Preparing for the exam. Solution of logarithmic and exponential inequalities by rationalization method. Manov's work "logarithmic inequalities in the exam" Solution of logarithmic inequalities exam profile level
Sections: Maths
Often, when deciding logarithmic inequalities, there are problems with a variable base of the logarithm. So, an inequality of the form
is a standard school inequality. As a rule, to solve it, a transition to an equivalent set of systems is applied:
The disadvantage of this method is the need to solve seven inequalities, not counting two systems and one set. Already with given quadratic functions, solving a set can be time-consuming.
An alternative, less laborious way of solving this standard inequality can be proposed. For this, we take into account the following theorem.
Theorem 1. Let a continuous increasing function on the set X. Then on this set the sign of the increment of the function will coincide with the sign of the increment of the argument, that is, where 
.
Note: if a continuous decreasing function on the set X, then.
Let's go back to inequality. Let's go to the decimal logarithm (you can go to any with a constant base greater than one).
Now you can use the theorem, noting in the numerator the increment of the functions 
and in the denominator. So it is true
As a result, the number of calculations leading to the answer is approximately halved, which not only saves time, but also allows you to potentially make fewer arithmetic and "inattention" errors.
Example 1.
Comparing with (1) we find 
, 
, .
Passing to (2) we will have:
Example 2.
Comparing with (1) we find,,.
Passing to (2) we will have:
Example 3.
Since the left side of the inequality is an increasing function for and 
, then the answer is set.
The set of examples in which Theorem 1 can be applied can be easily extended if Theorem 2 is taken into account.
Let on the set X functions,,, and on this set the signs and coincide, i.e. , then it will be fair.
Example 4.
Example 5.
With the standard approach, the example is solved according to the scheme: the product is less than zero, when the factors are of different signs. Those. the set of two systems of inequalities is considered, in which, as was indicated at the beginning, each inequality splits into seven more.
If we take into account Theorem 2, then each of the factors, taking into account (2), can be replaced by another function that has the same sign in this example O.D.Z.
The method of replacing the increment of a function with an increment of the argument, taking into account Theorem 2, turns out to be very convenient when solving typical problems C3 of the exam.
Example 6.
Example 7.
... Let us denote. We get
... Note that the replacement implies:. Returning to the equation, we get
.
Example 8.
In the theorems we use, there is no restriction on the classes of functions. In this article, for example, the theorems have been applied to the solution of logarithmic inequalities. The next few examples will demonstrate the promise of the method for solving other types of inequalities.
LOGARITHMIC INEQUALITIES IN THE USE
Sechin Mikhail Alexandrovich
Small Academy of Sciences for students of the Republic of Kazakhstan "Seeker"
MBOU "Sovetskaya secondary school №1", grade 11, town. Soviet Sovetsky District
Gunko Lyudmila Dmitrievna, teacher of MBOU "Soviet school №1"
Soviet district
Objective: investigation of the mechanism for solving logarithmic inequalities C3 using non-standard methods, identifying interesting facts 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 plan to enter a university, where profile subject is math. 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. Methods learned in school curriculum on this topic, do not provide a basis for solving C3 tasks. 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: are there logarithms in our life?
With this in mind, the topic was chosen:
"Logarithmic inequalities in the exam"
Objective: 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
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 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 \u200b\u200bof 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 comes down to the same thing, simply 1. This is how decimal logarithms appeared and the first logarithmic tables were printed. Later, Briggs tables were supplemented by the Dutch bookseller and lover of mathematics Andrian Flakk (1600-1667). 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".
The first logarithmic tables in Russian were published in 1703. But in all the logarithmic tables, errors were made in the calculation. The first error-free tables were published in Berlin in 1857, edited 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 technique" (1668) gives a series that gives the expansion of ln (x + 1) in
powers of x:
This expression exactly corresponds to the course of his thought, although, of course, he 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", read in 1907-1908, F. Klein proposed to use the formula as a starting point for the construction of the theory of logarithms.
Stage 3
Definition logarithmic function as a function of the inverse
exponential, logarithm as an indicator of the degree of a given base
was not immediately formulated. Composition 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 generalized interval method.
Equivalent transitions
if a\u003e 1
if 0 <
а <
1
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 
, and on the right 0.
2. Find the domain of the function .
3. Find the zeros of the function , that is, to solve the equation 
(and solving an equation is usually easier than solving an inequality).
4. Draw the domain and zeros of the function on the number line.
5. Determine the signs of the function at the intervals obtained.
6. Select intervals where the function takes the required values \u200b\u200band write down the answer.
Example 1.
Decision:
Let's apply the spacing method
from where
For these values, all expressions under the sign of the logarithms are positive.
Answer:
Example 2.
Decision:
1st way . ODZ is determined by the inequality x \u003e 3. Taking the logarithm for such x base 10, we get
The last inequality could be solved by applying 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, you can apply the method of intervals.
Function f(x) = 2x(x- 3,5) lgǀ x- 3ǀ is continuous at x \u003e 3 and vanishes at the points x 1 = 0, x 2 = 3,5, x 3 = 2, x 4 \u003d 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 \u003e 3 is equivalent to the inequality
or
The last inequality is solved by the method of intervals
Answer:
Example 3.
Decision:
Let's apply the spacing method
Answer:
Example 4.
Decision:
Since 2 x 2 - 3x + 3\u003e 0 for all real xthen
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 ythat satisfy the inequality -0.5< y < 1.
Where from, since
we obtain the inequality
which is carried out with those xfor 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.
Decision:
Inequality is equivalent to a set of systems
or
Let's apply the method of intervals or
Answer:
Example 6.
