Lesson objectives:
Didactic:
- Level 1 - to teach how to solve the simplest logarithmic inequalities using the definition of the logarithm, the properties of logarithms;
- Level 2 - solve logarithmic inequalities by choosing a solution method on your own;
- Level 3 - be able to apply knowledge and skills in non-standard situations.
Developing: develop memory, attention, logical thinking, comparison skills, be able to generalize and draw conclusions
Educational: to bring up accuracy, responsibility for the performed task, mutual assistance.
Teaching methods: verbal , pictorial , practical , partial search , self-government , control.
Forms of organizing the cognitive activity of students: frontal , individual , work in pairs.
Equipment: a set of test items, background notes, blank sheets for solutions.
Lesson type: learning new material.
During the classes
1. Organizational moment. The topic and goals of the lesson, the scheme of the lesson are announced: each student is given an assessment sheet, which the student fills out during the lesson; for each pair of students - printed materials with assignments, assignments must be completed in pairs; blank sheets for solutions; support sheets: definition of the logarithm; graph of a logarithmic function, its properties; properties of logarithms; algorithm for solving logarithmic inequalities.
All decisions after self-assessment are submitted to the teacher.
Student grade sheet
2. Updating knowledge.
Teacher instructions. Remember the definition of a logarithm, the graph of a logarithmic function and its properties. To do this, read the text on pp. 88–90, 98–101 of the textbook “Algebra and the beginnings of analysis 10–11” edited by Sh.A Alimov, Yu.M. Kolyagin and others.
Pupils are given sheets on which are written: the definition of the logarithm; shows a graph of a logarithmic function, its properties; properties of logarithms; algorithm for solving logarithmic inequalities, an example of solving a logarithmic inequality that reduces to a square one.
3. Learning new material.
The solution to logarithmic inequalities is based on the monotonicity of the logarithmic function.
Algorithm for solving logarithmic inequalities:
A) Find the domain of inequality (sub-logarithmic expression is greater than zero).
B) Present (if possible) the left and right sides of the inequality in the form of logarithms on the same base.
C) Determine whether the logarithmic function is increasing or decreasing: if t> 1, then it is increasing; if 0
D) Go to a simpler inequality (sub-logarithmic expressions), taking into account that the inequality sign will remain if the function increases, and will change if it decreases.
Learning element # 1.
Purpose: to fix the solution of the simplest logarithmic inequalities
The form of organizing the cognitive activity of students: individual work.
Self-study assignments for 10 minutes. For each inequality, there are several answer options, you need to choose the correct one and check by key.
KEY: 13321, maximum number of points - 6 pts.
Learning element # 2.
Purpose: to consolidate the solution of logarithmic inequalities, applying the properties of logarithms.
Teacher instructions. Remember the basic properties of logarithms. To do this, read the text of the textbook on pages 92, 103–104.
Self-study assignments for 10 minutes.
KEY: 2113, maximum number of points - 8 pts.
Learning element # 3.
Purpose: to study the solution of logarithmic inequalities by the method of reduction to the square.
Teacher's instructions: the method of reducing inequality to square is that it is necessary to transform the inequality to such a form that some logarithmic function is designated by a new variable, thus obtaining a square inequality with respect to this variable.
Let's apply the method of intervals.
You have passed the first level of assimilation of the material. Now you will have to independently choose a method for solving logarithmic equations, using all your knowledge and capabilities.
Learning element # 4.
Purpose: to consolidate the solution of logarithmic inequalities, choosing a rational way of solving independently.
Self-study assignments for 10 minutes
Learning element # 5.
Teacher instructions. Well done! You have mastered solving equations of the second level of difficulty. The purpose of your further work is to apply your knowledge and skills in more complex and non-standard situations.
Self-help tasks:
Teacher instructions. It is great if you have coped with the whole task. Well done!
The grade for the entire lesson depends on the number of points scored for all educational elements:
- if N ≥ 20, then you get the grade “5”,
- at 16 ≤ N ≤ 19 - rating “4”,
- at 8 ≤ N ≤ 15 - grade “3”,
- at N< 8 выполнить работу над ошибками к следующему уроку (решения можно взять у учителя).
Pass the assessment foxes to the teacher.
