Solving logarithmic inequalities with different bases. Complex logarithmic inequalities

Maintaining your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please review our privacy practices and let us know if you have any questions.

Collection and use of personal information

Personal information refers to data that can be used to identify or contact a specific person.

You may be asked to provide your personal information at any time when you contact us.

Below are some examples of the types of personal information we may collect and how we may use such information.

What personal information do we collect:

• When you submit an application on the site, we may collect various information, including your name, telephone number, address Email etc.

How we use your personal information:

• Collected by us personal information allows us to contact you and inform you about unique offers, promotions and other events and upcoming events.
• From time to time, we may use your personal information to send important notices and communications.
• We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
• If you participate in a prize draw, contest or similar promotion, we may use the information you provide to administer such programs.

Disclosure of information to third parties

We do not disclose the information received from you to third parties.

Exceptions:

• If necessary, in accordance with the law, judicial procedure, in legal proceedings, and/or based on public inquiries or requests from government agencies on the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public importance purposes.
• In the event of a reorganization, merger, or sale, we may transfer the personal information we collect to the applicable successor third party.

Protection of personal information

We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as unauthorized access, disclosure, alteration and destruction.

Respecting your privacy at the company level

To ensure that your personal information is secure, we communicate privacy and security standards to our employees and strictly enforce privacy practices.

Among all the diversity logarithmic inequalities Inequalities with variable bases 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 area 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 ​​is found, all that remains is to intersect it with the solution 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 the number equal to 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:

1. Any number can be represented as a logarithm with a given base;
2. 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, general scheme solutions to logarithmic inequalities are as follows:

1. Find the VA of each logarithm included in the inequality;
2. Reduce the inequality to a standard one using the formulas for adding and subtracting logarithms;
3. 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 of ODZ will be the same. 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:

1. ODZ: x ∈ (−∞ 2/3)∪(1; +∞);
2. 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.

With them are inside logarithms.

Examples:

\(\log_3⁡x≥\log_3⁡9\)
\(\log_3⁡ ((x^2-3))< \log_3⁡{(2x)}\)
\(\log_(x+1)⁡((x^2+3x-7))>2\)
\(\lg^2⁡((x+1))+10≤11 \lg⁡((x+1))\)

How to solve logarithmic inequalities:

We should strive to reduce any logarithmic inequality to the form \(\log_a⁡(f(x)) ˅ \log_a(⁡g(x))\) (the symbol \(˅\) means any of ). This type allows you to get rid of logarithms and their bases, making the transition 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 is a number and it is greater than 1, the inequality sign remains the same during the transition,
\(-\) if the base is a number greater than 0 but less than 1 (lies between zero and one), then the inequality sign should change to the opposite, i.e.

Very important! In any inequality, the transition from the form \(\log_a(⁡f(x)) ˅ \log_a⁡(g(x))\) to comparing expressions under logarithms can be done only if:

Example . Solve inequality: \(\log\)\(≤-1\)

Solution:

Example . Solve the inequality: \(\log^2_3⁡x-\log_3⁡x-2>0\)

Solution:

Logarithmic inequalities

In previous lessons, we got acquainted with logarithmic equations and now we know what they are and how to solve them. Today's lesson will be devoted to the study of logarithmic inequalities. What are these inequalities and what is the difference between solving a logarithmic equation and an inequality?

Logarithmic inequalities are inequalities that have a variable appearing under the logarithm sign or at its base.

Or, we can also say that a logarithmic inequality is an inequality in which its unknown value, as in a logarithmic equation, will appear under the sign of the logarithm.

The simplest logarithmic inequalities have the following form:

where f(x) and g(x) are some expressions that depend on x.

Let's look at this using this example: f(x)=1+2x+x2, g(x)=3x−1.

Solving logarithmic inequalities

Before solving logarithmic inequalities, it is worth noting that when solved they are similar to exponential inequalities, namely:

First, when moving from logarithms to expressions under the logarithm sign, we also need to compare the base of the logarithm with one;

Secondly, when solving a logarithmic inequality using a change of variables, we need to solve inequalities with respect to the change until we get the simplest inequality.

But you and I have considered similar aspects of solving logarithmic inequalities. Now let’s pay attention to a rather significant difference. You and I know that the logarithmic function has a limited domain of definition, therefore, when moving from logarithms to expressions under the logarithm sign, we need to take into account the range of permissible values ​​(ADV).

That is, it should be taken into account that when solving a logarithmic equation, you and I can first find the roots of the equation, and then check this solution. But solving a logarithmic inequality will not work this way, since moving from logarithms to expressions under the logarithm sign, it will be necessary to write down the ODZ of the inequality.

