Logarithmic inequalities are the simplest. Manov's work "logarithmic inequalities in the Unified State Examination". What is needed to solve logarithmic inequalities
Lesson objectives:
Didactic:
- Level 1 – teach how to solve the simplest logarithmic inequalities, using the definition of a logarithm and the properties of logarithms;
- Level 2 – solve logarithmic inequalities, choosing your own solution method;
- Level 3 – be able to apply knowledge and skills in non-standard situations.
Educational: develop memory, attention, logical thinking, comparison skills, be able to generalize and draw conclusions
Educational: cultivate accuracy, responsibility for the task being performed, and mutual assistance.
Teaching methods: verbal , visual , practical , partial-search , self-government , control.
Forms of organization of students’ cognitive activity: frontal , individual , work in pairs.
Equipment: a set of test tasks, reference notes, blank sheets for solutions.
Lesson type: learning new material.
Lesson progress
1. Organizational moment. The topic and goals of the lesson, the lesson plan are announced: each student is given an assessment sheet, which the student fills out during the lesson; for each pair of students - printed materials with tasks; tasks must be completed in pairs; blank solution sheets; support sheets: definition of logarithm; graph of a logarithmic function, its properties; properties of logarithms; algorithm for solving logarithmic inequalities.
All decisions after self-assessment are submitted to the teacher.
Student's score sheet
2. Updating knowledge.
Teacher's instructions. Recall the definition of a logarithm, the graph of a logarithmic function, and its properties. To do this, read the text on pp. 88–90, 98–101 of the textbook “Algebra and the beginnings of analysis 10–11” edited by Sh.A Alimov, Yu.M Kolyagin and others.
Students are given sheets on which are written: the definition of a logarithm; shows a graph of a logarithmic function and its properties; properties of logarithms; algorithm for solving logarithmic inequalities, an example of solving a logarithmic inequality that reduces to a quadratic one.
3. Studying new material.
Solving logarithmic inequalities is based on the monotonicity of the logarithmic function.
Algorithm for solving logarithmic inequalities:
A) Find the domain of definition of the inequality (the sublogarithmic expression is greater than zero).
B) Represent (if possible) the left and right sides of the inequality as logarithms to the same base.
C) Determine whether the logarithmic function is increasing or decreasing: if t>1, then increasing; if 0
D) Go to a simpler inequality (sublogarithmic expressions), taking into account that the sign of the inequality will remain the same if the function increases and will change if it decreases.
Learning element #1.
Goal: consolidate the solution to the simplest logarithmic inequalities
Form of organization of students' cognitive activity: individual work.
Tasks for independent work for 10 minutes. For each inequality there are several possible answers; you need to choose the correct one and check it using the key.
KEY: 13321, maximum number of points – 6 points.
Learning element #2.
Goal: consolidate the solution of logarithmic inequalities using the properties of logarithms.
Teacher's instructions. Remember the basic properties of logarithms. To do this, read the text of the textbook on pp. 92, 103–104.
Tasks for independent work for 10 minutes.
KEY: 2113, maximum number of points – 8 points.
Learning element #3.
Purpose: to study the solution of logarithmic inequalities by the method of reduction to quadratic.
Teacher's instructions: the method of reducing an inequality to a quadratic is to transform the inequality to such a form that a certain logarithmic function is denoted by a new variable, thereby obtaining a quadratic inequality with respect to this variable.
Let's use the interval method.
You have passed the first level of mastering the material. Now you will have to independently choose a method for solving logarithmic equations, using all your knowledge and capabilities.
Learning element #4.
Goal: consolidate the solution to logarithmic inequalities by independently choosing a rational solution method.
Tasks for independent work for 10 minutes
Learning element #5.
Teacher's instructions. Well done! You have mastered solving equations of the second level of complexity. The goal of your further work is to apply your knowledge and skills in more complex and non-standard situations.
Tasks for independent solution:
Teacher's instructions. It's great if you completed the whole task. Well done!
The grade for the entire lesson depends on the number of points scored for all educational elements:
- if N ≥ 20, then you get a “5” rating,
- for 16 ≤ N ≤ 19 – score “4”,
- for 8 ≤ N ≤ 15 – score “3”,
- at N< 8 выполнить работу над ошибками к следующему уроку (решения можно взять у учителя).
Submit the assessment papers to the teacher.
