Logarithmic inequalities
In the previous lessons, we met with logarithmic equations and now we know what it is 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 under the sign of the logarithm or at its base.
Or, we can also say that a logarithmic inequality is an inequality in which its unknown value, as in the logarithmic equation, will be under the sign of the logarithm.
The simplest logarithmic inequalities are as follows:
where f (x) and g (x) are some expressions that depend on x.
Let's look at this with an example: f (x) = 1 + 2x + x2, g (x) = 3x − 1.
Solving logarithmic inequalities
Before solving logarithmic inequalities, it is worth noting that when solving them, they resemble exponential 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 one;
Second, solving the logarithmic inequality using a change of variables, we need to solve the inequality about the change until we get the simplest inequality.
But it was we who considered similar aspects of solving logarithmic inequalities. And 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, passing from logarithms to expressions under the sign of the logarithm, it is necessary to take into account the range of permissible values (ADV).
That is, it should be borne in mind that when solving a logarithmic equation, you and I can first find the roots of the equation, and then check this solution. But solving the logarithmic inequality will not work that way, since passing from logarithms to expressions under the sign of the logarithm, it will be necessary to write down the ODZ of 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 it is necessary 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.
Solving inequalities with a variable, you need to find all its solutions. If two inequalities have one variable x, then such inequalities are equivalent, provided that their solutions coincide.
When performing tasks for solving logarithmic inequalities, it is necessary to 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.
Ways to solve logarithmic inequalities
Now let's look at some of the ways that take place when solving logarithmic inequalities. For a better understanding and assimilation, we will try to understand them with specific examples.
You and I know that the simplest logarithmic inequality has the following form:
In this inequality, V - is one of such inequality signs as:<,>, ≤ or ≥.
When the base of this logarithm is greater than one (a> 1), making the transition from logarithms to expressions under the sign of the logarithm, then in this version the inequality sign is preserved, and the inequality will look like this:
which is equivalent to such a 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 see more examples of solving the simplest logarithmic inequalities shown in the picture below: The task. Let's try to solve this inequality: Solution of the range of valid values. Now let's try to multiply its right side by: Let's see what we get: Now, let's move on with you to the transformation of sub-logarithmic 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 have obtained is wholly and completely owned by the GDZ and is a solution to such an inequality. Here's our answer: Now let's try to analyze what we need to successfully solve logarithmic inequalities? First, focus 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 prevent the expansion and contraction of the ODZ inequality, 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 inequality, while being guided by its ODV. Thirdly, in order to successfully solve such inequalities, each of you must perfectly know all the properties of elementary functions and clearly understand their meaning. These functions include not only logarithmic, but also rational, power, trigonometric, etc., in a word, all those that you studied during your school algebra studies. As you can see, having studied the topic of logarithmic inequalities, there is nothing difficult in solving these inequalities, provided you are attentive and persistent in achieving your goals. To avoid any problems in solving inequalities, you need to train as much as possible, solving various tasks and at the same time memorize the main ways of solving such inequalities and their systems. In case of unsuccessful solutions to logarithmic inequalities, you should carefully analyze your mistakes so as not to return to them again in the future. For a better understanding of the topic and consolidation of the passed material, solve the following inequalities: They are inside the logarithms. Examples: \ (\ log_3x≥ \ log_39 \) Any logarithmic inequality should be reduced to the form \ (\ log_a (f (x)) ˅ \ log_a (g (x)) \) (the symbol \ (˅ \) means any of). This form allows you to get rid of the logarithms and their bases by making the transition to the inequality of expressions under the logarithms, that is, to the form \ (f (x) ˅ g (x) \). But there is one very important subtlety when performing this transition: \ (\ log_2 ((8-x))<1\) Solution: \ (\ log \) \ (_ (0,5) \) \ ((2x-4) \) ≥ \ (\ log \) \ (_ (0,5) \) \ (((x + one))\) Solution: Very important! In any inequality, the transition from the form \ (\ log_a (f (x)) ˅ \ log_a (g (x)) \) to the comparison of expressions under logarithms can be done only if: Example
