Types of logarithmic inequalities. Manov's work "Logarithmic inequalities in the exam." Solving logarithmic inequalities

Objectives lesson:

Didactic:

• 1 level - to teach solving the simplest logarithmic inequalities, applying the definition of logarithm, the properties of logarithms;
• 2 level - to solve logarithmic inequalities, choosing an independent solution method;
• 3 level - to be able to apply knowledge and skills in non-standard situations.

Developing: develop memory, attention, logical thinking, comparison skills, be able to generalize and draw conclusions

Educational:educating accuracy, responsibility for the task performed, mutual assistance.

Teaching methods: verbal , visual , practical , partial search , self-government , control.

Forms of organization of cognitive activity of students: Frontal , individual , work in pairs.

Equipment: set of test tasks, reference abstract, clean sheets for solutions.

Type of lesson: Studying a new material.

During the classes

1. Organizational moment. The topic and objectives of the lesson are announced, the lesson scheme: each student is issued an estimated sheet, which the student fills in the course of the lesson; For each pair of students - printed materials with tasks, you need to perform tasks in pairs; Clean sheets for solutions; Support sheets: definition of logarithm; graph logarithmic function, its properties; properties of logarithms; Algorithm for solving logarithmic inequalities.

All decisions after self-esteem are handed over to the teacher.

Evaluation sheet student

2. Actualization of knowledge.

Instructions of the teacher. Recall the definition of logarithm, a graph of the logarithmic function and its properties. To do this, read the text on S.88-90, 98-101 of the textbook "Algebra and the beginning of Analysis of 10-11" edited by Sh.a Alimova, Yu.M Kolyagin, etc.

Pupils are distributed sheets on which they are recorded: the definition of logarithm; depicted graph of logarithmic function, its properties; properties of logarithms; The algorithm for solving logarithmic inequalities, an example of a solution of logarithmic inequality that boils into a square.

3. Studying a new material.

The solution of logarithmic inequalities is based on the logarithmic function monotony.

Logarithmic inequality algorithm:

A) Find the area of \u200b\u200bdefinition of inequality (the otogarithmic expression is greater than zero).
B) submit (if possible) the left and right portion of inequality in the form of logarithms on the same basis.
C) to determine whether the logarithmic function is increasing or decreasing: if T\u003e 1, then increasing; If 0 1, then decreasing.
D) Go to more simple inequality (fearful expressions), given that the inequality sign will be saved if the function increases and change if it decreases.

Training element number 1.

Purpose: consolidate the solution of the simplest logarithmic inequalities

Form of organization of cognitive activity of students: individual work.

Tasks for independent work for 10 minutes. For each inequality there are several options for answers, you need to choose the right and check by key.

Key: 13321, Maximum number of points - 6 b.

Training element number 2.

Purpose: consolidate the solution of logarithmic inequalities, applying the properties of logarithms.

Instructions of the teacher. Recall the basic properties of logarithms. To do this, read the text of the textbook on C.92, 103-104.

Tasks for independent work for 10 minutes.

The key: 2113, the maximum number of points - 8 b.

Training element number 3.

Purpose: to study the solution of logarithmic inequalities by the method of information to the square.

Teacher's instructions: The method of information of inequality to square is that it is necessary to transform inequality to this species so that some logarithmic function is to designate a new variable, having obtained a square inequality with respect to this variable.

Apply the interval method.

You have passed the first level of mastering the material. Now you have to independently choose the method of solving logarithmic equations using all your knowledge and opportunities.

Training element number 4.

Purpose: consolidate the solution of logarithmic inequalities by selecting an independently rational solution.

Tasks for independent work for 10 minutes

Training element number 5.

Instructions of the teacher. Well done! You have mastered the solution of equations of the second level of complexity. The purpose of your work further is to apply your knowledge and skills in more complex and non-standard situations.

Tasks for self solutions:

Instructions of the teacher. Wonderful if you coped with all the task. Well done!

The assessment for the entire lesson depends on the number of scored points for all educational elements:

• if N ≥ 20, then you get the rating "5",
• at 16 ≤ n ≤ 19 - rating "4",
• at 8 ≤ n ≤ 15 - rating "3",
• with N.< 8 выполнить работу над ошибками к следующему уроку (решения можно взять у учителя).

Estimated foxes to hand over to the teacher.

5. Homework: If you have scored no more than 15 b - work on errors (solutions can be taken from the teacher) if you scored more than 15 b - perform a creative task on the topic "Logarithmic inequalities".

Solving logarithmic inequalities, we use the property of the logarithmic function. We also use the definition of logarithm and the main logarithmic formulas.