5. Homework: if you scored no more than 15 points, work on your mistakes (solutions can be obtained from the teacher), if you scored more than 15 points, complete a creative task on the topic “Logarithmic inequalities.”
When studying the logarithmic function, we mainly considered inequalities of the form
log a x< 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 inequality log (x + 1) ≤ 2 (1).
Solution.
1) The right side of the inequality under consideration makes sense for all values of x, and the left side makes sense for x + 1 > 0, i.e. for x > -1.
2) The interval x > -1 is called the domain of definition of inequality (1). A logarithmic function with base 10 is increasing, therefore, provided x + 1 > 0, inequality (1) is satisfied if x + 1 ≤ 100 (since 2 = log 100). Thus, inequality (1) and the system of inequalities
(x > -1, (2)
(x + 1 ≤ 100,
are equivalent, in other words, the set of solutions to inequality (1) and the system of inequalities (2) are the same.
3) Solving system (2), we find -1< х ≤ 99.
Answer. -1< х ≤ 99.
Solve the inequality log 2 (x – 3) + log 2 (x – 2) ≤ 1 (3).
Solution.
1) The domain of definition of the logarithmic function under consideration is the set of positive values of the argument, therefore the left side of the inequality makes sense for x – 3 > 0 and x – 2 > 0.
Consequently, the domain of definition of this inequality is the interval x > 3.
2) According to the properties of the logarithm, inequality (3) for x > 3 is equivalent to the inequality log 2 (x – 3)(x – 2) ≤ log 2 (4).
3) The logarithmic function with base 2 is increasing. Therefore, for x > 3, inequality (4) is satisfied if (x – 3)(x – 2) ≤ 2.
4) Thus, the original inequality (3) is equivalent to the system of inequalities
((x – 3)(x – 2) ≤ 2,
(x > 3.
Solving the first inequality of this system, we obtain x 2 – 5x + 4 ≤ 0, whence 1 ≤ x ≤ 4. Combining this segment with the interval x > 3, we obtain 3< х ≤ 4.
Answer. 3< х ≤ 4.
Solve the inequality log 1/2 (x 2 + 2x – 8) ≥ -4. (5)
Solution.
1) The domain of definition of the inequality is found from the condition x 2 + 2x – 8 > 0.
2) Inequality (5) can be written as: 
log 1/2 (x 2 + 2x – 8) ≥ log 1/2 16.
3) Since the logarithmic function with base ½ is decreasing, then for all x from the entire domain of definition of the inequality we obtain:
x 2 + 2x – 8 ≤ 16.
Thus, the original equality (5) is equivalent to the system of inequalities
(x 2 + 2x – 8 > 0, or (x 2 + 2x – 8 > 0,
(x 2 + 2x – 8 ≤ 16, (x 2 + 2x – 24 ≤ 0.
Solving the first quadratic inequality, we get x< -4, х >2. Solving the second quadratic inequality, we obtain -6 ≤ x ≤ 4. Consequently, both inequalities of the system are satisfied simultaneously for -6 ≤ x< -4 и при 2 < х ≤ 4.
Answer. -6 ≤ x< -4; 2 < х ≤ 4.
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Logarithmic inequalities
In previous lessons, we got acquainted with logarithmic equations and now we know what they are and how to solve them. Today's lesson will be devoted to the study of logarithmic inequalities. What are these inequalities and what is the difference between solving a logarithmic equation and an inequality?
Logarithmic inequalities are inequalities that have a variable appearing under the logarithm sign or at its base.
Or, we can also say that a logarithmic inequality is an inequality in which its unknown value, as in a logarithmic equation, will appear under the sign of the logarithm.
The simplest logarithmic inequalities have the following form:
where f(x) and g(x) are some expressions that depend on x.
Let's look at this using this example: f(x)=1+2x+x2, g(x)=3x−1.
Solving logarithmic inequalities
Before solving logarithmic inequalities, it is worth noting that when solved they are similar to exponential inequalities, namely:
First, when moving from logarithms to expressions under the logarithm sign, we also need to compare the base of the logarithm with one;
Secondly, when solving a logarithmic inequality using a change of variables, we need to solve inequalities with respect to the change until we get the simplest inequality.
But you and I have considered similar aspects of solving logarithmic inequalities. Now let’s pay attention to a rather significant difference. You and I know that the logarithmic function has a limited domain of definition, therefore, when moving from logarithms to expressions under the logarithm sign, we need to take into account the range of permissible values (ADV).