... Solve inequality: \ (\ log \) \ (≤-1 \) Solution:
\ (\ log \) \ (_ (\ frac (1) (3)) (\ frac (3x-2) (2x-3)) \)\(≤-1\) Let's write out ODZ. ODZ: \ (\ frac (3x-2) (2x-3) \) \ (> 0 \) \ ( \ frac (3x-2-3 (2x-3)) (2x-3) \)\(≥\) \(0\) We open the brackets, we give. \ ( \ frac (-3x + 7) (2x-3) \) \ (≥ \) \ (0 \) We multiply the inequality by \ (- 1 \), not forgetting to reverse the comparison sign. \ ( \ frac (3x-7) (2x-3) \) \ (≤ \) \ (0 \) \ ( \ frac (3 (x- \ frac (7) (3))) (2 (x- \ frac (3) (2))) \)\(≤\) \(0\) Let's construct a numerical axis and mark on it the points \ (\ frac (7) (3) \) and \ (\ frac (3) (2) \). Please note that the dot from the denominator is punctured, despite the fact that the inequality is not strict. The point is that this point will not be a solution, since when substituted into an inequality, it will lead us to division by zero. Now, on the same numerical axis, we plot the ODZ and write in response the interval that falls into the ODZ. We write down the final answer. Example
... Solve the inequality: \ (\ log ^ 2_3x- \ log_3x-2> 0 \) Solution:
\ (\ log ^ 2_3x- \ log_3x-2> 0 \) Let's write out ODZ. ODZ: \ (x> 0 \) Let's get down to the solution. Solution: \ (\ log ^ 2_3x- \ log_3x-2> 0 \) We have before us a typical square-logarithmic inequality. We do it. \ (t = \ log_3x \) \ (D = 1 + 8 = 9 \) Now you need to go back to the original variable - x. To do this, go to one that has the same solution and make the reverse replacement. \ (\ left [\ begin (gathered) t> 2 \\ t<-1 \end{gathered} \right.\) \(\Leftrightarrow\) \(\left[ \begin{gathered} \log_3x>2 \\ \ log_3x<-1 \end{gathered} \right.\) Convert \ (2 = \ log_39 \), \ (- 1 = \ log_3 \ frac (1) (3) \). \ (\ left [\ begin (gathered) \ log_3x> \ log_39 \\ \ log_3x<\log_3\frac{1}{3} \end{gathered} \right.\) We make the transition to the comparison of arguments. The bases of logarithms are greater than \ (1 \), so the sign of the inequalities does not change. \ (\ left [\ begin (gathered) x> 9 \\ x<\frac{1}{3} \end{gathered} \right.\) Let us combine the solution of inequality and the DHS in one figure. Let's write down the answer. LOGARITHMIC INEQUALITIES IN THE USE
Sechin Mikhail Alexandrovich Small Academy of Sciences for students of the Republic of Kazakhstan "Seeker" MBOU "Soviet school №1", grade 11, town. Sovetsky Sovetsky district Gunko Lyudmila Dmitrievna, teacher of MBOU "Soviet school №1" Soviet district Purpose of work: investigation of the mechanism for solving logarithmic inequalities C3 using non-standard methods, revealing interesting facts of the logarithm. Subject of study:
3) Learn to solve specific logarithmic inequalities C3 using non-standard methods. Results:
Content
Introduction ………………………………………………………………………… .4
Chapter 1. Background ………………………………………………… ... 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. Trap Missions ………………………………………………… 27 Conclusion …………………………………………………………………… 30
Literature……………………………………………………………………. 31
Introduction
I am in 11th grade and am planning to enter a university, where the core subject is mathematics. Therefore, I work a lot with the problems of part C. In task C3, you need to solve a non-standard inequality or a system of inequalities, usually associated with logarithms. While preparing for the exam, I faced the problem of the lack of methods and techniques for solving the exam logarithmic inequalities offered in C3. The methods that are studied in the school curriculum on this topic do not provide a basis for solving tasks C3. The math teacher invited me to work with the C3 tasks on my own under her guidance. In addition, I was interested in the question: do logarithms occur in our life? With this in mind, the topic was chosen: "Logarithmic inequalities in the exam"
Purpose of work: investigation of the mechanism for solving C3 problems using non-standard methods, revealing interesting facts of the logarithm. Subject of study:
1) Find the necessary information about non-standard methods for solving logarithmic inequalities. 2) Find more information about logarithms. 3) Learn to solve specific C3 problems using non-standard methods. Results:
The practical significance lies in the expansion of the apparatus for solving C3 problems. This material can be used in some lessons, for circles, extracurricular activities in mathematics. The project product will be the collection “Logarithmic C3 inequalities with solutions”. Chapter 1. Background
During the 16th century, the number of approximate calculations increased rapidly, primarily in astronomy. Improving instruments, studying planetary movements and other work required colossal, sometimes many years, calculations. Astronomy was in real danger of drowning in unfulfilled calculations. Difficulties arose in other areas, for example, in the insurance business, tables of compound interest were needed for various values of interest. The main difficulty was represented by multiplication, division of multidigit numbers, especially trigonometric quantities. The discovery of logarithms was based on the well-known properties of progressions by the end of the 16th century. Archimedes spoke about the connection between the members of the geometric progression q, q2, q3, ... and the arithmetic progression of their exponents 1, 2, 3, ... in the Psalm. Another prerequisite was the extension of the concept of degree to negative and fractional indicators. Many authors have pointed out that multiplication, division, exponentiation and extraction of a root exponentially correspond in arithmetic - in the same order - addition, subtraction, multiplication and division. This was the idea behind the logarithm as an exponent. Several stages have passed in the history of the development of the doctrine of logarithms. 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 Burghi (1552-1632). Both wanted to give a new convenient means of arithmetic calculations, although they approached this task in different ways. Neper expressed kinematically the logarithmic function and thus entered a new field of function theory. Burghi remained on the basis of considering discrete progressions. However, the definition of the logarithm for both does not resemble the modern one. The term "logarithm" (logarithmus) belongs to Napier. It arose from a combination of 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), professor of mathematics at Gresch College in London, Napier proposed to take zero for the logarithm of unity, and 100 for the logarithm of ten, or, which amounts to the same thing, simply 1. This is how decimal logarithms appeared and the first logarithmic tables were printed. Later, the Dutch bookseller and mathematician Andrian Flakk (1600-1667) supplemented the Briggs tables. Napier and Briggs, although they came to logarithms earlier than anyone else, published their tables later than others - in 1620. The log and Log signs were introduced in 1624 by I. Kepler. The term "natural logarithm" was introduced by Mengoli in 1659, 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 title "New Logarithms". In Russian, the first logarithmic tables were published in 1703. But in all logarithmic tables, errors were made in the calculation. 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 analytic geometry and infinitesimal calculus. The establishment of a connection between the quadrature of an equilateral hyperbola and the natural logarithm dates back to that time. 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 the composition "Logarithmic engineering" (1668) gives a series that gives an expansion of ln (x + 1) in powers of x: This expression exactly corresponds to the course of his thought, although he, of course, did not use the signs d, ..., but more cumbersome symbols. 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 from the Highest Point of View", delivered in 1907-1908, F. Klein suggested using the formula as a starting point for constructing the theory of logarithms. Stage 3
Definition of a logarithmic function as a function of the inverse exponential, logarithm as an indicator of the degree of a given base was not immediately formulated. Writing by Leonard Euler (1707-1783) An Introduction to the Analysis of the Infinitesimal (1748) served as a further development of the theory of the logarithmic function. Thus, 134 years have passed since logarithms were first introduced (counting from 1614) before mathematicians came to the definition the concept of the 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 is the most versatile for solving inequalities of almost any type. The solution scheme looks like this: 1. Reduce the inequality to the form where the function is located on the left-hand side 2. Find the domain of the function 3. Find the zeros of the function 4. Draw the domain 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 spacing method where For these values, all expressions under the sign of the logarithms are positive. Answer:
Example 2.
