15 exam profile how to solve with logarithms. Manov's work "logarithmic inequalities in the exam"

“SOLUTION OF LOGARITHMIC INEQUALITIES (TASK №15 PROFILE USE). APPLICATION OF LOGARITHMS IN DIFFERENT SPHERES OF HUMAN LIFE "

The epigraph of the lesson will be the words of Maurice Kline “Music can elevate or pacify the soul, painting can please the eye, poetry can awaken feelings, philosophy can satisfy the needs of the mind, engineering can improve the material side of people's lives, andmathematics is capable of achieving all of these goals »

Now let's create the mood of success!

We will answer the following questions:

Verification practice examination papers, and I am a USE expert in mathematics since 2005 shows that the greatest difficulty for schoolchildren is the solution of transcendental inequalities, especially logarithmic inequalities with a variable base.

Therefore, I propose to consider, firstly, the rationalization method (the Modenov decomposition method), or otherwise called, the method of replacing the Golubev multipliers, which allows us to reduce complex, in particular, logarithmic inequalities, to a system of simpler rational inequalities.

So, for example, when solving the inequality
in the evaluative version, proposed to the examiners of the exam, the following solution was given:

I suggest using the rationalization method:

Solving the first inequality by the method of intervals and taking into account that we get

Solution of the following inequality

I saw it like this:

And I explained to the students that sometimes a graphical solution is simpler.

And as a result, the solution to this inequality has the form:

Consider the inequality

Solving this inequality, one can use the formula

but to go to the base is a number, and absolutely any:

and solve the resulting inequality by the method of intervals:

ODZ:

and solve the resulting inequality by the method of intervals

and taking into account the ODZ we get:

And, solving the next type of inequality, students, when writing down the answer, usually lose one of the solutions. You should definitely pay attention to this.

Let's find the ODZ:

and perform the replacement: we get:

I draw your attention to the fact that often students solving this, the resulting inequality, discard the denominator, thereby losing one of the solutions:

Taking into account ODZ we get: and

And at the end of the lesson, I offer students interesting facts about the application of logarithms in various fields.

Wherever there are processes that change over time, logarithms are used.

Logarithms are a mathematical concept that is used in all branches of science: chemistry, biology, physics, geography, computer science and many others, but the widest application of logarithms is found in economics.

The article is devoted to the analysis of tasks 15 of profile exam 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 open bank tasks of the USE in mathematics (profile), so that such inequalities are likely to come across you on the exam as task 15. Ideal for those who, in a short period of time, want to learn how to solve task 15 from the second part of the profile USE in mathematics in order to get more points on the exam.

Analysis of 15 tasks from the profile exam in mathematics

In the tasks of the 15 USE in mathematics (profile), logarithmic inequalities are often encountered. The solution of logarithmic inequalities begins with the definition of the area 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:

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The first inequality in the system is the square inequality. To solve it, we would really not hurt to decompose left side by factors. I think you know that anyone square trinomial of the kind 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, so 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 = "(! 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 range of admissible values ​​of 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 increase of the function, is equivalent to the inequality , the solution of which is the interval ... It remains to intersect it with the range of permissible values ​​of 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).

We begin the solution by determining the range of admissible values ​​of this inequality. At the base of each logarithm there should be positive number, which 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 ​​of this inequality, which is defined by the following system of inequalities:

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In the range of valid values, we can use the transformation formulas for the logarithms in order 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, as well as 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:

LOGARITHMIC INEQUALITIES IN THE USE

Sechin Mikhail Alexandrovich

Small Academy of Sciences of student youth of the Republic of Kazakhstan "Seeker"

MBOU "Sovetskaya secondary school No. 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, 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 that are 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: 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 necessary information on non-standard methods for solving logarithmic inequalities.

2) Find more information about logarithms.

3) Learn to solve specific tasks C3 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 conducting circles, extracurricular activities 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. The improvement of instruments, the study of 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 different meanings percent. The main difficulty was multiplication, division multi-digit numbers, especially trigonometric values.

The discovery of logarithms was based on the well-known properties of progressions by the end of the 16th century. About communication between members geometric progression q, q2, q3, ... and arithmetic progression their indicators 1, 2, 3, ... said in the "Psalm" Archimedes. Another prerequisite was the extension of the concept of degree to negative and fractional indicators. Many authors have pointed out that multiplication, division, raising to a power 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 problem in different ways. Napier kinematically expressed the logarithmic function and thus entered into new area 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 one, 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, 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 the calculus of infinitesimal. 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

"Logarithmology" (1668) gives a series that gives the expansion of ln (x + 1) in

powers of x:

This expression exactly corresponds to the line 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

if a> 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 is located on the left-hand side
, 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, 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 defined by the inequality x> 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 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 "new modern effective method solutions of exponential and logarithmic inequalities "(quote from the book of S. I. Kolesnikova)
And even if the teacher knew him, there was apprehension - but does he know 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."
The method is now widely promoted. And for experts there is guidelines related to this method, and in "The most complete editions standard options... "solution C3 uses this method.
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;