That is, it should be taken into account that when solving a logarithmic equation, you and I can first find the roots of the equation, and then check this solution. But solving a logarithmic inequality will not work this way, since when moving from logarithms to expressions under the logarithm sign, it will be necessary to write down the ODZ of the inequality.
In addition, it is worth remembering that the theory of inequalities consists of real numbers, which are positive and negative numbers, as well as the number 0.
For example, when the number “a” is positive, then you need to use the following notation: a >0. In this case, both the sum and the product of these numbers will also be positive.
The main principle for solving an inequality is to replace it with a simpler inequality, but the main thing is that it is equivalent to the given one. Further, we also obtained an inequality and again replaced it with one that has a simpler form, etc.
When solving inequalities with a variable, you need to find all its solutions. If two inequalities have the same variable x, then such inequalities are equivalent, provided that their solutions coincide.
When performing tasks on solving logarithmic inequalities, you must remember that when a > 1, then the logarithmic function increases, and when 0< a < 1, то такая функция имеет свойство убывать. Эти свойства вам будут необходимы при решении логарифмических неравенств, поэтому вы их должны хорошо знать и помнить.
Methods for solving logarithmic inequalities
Now let's look at some of the methods that take place when solving logarithmic inequalities. For better understanding and assimilation, we will try to understand them using specific examples.
We all know that the simplest logarithmic inequality has the following form:
In this inequality, V – is one of the following inequality signs:<,>, ≤ or ≥.
When the base of a given logarithm is greater than one (a>1), making the transition from logarithms to expressions under the logarithm sign, then in this version the inequality sign is preserved, and the inequality will have the following form:
which is equivalent to this system:
In the case when the base of the logarithm is greater than zero and less than one (0 This is equivalent to this system: Let's look at more examples of solving the simplest logarithmic inequalities shown in the picture below: Exercise. Let's try to solve this inequality: Solving the range of acceptable values. Now let's try to multiply its right side by: Let's see what we can come up with: Now, let's move on to converting sublogarithmic expressions. Due to the fact that the base of the logarithm is 0< 1/4 <1, то от сюда следует, что знак неравенства изменится на противоположный: 3x - 8 > 16; And from this it follows that the interval that we obtained entirely belongs to the ODZ and is a solution to such an inequality. Here's the answer we got: Now let's try to analyze what we need to successfully solve logarithmic inequalities? First, concentrate all your attention and try not to make mistakes when performing the transformations that are given in this inequality. Also, it should be remembered that when solving such inequalities, it is necessary to avoid expansions and contractions of the inequalities, which can lead to the loss or acquisition of extraneous solutions. Secondly, when solving logarithmic inequalities, you need to learn to think logically and understand the difference between concepts such as a system of inequalities and a set of inequalities, so that you can easily select solutions to the inequality, while being guided by its DL. Thirdly, to successfully solve such inequalities, each of you must perfectly know all the properties of elementary functions and clearly understand their meaning. Such functions include not only logarithmic, but also rational, power, trigonometric, etc., in a word, all those that you studied during school algebra. As you can see, having studied the topic of logarithmic inequalities, there is nothing difficult in solving these inequalities, provided that you are careful and persistent in achieving your goals. To avoid any problems in solving inequalities, you need to practice as much as possible, solving various tasks and at the same time remember the basic methods of solving such inequalities and their systems. If you fail to solve logarithmic inequalities, you should carefully analyze your mistakes so as not to return to them again in the future. To better understand the topic and consolidate the material covered, solve the following inequalities: When solving logarithmic inequalities, we use the monotonicity property of the logarithmic function. We also use the definition of logarithm and basic logarithmic formulas. Let's review what logarithms are: Logarithm a positive number to the base is an indicator of the power to which it must be raised to get . At the same time Basic logarithmic identity: Basic formulas for logarithms: (The logarithm of the product is equal to the sum of the logarithms) (The logarithm of the quotient is equal to the difference of the