Solution:
1st
way
.
ODZ is determined by the inequality x> 3. Taking the logarithm for such x base 10, we get The last inequality could be solved using the decomposition rules, i.e. comparing the factors to zero. However, in this case, it is easy to determine the intervals of constancy of the function therefore the spacing 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 define the intervals of constancy of the function f(x):
Answer: 2nd way
.
Let us apply the ideas of the method of intervals directly 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 for x> 3 is equivalent to the inequality or The last inequality is solved by the method of intervals Answer: Example 3.
Solution:
Let's apply the spacing method Answer: Example 4.
Solution:
Since 2 x 2 - 3x+ 3> 0 for all real x, then To solve the second inequality, we use the method of intervals In the first inequality, we make the replacement then we arrive at the inequality 2y 2 - y - 1 < 0 и, применив метод интервалов, получаем, что решениями будут те y that satisfy the inequality -0.5< y < 1.
Where, since we obtain the inequality which is carried out with those x for which 2 x 2 - 3x - 5 < 0. Вновь применим метод интервалов
Now, taking into account the solution of the second inequality of the system, we finally obtain Answer:
Example 5.
Solution:
Inequality is equivalent to a set of systems or Let's apply the method of intervals or Answer:
Example 6.
Solution:
Inequality is equivalent to the system Let be then y > 0,
and the first inequality system takes the form or by expanding square trinomial by factors, Applying the method of intervals 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, solutions to inequality are all 2.2. Method of rationalization. Previously, the method of rationalizing inequality was not solved, it was not known. This is "a new modern effective method for solving exponential and logarithmic inequalities" (quote from the book of 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 above reasoning is simple, but it considerably 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, 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 substitution y = 3 x -1; then this inequality takes the form Log 4 log 0.25 As log 0.25 We make the change 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 simplest inequalities Therefore, the original inequality is equivalent to a 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 is equivalent to the system The solution to the second inequality, which determines the DHS, will be the set of those x,
for whom x > 0.
To solve the first inequality, we make the substitution Then we obtain 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 Many of those x that 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. ODZ inequalities are 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 special methods for solving C3 problems from the 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 the ODZ. These methods are absent in the school curriculum. Using different methods, I solved 27 inequalities proposed in the exam in part C, namely C3. These inequalities with solutions by methods formed the basis of the collection "Logarithmic C3 inequalities with solutions", which became a project product of my work. The hypothesis that I posed at the beginning of the project was confirmed: the C3 tasks can be effectively solved, knowing these methods. In addition, I found interesting facts about logarithms. It was interesting for me to do it. My design products will be useful for both students and teachers. Conclusions:
Thus, the set goal of the project has been achieved, the problem has been resolved. And I got the most complete and versatile experience in project activities at all stages of work. In the course of 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, activity. A guarantee of success when creating a research project for I became: significant school experience, the ability to extract information from various sources, check its reliability, rank it by importance. In addition to direct subject knowledge in mathematics, he expanded his 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. In the course of the project activities, organizational, intellectual and communicative general educational skills and abilities were developed. Literature
1. Koryanov A. G., Prokofiev A. A. Systems of inequalities with one variable (typical tasks C3). 2. Malkova AG Preparation for the exam in mathematics. 3. Samarova SS Solution of logarithmic inequalities. 4. Mathematics. Collection of training works edited by A.L. Semyonov and I.V. Yashchenko. -M .: MTsNMO, 2009 .-- 72 p. -
An inequality is called logarithmic if it contains a logarithmic function. The methods for solving logarithmic inequalities are no different from, except for two things. First, when passing from a logarithmic inequality to an inequality of sub-logarithmic functions, it follows that watch the sign of the resulting inequality... He obeys the following rule. If the base of the logarithmic function is greater than $ 1 $, then when passing from the logarithmic inequality to the inequality of sub-logarithmic functions, the sign of the inequality is preserved, and if it is less than $ 1 $, then it changes to the opposite. Secondly, the solution of any inequality is an interval, and, therefore, at the end of the solution to the inequality of sub-logarithmic functions, it is necessary to compose a system of two inequalities: the first inequality of this system will be the inequality of sub-logarithmic functions, and the second is the interval of the domain of definition of logarithmic functions included in the logarithmic inequality. Let's solve the inequalities: 1.
