LOGARITHMIC INEQUALITIES IN THE USE
Sechin Mikhail Alexandrovich
Small Academy of Sciences for Students of the Republic of Kazakhstan "Seeker"
MBOU "Soviet secondary school No. 1", grade 11, town. Sovietsky Soviet District
Gunko Lyudmila Dmitrievna, teacher of MBOU "Soviet secondary school No. 1"
Sovietsky district
Objective: study of the mechanism for solving C3 logarithmic inequalities using non-standard methods, revealing interesting facts about the logarithm.
Subject of study:
3) Learn to solve specific logarithmic C3 inequalities 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. Tasks with traps…………………………………………………… 27
Conclusion…………………………………………………………………… 30
Literature……………………………………………………………………. 31
Introduction
I am in the 11th grade and I plan to enter a university where mathematics is a core subject. And that's why I work a lot with the tasks 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 encountered the problem of the lack of methods and techniques for solving the examination 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 suggested that I work with the C3 assignments on my own under her guidance. In addition, I was interested in the question: are there logarithms in our life?
With this in mind, the theme was chosen:
"Logarithmic inequalities in the exam"
Objective: study of the mechanism for solving C3 problems using non-standard methods, revealing 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 problems C3. This material can be used in some lessons, for conducting circles, optional classes 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. 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 also arose in other areas, for example, in the insurance business, tables of compound interest were needed for various percentage values. The main difficulty was multiplication, division of multi-digit 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 indicators 1, 2, 3, ... in the Psalmite. Another prerequisite was the extension of the concept of degree to negative and fractional exponents. Many authors have pointed out that multiplication, division, raising to a power, and extracting 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.
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 Burgi (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 thus entered a new field of function theory. Bürgi remained on the basis of consideration of 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 - "relationship" 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 for the logarithm of one, and 100 for the logarithm of ten, or, what amounts to the same, just 1. This is how decimal logarithms and The first logarithmic tables were printed. Later, the Briggs tables were supplemented by the Dutch bookseller and mathematician Andrian Flakk (1600-1667). Napier and Briggs, although they came to logarithms before anyone else, published their tables later than 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, followed by N. Mercator in 1668, and the London teacher John Spadel published tables of natural logarithms of numbers from 1 to 1000 under the name "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 in the processing of the German mathematician K. Bremiker (1804-1877).
Stage 2
Further development of the theory of logarithms is associated with a wider application of analytic geometry and infinitesimal calculus. By that time, the connection between the quadrature of an equilateral hyperbola and the natural logarithm was 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 his essay
"Logarithmotechnics" (1668) gives a series that gives the expansion of ln(x + 1) in terms of
powers x:
This expression corresponds exactly 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 a higher point of view", read 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 exponent of a given base
was not formulated immediately. The work of Leonhard Euler (1707-1783)
"Introduction to the analysis of infinitesimals" (1748) served as further
development of the theory of the logarithmic function. In this way,
134 years have passed since logarithms were first introduced
(counting from 1614) before mathematicians came up with a 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 universal in solving inequalities of almost any type. The solution scheme looks like this:
1. Bring the inequality to such a form, where the function is located on the left side 
, and 0 on the right.
2. Find the scope of the function .
3. Find the zeros of a 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 a real line.
5. Determine the signs of the function at the received intervals.
6. Select the intervals where the function takes the necessary values, and write down the answer.
Example 1
Solution:
Apply the interval method
where
For these values, all expressions under the signs of logarithms are positive.
Answer:
Example 2
Solution:
1st way . ODZ is determined by the inequality x> 3. Taking logarithms for such x in base 10, we get
The last inequality could be solved by applying the decomposition rules, i.e. comparing factors with zero. However, in this case it is easy to determine the intervals of constancy of the function
so the interval method can be applied.
Function f(x) = 2x(x- 3.5)lgǀ x- 3ǀ is continuous for 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 constancy of the function f(x):
Answer:
2nd way . Let us apply the ideas of the method of intervals directly to the original inequality.
For this, we 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 interval method
Answer:
Example 3
Solution:
Apply the interval 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 interval method
In the first inequality, we make the change
then we arrive at 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 with 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
Apply the interval method or
Answer:
Example 6
Solution:
Inequality is tantamount to a system
Let
then y > 0,
and the first inequality
system takes the form
or, expanding
square trinomial to factors,
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 of the inequality are all
2.2. rationalization method.
Previously, the method of rationalization of 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 by Kolesnikova S.I.)
And even if the teacher knew him, there was a fear - but does the USE 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.
THE METHOD IS GREAT!
"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 noticeably 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, we write (x-1-1) (x-1) instead of the denominator, and the product (x-1) (x-3-9 + x) instead of the numerator. 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 Let's make a replacement t =log 4 y and get 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 set of two exponential inequalities, The solution of the first inequality of this set is the interval 0<х≤1, решением второго – промежуток 2≤х<+ Example 8
Solution:
Inequality is tantamount to a system The solution of the second inequality, which determines the ODZ, will be the set of those x,
for which x > 0.
To solve the first inequality, we make the change Then we get the inequality or The set of solutions of the last inequality is found by the method intervals: -1< t < 2. Откуда, возвращаясь к переменной x, we get or Many 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 (2x +1-x 2)>log 2 (2x-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 of this collection is the intervals 0<у≤2 и 8≤у<+.
that is, 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 a large variety 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 offered at the USE in part C, namely C3. These inequalities with solutions by methods formed the basis of the collection "Logarithmic C3 Inequalities with Solutions", which became the project product of my activity. The hypothesis I put forward at the beginning of the project was confirmed: C3 problems can be effectively solved if these methods are known.
In addition, I discovered interesting facts about logarithms. It was interesting for me to do it. My project products will be useful for both students and teachers.
Conclusions:
Thus, the goal of the project is achieved, the problem is solved. 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, and activity.
A guarantee of success when creating a research project for I have become: significant school experience, the ability to extract information from various sources, check its reliability, rank it according to its significance.
In addition to directly 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 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 A. G. Preparing for the Unified State Examination in Mathematics.
3. S. S. Samarova, Solution of logarithmic inequalities.
4. Mathematics. Collection of training works edited by A.L. Semyonov 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 transformations needed to solve the inequality
log 1/5 (x - l) > - 2.
First you need to balance 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 the larger value of this function corresponds to the smaller value of the argument. 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 can only be positive. It turns out that this inequality is identical to the system of two linear inequalities. Given that the base of the logarithm is less than one, in an identical system, the inequality sign is reversed:
Solving which we see that:
1 < х < 26.
It is of great importance not to forget the condition x-1 > 0, otherwise the conclusion will not be correct: x< 26. Тогда бы в эти «решения» входило бы и значение х = 0, при котором левая часть первоначального неравенства не существует.
With them are inside 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 \(\log_a(f(x)) ˅ \log_a(g(x))\) (symbol \(˅\) means any of ). This form allows us to get rid of logarithms and their bases by passing to the inequality of expressions under logarithms, that is, to the form \(f(x) ˅ g(x)\).
But when making this transition, there is one very important subtlety:
\(-\) if - 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 (between zero and one), then the inequality sign must be reversed, i.e.
|
\(\log_2((8-x))<1\) Solution: |
\(\log\)\(_(0.5)\) \((2x-4)\)≥\(\log\)\(_(0.5)\) \(((x+ one))\) Solution: |
Very important! In any inequality, the transition from the form \(\log_a(f(x)) ˅ \log_a(g(x))\) to comparing expressions under logarithms can only be done if:
Example . Solve the inequality: \(\log\)\(≤-1\)
Solution:
|
\(\log\) \(_(\frac(1)(3))(\frac(3x-2)(2x-3))\)\(≤-1\) |
Let's write out the ODZ. |
|
ODZ: \(\frac(3x-2)(2x-3)\) \(>0\) |
|
|
\(\frac(3x-2-3(2x-3))(2x-3)\)\(≥\) \(0\) |
We open the brackets, give . |
|
\(\frac(-3x+7)(2x-3)\) \(≥\) \(0\) |
We multiply the inequality by \(-1\), remembering to reverse the comparison sign. |
|
\(\frac(3x-7)(2x-3)\) \(≤\) \(0\) |
|
|
\(\frac(3(x-\frac(7)(3)))(2(x-\frac(3)(2)))\)\(≤\) \(0\) |
