Preparation for the Unified State Exam. Solving logarithmic and exponential inequalities using the rationalization method. Manovskaya work "logarithmic inequalities in the Unified State Exam" Unified State Examination profile solution of logarithmic inequalities

Sections: Mathematics

Often, when deciding logarithmic inequalities, there are problems with a variable logarithm base. Thus, 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 used:

The disadvantage of this method is the need to solve seven inequalities, not counting two systems and one population. Already with these quadratic functions, solving the population can take a lot of time.

It is possible to propose an alternative, less time-consuming way to solve this standard inequality. To do this, we take into account the following theorem.

Theorem 1. Let there be a continuous increasing function on a 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, i.e. , Where .

Note: if a continuous decreasing function on a set X, then .

Let's return to inequality. Let's move on to the decimal logarithm (you can move on to any with a constant base greater than one).

Now you can use the theorem, noticing the increment of functions in the numerator and in the denominator. So it's true

As a result, the number of calculations leading to the answer is reduced by approximately half, which saves not only time, but also allows you to potentially make fewer arithmetic and careless errors.

Example 1.

Comparing with (1) we find , , .

Moving on to (2) we will have:

Example 2.

Comparing with (1) we find , , .

Moving on to (2) we will have:

Example 3.

Since the left side of the inequality is an increasing function as and , then the answer will be many.

The many examples in which Theme 1 can be applied can easily be expanded by taking into account Theme 2.

Let on the set X the functions , , , are defined, 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 following scheme: the product is less than zero when the factors are of different signs. Those. a set of two systems of inequalities is considered, in which, as indicated at the beginning, each inequality breaks down 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 argument, taking into account Theorem 2, turns out to be very convenient when solving standard C3 Unified State Examination problems.

Example 6.

Example 7.

. Let's denote . We get

. Note that the replacement implies: . Returning to the equation, we get .

Example 8.

In the theorems we use there are no restrictions on classes of functions. In this article, as an example, the theorems were applied to solving logarithmic inequalities. The following several 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 “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

Goal of the work: study 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. 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. The methods that are studied in the 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”

Goal of the work: study of the mechanism for solving C3 problems using non-standard methods, identifying interesting facts about the logarithm.

Subject of study:

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:

The practical significance lies in the expansion of 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 arose in other areas, for example, in the insurance business, compound interest tables were needed for various interest rates. 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. On the connection between the terms of the 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 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), a professor of mathematics at Gresh College in London, Napier suggested taking zero as the logarithm of one, and 100 as the logarithm of ten, or, what amounts to the same thing, just 1. This is how 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 quadrature of an equilateral hyperbola and the natural logarithm had been established. 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 from a Higher Point of View,” given in 1907-1908, F. Klein proposed using the formula as the starting point for constructing the theory of logarithms.

Stage 3

Definition of a 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

, if a > 1

, if 0 < а < 1

Generalized interval method

This method is the 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
, and on the right 0.

2. Find the domain of the function
.

3. Find the zeros of the function
, that is, solve the equation
(and solving an equation is usually easier than solving an inequality).

4. Draw the domain of definition and zeros of the function on the number line.

5. Determine the signs of the function
on the obtained intervals.

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 factored,

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)
And even if the teacher knew him, there was a fear - does the Unified State Exam expert know him, and why don’t they give him 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 are guidelines associated with this method, and in the “Most Complete Editions of Standard Options...” in Solution C3 this method is used.
WONDERFUL METHOD!

"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:

Answer. (0; 0.5)U.

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).

Answer : (3;6)

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 = -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 ≤.

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
The solution to this set is the intervals 0<у≤2 и 8≤у<+.

Therefore, the original inequality is equivalent to the set of two exponential inequalities,
that is, aggregates

The solution to the first inequality of this set is the interval 0<х≤1, решением второго – промежуток 2≤х<+. Thus, the original inequality is satisfied for all values ​​of x from the intervals 0<х≤1 и 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 . Therefore, all x are from the interval 0

Example 2.

log 2 (2 x +1-x 2)>log 2 (2 x-1 +1-x)+1.. ? The point is that the second number is obviously greater than

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.-

The article is devoted to the analysis of tasks 15 from the profile Unified State Examination in mathematics for 2017. In this task, schoolchildren are asked to solve inequalities, most often logarithmic ones. Although there may be indicative ones. This article provides an analysis of examples of logarithmic inequalities, including those containing a variable in the base of the logarithm. All examples are taken from the open bank of Unified State Examination tasks in mathematics (profile), so such inequalities are likely to come across in the exam as task 15. Ideal for those who want to learn how to solve task 15 from the second part of the profile Unified State Exam in a short period of time in mathematics to get more marks in the exam.

Analysis of tasks 15 from the profile Unified State Examination in mathematics

Example 1. Solve the inequality:

In tasks 15 of the Unified State 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 in the base of both logarithms, there is only the number 11, which greatly simplifies the problem. So the only limitation we have here is that both expressions under the logarithm sign are positive:

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The first inequality in the system is the quadratic inequality. To solve it, we would really like to factorize the left-hand side. I think you know that any quadratic 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 numerical coefficient in front of ). The coefficient is also equal to 1, and the coefficient is the dummy term, it is equal to -20. The roots of a trinomial are most easily determined using Vieta's theorem. The equation we have given means 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's easy to guess that the roots will be -5 and 4.

Now the left side of the inequality can be factorized: title="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. This means that the required solution to the inequality is the interval . For those who do not understand what is written here, you can watch 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 resolved. 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 permissible values ​​of the inequality.

So, taking into account factorization, the original inequality takes the form:

Using the formula, we add 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, changing its sign to the opposite:

After reduction we get:

The last inequality, due to the increase of the function, is equivalent to the inequality , whose solution is the interval . All that remains is to intersect it with the region of acceptable values ​​of the inequality, and this will be the answer to the entire task.

So, the required answer to the task looks like:

We have dealt with this task, now we move on to the next example of task 15 of the Unified State Exam in mathematics (profile).

Example 2. Solve the inequality:

We begin the solution by determining the range of acceptable values ​​of this inequality. At the base of each logarithm there must be a positive number that is not equal to 1. All expressions under the sign of the logarithm must be positive. The denominator of the fraction must not contain zero. The last condition is equivalent to the fact that , since only otherwise both logarithms in the denominator vanish. All these conditions determine the range of permissible values ​​of this inequality, given by the following system of inequalities:

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In the range of acceptable values, we can use logarithm conversion formulas to simplify the left side of the inequality. Using formula we get rid of the denominator:

Now we only have logarithms with a base. 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 was in the range of acceptable values. Using the substitution we arrive at the expression:

Let's use one more replacement: . As a result, we arrive at the following result:

So, we gradually return to the original variables. First to the variable: