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. It often comes up in tasks for the Unified State Exam. this wording. ODZ will be useful to you not only in case 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 Examination 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 values 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 with a solution to this simple equation you won't have any problems. 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.
Apply again interval method in the decision. 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 for different reasons presupposes an initial reduction to one base. Next, use the method described above. But there is more difficult case. Let's consider one of the most complex species 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 benefit 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.
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 highest score. Good luck to you in your difficult task!
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 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: inequality solution in mode online solution almost any given inequality online. Mathematical inequalities online to solve mathematics. Find quickly inequality solution in mode online. The website www.site allows you to find solution almost any given algebraic, trigonometric or transcendental inequality online. 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Solving inequalities online yourself, it is useful to check the received answer using online solution of inequalities on the website www.site. You need to write the inequality correctly and instantly get online solution, after which all that remains is to compare the answer with your solution to the inequality. Checking the answer will take no more than a minute, it’s enough solve inequality online and compare the answers. This will help you avoid mistakes in decision and correct the answer in time when solving inequalities online be it algebraic, trigonometric, transcendental or inequality with unknown parameters. We looked at solving the simplest logarithmic inequalities and inequalities where the base of the logarithm is fixed in the last lesson. But what if there is a variable at the base of the logarithm? Then it will come to our aid rationalization of inequalities. To understand how this works, let's consider, for example, the inequality: $$\log_(2x) x^2 > \log_(2x) x.$$ As expected, let's start with ODZ. $$\left[ \begin(array)(l)x>0,\\ 2x ≠ 1. \end(array)\right.$$ Let's reason as if we were solving an inequality with a fixed base. If the base is greater than one, we get rid of logarithms, and the inequality sign does not change; if it is less than one, it changes. Let's write this as a system: $$\left[ \begin(array)(l) \left\( \begin(array)(l)2x>1,\\ x^2 > x; \end(array)\right. \\ \left\ ( \begin(array)(l)2x<1,\\ x^2 < x; \end{array}\right. \end{array} \right.$$ For further reasoning, let us move all the right-hand sides of the inequalities to the left. $$\left[ \begin(array)(l) \left\( \begin(array)(l)2x-1>0,\\ x^2 -x>0; \end(array)\right. \ \ \left\( \begin(array)(l)2x-1<0,\\ x^2 -x<0; \end{array}\right. \end{array} \right.$$ What did we get? It turns out that we need the expressions `2x-1` and `x^2 - x` to be either positive or negative at the same time. The same result will be obtained if we solve the inequality: $$(2x-1)(x^2 - x) >0.$$ This inequality, like the original system, is true if both factors are either positive or negative. It turns out that you can move from a logarithmic inequality to a rational one (taking into account the ODZ). Let's formulate method for rationalizing logarithmic inequalities$$\log_(f(x)) g(x) \vee \log_(f(x)) h(x) \Leftrightarrow (f(x) - 1)(g(x)-h(x)) \ vee 0,$$ where `\vee` is any inequality sign. (For the `>` sign, we have just checked the validity of the formula. For the rest, I suggest you check it yourself - it will be remembered better). Let's return to solving our inequality. Expanding it into brackets (to make the zeros of the function easier to see), we get $$(2x-1)x(x - 1) >0.$$ The interval method will give the following picture: (Since the inequality is strict and we are not interested in the ends of the intervals, they are not shaded.) As can be seen, the resulting intervals satisfy the ODZ. We received the answer: `(0,\frac(1)(2)) \cup (1,∞)`. $$\log_(2-x) 3 \leqslant \log_(2-x) x.$$ $$\left\(\begin(array)(l)2-x > 0,\\ 2-x ≠ 1,\\ x > 0. \end(array)\right.$$ $$\left\(\begin(array)(l)x< 2,\\ x ≠ 1, \\ x >0.