5. Homework: if you scored no more than 15 p - complete the work on the mistakes (solutions can be taken from the teacher), if you scored more than 15 p - complete the creative task on the topic “Logarithmic inequalities”.
They are inside the logarithms.
Examples:
\ (\ log_3x≥ \ log_39 \)
\ (\ 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:
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:
\ (- \) 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 sign of the inequality must be reversed, i.e.
|
\ (\ log_2 ((8-x))<1\) Solution: |
\ (\ log \) \ (_ (0,5) \) \ ((2x-4) \) ≥ \ (\ log \) \ (_ (0,5) \) \ (((x + 1))\) 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:
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\ (\ log \) \ (_ (\ frac (1) (3)) (\ frac (3x-2) (2x-3)) \)\(≤-1\) |
Let's write out the ODZ. |
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ODZ: \ (\ frac (3x-2) (2x-3) \) \ (> 0 \) |
|
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\ ( \ frac (3x-2-3 (2x-3)) (2x-3) \)\(≥\) \(0\) |
We open the brackets, we give. |
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\ ( \ frac (-3x + 7) (2x-3) \) \ (≥ \) \ (0 \) |
We multiply the inequality by \ (- 1 \), not forgetting to reverse the comparison sign. |
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\ ( \ frac (3x-7) (2x-3) \) \ (≤ \) \ (0 \) |
|
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\ ( \ frac (3 (x- \ frac (7) (3))) (2 (x- \ frac (3) (2))) \)\(≤\) \(0\) |
Let's construct a numerical axis and mark the points \ (\ frac (7) (3) \) and \ (\ frac (3) (2) \) on it. 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 inequality, it will lead us to division by zero. |
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Now, on the same numerical axis, we plot the ODZ and write in response the interval that falls into the ODZ. |
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We write down the final answer. |
Example ... Solve the inequality: \ (\ log ^ 2_3x- \ log_3x-2> 0 \)
Solution:
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\ (\ log ^ 2_3x- \ log_3x-2> 0 \) |
Let's write out the ODZ. |
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ODZ: \ (x> 0 \) |
Let's get down to the solution. |
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Solution: \ (\ log ^ 2_3x- \ log_3x-2> 0 \) |
We have before us a typical square-logarithmic inequality. We do it. |
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\ (t = \ log_3x \) |
Expand the left side of the inequality into. |
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\ (D = 1 + 8 = 9 \) |
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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. |
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\ (\ 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) \). |
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\ (\ 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. |
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\ (\ 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. |
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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 mathematics is a specialized subject. 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 was faced with 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 as well, 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, ... 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.
Here was the idea of 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 kinematically expressed the logarithmic function and thus entered a new area of the theory of functions. 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
"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, 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", 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 
, 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 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 method of spacing 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)
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!
"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 Because 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 the collection 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.
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
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. -
Among all the variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely told at school:
log k (x) f (x) ∨ log k (x) g (x) ⇒ (f (x) - g (x)) (k (x) - 1) ∨ 0
Instead of the "∨" checkbox, you can put any inequality sign: more or less. The main thing is that in both inequalities the signs are the same.
So we get rid of logarithms and reduce the problem to rational inequality. The latter is much easier to solve, but when dropping logarithms, unnecessary roots may appear. To cut them off, it is enough to find the range of acceptable values. If you have forgotten the ODZ of the logarithm, I strongly recommend repeating it - see "What is a logarithm".
Everything related to the range of permissible values must be written out and solved separately:
f (x)> 0; g (x)> 0; k (x)> 0; k (x) ≠ 1.
These four inequalities constitute a system and must be fulfilled simultaneously. When the range of acceptable values is found, it remains to cross it with the solution of rational inequality - and the answer is ready.
Task. Solve the inequality:
To begin with, let's write out the ODZ of the logarithm:
The first two inequalities are fulfilled automatically, and the last one will have to be described. Since the square of a number is zero if and only if the number itself is zero, we have:
x 2 + 1 ≠ 1;
x 2 ≠ 0;
x ≠ 0.
It turns out that the ODZ of the logarithm is all numbers except zero: x ∈ (−∞ 0) ∪ (0; + ∞). Now we solve the main inequality:
We carry out the transition from a logarithmic inequality to a rational one. In the original inequality there is a “less” sign, which means that the resulting inequality must also be with a “less” sign. We have:
(10 - (x 2 + 1)) (x 2 + 1 - 1)< 0;
(9 - x 2) x 2< 0;
(3 - x) (3 + x) x 2< 0.