Decision:
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 \u003e 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)
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 model variants ..." in the solution C3 this method is used.
WONDERFUL METHOD!
"Magic table"
In other sources
if a a\u003e 1 and b\u003e 1, then log a b\u003e 0 and (a -1) (b -1)\u003e 0;
if a a\u003e 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)\u003e 0. The above reasoning is simple, but it considerably simplifies the solution of logarithmic inequalities. Example 4.
log x (x 2 -3)<0
Decision:
Example 5.
log 2 x (2x 2 -4x +6) ≤log 2 x (x 2 + x) Decision: 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. Nonstandard 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 \u003d 3 x -1; then this inequality takes the form Log 4 log 0.25 As log 0.25 We make the change t \u003d log 4 y and obtain the inequality t 2 -2t + ≥0, the solution of which is the intervals - Thus, to find the values \u200b\u200bof 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.
Decision:
Inequality is equivalent to the system The solution to the second inequality, which determines the DHS, is the set of those x,
for whom x > 0.
To solve the first inequality, we make the change 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 xthat satisfy the last inequality belongs to ODZ ( x \u003e 0), therefore, is a solution to the system and hence the original inequality. Answer: 2.4. Trap quests. Example 1.
Decision. ODZ inequalities are all x satisfying the condition 0 Example 2.
log 2 (2 x + 1-x 2)\u003e log 2 (2 x-1 + 1-x) +1.
Answer... (0; 0.5) U.
Answer :
(3;6)
.
\u003d -log 4 \u003d - (log 4 y -log 4 16) \u003d 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 totality 
... Thus, the original inequality holds for all values \u200b\u200bof x from the intervals 0<х≤1 и 2≤х<+.
.
... Therefore, all x from the interval 0
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 the 27 inequalities proposed on 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 solved. And I got the most complete and versatile experience in project activities at all stages of work. In the course of work 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. -
The article is devoted to the analysis of 15 tasks from the profile USE in mathematics for 2017. In this task, students are offered to solve inequalities, most often logarithmic ones. Although there may be indicative. This article provides an analysis of examples of logarithmic inequalities, including those containing a variable at the base of the logarithm. All examples are taken from the open bank of USE tasks in mathematics (profile), so such inequalities are likely to come across you on the exam as task 15. Ideal for those who want to learn how to solve task 15 from the second part of the profile USE in a short period of time in math to get more points on the exam.
Analysis of 15 tasks from the profile exam in mathematics
| Example 1. Solve the inequality: |
In the tasks of the 15th exam in mathematics (profile), logarithmic inequalities are often encountered. Solving logarithmic inequalities begins with determining the range of acceptable values. In this case, there is no variable at the base of both logarithms, there is only the number 11, which greatly simplifies the task. Therefore, the only limitation we have here is that both expressions under the sign of the logarithm are positive:
Title \u003d "(! LANG: Rendered by QuickLaTeX.com">!}
The first inequality in the system is the square inequality. To solve it, we would really not hurt to factor the left side into factors. I think you know that any square trinomial of the form 
is factorized as follows:
where and are the roots of the equation. In this case, the coefficient is 1 (this is the numeric coefficient in front of). The coefficient is also 1, and the coefficient is an intercept, it is -20. The roots of a trinomial are most easily determined by Vieta's theorem. The equation we have given, then the sum of the roots will be equal to the coefficient with the opposite sign, that is, -1, and the product of these roots will be equal to the coefficient, that is, -20. It is easy to guess that the roots will be -5 and 4.
Now the left side of the inequality can be factorized: title \u003d "(! LANG: Rendered by QuickLaTeX.com" height="20" width="163" style="vertical-align: -5px;"> Решаем это неравенство. График соответствующей функции — это парабола, ветви которой направлены вверх. Эта парабола пересекает ось !} X at points -5 and 4. Hence, the desired solution to the inequality is an interval. For those who do not understand what is written here, you can see the details in the video, starting from this moment. There you will also find a detailed explanation of how the second inequality of the system is solved. It is being solved. Moreover, the answer is exactly the same as for the first inequality of the system. That is, the set written above is the region of admissible values \u200b\u200bof inequality.
So, taking into account the factorization, the original inequality takes the form:
Using the formula, we bring 11 to the power of the expression under the sign of the first logarithm, and move the second logarithm to the left side of the inequality, while changing its sign to the opposite:
After reduction we get:
The last inequality, due to the increasing function, is equivalent to the inequality 
, the solution of which is the interval 
... It remains to intersect it with the range of admissible values \u200b\u200bof inequality, and this will be the answer to the entire task.
So, the desired answer to the task is:
We figured out this task, now we turn to the next example of the 15 USE task in mathematics (profile).
| Example 2. Solve the inequality:
|
We begin the solution by determining the range of admissible values \u200b\u200bof this inequality. At the base of each logarithm must be a positive number that is not equal to 1. All expressions under the sign of the logarithm must be positive. There should be no zero in the denominator of the fraction. The last condition is equivalent to that, since only otherwise both logarithms in the denominator vanish. All these conditions determine the range of admissible values \u200b\u200bof this inequality, which is defined by the following system of inequalities:
Title \u003d "(! LANG: Rendered by QuickLaTeX.com">!}
In the range of valid values, we can use the logarithm transformation formulas to simplify the left side of the inequality. Using the formula 
get rid of the denominator:
Now we have only base logarithms. This is already more convenient. Next, we use the formula, and also the formula in order to bring the expression worth glory to the following form:
In the calculations, we used what is in the range of acceptable values. Using the replacement, we arrive at the expression:
We use one more replacement:. As a result, we come to the following result:
So, we gradually return to the original variables. First to the variable:
