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 about the 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. Archimedes spoke about the connection between the terms 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 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 “a new modern effective method for solving 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: 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). 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 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 Therefore, the original inequality is equivalent to the set of two exponential inequalities, The solution to the first inequality of this set is the interval 0<х≤1, решением второго – промежуток 2≤х<+ Example 8.
Solution:
Inequality 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 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 we 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, aggregates 
. Thus, the original inequality is satisfied for all values of x from the intervals 0<х≤1 и 2≤х<+.
.
. Therefore, all x are from the interval 0
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.-
When deciding logarithmic inequalities we take as a basis properties of logarithmic functions. Namely, that the function at=log a x at A> 1 will be monotonically increasing, and at 0< A< 1 - монотонно убывающей.
Let's analyze transformation necessary to solve inequalities
log 1/5 (x - l) > - 2.
Initially, you need to equalize bases of logarithms, in this case, show the right side in the form of a logarithm with the necessary basis. Let's transform -2=-2 log 1/5 1/5= log 1/5 1/5 -2 = log 1/5 25, then we indicate the chosen inequality in the form:
log 1/5 (x- l) > log 1/5 25.
Function at= log 1/5 x will be monotonically decreasing. It turns out that a larger value of this function corresponds to a smaller argument value. And accordingly we have, X—1 < 25. К указанному неравенству требуется добавить еще неравенство X- 1 > 0, corresponding to the fact that under the sign logarithm there can only be a positive value. It turns out that this inequality is identical to the system of two linear inequalities. Considering that the base of the logarithm is less than one, in an identical system the sign of inequality is reversed:
Having solved which we see that:
1 < х < 26.
It is of great importance not to forget the condition x- 1 > 0, otherwise the conclusion will be incorrect: x< 26. Тогда бы в эти «решения» входило бы и значение х = 0, при котором левая часть первоначального неравенства не существует.
Among the whole 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 taught in 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.
This way we get rid of logarithms and reduce the problem to a rational inequality. The latter is much easier to solve, but when discarding logarithms, extra roots may appear. To cut them off, it is enough to find the range of acceptable values. If you have forgotten the ODZ of a logarithm, I strongly recommend repeating it - see “What is a logarithm”.
Everything related to the range of acceptable 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 satisfied simultaneously. When the range of acceptable values has been found, all that remains is to intersect it with the solution of the rational inequality - and the answer is ready.
Task. Solve the inequality:
First, let’s write out the logarithm’s ODZ:
The first two inequalities are satisfied automatically, but the last one will have to be written out. Since the square of a number is zero if and only if the number itself is zero, we have:
x 2 + 1 ≠ 1;
x2 ≠ 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 make the transition from logarithmic inequality to rational one. The original inequality has a “less than” sign, which means the resulting inequality must also have a “less than” 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 are: 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 this is the answer.
Converting logarithmic inequalities
Often the original inequality is different from the one above. This can be easily corrected using 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 by one logarithm.
Separately, I would like to remind you about the range of acceptable values. Since there may be several logarithms in the original inequality, it is required to find the VA of each of them. Thus, the general scheme for solving logarithmic inequalities is as follows:
- Find the VA of each logarithm included in the inequality;
- Reduce the inequality to a standard one using the formulas for adding and subtracting logarithms;
- Solve the resulting inequality using the scheme given above.
Task. Solve the inequality:
Let's find the domain of definition (DO) of the first logarithm:
We solve using the interval method. Finding the zeros of the numerator:
3x − 2 = 0;
x = 2/3.
Then - the zeros of the denominator:
x − 1 = 0;
x = 1.
We mark zeros and signs on the coordinate arrow:
We get x ∈ (−∞ 2/3)∪(1; +∞). The second logarithm will have the same VA. If you don't believe it, you can check it. Now we transform the second logarithm so that the base is two:
As you can see, the threes at the base and in front of the logarithm have been reduced. We got two logarithms with the same base. Let's add them up:
log 2 (x − 1) 2< 2;
log 2 (x − 1) 2< log 2 2 2 .
We obtained the standard logarithmic inequality. We get rid of logarithms using 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 intersect these sets - we get the real answer:
We are interested in the intersection of sets, so we select intervals that are shaded on both arrows. We get x ∈ (−1; 2/3)∪(1; 3) - all points are punctured.
Lesson objectives:
Didactic:
- Level 1 – teach how to solve the simplest logarithmic inequalities, using the definition of a logarithm and the properties of logarithms;
- Level 2 – solve logarithmic inequalities, choosing your own solution method;
- Level 3 – be able to apply knowledge and skills in non-standard situations.
Educational: develop memory, attention, logical thinking, comparison skills, be able to generalize and draw conclusions
Educational: cultivate accuracy, responsibility for the task being performed, and mutual assistance.
Teaching methods: verbal , visual , practical , partial-search , self-government , control.
Forms of organization of students’ cognitive activity: frontal , individual , work in pairs.
Equipment: a set of test tasks, reference notes, blank sheets for solutions.
Lesson type: learning new material.
During the classes
1. Organizational moment. The topic and goals of the lesson, the lesson plan 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 tasks; tasks must be completed in pairs; blank solution sheets; support sheets: definition of 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's score sheet
2. Updating knowledge.
Teacher's instructions. Recall the definition of logarithm, the graph of the 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.
Students are given sheets on which are written: the definition of a logarithm; shows a graph of a logarithmic function and its properties; properties of logarithms; algorithm for solving logarithmic inequalities, an example of solving a logarithmic inequality that reduces to a quadratic one.
3. Studying new material.
Solving logarithmic inequalities is based on the monotonicity of the logarithmic function.
