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Do you think that there is still time before the Unified State Exam and you will have time to prepare? Perhaps this is so. But in any case, the earlier a student begins preparation, the more successfully he passes the exams. Today we decided to devote an article to logarithmic inequalities. This is one of the tasks, which means an opportunity to get extra credit.
Do you already know what a logarithm is? We really hope so. But even if you don't have an answer to this question, it's not a problem. Understanding what a logarithm is is very simple.
Why 4? You need to raise the number 3 to this power to get 81. Once you understand the principle, you can proceed to more complex calculations.
You went through inequalities a few years ago. And since then you have constantly encountered them in mathematics. If you have problems solving inequalities, check out the appropriate section.
Now that we have become familiar with the concepts individually, let's move on to considering them in general.
The simplest logarithmic inequality.
The simplest logarithmic inequalities are not limited to this example; there are three more, only with different signs. Why is this necessary? To better understand how to solve inequalities with logarithms. Now let's give a more applicable example, still quite simple; we'll leave complex logarithmic inequalities for later.
How to solve this? It all starts with ODZ. It’s worth knowing more about it if you want to always easily solve any inequality.
What is ODZ? ODZ for logarithmic inequalities
The abbreviation stands for the range of acceptable values. This formulation often comes up in tasks for the Unified State Exam. ODZ will be useful to you not only in the case of logarithmic inequalities.
Look again at the above example. We will consider the ODZ based on it, so that you understand the principle, and solving logarithmic inequalities does not raise questions. From the definition of a logarithm it follows that 2x+4 must be greater than zero. In our case this means the following.
This number, by definition, must be positive. Solve the inequality presented above. This can even be done orally; here it is clear that X cannot be less than 2. The solution to the inequality will be the definition of the range of acceptable values.
Now let's move on to solving the simplest logarithmic inequality.
We discard the logarithms themselves from both sides of the inequality. What are we left with as a result? Simple inequality.
It's not difficult to solve. X must be greater than -0.5. Now we combine the two obtained values into a system. Thus,
This will be the range of acceptable values for the logarithmic inequality under consideration.
Why do we need ODZ at all? This is an opportunity to weed out incorrect and impossible answers. If the answer is not within the range of acceptable values, then the answer simply does not make sense. This is worth remembering for a long time, since in the Unified State Exam there is often a need to search for ODZ, and it concerns not only logarithmic inequalities.
Algorithm for solving logarithmic inequality
The solution consists of several stages. First, you need to find the range of acceptable values. There will be two meanings in the ODZ, we discussed this above. Next, you need to solve the inequality itself. The solution methods are as follows:
- multiplier replacement method;
- decomposition;
- rationalization method.
Depending on the situation, it is worth using one of the above methods. Let's move directly to the solution. Let us reveal the most popular method, which is suitable for solving Unified State Examination tasks in almost all cases. Next we will look at the decomposition method. It can help if you come across a particularly tricky inequality. So, an algorithm for solving logarithmic inequality.
Examples of solutions :
It’s not for nothing that we took exactly this inequality! Pay attention to the base. Remember: if it is greater than one, the sign remains the same when finding the range of acceptable values; otherwise, you need to change the inequality sign.
As a result, we get the inequality:
Now we reduce the left side to the form of the equation equal to zero. Instead of the “less than” sign we put “equals” and solve the equation. Thus, we will find the ODZ. We hope that you will not have problems solving such a simple equation. The answers are -4 and -2. That's not all. You need to display these points on the graph, placing “+” and “-”. What needs to be done for this? Substitute the numbers from the intervals into the expression. Where the values are positive, we put “+” there.
Answer: x cannot be greater than -4 and less than -2.
We have found the range of acceptable values only for the left side; now we need to find the range of acceptable values for the right side. This is much easier. Answer: -2. We intersect both resulting areas.
And only now are we beginning to address the inequality itself.
Let's simplify it as much as possible to make it easier to solve.
We again use the interval method in the solution. Let’s skip the calculations; everything is already clear with it from the previous example. Answer.
But this method is suitable if the logarithmic inequality has the same bases.
Solving logarithmic equations and inequalities with different bases requires an initial reduction to the same base. Next, use the method described above. But there is a more complicated case. Let's consider one of the most complex types of logarithmic inequalities.