The zeros of this expression: x = 3; x = −3; x = 0. Moreover, x = 0 is a root of the second multiplicity, which means that when passing through it, the sign of the function does not change. We have:
We get x ∈ (−∞ −3) ∪ (3; + ∞). This set is completely contained in the ODZ of the logarithm, which means that this is the answer.
Transforming Logarithmic Inequalities
Often the original inequality differs from the one above. It is easy to fix it according to the standard rules for working with logarithms - see "Basic properties of logarithms". Namely:
- Any number can be represented as a logarithm with a given base;
- The sum and difference of logarithms with the same bases can be replaced with one logarithm.
I would also like to remind you about the range of acceptable values. Since the original inequality may contain several logarithms, it is required to find the ODV for each of them. Thus, the general scheme for solving logarithmic inequalities is as follows:
- Find the ODV of each logarithm included in the inequality;
- Reduce inequality to the standard one according to the formulas for addition and subtraction of logarithms;
- Solve the resulting inequality according to the scheme given above.
Task. Solve the inequality:
Let's find the domain of definition (ODZ) of the first logarithm:
We solve by the method of intervals. Find the zeros of the numerator:
3x - 2 = 0;
x = 2/3.
Then the zeros of the denominator:
x - 1 = 0;
x = 1.
We mark the zeros and signs on the coordinate arrow:
We get x ∈ (−∞ 2/3) ∪ (1; + ∞). The second logarithm of ODV will be the same. If you don't believe it, you can check it out. Now we transform the second logarithm so that there is a two at the base:
As you can see, the triplets at the base and in front of the logarithm have contracted. Received two logarithms with the same base. We add them:
log 2 (x - 1) 2< 2;
log 2 (x - 1) 2< log 2 2 2 .
Received the standard logarithmic inequality. We get rid of the logarithms by the formula. Since the original inequality contains a less than sign, the resulting rational expression must also be less than zero. We have:
(f (x) - g (x)) (k (x) - 1)< 0;
((x - 1) 2 - 2 2) (2 - 1)< 0;
x 2 - 2x + 1 - 4< 0;
x 2 - 2x - 3< 0;
(x - 3) (x + 1)< 0;
x ∈ (−1; 3).
We got two sets:
- ODZ: x ∈ (−∞ 2/3) ∪ (1; + ∞);
- Candidate answer: x ∈ (−1; 3).
It remains to cross these sets - we get the real answer:
We are interested in the intersection of sets, so we select the intervals filled in on both arrows. We get x ∈ (−1; 2/3) ∪ (1; 3) - all points are punctured.
Solving logarithmic inequalities, we use the property of monotonicity of the logarithmic function. We also use the definition of a logarithm and basic logarithmic formulas.
Let's recap what logarithms are:
Logarithm a positive base number is an indicator of the degree to which you need to raise in order to get.
Wherein
Basic logarithmic identity:
Basic formulas for logarithms:
(The logarithm of the product is equal to the sum of the logarithms)
(The logarithm of the quotient is equal to the difference of the logarithms)
(Formula for logarithm of power)
The formula for the transition to a new base:
Algorithm for solving logarithmic inequalities
We can say that logarithmic inequalities are solved according to a certain algorithm. We need to write down the range of acceptable values (ADV) of the inequality. Reduce the inequality to the form The sign here can be any: It is important that the left and right in the inequality are the logarithms on the same base.
And after that we "discard" the logarithms! Moreover, if the base is the degree, the inequality sign remains the same. If the basis is such that the sign of inequality is reversed.
Of course, we don't just “drop” the logarithms. We use the property of monotonicity of the logarithmic function. If the base of the logarithm is greater than one, the logarithmic function increases monotonically, and then a larger value of x corresponds to a larger value of the expression.
If the base is greater than zero and less than one, the logarithmic function decreases monotonically. A larger value of the argument x will correspond to a smaller value
Important note: it is best to write the solution as a chain of equivalent transitions.
Let's move on to practice. As always, let's start with the simplest inequalities.