logarithms) (Formula for logarithm of power) Formula for moving to a new base: Algorithm for solving logarithmic inequalities We can say that logarithmic inequalities are solved using a specific algorithm. We need to write down the range of acceptable values (APV) of the inequality. Reduce the inequality to the form The sign here can be anything: It is important that on the left and on the right in the inequality there are logarithms to the same base. And after that we “discard” the logarithms! Moreover, if the base is a degree , the inequality sign remains the same. If the base is such that the sign of inequality changes to the opposite. Of course, we don't just "throw away" logarithms. We use the monotonicity property of a logarithmic function. If the base of the logarithm is greater than one, the logarithmic function increases monotonically, and then a larger value of x corresponds to a larger value of the expression. If the base is greater than zero and less than one, the logarithmic function decreases monotonically. A larger value of the argument x will correspond to a smaller value Important note: it is best to write the solution in the form of a chain of equivalent transitions. Let's move on to practice. As always, let's start with the simplest inequalities. 1. Consider the inequality log 3 x > log 3 5. Well, this formulation sounds dashing and is easy to remember. But why can we still do this? We are people, we have intelligence. Our mind is designed in such a way that everything that is logical, understandable, and has an internal structure is remembered and applied much better than random and unrelated facts. That’s why it’s important not to mechanically memorize the rules like a trained math dog, but to act consciously. So why do we still “drop logarithms”? The answer is simple: if the base is greater than one (as in our case), the logarithmic function increases monotonically, which means that a larger value of x corresponds to a larger value of y and from the inequality log 3 x 1 > log 3 x 2 it follows that x 1 > x 2. So x > 5. The following logarithmic inequality is also simple. 2. log 5 (15 + 3x) > log 5 2x Let's start with the range of acceptable values. Logarithms are only defined for positive numbers, so Solving this system, we get: x > 0. Now let’s move from the logarithmic inequality to the algebraic one - “discard” the logarithms. Since the base of the logarithm is greater than one, the inequality sign remains the same. 15 + 3x > 2x. We get: x > −15. Answer: x > 0. But what happens if the base of the logarithm is less than one? It is easy to guess that in this case, when moving to an algebraic inequality, the sign of the inequality will change. Let's give an example. Let's write down the ODZ. The expressions from which logarithms are taken must be positive, that is Solving this system, we get: x > 4.5. Since , a logarithmic function with a base decreases monotonically. This means that a larger value of the function corresponds to a smaller value of the argument: We get that x ≤ 9. Considering that x > 4.5, we write the answer: In the next problem, the exponential inequality is reduced to a quadratic inequality. So we recommend repeating the topic “quadratic inequalities”. Now for more complex inequalities: 4. Solve the inequality 5. Solve the inequality If, then. We're lucky! We know that the base of the logarithm is greater than one for all values of x included in the ODZ. Let's make a replacement Note that we first solve the inequality completely with respect to the new variable t. And only after that we return to the variable x. Remember this and don’t make mistakes in the exam! Let us remember the rule: if an equation or inequality contains roots, fractions or logarithms, the solution must begin from the range of acceptable values. Since the base of the logarithm must be positive and not equal to one, we obtain a system of conditions: Let's simplify this system: This is the range of acceptable values of inequality. We see that the variable is contained in the base of the logarithm. Let's move on to the permanent base. Let us remind you that In this case, it is convenient to go to base 4. Let's simplify the inequality and solve it using the interval method: Let's return to the variable x: 7. The following problem can also be solved using the interval method As always, we start solving a logarithmic inequality from the range of acceptable values. In this case This condition must be met, and we will return to it. Let's look at the inequality itself for now. Let's write the left side as a logarithm to base 3: The right-hand side can also be written as a logarithm to base 3, and then move on to the algebraic inequality: We see that the condition (that is, the ODZ) is now fulfilled automatically. Well, this makes solving the inequality easier. We solve the inequality using the interval method: Did it work? Well, let's increase the difficulty level: 8. Solve the inequality: Inequality is equivalent to the system: 9. Solve the inequality: Expression 5 - x 2 is compulsively repeated in the problem