Logarithmic inequalities with variable base
How to solve inequalities with such characteristics? Yes, and such people can be found in the Unified State Examination. Solving inequalities in the following way will also have a beneficial effect on your educational process. Let's look at the issue in detail. Let's discard theory and go straight to practice. To solve logarithmic inequalities, it is enough to familiarize yourself with the example once.
To solve a logarithmic inequality of the form presented, it is necessary to reduce the right-hand side to a logarithm with the same base. The principle resembles equivalent transitions. As a result, the inequality will look like this.
Actually, all that remains is to create a system of inequalities without logarithms. Using the rationalization method, we move on to an equivalent system of inequalities. You will understand the rule itself when you substitute the appropriate values and track their changes. The system will have the following inequalities.
When using the rationalization method when solving inequalities, you need to remember the following: one must be subtracted from the base, x, by definition of the logarithm, is subtracted from both sides of the inequality (right from left), two expressions are multiplied and set under the original sign in relation to zero.
The further solution is carried out using the interval method, everything is simple here. It is important for you to understand the differences in solution methods, then everything will start to work out easily.
There are many nuances in logarithmic inequalities. The simplest of them are quite easy to solve. How can you solve each of them without problems? You have already received all the answers in this article. Now you have a long practice ahead of you. Constantly practice solving a variety of problems in the exam and you will be able to get the highest score. Good luck to you in your difficult task!
Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely taught in school. The presentation presents solutions to tasks C3 of the Unified State Exam - 2014 in mathematics.
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Solving logarithmic inequalities containing a variable in the base of the logarithm: methods, techniques, equivalent transitions, mathematics teacher, Secondary School No. 143 Knyazkina T.V.
Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely taught in school: log k (x) f (x) ∨ log k (x) g (x) ⇒ (f (x) − g (x)) (k ( x) − 1) ∨ 0 Instead of the “∨” checkbox, you can put any inequality sign: more or less. The main thing is that in both inequalities the signs are the same. This way we get rid of logarithms and reduce the problem to a rational inequality. The latter is much easier to solve, but when discarding logarithms, extra roots may appear. To cut them off, it is enough to find the range of acceptable values. Don't forget the ODZ of the logarithm! Everything related to the range of acceptable values must be written out and solved separately: f (x) > 0; g(x) > 0; k(x) > 0; k(x) ≠ 1. These four inequalities constitute a system and must be satisfied simultaneously. When the range of acceptable values has been found, all that remains is to intersect it with the solution of the rational inequality - and the answer is ready.
Solve the inequality: Solution First, let's write out the OD of the logarithm. The first two inequalities are satisfied automatically, but the last one will have to be written down. Since the square of a number is equal to zero if and only if the number itself is equal to zero, we have: x 2 + 1 ≠ 1; x 2 ≠ 0; x ≠ 0. It turns out that the ODZ of a logarithm is all numbers except zero: x ∈ (−∞0)∪(0 ;+ ∞). Now we solve the main inequality: We make the transition from the logarithmic inequality to the rational one. The original inequality has a “less than” sign, which means the resulting inequality must also have a “less than” sign.
We have: (10 − (x 2 + 1)) · (x 2 + 1 − 1)
Transforming Logarithmic Inequalities Often the original inequality is different from the one above. This can be easily corrected using standard rules for working with logarithms. Namely: Any number can be represented as a logarithm with a given base; The sum and difference of logarithms with the same bases can be replaced by one logarithm. Separately, I would like to remind you about the range of acceptable values. Since there may be several logarithms in the original inequality, it is required to find the VA of each of them. Thus, the general scheme for solving logarithmic inequalities is as follows: Find the VA of each logarithm included in the inequality; Reduce the inequality to a standard one using the formulas for adding and subtracting logarithms; Solve the resulting inequality using the scheme given above.