In addition, it is worth remembering that the theory of inequalities consists of real numbers, which are positive and negative numbers, as well as the number 0.

For example, when the number “a” is positive, then you need to use the following notation: a >0. In this case, both the sum and the product of these numbers will also be positive.

The main principle for solving an inequality is to replace it with a simpler inequality, but the main thing is that it is equivalent to the given one. Further, we also obtained an inequality and again replaced it with one that has a simpler form, etc.

When solving inequalities with a variable, you need to find all its solutions. If two inequalities have the same variable x, then such inequalities are equivalent, provided that their solutions coincide.

When performing tasks on solving logarithmic inequalities, you must remember that when a > 1, then the logarithmic function increases, and when 0< a < 1, то такая функция имеет свойство убывать. Эти свойства вам будут необходимы при решении логарифмических неравенств, поэтому вы их должны хорошо знать и помнить.

Methods for solving logarithmic inequalities

Now let's look at some of the methods that take place when solving logarithmic inequalities. For better understanding and assimilation, we will try to understand them using specific examples.

We all know that the simplest logarithmic inequality has the following form:

In this inequality, V – is one of the following inequality signs:<,>, ≤ or ≥.

When the base of a given logarithm is greater than one (a>1), making the transition from logarithms to expressions under the logarithm sign, then in this version the inequality sign is preserved, and the inequality will have the following form:

which is equivalent to this system:

In the case when the base of the logarithm is greater than zero and less than one (0

This is equivalent to this system:

Let's look at more examples of solving the simplest logarithmic inequalities shown in the picture below:

Solving Examples

Exercise. Let's try to solve this inequality:

Solving the range of acceptable values.

Now let's try to multiply its right side by:

Let's see what we can come up with:

Now, let's move on to converting sublogarithmic expressions. Due to the fact that the base of the logarithm is 0< 1/4 <1, то от сюда следует, что знак неравенства изменится на противоположный:

3x - 8 > 16;
3x > 24;
x > 8.

And from this it follows that the interval that we obtained entirely belongs to the ODZ and is a solution to such an inequality.

Here's the answer we got:

What is needed to solve logarithmic inequalities?

Now let's try to analyze what we need to successfully solve logarithmic inequalities?

First, concentrate all your attention and try not to make mistakes when performing the transformations that are given in this inequality. Also, it should be remembered that when solving such inequalities, it is necessary to avoid expansions and contractions of the inequalities, which can lead to the loss or acquisition of extraneous solutions.

Secondly, when solving logarithmic inequalities, you need to learn to think logically and understand the difference between concepts such as a system of inequalities and a set of inequalities, so that you can easily select solutions to the inequality, while being guided by its DL.

Thirdly, to successfully solve such inequalities, each of you must perfectly know all the properties of elementary functions and clearly understand their meaning. Such functions include not only logarithmic, but also rational, power, trigonometric, etc., in a word, all those that you have studied throughout schooling algebra.

As you can see, having studied the topic of logarithmic inequalities, there is nothing difficult in solving these inequalities, provided that you are careful and persistent in achieving your goals. To avoid any problems in solving inequalities, you need to practice as much as possible, solving various tasks and at the same time remember the basic methods of solving such inequalities and their systems. If you fail to solve logarithmic inequalities, you should carefully analyze your mistakes so as not to return to them again in the future.

Homework

To better understand the topic and consolidate the material covered, solve the following inequalities:

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 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 generalized interval method…………… 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 specialized 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. Methods that are studied in 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 necessary information O non-standard methods solutions to logarithmic inequalities.

2) Find additional information about logarithms.

3) Learn to decide specific tasks C3 using non-standard methods.

Results:

The practical significance lies in the expansion of the apparatus for solving C3 problems. This material can be used in some lessons, for 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 to different meanings percent. 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 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 most universal when solving inequalities of almost any type. The solution diagram looks like in the following way:

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 by 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 to the inequality are all

2.2. Rationalization method.

Previously, inequality was not solved using the rationalization method; it was not known. This is the "new modern" effective method solutions to 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 is guidelines, associated with this method, and in the "Most complete editions typical options..." Solution C3 uses this method.
WONDERFUL METHOD!

"Magic Table"

In other sources

If a >1 and b >1, then log a b >0 and (a -1)(b -1)>0;

If a >1 and 0

if 0<a<1 и b >1, then log a b<0 и (a -1)(b -1)<0;