulas carefully, you will notice that the sign of the difference g(x) – h(x) matches 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)

Consequently, 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 definition of the expression A.

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

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

Example 2. Solve inequality with rationalization method:

The domain of inequality is defined by the conditions:

We get:

It remains to write the inequality (5)

Taking into account the domain of definition

Answer: (3; 5)

5. Consolidation of the studied material

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

II. Imagine the right-hand side of the inequality as the logarithm to the base required and go to the equivalent system:

The teacher calls the students who have written down the systems from groups I and II to the blackboard and offers one of the strongest students to solve domestic inequality (No. 11.81 * a) by the method of rationalization.

6. Verification work

Option 1

Option 2

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

2. Solve inequality by rationalization

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 more efficiently use your time on the exam?

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

Bibliography:

1. Algebra and the beginning of analysis: Textbook. For 11 cl. 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... Course materials "Preparing good students for the Unified State Exam": lectures 1-4. - M .: Pedagogical University "September First", 2012.

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

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

Equivalence

Equivalent or equivalent are equations (inequalities) whose root sets coincide. Equations (inequalities) that have no 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. Inequalities and are equivalent, since the solution to both is many.

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

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

That is, equivalent inequalities (equations) in appearance may be quite far from similarity.

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

Example 6. Let's remember how we solved inequality before getting to know the interval method... We 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 our hands the aggregate

replace it with an inequality, which can be solved in no time by the method of intervals.

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

Rationalization method in logarithmic inequalities

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, a combination (union) of two systems arises:

But, ATTENTION, this system should be solved taking into account the OHS! 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 transfer everything in each line of inequality to the left side):

Does this remind you of anything? By analogy with example 6 we replace this set of systems with the inequality:

Having solved this inequality on the ODZ, we will obtain a solution to the inequality.

Let us first find the ODV of the original inequality:

Now let's decide

Solution of the last inequality, taking into account the DHS:

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 signs

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

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 from, - function or number, - one of the symbols The table works on condition. Also in the third, fourth lines - additionally -

Again, in fact, you need to memorize 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.

Method of rationalization in inequalities containing modulus

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

Let's solve the inequality ”.

A here I suggest more consider some examples on the topic “Rationalizing inequalities”.

Municipal Autonomous Educational Institution "Yarkovskaya Secondary School"

Study project

Solving logarithmic inequalities by rationalization method

MAOU "Yarkovskaya secondary school"

Shanskikh Daria

Supervisor: 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) Appendices ………………………………………………… 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. For example, here's the standard inequality:

As a rule, the classical method is used to solve such tasks, 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 opposite signs. That is, a set of two systems of inequalities is considered, in which each inequality splits into seven more. Therefore, a less laborious method for solving this standard inequality can be proposed. This is a rationalization method known in the mathematical literature as decomposition.

When completing the project, I set the following goals :

1) Master this decision technique

2) To practice the skills of solving on tasks C3 from the training and diagnostic works in 2013.

The task of the projectis the study of the theoretical basis of the method of rationalization.

Relevanceof the work lies in the fact that this method allows you to successfully solve the logarithmic inequalities of the C3 part of the exam 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 two cases into the range of acceptable values ​​of the inequality.

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

font-size: 14.0pt; line-height: 150% ">, the inequality sign is drawn: 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, we will 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 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 inequality system :

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 initial 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 canceling by it, you will have to change the sign of the inequality to the opposite, then you get the inequality .

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

Terem is proven.

The main provisions of the theory of the method of rationalization.

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

Let's highlight 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).

	Expression F	Expression G
		(a –1) ( v - φ)
1 b
		)
2 b

Proof

1. Let logav - logaφ> 0, that is logav> logaφ, moreover 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

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, then v > φ . Therefore, the inequality ( a – 1)( v – φ )> 0. Conversely, if the inequality ( a – 1)( v – φ )> 0 on the range of admissible values ( a > 0, a ≠ 1, v> 0, φ> 0),then in this area it is equivalent to a 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

logu 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 a> 1, thenloga uv > logauφ or

( u – φ) loga u > 0.