Solve the inequality: Solution Let's find the domain of definition (DO) of the first logarithm: Solve by the method of intervals. Find the zeros of the numerator: 3 x − 2 = 0; x = 2/3. Then - the zeros of the denominator: x − 1 = 0; x = 1. Mark zeros and signs on the coordinate line:
We get x ∈ (−∞ 2/3) ∪ (1; +∞). The second logarithm will have the same VA. If you don't believe it, you can check it. Now let's transform the second logarithm so that there is a two at the base: As you can see, the threes at the base and in front of the logarithm have been canceled. We got two logarithms with the same base. Add them up: log 2 (x − 1) 2
(f (x) − g (x)) (k (x) − 1)
We are interested in the intersection of sets, so we select intervals that are shaded on both arrows. We get: x ∈ (−1; 2/3) ∪ (1; 3) - all points are punctured. Answer: x ∈ (−1; 2/3)∪(1; 3)
Solving USE-2014 tasks type C3
Solve the system of inequalities. Solution. ODZ: 1) 2)
Solve the system of inequalities 3) -7 -3 - 5 x -1 + + + − − (continued)
Solve the system of inequalities 4) General solution: and -7 -3 - 5 x -1 -8 7 log 2 129 (continued)
Solve the inequality (continued) -3 3 -1 + − + − x 17 + -3 3 -1 x 17 -4
Solve the inequality Solution. ODZ:
Solve the inequality (continued)
Solve the inequality Solution. ODZ: -2 1 -1 + − + − x + 2 -2 1 -1 x 2
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 out and solved separately:
f(x) > 0; g(x) > 0; k(x) > 0; k(x) ≠ 1.
These four inequalities constitute a system and must be satisfied simultaneously. When the range of acceptable values 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;
x2 ≠ 0;
x ≠ 0.
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;
(9 − x 2) x 2< 0;
(3 − x) · (3 + x) · x 2< 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.
Converting logarithmic inequalities
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:
- Any number can be represented as a logarithm with a given base;
- The sum and difference of logarithms with the same bases can be replaced by one logarithm.
Separately, I would like to remind you about the range of acceptable values. Since there may be several logarithms in the original inequality, it is required to find the VA of each of them. Thus, the general scheme for solving logarithmic inequalities is as follows:
- Find the VA of each logarithm included in the inequality;
- Reduce the inequality to a standard one using the formulas for adding and subtracting logarithms;
- Solve the resulting inequality using the scheme given above.
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;
x = 2/3.
Then - the zeros of the denominator:
x − 1 = 0;
x = 1.
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 it, 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;
log 2 (x − 1) 2< log 2 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;
((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).
We got two sets:
- ODZ: x ∈ (−∞ 2/3)∪(1; +∞);
- Candidate answer: x ∈ (−1; 3).
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.
A lesson on one inequality develops research skills, awakens students' thoughts, develops intelligence, and increases students' interest in work. It is best to conduct it when students have mastered the necessary concepts and have analyzed a number of particular techniques for solving logarithmic inequalities. In this lesson, students are active participants in finding a solution.
Lesson type
. A lesson in applying knowledge, skills, abilities in a new situation. (Lesson of systematization and generalization of the studied material).Lesson Objectives
:- educational : to develop skills and abilities to solve logarithmic inequalities of the specified type in different ways; teach how to independently acquire knowledge (students’ own activities in studying and mastering the content of educational material);
- developing : work on speech development; teach to analyze, highlight the main thing, prove and disprove logical conclusions;
- educational : formation of moral qualities, humane relations, accuracy, discipline, self-esteem, responsible attitude towards achieving the goal.
During the classes.
1. Organizational moment.
Oral work.
2. Checking homework.
Write down the following sentences in mathematical language: “The numbers a and b are on the same side of one,” “The numbers a and b are on opposite sides of the unit,” and prove the resulting inequalities. (One of the students prepared a solution in advance on the board).
3. Report the topic of the lesson, its goals and objectives.
Analyzing the options for entrance exams in mathematics, one can notice that from the theory of logarithms in exams one often encounters logarithmic inequalities containing a variable under the logarithm and at the base of the logarithm.
Our lesson is lesson of one inequality, containing a variable under the logarithm and at the base of the logarithm, solved in different ways. They say that it is better to solve one inequality, but in different ways, than several inequalities in the same way. Indeed, you should be able to check your decisions. There is no better test than solving a problem in a different way and getting the same answer (you can arrive at the same systems, the same inequalities, equations in different ways). But not only this goal is pursued when solving tasks in different ways. The search for different solutions, consideration of all possible cases, critical assessment of them in order to highlight the most rational and beautiful, is an important factor in the development of mathematical thinking, and leads away from the template. Therefore, today we will solve only one inequality, but we will try to find several ways to solve it.
