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How to solve inequalities with decimal logarithms. Solving simple logarithmic inequalities

Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely taught in school:

log k (x) f (x) ∨ log k (x) g (x) ⇒ (f (x) − g (x)) (k (x) − 1) ∨ 0

Instead of the “∨” checkbox, you can put any inequality sign: more or less. The main thing is that in both inequalities the signs are the same.

This way we get rid of logarithms and reduce the problem to a rational inequality. The latter is much easier to solve, but when discarding logarithms, extra roots may appear. To cut them off, it is enough to find the range of acceptable values. If you have forgotten the ODZ of a logarithm, I strongly recommend repeating it - see “What is a logarithm".

Everything related to the range of acceptable values must be written down and solved separately:

f(x) > 0; g(x) > 0; k(x) > 0; k(x) ≠ 1.

These four inequalities constitute a system and must be satisfied simultaneously. When the range of acceptable values has been found, all that remains is to intersect it with the solution of the rational inequality - and the answer is ready.

Task. Solve the inequality:

First, let’s write out the logarithm’s ODZ:

The first two inequalities are satisfied automatically, but the last one will have to be written out. Since the square of a number is zero if and only if the number itself is zero, we have:

x 2 + 1 ≠ 1;
x 2 ≠ 0;
x ≠ 0.

It turns out that the ODZ of the logarithm is all numbers except zero: x ∈ (−∞ 0)∪(0; +∞). Now we solve the main inequality:

We 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 is easy to fix by standard rules working with logarithms - see “Basic properties of logarithms”. Namely:

Any number can be represented as a logarithm with a given base;
The sum and difference of logarithms with the same bases can be replaced by one logarithm.

Separately, I would like to remind you about the range of acceptable values. Since there may be several logarithms in the original inequality, it is required to find the VA of each of them. Thus, general scheme solutions to logarithmic inequalities are as follows:

Find the VA of each logarithm included in the inequality;
Reduce the inequality to a standard one using the formulas for adding and subtracting logarithms;
Solve the resulting inequality according to the scheme given above.

Task. Solve the inequality:

Let's find the domain of definition (DO) of the first logarithm:

We solve using the interval method. Finding the zeros of the numerator:

3x − 2 = 0;
x = 2/3.

Then - the zeros of the denominator:

x − 1 = 0;
x = 1.

We mark zeros and signs on the coordinate arrow:

We get x ∈ (−∞ 2/3)∪(1; +∞). The second logarithm will have the same VA. If you don't believe me, 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 basis. 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 should be too less than zero. We have:

(f (x) − g (x)) (k (x) − 1)< 0;
((x − 1) 2 − 2 2)(2 − 1)< 0;
x 2 − 2x + 1 − 4< 0;
x 2 − 2x − 3< 0;
(x − 3)(x + 1)< 0;
x ∈ (−1; 3).

We got two sets:

ODZ: x ∈ (−∞ 2/3)∪(1; +∞);
Candidate answer: x ∈ (−1; 3).

It remains to intersect these sets - we get the real answer:

We are interested in the intersection of sets, so we select intervals that are shaded on both arrows. We get x ∈ (−1; 2/3)∪(1; 3) - all points are punctured.

There is less and less time left before passing the Unified State Examination in mathematics. The situation is heating up, the nerves of schoolchildren, parents, teachers and tutors are becoming increasingly strained. Take off nervous tension Daily in-depth math classes will help you. After all, nothing, as we know, charges you with positivity and helps you pass exams like confidence in your abilities and knowledge. Today, a mathematics tutor will tell you about solving systems of logarithmic and exponential inequalities, tasks that traditionally cause difficulties for many modern high school students.

In order to learn how to solve C3 problems from the Unified State Examination in mathematics as a mathematics tutor, I recommend that you pay attention to the following important points.

1. Before you begin solving systems of logarithmic and exponential inequalities, you need to learn how to solve each of these types of inequalities separately. In particular, to understand how the range of acceptable values is located, equivalent transformations of logarithmic and exponential expressions are carried out. You can understand some of the secrets related to this by studying the articles “” and “”.

2. At the same time, it is necessary to realize that solving a system of inequalities does not always come down to solving each inequality separately and intersecting the resulting intervals. Sometimes, knowing the solution to one inequality of the system, the solution to the second becomes much simpler. As a mathematics tutor preparing schoolchildren to take final exams in Unified State Exam format, I will reveal a couple of secrets related to this in this article.

3. It is necessary to clearly understand the difference between the intersection and union of sets. This is one of the most important mathematical knowledge that an experienced professional tutor tries to give to his student from the very first lessons. A visual representation of the intersection and union of sets is given by the so-called “Euler circles”.

Intersection of sets is a set that contains only those elements that each of these sets has.

intersection

Representation of the intersection of sets using “Eulerian circles”

Explanation at your fingertips. Diana has a “set” in her purse consisting of ( pens, pencil, rulers, notebooks, combs). Alice has a “set” in her purse consisting of ( notebook , pencil, mirrors, notebooks, Chicken Kiev). The intersection of these two “sets” will be the “set” consisting of ( pencil, notebooks), since both Diana and Alice have both of these “elements”.