\end(array)\right.$$ According to the rule we just received rationalization of logarithmic inequalities, we find that this inequality is identical (taking into account the ODZ) to the following: $$(2-x -1) (3-x) \leqslant 0.$$ $$(1-x) (3-x) \leqslant 0.$$ Combining this solution with the ODZ, we get the answer: `(1,2)`. $$\log_x\frac(4x+5)(6-5x) \leqslant -1.$$ $$\left\(\begin(array)(l) \dfrac(4x+5)(6-5x)>0, \\ x>0,\\ x≠ 1.\end(array) \right.$ $ Since the system is relatively complex, let's immediately plot the solution to the inequalities on the number line: Thus, ODZ: `(0,1)\cup \left(1,\frac(6)(5)\right)`. Let's represent `-1` as a logarithm with the base `x`. $$\log_x\frac(4x+5)(6-5x) \leqslant \log_x x^(-1).$$ By using rationalization of logarithmic inequality we get a rational inequality: $$(x-1)\left(\frac(4x+5)(6-5x) -\frac(1)(x)\right)\leqslant0,$$ $$(x-1)\left(\frac(4x^2+5x - 6+5x)(x(6-5x))\right)\leqslant0,$$ $$(x-1)\left(\frac(2x^2+5x - 3)(x(6-5x))\right)\leqslant0.$$ LOGARITHMIC INEQUALITIES IN THE USE
Sechin Mikhail Alexandrovich Small Academy sciences for students of the Republic of Kazakhstan "Iskatel" MBOU "Sovetskaya Secondary School No. 1", 11th grade, town. Sovetsky Sovetsky district Gunko Lyudmila Dmitrievna, teacher of the Municipal Budgetary Educational Institution “Sovetskaya Secondary School No. 1” Sovetsky district Purpose of the work: study of the mechanism for solving logarithmic inequalities C3 using non-standard methods, identifying interesting facts logarithm Subject of research:
3) Learn to solve specific logarithmic inequalities C3 using non-standard methods. Results:
Content
Introduction………………………………………………………………………………….4
Chapter 1. History of the issue……………………………………………………...5
Chapter 2. Collection of logarithmic inequalities ………………………… 7
2.1. Equivalent transitions and the generalized method of intervals…………… 7 2.2. Rationalization method……………………………………………………………… 15 2.3. Non-standard substitution……………….................................................... ..... 22 2.4. Tasks with traps……………………………………………………27 Conclusion……………………………………………………………………………… 30
Literature……………………………………………………………………. 31
Introduction
I am in 11th grade and plan to enter a university where the core subject is mathematics. That’s why I work a lot with problems in part C. In task C3, I need to solve a non-standard inequality or system of inequalities, usually related to logarithms. When preparing for the exam, I was faced with the problem of a shortage of methods and techniques for solving exam logarithmic inequalities offered in C3. Methods that are studied in school curriculum on this topic, do not provide a basis for solving C3 tasks. The math teacher suggested that I work on C3 assignments independently under her guidance. In addition, I was interested in the question: do we encounter logarithms in our lives? With this in mind, the topic was chosen: “Logarithmic inequalities in the Unified State Exam”
Purpose of the work: study of the mechanism for solving C3 problems using non-standard methods, identifying interesting facts about the logarithm. Subject of research:
1) Find the necessary information about non-standard methods for solving logarithmic inequalities. 2) Find additional information about logarithms. 3) Learn to solve specific C3 problems using non-standard methods. Results:
Practical significance consists in expanding the apparatus for solving C3 problems. This material can be used in some lessons, for clubs, and elective classes in mathematics. The project product will be the collection “C3 Logarithmic Inequalities with Solutions.” Chapter 1. Background
Throughout the 16th century, the number of approximate calculations increased rapidly, primarily in astronomy. Improving instruments, studying planetary movements and other work required colossal, sometimes multi-year, calculations. Astronomy was in real danger of drowning in unfulfilled calculations. Difficulties also arose in other areas, for example in insurance business Compound interest tables were needed for various percentage values. The main difficulty was multiplication and division of multi-digit numbers, especially trigonometric quantities. The discovery of logarithms was based on the properties of progressions that were well known by the end of the 16th century. About the connection between members geometric progression q, q2, q3, ... and arithmetic progression their indicators are 1, 2, 3,... Archimedes spoke in his “Psalmitis”. Another prerequisite was the extension of the concept of degree to negative and fractional exponents. Many authors have pointed out that multiplication, division, exponentiation and root extraction in geometric progression correspond in arithmetic - in the same order - addition, subtraction, multiplication and division. Here was the idea of the logarithm as an exponent. In the history of the development of the doctrine of logarithms, several stages have passed. Stage 1