Let's repeat what logarithms are:

Logarithm A positive number based on the ground is an indicator of the degree in which you need to build to get.

Wherein

Basic logarithmic identity:

Basic formulas for logarithms:

(The logarithm of the work is equal to the sum of logarithms)

(Private logarithm is equal to the difference in logarithms)

(Formula for logarithm degree)

Formula of the transition to a new base:

Algorithm Solutions of Logarithmic Inequalities

It can be said that logarithmic inequalities are solved according to a specific algorithm. We need to record the area of \u200b\u200bpermissible values \u200b\u200b(OTZ) inequalities. Certain the inequality to mind the sign here can be any: it is important that logarithms were on the left and right in the inequality on the same basis.

And then "cast" logarithm! At the same time, if the foundation of the degree, the sign of inequality remains the same. If the basis is such that the sign of inequality changes to the opposite.

Of course, we are not just "throwing out" logarithms. We use the property of the logarithmic function. If the base of the logarithm is greater than the unit, the logarithmic function increases monotonously, and then the greater value of x corresponds to the greater value of the expression.

If the base is greater than zero and less unit, the logarithmic function monotonously decreases. The greater value of the argument will correspond to less

Important note: It is best to record a solution in the form of a chain of equivalent transitions.

Let's go to practice. As always, let's start with the most simple inequalities.

1. Consider the inequality of LOG 3 X\u003e LOG 3 5.
Since logarithms are defined only for positive numbers, it is necessary that X be positive. Condition X\u003e 0 is called an area of \u200b\u200bpermissible values \u200b\u200b(odb) of this inequality. Only with such x inequality makes sense.

Well, this formulation famously sounds and easily remembered. But why do we still do it?

We are people, we possess intelligence. Our mind is arranged in such a way that everything is logical, understandable, having an internal structure is remembered and is applied much better than random and non-related facts. That is why it is important not to mechanically to drive the rules as a trained dog-mathematician, but to act consciously.

So why are we still "throwing out logarithms"?

The answer is simple: if the base is greater than one (as in our case), the logarithmic function increases monotonically, it means that the greater value of X corresponds to the greater value of Y and from the inequality of Log 3 x 1\u003e Log 3 x 2, it follows that x 1\u003e x 2.


Please note, we switched to algebraic inequality, and the sign of inequality is preserved.

So, x\u003e 5.

The following logarithmic inequality is also simple.

2. Log 5 (15 + 3x)\u003e Log 5 2X

Let's start with the area of \u200b\u200bpermissible values. Logarithms are defined only for positive numbers, so

Solving this system, we get: x\u003e 0.

Now from the logarithmic inequality, we turn to the algebraic - "throw" logarithm. Since the base of the logarithm is greater than the unit, the sign of inequality is preserved.

15 + 3x\u003e 2x.

We get: x\u003e -15.

Answer: x\u003e 0.

And what will happen if the basis of the logarithm is less than one? It is easy to guess that in this case, when moving to algebraic inequality, the inequality sign will change.

Let us give an example.

We write OTZ. Expressions from which logarithms are taken should be positively, that is

Solving this system, we get: x\u003e 4.5.

Because the logarithmic function with the base monotonically decreases. This means that the smaller value of the argument is responsible for greater value:


And if, then
2x - 9 ≤ x.

We get that x ≤ 9.

Considering that X\u003e 4.5, we write the answer:

In the following task, the indicative inequality is reduced to the square. So the topic "Square inequalities" is recommended to repeat.

Now more complex inequalities:

4. Decide inequality

5. Decide inequality

If, then. We were lucky! We know that the base of the logarithm is greater than the unit for all the values \u200b\u200bof x included in the OTZ.

We will replace

Please note that at first we fully solve the inequality relative to the new variable t. And only after that we return to the variable x. Remember this and do not butt on the exam!

We remember the rule: if roots or logarithms are present in the equation or inequality - the solution should be started with the area of \u200b\u200bpermissible values. Since the base of the logarithm must be positive and not equal to one, we obtain the system of conditions:

Simplify this system:

This is the area of \u200b\u200bpermissible inequality values.

We see that the variable is contained at the base of the logarithm. Let us turn to a constant basis. Recall that

In this case, it is convenient to go to the base 4.

We will replace

We simplify inequality and solve it by intervals:

Let's return to the variable x.:

We have added a condition x. \u003e 0 (from OTZ).

7. The following task is also solved using the interval method.

As always, the solution of logarithmic inequality is starting with the area of \u200b\u200bpermissible values. In this case

This condition must be performed, and we will return to it. Consider while inequality itself. We write down the left side as a logarithm based on 3:

The right-hand side can also be written as a logarithm based on 3, and then move to algebraic inequality:

We see that the condition (that is, the OTZ) is now performed automatically. Well, it simplifies the decision of inequality.

