Logarithmic inequalities with fractional bases. Inequalities. Logarithmic inequalities. What is ODZ? ODZ for logarithmic inequalities
Logarithmic equations and inequalities in the Unified State Examination in mathematics it is devoted to problem C3 . Every student must learn to solve C3 tasks from the Unified State Exam in mathematics if he wants to pass the upcoming exam with “good” or “excellent”. This article provides a brief overview of commonly encountered logarithmic equations and inequalities, as well as basic methods for solving them.
So, let's look at a few examples today. logarithmic equations and inequalities, which were offered to students in the Unified State Examination in mathematics of previous years. But it will begin with a brief summary of the main theoretical points that we will need to solve them.
Logarithmic function
Definition
Function of the form
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called logarithmic function.
Basic properties
Basic properties of the logarithmic function y= log a x:
The graph of a logarithmic function is logarithmic curve:
Properties of logarithms
Logarithm of the product two positive numbers is equal to the sum of the logarithms of these numbers:
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Logarithm of the quotient two positive numbers is equal to the difference between the logarithms of these numbers:
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If a And b a≠ 1, then for any number r equality is true:
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Equality log a t= log a s, Where a > 0, a ≠ 1, t > 0, s> 0, valid if and only if t = s.
If a, b, c are positive numbers, and a And c are different from unity, then the equality ( formula for moving to a new logarithm base):
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Theorem 1. If f(x) > 0 and g(x) > 0, then the logarithmic equation log a f(x) = log a g(x) (Where a > 0, a≠ 1) is equivalent to the equation f(x) = g(x).
Solving logarithmic equations and inequalities
Example 1. Solve the equation:
Solution. The range of acceptable values includes only those x, for which the expression under the logarithm sign is greater than zero. These values are determined by the following system of inequalities:
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Considering that
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we obtain the interval that defines the range of permissible values of this logarithmic equation:
Based on Theorem 1, all conditions of which are satisfied here, we proceed to the following equivalent quadratic equation:
The range of acceptable values includes only the first root.
Answer: x = 7.
Example 2. Solve the equation:
Solution. The range of acceptable values of the equation is determined by the system of inequalities:
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Solution. The range of acceptable values of the equation is determined here easily: x > 0.
We use substitution:
The equation becomes:
Reverse substitution:
Both answer are within the range of acceptable values of the equation because they are positive numbers.
Example 4. Solve the equation:
Solution. Let's start the solution again by determining the range of acceptable values of the equation. It is determined by the following system of inequalities:
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The bases of the logarithms are the same, so in the range of acceptable values we can proceed to the following quadratic equation:
The first root is not within the range of acceptable values of the equation, the second is.
Answer: x = -1.
Example 5. Solve the equation:
Solution. We will look for solutions in between x > 0, x≠1. Let's transform the equation to an equivalent one:
Both answer are within the range of acceptable values of the equation.
Example 6. Solve the equation:
Solution. The system of inequalities defining the range of permissible values of the equation this time has the form:
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Using the properties of the logarithm, we transform the equation to an equation that is equivalent in the range of acceptable values:
Using the formula for moving to a new logarithm base, we get:
The range of acceptable values includes only one answer: x = 4.
Let's now move on to logarithmic inequalities . This is exactly what you will have to deal with on the Unified State Exam in mathematics. To solve further examples we need the following theorem:
Theorem 2. If f(x) > 0 and g(x) > 0, then:
at a> 1 logarithmic inequality log a f(x) > log a g(x) is equivalent to an inequality of the same meaning: f(x) > g(x);
at 0< a < 1 логарифмическое неравенство log a f(x) > log a g(x) is equivalent to an inequality with the opposite meaning: f(x) < g(x).
Example 7. Solve the inequality:
Solution. Let's start by defining the range of acceptable values of the inequality. The expression under the sign of the logarithmic function must take only positive values. This means that the desired range of acceptable values is determined by the following system of inequalities:
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Since the base of the logarithm is a number less than one, the corresponding logarithmic function will be decreasing, and therefore, according to Theorem 2, the transition to the following quadratic inequality will be equivalent:
Finally, taking into account the range of acceptable values, we obtain answer:
Example 8. Solve the inequality:
Solution. Let's start again by defining the range of acceptable values:
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On the set of admissible values of the inequality we carry out equivalent transformations:
After reduction and transition to the inequality equivalent by Theorem 2, we obtain:
Taking into account the range of acceptable values, we obtain the final answer:
Example 9. Solve logarithmic inequality:
Solution. The range of acceptable values of inequality is determined by the following system:
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It can be seen that in the range of acceptable values, the expression at the base of the logarithm is always greater than one, and therefore, according to Theorem 2, the transition to the following inequality will be equivalent:
Taking into account the range of acceptable values, we obtain the final answer:
Example 10. Solve the inequality:
Solution.