Logarithms were invented no later than 1594 independently by the Scottish Baron Napier (1550-1617) and ten years later by the Swiss mechanic Bürgi (1552-1632). Both wanted to provide a new, convenient means of arithmetic calculations, although they approached this problem in different ways. Napier kinematically expressed the logarithmic function and thereby entered a new field of function theory. Bürgi remained on the basis of considering discrete progressions. However, the definition of the logarithm for both is not similar to the modern one. The term "logarithm" (logarithmus) belongs to Napier. It arose from a combination Greek words: logos - “relation” and ariqmo - “number”, which meant “number of relations”. Initially, Napier used a different term: numeri artificiales - “artificial numbers”, as opposed to numeri naturalts - “natural numbers”. In 1615, in a conversation with Henry Briggs (1561-1631), a professor of mathematics at Gresh College in London, Napier proposed taking zero as the logarithm of one, and 100 as the logarithm of ten, or, what amounts to the same thing, simply 1. This is how they appeared decimal logarithms and the first logarithmic tables were printed. Later, Briggs' tables were supplemented by the Dutch bookseller and mathematics enthusiast Adrian Flaccus (1600-1667). Napier and Briggs, although they came to logarithms earlier than everyone else, published their tables later than the others - in 1620. The signs log and Log were introduced in 1624 by I. Kepler. The term “natural logarithm” was introduced by Mengoli in 1659 and followed by N. Mercator in 1668, and the London teacher John Speidel published tables of natural logarithms of numbers from 1 to 1000 under the name “New Logarithms”. The first logarithmic tables were published in Russian in 1703. But in all logarithmic tables there were calculation errors. The first error-free tables were published in 1857 in Berlin, processed by the German mathematician K. Bremiker (1804-1877). Stage 2
Further development of the theory of logarithms is associated with a wider application of analytical geometry and infinitesimal calculus. By that time, the connection between the squaring of an equilateral hyperbola and natural logarithm. The theory of logarithms of this period is associated with the names of a number of mathematicians. German mathematician, astronomer and engineer Nikolaus Mercator in an essay "Logarithmotechnics" (1668) gives a series giving the expansion of ln(x+1) in powers of x: This expression exactly corresponds to his train of thought, although, of course, he did not use the signs d, ..., but more cumbersome symbolism. With the discovery of the logarithmic series, the technique for calculating logarithms changed: they began to be determined using infinite series. In his lectures "Elementary Mathematics with highest point vision", read in 1907-1908, F. Klein proposed using the formula as the starting point for constructing the theory of logarithms. Stage 3
Definition logarithmic function as an inverse function exponential, logarithm as an exponent of a given base was not formulated immediately. Essay by Leonhard Euler (1707-1783) "An Introduction to the Analysis of Infinitesimals" (1748) served to further development of the theory of logarithmic functions. Thus, 134 years have passed since logarithms were first introduced (counting from 1614), before mathematicians came to the definition the concept of logarithm, which is now the basis of the school course. Chapter 2. Collection of logarithmic inequalities
2.1. Equivalent transitions and the generalized method of intervals. Equivalent transitions
Generalized interval method
This method most universal for solving inequalities of almost any type. The solution diagram looks like this: 1. Bring the inequality to a form where the function on the left side is 2. Find the domain of the function 3. Find the zeros of the function 4. Draw the domain of definition and zeros of the function on the number line. 5. Determine the signs of the function 6. Select intervals where the function takes the required values and write down the answer. Example 1.