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

( v – φ)( a – 1)( u – 1) > 0, ( v – φ)( u – 1) > 0. Similarly, 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 inequalities | p | > | q | and p 2> q 2

(| p |< | q | и p 2 < q 2 ).

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

3. Conclusion

I believe that the tasks that I set for myself while doing the 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 approximately halved, which not only saves time, but also allows you to potentially make fewer arithmetic and "inattention" 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. - A guide to mathematics. - 1972.

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

Sections: Maths

Often, when solving logarithmic inequalities, problems with a variable base of the logarithm are encountered. 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 applied:

The disadvantage of this method is the need to solve seven inequalities, not counting two systems and one set. Already with given quadratic functions, solving a set can be time-consuming.

An alternative, less laborious way of solving this standard inequality can be proposed. For this, we take into account the following theorem.

Theorem 1. Let a continuous increasing function on the 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, that is, , where .

Note: if a continuous decreasing function on the set X, then.

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

Now you can use the theorem, noting in the numerator the increment of the functions and in the denominator. So it is true

As a result, the number of calculations leading to the answer is approximately halved, which not only saves time, but also allows you to potentially make fewer arithmetic and “carelessness” 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-hand side of the inequality is an increasing function for and , then the answer is set.

The set of examples in which Theorem 1 can be applied can be easily extended if Theorem 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 opposite signs. Those. the set of two systems of inequalities is considered, in which, as indicated at the beginning, each inequality splits 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 the argument, taking into account Theorem 2, turns out to be very convenient when solving typical problems C3 of the exam.

Example 6.

Example 7.

... Let us denote. We 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, for example, the theorems have been applied to the solution of logarithmic inequalities. The next few examples will demonstrate the promise of the method for solving other types of inequalities.

Yezhova Elena Sergeevna
Position: mathematic teacher
Educational institution: MOU "Secondary School No. 77"
Locality: Saratov
Material name: methodical development
Theme: The rationalization method for solving inequalities in preparation for the exam "
Date of publication: 16.05.2018
Chapter: complete education

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

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

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

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

solution of logarithmic and exponential inequalities, as well as inequalities containing

variable under the module sign.

The main idea of ​​the method.

The method of replacing factors is used to solve inequalities that are reduced to the form

Where 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 the 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 domain of inequality definition and having in this domain

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 if the inequality

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

) matches

sign with the difference (f (t

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

(↔ means coincidence)

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

any values ​​of t

from the domain of the function, the difference (t

) matches

sign with the difference (f (t

)), that is, f ↓<=>(t

The substantiation of these statements follows directly from the definition of strictly

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

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

the product of the difference between the indicators of these degrees by the deviation of the base from one,

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

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

The fact that the difference of non-negative quantities coincides in sign with the difference

squares of these quantities, allows the following substitutions:

Solve inequality

Solution.

Let's move on to an equivalent system:

From the first inequality we obtain

The second inequality holds for all

From the third inequality we obtain

Thus, the set of solutions to the original inequality:

Solve inequality

Solution.

Let's solve the inequality:

Answer: (−4; −3)

Solve inequality

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

Replace the difference in the values ​​of the logarithmic function with the difference in the values ​​of the argument. V

the function is increasing in the numerator, and decreasing in the denominator, therefore the inequality sign

will change to the opposite. It is important not to forget to consider the scope of definition

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

Numerator roots: 8; eight;

Denominator root: 1

Solve inequality

We replace in the numerator the difference of the absolute values ​​of two functions with the difference of their squares, and in

the denominator is the difference of the values ​​of the logarithmic function by the difference of the arguments.

In the denominator, the function is decreasing, which means that the sign of the inequality 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 method of intervals.

Numerator roots:

Denominator roots:

Solve inequality

We 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 monotony.

Numerator roots:

Denominator roots:

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

a) Replacement of constant sign factors.

b) Replacement of non-constant multipliers with the module.

c) Replacement of 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 quantities, since 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, the rule of signs) is to replace 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).