The range of acceptable values of inequality is determined by the system of inequalities:
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Method I Let us use the formula for transition to a new base of the logarithm and move on to an inequality that is equivalent in the range of acceptable values.
When solving logarithmic inequalities, we use the monotonicity property of the logarithmic function. We also use the definition of logarithm and basic logarithmic formulas.
Let's review what logarithms are:
Logarithm a positive number to the base is an indicator of the power to which it must be raised to get .
At the same time
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)
Formula for moving to a new base:
Algorithm for solving logarithmic inequalities
We can say that logarithmic inequalities are solved using a specific algorithm. We need to write down the range of acceptable values (APV) of the inequality. Reduce the inequality to the form The sign here can be anything: It is important that on the left and on the right in the inequality there are logarithms to the same base.
And after that we “discard” the logarithms! Moreover, if the base is a degree , the inequality sign remains the same. If the base is such that the sign of inequality changes to the opposite.
Of course, we don't just "throw away" logarithms. We use the monotonicity property of a 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 in the form of 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 defined only for positive numbers, it is necessary that x be positive. The condition x > 0 is called the range of permissible values (APV) of this inequality. Only for such x does the inequality make sense.
Well, this formulation sounds dashing and is easy to remember. But why can we still do this?
We are people, 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’s why it’s important not to mechanically memorize the rules like a trained math dog, but to act consciously.
So why do we still “drop logarithms”?
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 moved on to an algebraic inequality, and the inequality sign remains the same.
So x > 5.
The following logarithmic inequality is also simple.
2. log 5 (15 + 3x) > log 5 2x
Let's start with the range of acceptable values. Logarithms are only defined for positive numbers, so
Solving this system, we get: x > 0.
Now let’s move from the logarithmic inequality to the algebraic one - “discard” the logarithms. Since the base of the logarithm is greater than one, the inequality sign remains the same.
15 + 3x > 2x.
We get: x > −15.
Answer: x > 0.
But what happens if the base of the logarithm is less than one? It is easy to guess that in this case, when moving 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 logarithms are taken must be positive, that is
Solving this system, we get: x > 4.5.
Since , a logarithmic function with a 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 that x ≤ 9.
Considering that x > 4.5, we write the answer:
In the next problem, the exponential inequality is reduced to a quadratic inequality. So we recommend repeating the topic “quadratic inequalities”.
Now for more complex inequalities:
4. Solve the inequality
5. Solve the inequality
If, then. We're lucky! We know that the base of the logarithm is greater than one for all values of x included in the ODZ.
Let's make a replacement
Note that we first solve the inequality completely with respect to the new variable t. And only after that we return to the variable x. Remember this and don’t make mistakes in the exam!
Let us remember the rule: if an equation or inequality contains roots, fractions or logarithms, the solution must begin from the range of acceptable values. Since the base of the logarithm must be positive and not equal to one, we obtain a system of conditions:
Let's simplify this system:
This is the range of acceptable values of inequality.
We see that the variable is contained in the base of the logarithm. Let's move on to the permanent base. Let us remind you that
In this case, it is convenient to go to base 4.
Let's make a replacement
Let's simplify the inequality and solve it using the interval method:
Let's return to the variable x:
We have added a condition x> 0 (from ODZ).
7. The following problem can also be solved using the interval method
As always, we start solving a logarithmic inequality from the range of acceptable values. In this case
This condition must be met, and we will return to it. Let's look at the inequality itself for now. Let's write the left side as a logarithm to base 3:
The right-hand side can also be written as a logarithm to base 3, and then move on to the algebraic inequality:
We see that the condition (that is, the ODZ) is now fulfilled automatically. Well, this makes solving the inequality easier.
We solve the inequality using the interval method:
Answer:
Did it work? Well, let's increase the difficulty level:
8. Solve the inequality:
Inequality is equivalent to the system:
9. Solve the inequality:
Expression 5 - x 2 is compulsively 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:
Already better. Let's find the range of acceptable 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 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 625 t− 2 ≠ 0, that is
Let's carefully write down the ODZ
and solve the resulting system using the interval method.
So,
Well, half the battle is done - we sorted out the ODZ. We solve the inequality itself. Let us represent the sum of logarithms on the left side as the logarithm of the product.