Solution:
Let's apply the interval method where For these values, all expressions under the logarithmic signs are positive. Answer:
Example 2.
Solution:
1st
way
.
ADL is determined by inequality x> 3. Taking logarithms for such x in base 10, we get The last inequality could be solved by applying expansion rules, i.e. comparing factors to zero. However, in this case it is easy to determine the intervals of constant sign of the function therefore, the interval method can be applied. Function f(x) = 2x(x- 3.5)lgǀ x- 3ǀ is continuous at x> 3 and vanishes at points x 1 = 0, x 2 = 3,5, x 3 = 2, x 4 = 4. Thus, we determine the intervals of constant sign of the function f(x):
Answer: 2nd method
.
Let us directly apply the ideas of the interval method to the original inequality. To do this, recall that the expressions a b- a c and ( a - 1)(b- 1) have one sign. Then our inequality at x> 3 is equivalent to inequality or The last inequality is solved using the interval method Answer: Example 3.
Solution:
Let's apply the interval method Answer: Example 4.
Solution:
Since 2 x 2 - 3x+ 3 > 0 for all real x, That To solve the second inequality we use the interval method In the first inequality we make the replacement then we come to the inequality 2y 2 - y - 1 < 0 и, применив метод интервалов, получаем, что решениями будут те y, which satisfy the inequality -0.5< y < 1.
From where, because we get the inequality which is carried out when x, for which 2 x 2 - 3x - 5 < 0. Вновь применим метод интервалов
Now, taking into account the solution to the second inequality of the system, we finally obtain Answer:
Example 5.
Solution:
Inequality is equivalent to a collection of systems or Let's use the interval method or Answer:
Example 6.
Solution:
Inequality equals system Let Then y > 0,
and the first inequality system takes the form or, unfolding quadratic trinomial by factors, Applying the interval method to the last inequality, we see that its solutions satisfying the condition y> 0 will be all y > 4.
Thus, the original inequality is equivalent to the system: So, the solutions to the inequality are all 2.2. Rationalization method. Previously, inequality was not solved using the rationalization method; it was not known. This is the "new modern" effective method solutions to exponential and logarithmic inequalities" (quote from the book by S.I. Kolesnikova) "Magic Table"
In other sources
If a >1 and b >1, then log a b >0 and (a -1)(b -1)>0; If a >1 and 0 if 0<a<1 и b
>1, then log a b<0 и (a
-1)(b
-1)<0;
if 0<a<1 и 00 and (a -1)(b -1)>0. The reasoning carried out is simple, but significantly simplifies the solution of logarithmic inequalities. Example 4.
log x (x 2 -3)<0
Solution:
Example 5.
log 2 x (2x 2 -4x +6)≤log 2 x (x 2 +x ) Solution: Example 6.
To solve this inequality, instead of the denominator, we write (x-1-1)(x-1), and instead of the numerator, we write the product (x-1)(x-3-9 + x). Example 7.
Example 8.
2.3. Non-standard substitution. Example 1.
Example 2.
Example 3.
Example 4.
Example 5.
Example 6.
Example 7.
log 4 (3 x -1)log 0.25 Let's make the replacement y=3 x -1; then this inequality will take the form Log 4 log 0.25 Because log 0.25 Let us make the replacement t =log 4 y and obtain the inequality t 2 -2t +≥0, the solution of which is the intervals - Thus, to find the values of y we have a set of two simple inequalities Therefore, the original inequality is equivalent to the set of two exponential inequalities, The solution to the first inequality of this set is the interval 0<х≤1, решением второго – промежуток 2≤х<+ Example 8.