Let's build a number line and mark the points \(\frac(7)(3)\) and \(\frac(3)(2)\) on it. Note that the point from the denominator is punctured, despite the fact that the inequality is not strict. The fact is that this point will not be a solution, since when substituting into an inequality, it will lead us to division by zero. |
|
|
Now we plot the ODZ on the same numerical axis and write down in response the interval that falls into the ODZ. |
|
|
Write down the final answer. |
Example . Solve the inequality: \(\log^2_3x-\log_3x-2>0\)
Solution:
|
\(\log^2_3x-\log_3x-2>0\) |
Let's write out the ODZ. |
|
ODZ: \(x>0\) |
Let's get to the decision. |
|
Solution: \(\log^2_3x-\log_3x-2>0\) |
Before us is a typical square-logarithmic inequality. We do. |
|
\(t=\log_3x\) |
Expand the left side of the inequality into . |
|
\(D=1+8=9\) |
|
|
Now you need to return to the original variable - x. To do this, we pass to , which has the same solution, and make the reverse substitution. |
|
|
\(\left[ \begin(gathered) t>2 \\ t<-1 \end{gathered} \right.\) \(\Leftrightarrow\) \(\left[ \begin{gathered} \log_3x>2 \\ \log_3x<-1 \end{gathered} \right.\) |
Transform \(2=\log_39\), \(-1=\log_3\frac(1)(3)\). |
|
\(\left[ \begin(gathered) \log_3x>\log_39 \\ \log_3x<\log_3\frac{1}{3} \end{gathered} \right.\) |
Let's move on to comparing arguments. The bases of logarithms are greater than \(1\), so the sign of the inequalities does not change. |
|
\(\left[ \begin(gathered) x>9 \\ x<\frac{1}{3} \end{gathered} \right.\) |
Let's combine the solution of the inequality and the ODZ in one figure. |
|
|
Let's write down the answer. |
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 lg (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 - for x + 1 > 0, i.e. for x > -1.
2) The interval x\u003e -1 is called the domain of definition of inequality (1). The logarithmic function with base 10 is increasing, therefore, under the condition x + 1 > 0, inequality (1) is satisfied if x + 1 ≤ 100 (since 2 = lg 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. -one< х ≤ 99.
Solve the inequality log 2 (x - 3) + log 2 (x - 2) ≤ 1 (3).
Solution.
1) The domain of the considered logarithmic function 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.
Therefore, the domain of this inequality is the interval x > 3.
2) According to the properties of the logarithm, inequality (3) for х > 3 is equivalent to the inequality log 2 (х – 3)(х – 2) ≤ log 2 (4).
3) The base 2 logarithmic function is increasing. Therefore, for х > 3, inequality (4) is satisfied if (х – 3)(х – 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 get x 2 - 5x + 4 ≤ 0, whence 1 ≤ x ≤ 4. Combining this segment with the interval x > 3, we get 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 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 the inequality we get:
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 get -6 ≤ x ≤ 4. Therefore, both inequalities of the system are fulfilled simultaneously at -6 ≤ x< -4 и при 2 < х ≤ 4.
Answer. -6 ≤ x< -4; 2 < х ≤ 4.
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Do you think that there is still time before the exam, and you will have time to prepare? Perhaps this is so. But in any case, the earlier the student begins training, the more successfully he passes the exams. Today we decided to dedicate an article to logarithmic inequalities. This is one of the tasks, which means an opportunity to get an extra point.
Do you already know what a logarithm (log) is? We really hope so. But even if you don't have an answer to this question, it's not a problem. It is very easy to understand what a logarithm is.
Why exactly 4? You need to raise the number 3 to such a power to get 81. When you understand the principle, you can proceed to more complex calculations.
You went through the inequalities a few years ago. And since then, you constantly meet them in mathematics. If you're having trouble solving inequalities, check out the appropriate section.
Now, when we have got acquainted with concepts separately, we will pass to their consideration in general.
The simplest logarithmic inequality.
The simplest logarithmic inequalities are not limited to this example, there are three more, only with different signs. Why is this needed? To better understand how to solve inequality with logarithms. Now we give a more applicable example, still quite simple, we leave complex logarithmic inequalities for later.
How to solve it? It all starts with ODZ. You should know more about it if you want to always easily solve any inequality.
What is ODZ? DPV for logarithmic inequalities
The abbreviation stands for the range of valid values. In assignments for the exam, this wording often pops up. DPV is useful to you not only in the case of logarithmic inequalities.