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
Purpose of the work: study of the mechanism for solving logarithmic inequalities C3 using non-standard methods, identifying interesting facts about the logarithm.
Subject of research:
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”
Purpose of the work: study of the mechanism for solving C3 problems using non-standard methods, identifying interesting facts about the logarithm.
Subject of research:
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 many years, 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 the course of his thought, although he, of course, 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)
"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 the 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 the 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's make the 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 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.-
We looked at solving the simplest logarithmic inequalities and inequalities where the base of the logarithm is fixed in the last lesson.
But what if there is a variable at the base of the logarithm?
Then it will come to our aid rationalization of inequalities. To understand how this works, let's consider, for example, the inequality:
$$\log_(2x) x^2 > \log_(2x) x.$$
As expected, let's start with ODZ.
ODZ
$$\left[ \begin(array)(l)x>0,\\ 2x ≠ 1. \end(array)\right.$$
Solution to inequality
Let's reason as if we were solving an inequality with a fixed base. If the base is greater than one, we get rid of logarithms, and the inequality sign does not change; if it is less than one, it changes.
Let's write this as a system:
$$\left[ \begin(array)(l) \left\( \begin(array)(l)2x>1,\\ x^2 > x; \end(array)\right. \\ \left\ ( \begin(array)(l)2x<1,\\ x^2 < x; \end{array}\right. \end{array} \right.$$
For further reasoning, let us move all the right-hand sides of the inequalities to the left.
$$\left[ \begin(array)(l) \left\( \begin(array)(l)2x-1>0,\\ x^2 -x>0; \end(array)\right. \ \ \left\( \begin(array)(l)2x-1<0,\\ x^2 -x<0; \end{array}\right. \end{array} \right.$$
What did we get? It turns out that we need the expressions `2x-1` and `x^2 - x` to be either positive or negative at the same time. The same result will be obtained if we solve the inequality:
$$(2x-1)(x^2 - x) >0.$$
This inequality, like the original system, is true if both factors are either positive or negative. It turns out that you can move from a logarithmic inequality to a rational one (taking into account the ODZ).
Let's formulate method for rationalizing logarithmic inequalities$$\log_(f(x)) g(x) \vee \log_(f(x)) h(x) \Leftrightarrow (f(x) - 1)(g(x)-h(x)) \ vee 0,$$ where `\vee` is any inequality sign. (For the `>` sign, we have just checked the validity of the formula. For the rest, I suggest you check it yourself - it will be remembered better).
Let's return to solving our inequality. Expanding it into brackets (to make the zeros of the function easier to see), we get
$$(2x-1)x(x - 1) >0.$$
The interval method will give the following picture:
(Since the inequality is strict and we are not interested in the ends of the intervals, they are not shaded.) As can be seen, the resulting intervals satisfy the ODZ. We received the answer: `(0,\frac(1)(2)) \cup (1,∞)`.
Example two. Solving a logarithmic inequality with a variable base
$$\log_(2-x) 3 \leqslant \log_(2-x) x.$$
ODZ
$$\left\(\begin(array)(l)2-x > 0,\\ 2-x ≠ 1,\\ x > 0. \end(array)\right.$$
$$\left\(\begin(array)(l)x< 2,\\ x ≠ 1, \\ x >0.\end(array)\right.$$
Solution to inequality
According to the rule we just received rationalization of logarithmic inequalities, we find that this inequality is identical (taking into account the ODZ) to the following:
$$(2-x -1) (3-x) \leqslant 0.$$
$$(1-x) (3-x) \leqslant 0.$$
Combining this solution with the ODZ, we get the answer: `(1,2)`.
Third example. Logarithm of a fraction
$$\log_x\frac(4x+5)(6-5x) \leqslant -1.$$
ODZ
$$\left\(\begin(array)(l) \dfrac(4x+5)(6-5x)>0, \\ x>0,\\ x≠ 1.\end(array) \right.$ $
Since the system is relatively complex, let's immediately plot the solution to the inequalities on the number line:
Thus, ODZ: `(0,1)\cup \left(1,\frac(6)(5)\right)`.