statement. This means that you can make a replacement: Since the exponential function only takes positive values, t> 0. Then The inequality will take the form: Already better. Let's find the range of acceptable values of the inequality. We have already said that t> 0. In addition, ( t− 3) (5 9 · t − 1) > 0 If this condition is met, then the quotient will be positive. And the expression under the logarithm on the right side of the inequality must be positive, that is (625 t − 2) 2 . This means 625 t− 2 ≠ 0, that is Let's carefully write down the ODZ and solve the resulting system using the interval method. Well, half the battle is done - we sorted out the ODZ. We solve the inequality itself. Let us represent the sum of logarithms on the left side as the logarithm of the product. Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely taught in school: log k (x) f (x) ∨ log k (x) g (x) ⇒ (f (x) − g (x)) (k (x) − 1) ∨ 0 Instead of the “∨” checkbox, you can put any inequality sign: more or less. The main thing is that in both inequalities the signs are the same. This way we get rid of logarithms and reduce the problem to a rational inequality. The latter is much easier to solve, but when discarding logarithms, extra roots may appear. To cut them off, it is enough to find the range of acceptable values. If you have forgotten the ODZ of a logarithm, I strongly recommend repeating it - see “What is a logarithm”. Everything related to the range of acceptable values must be written down and solved separately: f(x) > 0; g(x) > 0; k(x) > 0; k(x) ≠ 1. These four inequalities constitute a system and must be satisfied simultaneously. When the range of acceptable values has been found, all that remains is to intersect it with the solution of the rational inequality - and the answer is ready. Task. Solve the inequality: First, let’s write out the logarithm’s ODZ: The first two inequalities are satisfied automatically, but the last one will have to be written out. Since the square of a number is zero if and only if the number itself is zero, we have: x 2 + 1 ≠ 1; It turns out that the ODZ of the logarithm is all numbers except zero: x ∈ (−∞ 0)∪(0; +∞). Now we solve the main inequality: We make the transition from logarithmic inequality to rational one. The original inequality has a “less than” sign, which means the resulting inequality must also have a “less than” sign. We have: (10 − (x 2 + 1)) · (x 2 + 1 − 1)< 0; The zeros of this expression are: x = 3; x = −3; x = 0. Moreover, x = 0 is a root of the second multiplicity, which means that when passing through it, the sign of the function does not change. We have: We get x ∈ (−∞ −3)∪(3; +∞). This set is completely contained in the ODZ of the logarithm, which means this is the answer. Often the original inequality is different from the one above. This can be easily corrected using the standard rules for working with logarithms - see “Basic properties of logarithms”. Namely: Separately, I would like to remind you about the range of acceptable values. Since there may be several logarithms in the original inequality, it is required to find the VA of each of them. Thus, the general scheme for solving logarithmic inequalities is as follows: Task. Solve the inequality: Let's find the domain of definition (DO) of the first logarithm: We solve using the interval method. Finding the zeros of the numerator: 3x − 2 = 0; Then - the zeros of the denominator: x − 1 = 0; We mark zeros and signs on the coordinate arrow: We get x ∈ (−∞ 2/3)∪(1; +∞). The second logarithm will have the same VA. If you don't believe me, you can check it. Now we transform the second logarithm so that the base is two: As you can see, the threes at the base and in front of the logarithm have been reduced. We got two logarithms with the same base. Let's add them up: log 2 (x − 1) 2< 2; We obtained the standard logarithmic inequality. We get rid of logarithms using the formula. Since the original inequality contains a “less than” sign, the resulting rational expression must also be less than zero. We have: (f (x) − g (x)) (k (x) − 1)< 0; We got two sets: It remains to intersect these sets - we get the real answer: We are interested in the intersection of sets, so we select intervals that are shaded on both arrows. We get x ∈ (−1; 2/3)∪(1; 3) - all points are punctured.
Solving Examples
3x > 24;
x > 8. 
What is needed to solve logarithmic inequalities?
Homework
Since logarithms are defined only for positive numbers, it is necessary that x be positive. The condition x > 0 is called the range of permissible values (APV) of this inequality. Only for such x does the inequality make sense.
Please note that we have moved on to an algebraic inequality, and the inequality sign remains the same.
And if then
2x − 9 ≤ x.
Let's make a replacement
We have added a condition x> 0 (from ODZ).
Answer:
So,
x2 ≠ 0;
x ≠ 0.
(9 − x 2) x 2< 0;
(3 − x) · (3 + x) · x 2< 0.
Converting logarithmic inequalities
x = 2/3.
x = 1.
log 2 (x − 1) 2< log 2 2 2 .
((x − 1) 2 − 2 2)(2 − 1)< 0;
x 2 − 2x + 1 − 4< 0;
x 2 − 2x − 3< 0;
(x − 3)(x + 1)< 0;
x ∈ (−1; 3).