1. Consider the inequality log 3 x> log 3 5.
Since logarithms are only defined for positive numbers, x must be positive. The condition x> 0 is called the range of admissible values (ADV) of this inequality. Only for such x does the inequality make sense.
Well, this wording sounds dashing and easy to remember. But why can we do it anyway?
We are human beings, we have intelligence. Our mind is designed in such a way that everything that is logical, understandable, and has an internal structure is remembered and applied much better than random and unrelated facts. That is why it is important not to mechanically memorize the rules, like a trained mathematician dog, but to act consciously.
So why are we “dropping logarithms” anyway?
The answer is simple: if the base is greater than one (as in our case), the logarithmic function increases monotonically, which means that a larger value of x corresponds to a larger value of y, and from the inequality log 3 x 1> log 3 x 2 it follows that x 1> x 2. 
Please note that we have passed to an algebraic inequality, and the inequality sign remains the same.
So x> 5.
The next logarithmic inequality is also simple.
2.log 5 (15 + 3x)> log 5 2x
Let's start with the range of valid values. Logarithms are only defined for positive numbers, so
Solving this system, we get: x> 0.
Now we will pass from the logarithmic inequality to the algebraic one - we will "discard" the logarithms. Since the base of the logarithm is greater than one, the sign of the inequality is preserved.
15 + 3x> 2x.
We get: x> −15.
Answer: x> 0.
What happens if the base of the logarithm is less than one? It is easy to guess that in this case, when passing to an algebraic inequality, the sign of the inequality will change.
Let's give an example.
Let's write down the ODZ. The expressions from which the logarithms are taken must be positive, that is,
Solving this system, we get: x> 4.5.
Since, the logarithmic function with the base decreases monotonically. This means that a larger value of the function corresponds to a smaller value of the argument: 
And if, then
2x - 9 ≤ x.
We get x ≤ 9.
Considering that x> 4.5, we write the answer:
In the next problem, the exponential inequality is reduced to a square one. So we recommend repeating the topic "square inequalities".
Now for more complex inequalities:
4. Solve the inequality
5. Solve the inequality
If, then. We were lucky! We know that the base of the logarithm is greater than one for all values of x included in the ODV.
Let's make a replacement
Note that we first completely solve the inequality with respect to the new variable t. And only after that we return to the variable x. Remember this and do not make mistakes on the exam!
Let us remember the rule: if there are roots, fractions or logarithms in an equation or inequality, the solution must be started from the range of permissible values. Since the base of the logarithm must be positive and not equal to one, we get a system of conditions:
Let's simplify this system:
This is the range of valid values for inequality.
We see that the variable is contained in the base of the logarithm. Let's move on to the permanent base. Recall that
In this case, it is convenient to go to base 4.
Let's make a replacement
Let us simplify the inequality and solve it by the method of intervals:
Let's go back to the variable x:
We have added the condition x> 0 (from ODZ).
7. The next problem is also solved using the method of intervals
As always, we start solving the logarithmic inequality from the range of admissible values. In this case
This condition must be met, and we will return to it. Consider for now the inequality itself. Let's write the left side as logarithm base 3:
The right side can also be written as a base 3 logarithm, and then go to the algebraic inequality:
We see that the condition (that is, ODZ) is now met automatically. Well, this makes it easier to solve inequality.
We solve the inequality by the method of intervals:
Answer:
Happened? Well, we increase the level of difficulty:
8. Solve the inequality:
Inequality is equivalent to the system:
9. Solve the inequality:
Expression 5 - x 2 is obtrusively repeated in the problem statement. This means that you can make a replacement:
Since the exponential function only takes positive values, t> 0. Then
The inequality will take the form:
Better now. Let us find the range of admissible values of the inequality. We have already said that t> 0. In addition, ( t- 3) (5 9 t − 1) > 0
If this condition is met, then the quotient will also be positive.
And the expression under the logarithm on the right side of the inequality must be positive, that is, (625 t − 2) 2 .
This means that 625 t- 2 ≠ 0, that is
We will carefully write down the ODZ
and solve the resulting system using the method of intervals.
So,
Well, half the battle is done - we have dealt with the ODZ. Solving the inequality itself. The sum of the logarithms on the left-hand side is represented as the logarithm of the product.