Solution to inequality
Let's represent `-1` as a logarithm with the base `x`.
$$\log_x\frac(4x+5)(6-5x) \leqslant \log_x x^(-1).$$
By using rationalization of logarithmic inequality we get a rational inequality:
$$(x-1)\left(\frac(4x+5)(6-5x) -\frac(1)(x)\right)\leqslant0,$$
$$(x-1)\left(\frac(4x^2+5x - 6+5x)(x(6-5x))\right)\leqslant0,$$
$$(x-1)\left(\frac(2x^2+5x - 3)(x(6-5x))\right)\leqslant0.$$
Do you think that there is still time before the Unified State Exam and you will have time to prepare? Perhaps this is so. But in any case, the earlier a student begins preparation, the more successfully he passes the exams. Today we decided to devote an article to logarithmic inequalities. This is one of the tasks, which means an opportunity to get extra credit.
Do you already know what a logarithm is? We really hope so. But even if you don't have an answer to this question, it's not a problem. Understanding what a logarithm is is very simple.
Why 4? You need to raise the number 3 to this power to get 81. Once you understand the principle, you can proceed to more complex calculations.
You went through inequalities a few years ago. And since then you have constantly encountered them in mathematics. If you have problems solving inequalities, check out the appropriate section.
Now that we have become familiar with the concepts individually, let's move on to considering them 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 necessary? To better understand how to solve inequalities with logarithms. Now let's give a more applicable example, still quite simple; we'll leave complex logarithmic inequalities for later.
How to solve this? It all starts with ODZ. It’s worth knowing more about it if you want to always easily solve any inequality.
What is ODZ? ODZ for logarithmic inequalities
The abbreviation stands for the range of acceptable values. This formulation often comes up in tasks for the Unified State Exam. ODZ will be 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 solving logarithmic inequalities does not raise questions. From the definition of a logarithm it follows that 2x+4 must be greater than zero. In our case this means the following.
This number, by definition, must be positive. Solve the inequality presented above. This can even be done orally; here it is clear that X cannot be less than 2. The solution to 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 sides of the inequality. What are we left with as a result? Simple inequality.
It's not difficult to solve. X must be greater than -0.5. Now we combine the two obtained values into a system. Thus,
This will be the range of acceptable values for the logarithmic inequality under consideration.
Why do we need ODZ 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 Unified State Examination 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 stages. First, you need to find the range of acceptable values. There will be two meanings in the ODZ, we discussed this above. Next we need to solve the inequality itself. The solution methods are as follows:
- multiplier replacement method;
- decomposition;
- rationalization method.
Depending on the situation, it is worth using one of the above methods. Let's move directly to the solution. Let us reveal the most popular method, which is suitable for solving Unified State Examination tasks in almost all cases. Next we will look at the decomposition method. It can help if you come across a particularly tricky inequality. So, an algorithm for solving logarithmic inequality.
Examples of solutions :
It’s not for nothing that we took exactly this inequality! Pay attention to the base. Remember: if it is greater than one, the sign remains the same when finding the range of acceptable values; otherwise, you need to change the inequality sign.
As a result, we get the inequality:
Now we reduce the left side to the form of the equation equal to zero. Instead of the “less than” sign we put “equals” and solve the equation. Thus, we will find the ODZ. We hope that you will not have problems solving such a simple equation. The answers are -4 and -2. That's not all. You need to display these points on the graph, placing “+” and “-”. What needs to be done for this? Substitute the 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 have found the range of acceptable values only for the left side; now we need to find the range of acceptable values for the right side. This is much easier. Answer: -2. We intersect both resulting areas.
And only now are we beginning to address the inequality itself.
Let's simplify it as much as possible to make it easier to solve.
We again use the interval method in the solution. Let’s skip the calculations; everything is already clear with it 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 requires an initial reduction to the same base. Next, use the method described above. But there is a more complicated case. Let's 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 people can be found in the Unified State Examination. 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 discard theory and go straight to practice. To solve logarithmic inequalities, it is enough to familiarize yourself with the example once.
To solve a logarithmic inequality of the form presented, it is necessary to reduce the right-hand side to a logarithm with the same base. The principle resembles equivalent transitions. As a result, the inequality will look like this.
Actually, all that remains is to create a system of inequalities without logarithms. Using the rationalization method, we move on to an equivalent system of inequalities. You will understand the rule itself when you substitute the appropriate values and track their changes. The system will have the following inequalities.
When using the rationalization method when solving inequalities, you need to remember the following: one must be subtracted from the base, x, by definition of the logarithm, is subtracted from both sides of the inequality (right from left), two expressions are multiplied and set under the original sign in relation to zero.
Further solution is carried out using the interval method, everything is simple here. It is important for you to understand the differences in solution methods, then everything will start to work out easily.
There are many nuances in logarithmic inequalities. The simplest of them are quite easy to solve. How can you 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 a variety of problems in the exam and you will be able to get the highest score. Good luck to you in your difficult task!