We solve the inequality of intervals:

Answer:

Happened? What, increase the level of complexity:

8. Decide inequality:

Inequality is equivalent to the system:

9. Decide inequality:

Expression 5 - x. 2 is obsessively repeated in the condition of the problem. This means that you can make a replacement:

Since the indicative function takes only positive values, t. \u003e 0. then

The inequality will take the form:

Already better. Find the area of \u200b\u200bpermissible values \u200b\u200bof inequality. We have already said that t. \u003e 0. In addition, t. - 3) (5 9 · t. − 1) > 0

If this condition is fulfilled, the private will be positive.

And the expression under the logarithm in the right side of the inequality should be positive, that is (625 t. − 2) 2 .

This means that 625 t. - 2 ≠ 0, that is

Gently write OTZ

and solve the resulting system, applying the interval method.

So,

Well, half the way was done - dealt with OTZ. We solve the inequality itself. The sum of logarithms in the left side will represent as a logarithm of the work.

Logarithmic inequalities

In previous lessons, we met with logarithmic equations and now we know what it is and how to solve them. And today's lesson will be devoted to the study of logarithmic inequalities. What are these inequalities and what is the difference between the solution of the logarithmic equation and inequality?

Logarithmic inequalities are inequalities that have a variable standing under the logarithm sign or in its foundation.

Or, one can still say that logarithmic inequality is such an inequality in which its unknown value, as in the logarithmic equation, will stand under the sign of the logarithm.

The simplest logarithmic inequalities have this kind:

where f (x) and g (x) are some expressions that depend on x.

Let's look at this example: f (x) \u003d 1 + 2x + x2, g (x) \u003d 3x-1.

Solving logarithmic inequalities

Before the solution of logarithmic inequalities, it is worth noting that, when deciding, they have similarities with indicative inequalities, namely:

First, when moving from logarithms to expressions under the sign of the logarithm, we also need to compare the base of the logarithm with unit;

Secondly, solving logarithmic inequality, using the replacement of variables, we need to solve inequalities regarding replacement until we get the simplest inequality.

But we looked at the similar issues of solving logarithmic inequalities. And now we will pay attention to a rather significant difference. We are well aware that the logarithmic function has a limited area of \u200b\u200bdefinition, so moving from logarithms to expressions under the logarithm sign, you need to take into account the area of \u200b\u200bpermissible values \u200b\u200b(OTZ).

That is, it should be borne in mind that solving the logarithmic equation, we can first find the roots of the equation, and then check this solution. But it will not be possible to solve the logarithmic inequality, because moving from logarithms to expressions under the logarithm sign, it will be necessary to record odd inequality.

In addition, it is worth remembering that the theory of inequality consists of valid numbers that are positive and negative numbers, as well as the number 0.

For example, when the number "A" is positive, then you must use such an entry: a\u003e 0. In this case, both the amount and the product of these numbers will also be positive.

The basic principle of the solution of inequality is its replacement for a simpler inequality, but the main thing is that it is equivalent to this. Further, we also got inequality and again replaced it to the one that has a simpler form, etc.

Solving inequalities with a variable need to find all its solutions. If two inequalities have one variable x, then such inequalities are equivalent, provided that their solutions coincide.

Performing tasks to solve logarithmic inequalities, it is necessary to remember that when a\u003e 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.

Ways to solve logarithmic inequalities

Now consider some methods that take place when solving logarithmic inequalities. For a better understanding and assimilation, we will try to deal with specific examples.

We are well aware that the simplest logarithmic inequality has this kind:

In this inequality V - is one of the inequality signs as:<,>, ≤ or ≥.

When the basis of this logarithm is greater than the unit (A\u003e 1), carrying out the transition from logarithms to expressions under the sign of the logarithm, then in this embodiment the sign of inequalities is preserved, and inequality will have this kind:

what is equivalent to such a system:

In the case when the base of the logarithm is greater than zero and less than one (0

It is equivalent to this system:

Let's see more examples of solving the simplest logarithmic inequalities shown in the picture below:

Solution of examples

The task. Let's try to solve such an inequality:

Solving the area of \u200b\u200bpermissible values.

Now let's try to multiply its right side on:

We look at what we do:

Now, let's go to the conversion of fearful expressions. Due to the fact that the foundation of logarithm 0< 1/4 <1, то от сюда следует, что знак неравенства изменится на противоположный:

3x - 8\u003e 16;
3x\u003e 24;
x\u003e 8.