4. Creative application and acquisition of knowledge, mastering methods of activity by solving problematic problems built on the basis of previously acquired knowledge and skills in solving the inequality log x (x 2 – 2x – 3)< 0.
Here is the solution to this inequality, taken from one exam paper. Look at it carefully and try to analyze the solution. (The solution to the inequality is written down on the board in advance)
log x (x 2 – 2x – 3)< log x 1;
a) x 2 – 2x – 3 > 0; b) x 2 – 2x – 3< 1;
x 2 – 2x – 3 = 0; x 2 – 2x – 4< 0;
x 1 = - 1, x 2 = 3; x 2 – 2x – 4 = 0;
c) solution of the system
Possible student explanations:
This is not an equation, but an inequality, therefore, when moving from a logarithmic inequality to a rational one, the sign of the inequality will depend on the base of the logarithm and the monotonicity of the logarithmic function.
With such a decision, it is possible to acquire extraneous solutions, or to lose solutions, and it is possible that with an incorrect decision, the correct answer will be obtained.
So how was it necessary to solve this inequality, in which the variable is under the sign of the logarithm and in the base of the logarithm?!
This inequality is equivalent to a combination of two systems of inequalities.
The first system of inequalities has no solutions.
The solution to the system of inequalities will be
In the proposed solution to the inequality from the exam paper, the answer was correct. Why?
Possible student answers:
Since the domain of definition of the function on the left side of the inequality consists of numbers greater than 3, therefore, the function y = log x t is increasing. Therefore, the answer turned out to be correct.
How was it possible to write down a mathematically correct solution in an exam paper?
II method.
Let's find the domain of definition of the function on the left side of the inequality, and then, taking into account the domain of definition, consider only one case
How else can this inequality be resolved? What formulas can be used?
Formula for moving to a new base a > 0, a 1
III method.
IV method.
Is it possible to apply to the inequality itself the fact that the logarithm is less than zero?
Yes. The expression under the logarithm and the base of the logarithm are on opposite sides of one, but are positive!
That is, we again obtain the same set of two systems of inequalities:
All considered methods lead to a combination of two systems of inequalities. In all cases the same answer is obtained. All methods are theoretically justified.
Question to students: why do you think a question was asked in the homework that was not related to the material studied in grade 11?
Knowing the properties of the logarithm that log a b< 0 , If a And b on opposite sides of 1,
log a b > 0 if a And b on one side of 1, you can get a very interesting and unexpected way to solve the inequality. This method is written about in the article “Some useful logarithmic relationships” in the magazine “Quantum” No. 10 for 1990.
log g(x) f(x) > 0 if
log g(x) f(x)< 0, если
(Why condition g(x) 1 is not necessary to write?)
Solution to inequality log x (x 2 – 2x – 3)< 0 looks like that:
a) x 2 – 2x – 3 > 0; b) (x – 1)(x 2 – 2x – 4)< 0;
c) solution to the system of inequality
VI method.
Interval method. (“Solving logarithmic inequalities using the interval method” is the topic of the next lesson).
5. The result of the work done.
1. In what ways was inequality resolved? How many ways to solve this
Did we find any inequalities?
2. Which one is the most rational? Beautiful?
3. What was the solution to the inequality based on in each case?
4. Why is this inequality interesting?
Qualitative characteristics of the teacher’s work in the classroom.
6. Generalization of the studied material.
Is it possible to consider this inequality as a special case of a more general problem?
Inequality of the form log g(x) f(x)<(>) log g(x) h(x) can be reduced to inequality log g(x) p(x)<(>) 0 using the properties of logarithms and the properties of inequalities.
Solve inequality
log x (x 2 + 3x – 3) > 1
by any of the methods considered.
7. Homework, instructions on how to complete it
.1. Solve the inequalities (from the options for entrance exams in mathematics):
2. In the next lesson we will consider logarithmic inequalities that are solved by the interval method. Repeat the algorithm for solving inequalities using the interval method.
3. Arrange the numbers in ascending order (explain why this arrangement):
log 0.3 5; ; ; log 0.5 3 (repeat for next lesson).