Algorithm for solving logarithmic inequalities:
A) Find the domain of definition of the inequality (the sublogarithmic expression is greater than zero).
B) Represent (if possible) the left and right sides of the inequality as logarithms to the same base.
C) Determine whether the logarithmic function is increasing or decreasing: if t>1, then increasing; if 0
D) Go to a simpler inequality (sublogarithmic expressions), taking into account that the sign of the inequality will remain the same if the function increases and will change if it decreases.
Learning element #1.
Goal: consolidate the solution to the simplest logarithmic inequalities
Form of organization of students' cognitive activity: individual work.
Tasks for independent work for 10 minutes. For each inequality there are several possible answers; you need to choose the correct one and check it using the key.
KEY: 13321, maximum number of points – 6 points.
Learning element #2.
Goal: consolidate the solution of logarithmic inequalities using the properties of logarithms.
Teacher's instructions. Remember the basic properties of logarithms. To do this, read the text of the textbook on pp. 92, 103–104.
Tasks for independent work for 10 minutes.
KEY: 2113, maximum number of points – 8 points.
Learning element #3.
Purpose: to study the solution of logarithmic inequalities by the method of reduction to quadratic.
Teacher's instructions: the method of reducing an inequality to a quadratic is to transform the inequality to such a form that a certain logarithmic function is denoted by a new variable, thereby obtaining a quadratic inequality with respect to this variable.
Let's use the interval method.
You have passed the first level of mastering the material. Now you will have to independently choose a method for solving logarithmic equations, using all your knowledge and capabilities.
Learning element #4.
Goal: consolidate the solution to logarithmic inequalities by independently choosing a rational solution method.
Tasks for independent work for 10 minutes
Learning element #5.
Teacher's instructions. Well done! You have mastered solving equations of the second level of complexity. The goal of your further work is to apply your knowledge and skills in more complex and non-standard situations.
Tasks for independent solution:
Teacher's instructions. It's great if you completed 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 a “5” rating,
- for 16 ≤ N ≤ 19 – score “4”,
- for 8 ≤ N ≤ 15 – score “3”,
- at N< 8 выполнить работу над ошибками к следующему уроку (решения можно взять у учителя).
Submit the assessment papers to the teacher.
5. Homework: if you scored no more than 15 points, work on your mistakes (solutions can be obtained from the teacher), if you scored more than 15 points, complete a creative task on the topic “Logarithmic inequalities.”
When studying the logarithmic function, we mainly considered inequalities of the form
log a x< b и log а х ≥ b. Рассмотрим решение более сложных логарифмических неравенств. Обычным способом решения таких неравенств является переход от данного неравенства к более простому неравенству или системе неравенств, которая имеет то же самое множество решений.
Solve the inequality log (x + 1) ≤ 2 (1).
Solution.
1) The right side of the inequality under consideration makes sense for all values of x, and the left side makes sense for x + 1 > 0, i.e. for x > -1.
2) The interval x > -1 is called the domain of definition of inequality (1). A logarithmic function with base 10 is increasing, therefore, provided x + 1 > 0, inequality (1) is satisfied if x + 1 ≤ 100 (since 2 = log 100). Thus, inequality (1) and the system of inequalities
(x > -1, (2)
(x + 1 ≤ 100,
are equivalent, in other words, the set of solutions to inequality (1) and the system of inequalities (2) are the same.
3) Solving system (2), we find -1< х ≤ 99.
Answer. -1< х ≤ 99.
Solve the inequality log 2 (x – 3) + log 2 (x – 2) ≤ 1 (3).
Solution.
1) The domain of definition of the logarithmic function under consideration is the set of positive values of the argument, therefore the left side of the inequality makes sense for x – 3 > 0 and x – 2 > 0.
Consequently, the domain of definition of this inequality is the interval x > 3.
2) According to the properties of the logarithm, inequality (3) for x > 3 is equivalent to the inequality log 2 (x – 3)(x – 2) ≤ log 2 (4).
3) The logarithmic function with base 2 is increasing. Therefore, for x > 3, inequality (4) is satisfied if (x – 3)(x – 2) ≤ 2.
4) Thus, the original inequality (3) is equivalent to the system of inequalities
((x – 3)(x – 2) ≤ 2,
(x > 3.
Solving the first inequality of this system, we obtain x 2 – 5x + 4 ≤ 0, whence 1 ≤ x ≤ 4. Combining this segment with the interval x > 3, we obtain 3< х ≤ 4.
Answer. 3< х ≤ 4.
Solve the inequality log 1/2 (x 2 + 2x – 8) ≥ -4. (5)
Solution.
1) The domain of definition of the inequality is found from the condition x 2 + 2x – 8 > 0.
2) Inequality (5) can be written as: 
log 1/2 (x 2 + 2x – 8) ≥ log 1/2 16.
3) Since the logarithmic function with base ½ is decreasing, then for all x from the entire domain of definition of the inequality we obtain:
x 2 + 2x – 8 ≤ 16.
Thus, the original equality (5) is equivalent to the system of inequalities
(x 2 + 2x – 8 > 0, or (x 2 + 2x – 8 > 0,
(x 2 + 2x – 8 ≤ 16, (x 2 + 2x – 24 ≤ 0.
Solving the first quadratic inequality, we get x< -4, х >2. Solving the second quadratic inequality, we obtain -6 ≤ x ≤ 4. Consequently, both inequalities of the system are satisfied simultaneously for -6 ≤ x< -4 и при 2 < х ≤ 4.
Answer. -6 ≤ x< -4; 2 < х ≤ 4.
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