Solution:
Inequality equals system The solution to the second inequality defining the ODZ will be the set of those x,
for which x > 0.
To solve the first inequality we make the substitution Then we get the inequality or The set of solutions to the last inequality is found by the method intervals: -1< t < 2. Откуда, возвращаясь к переменной x, we get or Lots of those x, which satisfy the last inequality belongs to ODZ ( x> 0), therefore, is a solution to the system, and hence the original inequality. Answer: 2.4. Tasks with traps. Example 1.
Solution. The ODZ of the inequality is all x satisfying the condition 0 Example 2.
log 2 (2 x +1-x 2)>log 2 (2 x-1 +1-x)+1. Conclusion
It was not easy to find specific methods for solving C3 problems from a large abundance of different educational sources. In the course of the work done, I was able to study non-standard methods for solving complex logarithmic inequalities. These are: equivalent transitions and the generalized method of intervals, the method of rationalization ,
non-standard substitution ,
tasks with traps on ODZ. These methods are not included in the school curriculum. Using different methods, I solved 27 inequalities proposed on the Unified State Exam in part C, namely C3. These inequalities with solutions by methods formed the basis of the collection “C3 Logarithmic Inequalities with Solutions,” which became a project product of my activity. The hypothesis I posed at the beginning of the project was confirmed: C3 problems can be effectively solved if you know these methods. In addition, I discovered interesting facts about logarithms. It was interesting for me to do this. My project products will be useful for both students and teachers. Conclusions:
Thus, the project goal has been achieved and the problem has been solved. And I received the most complete and varied experience of project activities at all stages of work. While working on the project, my main developmental impact was on mental competence, activities related to logical mental operations, the development of creative competence, personal initiative, responsibility, perseverance, and activity. A guarantee of success when creating a research project for I gained: significant school experience, the ability to obtain information from various sources, check its reliability, and rank it by importance. In addition to direct subject knowledge in mathematics, I expanded my practical skills in the field of computer science, gained new knowledge and experience in the field of psychology, established contacts with classmates, and learned to cooperate with adults. During the project activities, organizational, intellectual and communicative general educational skills were developed. Literature
1. Koryanov A. G., Prokofiev A. A. Systems of inequalities with one variable (standard tasks C3). 2. Malkova A. G. Preparation for the Unified State Exam in Mathematics. 3. Samarova S. S. Solving logarithmic inequalities. 4. Mathematics. Collection of training works edited by A.L. Semenov and I.V. Yashchenko. -M.: MTsNMO, 2009. - 72 p.-
Solving Examples
3x > 24;
x > 8. 
What is needed to solve logarithmic inequalities?
Homework
ODZ
Solution to inequality
Example two. Solving logarithmic inequality with variable base
ODZ
Solution to inequality
Third example. Logarithm of a fraction
ODZ
Solution to inequality
, if a > 1
, if 0 <
а <
1
, and on the right 0..
, that is, solve the equation 
(and solving an equation is usually easier than solving an inequality).on the obtained intervals.
And even if the teacher knew him, there was a fear - did he know him? Unified State Exam expert, why don’t they give it at school? There were situations when the teacher said to the student: “Where did you get it? Sit down - 2.”
Now the method is being promoted everywhere. And for experts there is guidelines, associated with this method, and in the "Most Complete Editions of Model Options..." solution C3 uses this method.
WONDERFUL METHOD!
Answer. (0; 0.5)U.
Answer :
(3;6)
.
= -log 4 = -(log 4 y -log 4 16)=2-log 4 y , then we rewrite the last inequality as 2log 4 y -log 4 2 y ≤.
The solution to this set is the intervals 0<у≤2 и 8≤у<+.
that is, aggregates 
. Thus, the original inequality is satisfied for all values of x from the intervals 0<х≤1 и 2≤х<+.
.
. Therefore, all x are from the interval 0