$ \ log_ (2) ((x + 3)) \ geq 3. $ $ D (y): \ x + 3> 0. $ $ x \ in (-3; + \ infty) $ The base of the logarithm is $ 2> 1 $, so the sign does not change. Using the definition of the logarithm, we get: $ x + 3 \ geq 2 ^ (3), $ $ x \ in)
Solution examples
3x> 24;
x> 8. 
What is needed to solve logarithmic inequalities?
Homework
\ (\ log_3 ((x ^ 2-3))< \log_3{(2x)}\)
\ (\ log_ (x + 1) ((x ^ 2 + 3x-7))> 2 \)
\ (\ lg ^ 2 ((x + 1)) + 10≤11 \ lg ((x + 1)) \)How to solve logarithmic inequalities:
\ (- \) if is a number and it is greater than 1, the inequality sign remains the same during the transition,
\ (- \) if the base is a number greater than 0, but less than 1 (lies between zero and one), then the inequality sign should be reversed, i.e.
ODZ: \ (8-x> 0 \)
\ (- x> -8 \)
\ (x<8\)
\ (\ log \) \ (_ 2 \) \ ((8-x)<\log\)\(_2\)
\({2}\)
\ (8-x \) \ (<\)
\(2\)
\(8-2
Answer: \ ((6; 8) \)
ODZ: \ (\ begin (cases) 2x-4> 0 \\ x + 1> 0 \ end (cases) \)
\ (\ begin (cases) 2x> 4 \\ x> -1 \ end (cases) \) \ (\ Leftrightarrow \) \ (\ begin (cases) x> 2 \\ x> -1 \ end (cases) \) \ (\ Leftrightarrow \) \ (x \ in (2; \ infty) \)
\ (2x-4 \) \ (≤ \) \ (x + 1 \)
\ (2x-x≤4 + 1 \)
\ (x≤5 \)
Answer: \ ((2; 5] \)
Answer:
\ (x∈ (\) \ (\ frac (3) (2) \) \ (; \) \ (\ frac (7) (3)] \)
\ (x∈ (\) \ (\ frac (3) (2) \) \ (; \) \ (\ frac (7) (3)] \)
Answer:
\ ((0; \ frac (1) (3)) ∪ (9; ∞) \)
\ (t ^ 2-t-2> 0 \) Expand the left side of the inequality into.
\ (t_1 = \ frac (1 + 3) (2) = 2 \)
\ (t_2 = \ frac (1-3) (2) = - 1 \)
\ ((t + 1) (t-2)> 0 \)
if a> 1
if 0 <
а <
1
, and on the right 0..
, that is, to solve the equation 
(and solving an equation is usually easier than solving an inequality).at the intervals obtained.
And even if the teacher knew him, there was apprehension - does the examiner know him, and why is he not given at school? There were situations when the teacher said to the student: "Where did you get it? Sit down - 2."
Now the method is widely promoted. And for experts there are guidelines associated with this method, and in the "Most complete editions of the standard options ..." in the solution C3 this method is used.
WONDERFUL METHOD!
Answer... (0; 0.5) U.
Answer :
(3;6)
.
= -log 4 = - (log 4 y -log 4 16) = 2-log 4 y, then 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, the aggregates 
... Thus, the original inequality holds for all values of x from the intervals 0<х≤1 и 2≤х<+.
.
... Therefore, all x from the interval 0 Practice.