Look again at the above example. We will consider the ODZ based on it, so that you understand the principle, and the solution of logarithmic inequalities does not raise questions. It follows from the definition of the logarithm that 2x+4 must be greater than zero. In our case, this means the following.
This number must be positive by definition. Solve the inequality presented above. This can even be done orally, here it is clear that X cannot be less than 2. The solution of the inequality will be the definition of the range of acceptable values.
Now let's move on to solving the simplest logarithmic inequality.
We discard the logarithms themselves from both parts of the inequality. What is left for us as a result? simple inequality.
It's easy to solve. X must be greater than -0.5. Now we combine the two obtained values into the system. In this way,
This will be the region of admissible values for the considered logarithmic inequality.
Why is ODZ needed at all? This is an opportunity to weed out incorrect and impossible answers. If the answer is not within the range of acceptable values, then the answer simply does not make sense. This is worth remembering for a long time, since in the exam there is often a need to search for ODZ, and it concerns not only logarithmic inequalities.
Algorithm for solving logarithmic inequality
The solution consists of several steps. First, it is necessary to find the range of acceptable values. There will be two values in the ODZ, we considered this above. The next step is to solve the inequality itself. The solution methods are as follows:
- multiplier replacement method;
- decomposition;
- rationalization method.
Depending on the situation, one of the above methods should be used. Let's go straight to the solution. We will reveal the most popular method that is suitable for solving USE tasks in almost all cases. Next, we will consider the decomposition method. It can help if you come across a particularly "tricky" inequality. So, the algorithm for solving the logarithmic inequality.
Solution examples :
It is not in vain that we took precisely such an inequality! Pay attention to the base. Remember: if it is greater than one, the sign remains the same when finding the range of valid values; otherwise, the inequality sign must be changed.
As a result, we get the inequality:
Now we bring the left side to the form of the equation equal to zero. Instead of the “less than” sign, we put “equal”, we solve the equation. Thus, we will find the ODZ. We hope that you will have no problems with solving such a simple equation. The answers are -4 and -2. That's not all. You need to display these points on the chart, place "+" and "-". What needs to be done for this? Substitute numbers from the intervals into the expression. Where the values are positive, we put "+" there.
Answer: x cannot be greater than -4 and less than -2.
We found the range of valid values only for the left side, now we need to find the range of valid values for the right side. This is by no means easier. Answer: -2. We intersect both received areas.
And only now we begin to solve the inequality itself.
Let's simplify it as much as possible to make it easier to decide.
We again use the interval method in the solution. Let's skip the calculations, with him everything is already clear from the previous example. Answer.
But this method is suitable if the logarithmic inequality has the same bases.
Solving logarithmic equations and inequalities with different bases involves initial reduction to one base. Then use the above method. But there is also a more complicated case. Consider one of the most complex types of logarithmic inequalities.
Logarithmic inequalities with variable base
How to solve inequalities with such characteristics? Yes, and such can be found in the exam. Solving inequalities in the following way will also have a beneficial effect on your educational process. Let's look at the issue in detail. Let's put theory aside and go straight to practice. To solve logarithmic inequalities, it is enough to once familiarize yourself with the example.
To solve the logarithmic inequality of the presented form, it is necessary to reduce the right side to the logarithm with the same base. The principle resembles equivalent transitions. As a result, the inequality will look like this.
Actually, it remains to create a system of inequalities without logarithms. Using the rationalization method, we pass to an equivalent system of inequalities. You will understand the rule itself when you substitute the appropriate values and follow their changes. The system will have the following inequalities.
Using the rationalization method when solving inequalities, you need to remember the following: you need to subtract one from the base, x, by definition of the logarithm, is subtracted from both parts of the inequality (the right from the left), the two expressions are multiplied and set under the original sign relative to zero.
The further solution is carried out by the interval method, everything is simple here. It is important for you to understand the differences in the solution methods, then everything will start to work out easily.
There are many nuances in logarithmic inequalities. The simplest of them are easy enough to solve. How to make it so that to solve each of them without problems? You have already received all the answers in this article. Now you have a long practice ahead of you. Constantly practice solving various problems within the exam and you will be able to get the highest score. Good luck in your difficult work!