And from this it follows that the interval that we received is entirely and fully belongs to OTZ and is the solution of such inequality.

This is what the answer we turned out:

What is needed to solve logarithmic inequalities?

And now let's try to analyze what we need to successfully solve logarithmic inequalities?

First, focus all your attention and try not to allow errors when performing transformations that are given in this inequality. Also, it should be remembered that in solving such inequality, it is necessary to prevent expansion and narrowings of the OI inequality, which may lead to the loss or acquisition of foreign solutions.

Secondly, in solving logarithmic inequalities, it is necessary to learn to think logically and understand the difference between such concepts as a system of inequality and a set of inequalities so that you can easily carry out the selection of inequality decisions, while guided by the OTZ.

Thirdly, for the successful solution of such inequalities, each of you should know perfectly to 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 one word, all those that you studied throughout school learning algebra.

As you can see, having studied the theme about logarithmic inequalities, there is nothing difficult in solving these inequalities, provided if you are attentive and persistent in achieving the goals. To solve inequalities, no problems arise in the decision, you need to train as much as possible, solving various tasks and at the same time to memorize the main ways to solve such inequalities and their systems. With unsuccessful solutions of logarithmic inequalities, you should carefully analyze your mistakes so that in the future it is not to return to them again.

Homework

For better assimilation of the theme and consolidation of the material passed, decide the following inequalities:

When studying the logarithmic function, we considered mainly inequalities of the form
log A H.< b и log а х ≥ b. Рассмотрим решение более сложных логарифмических неравенств. Обычным способом решения таких неравенств является переход от данного неравенства к более простому неравенству или системе неравенств, которая имеет то же самое множество решений.

Solve the LG inequality (x + 1) ≤ 2 (1).

Decision.

1) The right side of the inequality considered meaning is at all values, and the left side is at x + 1\u003e 0, i.e. at x\u003e -1.

2) the gap of x\u003e -1 is called an area of \u200b\u200bdefinition of inequality (1). The logarithmic function with base 10 is increasing, therefore, under the condition x + 1\u003e 0, the inequality (1) is performed if X + 1 ≤ 100 (since 2 \u003d lg 100). Thus, inequality (1) and inequality system

(x\u003e -1, (2)
(x + 1 ≤ 100,

tale, in other words, many solutions of inequality (1) and inequality systems (2) is the same.

3) solving the system (2), we find -1< х ≤ 99.

Answer. -one< х ≤ 99.

Solve inequality log 2 (x - 3) + log 2 (x - 2) ≤ 1 (3).

Decision.

1) The area of \u200b\u200bdefinition of the logarithmic function under consideration is the set of positive values \u200b\u200bof the argument, so the left part of inequality meaning has at x - 3\u003e 0 and x - 2\u003e 0.

Consequently, the area of \u200b\u200bdetermining this inequality is the range of X\u003e 3.

2) according to the properties of the logarithm inequality (3) at x\u003e 3 is equivalent to the inequality of Log 2 (x - 3) (x - 2) ≤ log 2 (4).

3) Logarithmic function with base 2 is increasing. Therefore, at x\u003e 3, inequality (4) is performed if (x - 3) (x - 2) ≤ 2.

4) Thus, the initial inequality (3) is equivalent to the inequality system

((x - 3) (x - 2) ≤ 2,
(x\u003e 3.

Solving the first inequality of this system, we obtain x 2 - 5x + 4 ≤ 0, from where 1 ≤ x ≤ 4. Combining this segment with a gap\u003e 3, we get 3< х ≤ 4.

Answer. 3.< х ≤ 4.

Solve the inequality of Log 1/2 (x 2 + 2x - 8) ≥ -4. (five)

Decision.

1) The region of determining the inequality is found from the condition x 2 + 2x - 8\u003e 0.

2) inequality (5) can be written in the form:

log 1/2 (x 2 + 2x - 8) ≥ Log 1/2 16.

3) Since the logarithmic function with the base is ½ decreasing, then for all x from the entire region of determining the inequality, we get:

x 2 + 2x - 8 ≤ 16.

Thus, the initial equality (5) is equivalent to the inequality system

(x 2 + 2x - 8\u003e 0, or (x 2 + 2x - 8\u003e 0,
(x 2 + 2x - 8 ≤ 16, (x 2 + 2x - 24 ≤ 0.

Solving the first square inequality, we get x< -4, х > 2. Solving a second square inequality, we obtain -6 ≤ x ≤ 4. Therefore, both inequalities of the system are performed simultaneously at -6 ≤ x< -4 и при 2 < х ≤ 4.

Answer. -6 ≤ x.< -4; 2 < х ≤ 4.

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