Important to remember! If the solution to an inequality is an interval and the solution to an inequality is an interval, then the solution to the systems is:

is the interval that is intersection original intervals. Here and belowmeans any of the signs title="Rendered by QuickLaTeX.com" height="17" width="93" style="vertical-align: -4px;">!} and under - it is the opposite sign.

Union of sets is a set that consists of all elements of the original sets.

In other words, if two sets are given and then their unification will be a set of the following form:

Depiction of set union using “Eulerian circles”

Explanation at your fingertips. The union of the “sets” taken in the previous example will be the “set” consisting of ( pens, pencil, rulers, notebooks, combs, notebook, mirrors, Chicken Kiev), since it consists of all the elements of the original “sets”. One clarification that may not be superfluous. Many can't contain identical elements.

Important to remember! If the solution to an inequality is an interval and the solution to an inequality is an interval, then the solution to the population is:

is the interval that is association original intervals.

Let's move directly to the examples.

Example 1. Solve the system of inequalities:

Solution to problem C3.

1. Let us first solve the first inequality. Using the substitution we go to the inequality:

2. Let us now solve the second inequality. The range of its permissible values is determined by the inequality:

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In the range of acceptable values, taking into account that the base of the logarithm title="Rendered by QuickLaTeX.com" height="18" width="52" style="vertical-align: -4px;"> переходим к равносильному неравенству:!}

Excluding solutions that are not within the range of acceptable values, we obtain the interval

3. Reply to system there will be inequalities intersection

The resulting intervals on the number line. The solution is their intersection

Example 2. Solve the system of inequalities:

Solution to problem C3.

1. Let's solve the first inequality first. Multiply both parts by title="Rendered by QuickLaTeX.com" height="14" width="55" style="vertical-align: 0px;"> и делаем замену в результате чего приходим к неравенству:!}

Let's move on to reverse substitution:

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Graphic representation of the resulting interval. The solution to the system is their intersection

Example 3. Solve the system of inequalities:

Solution to problem C3.

1. Let's solve the first inequality first. Multiply both parts by title="Rendered by QuickLaTeX.com" height="18" width="61" style="vertical-align: -4px;"> после чего получаем неравенство:!}

Using substitution we go to the following inequality:

Let's move on to reverse substitution:

2. Let us now solve the second inequality. Let us first determine the range of acceptable values of this inequality:

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Please note that

Then, taking into account the range of acceptable values, we obtain:

3. We find a general solution to the inequalities. Comparison of the obtained irrational values of nodal points is a task in in this example is by no means trivial. This can be done as follows. Because

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That and the final response to the system looks like:

Example 4. Solve the system of inequalities:

Solution of problem C3.

1. Let's solve the second inequality first:

2. The first inequality of the original system is a logarithmic inequality with a variable base. Convenient way solutions to such inequalities are described in the article “Complex logarithmic inequalities”; it is based on a simple formula:

Any inequality sign can be substituted for the sign, the main thing is that it is the same in both cases. Using this formula greatly simplifies solving the inequality:

Let us now determine the range of acceptable values of this inequality. It is set by the following system:

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It is easy to see that at the same time this interval will also be a solution to our inequality.

3. The final answer to the original systems there will be inequalities intersection the resulting intervals, that is

Example 5. Solve the system of inequalities:

Solution to task C3.

1. Let's solve the first inequality first. We use substitution. We proceed to the following quadratic inequality:

2. Let us now solve the second inequality. The range of its permissible values is determined by the system:

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This inequality is equivalent to the following mixed system:

In the range of acceptable values, that is, with title="Rendered by QuickLaTeX.com" height="18" width="53" style="vertical-align: -4px;"> используя равносильные преобразования переходим к следующей смешанной системе:!}

Taking into account the range of acceptable values, we obtain:

3. The final decision of the original systems is

Solution to problem C3.

1. Let's solve the first inequality first. By equivalent transformations we bring it to the form:

2. Let us now solve the second inequality. The range of its valid values is determined by the interval: title="Rendered by QuickLaTeX.com" height="14" width="68" style="vertical-align: 0px;"> Используя замену переменной переходим к следующему квадратичному неравенству:!}

This answer entirely belongs to the range of acceptable values of inequality.

3. By intersecting the intervals obtained in the previous paragraphs, we obtain the final answer to the system of inequalities:

Today we solved systems of logarithmic and exponential inequalities. Tasks of this kind were offered in trial options Unified State Examination in mathematics throughout the current academic year. However, as a mathematics tutor with experience in preparing for the Unified State Exam, I can say that this does not mean that similar tasks will be real options Unified State Examination in mathematics in June.

Let me express one warning, addressed primarily to tutors and school teachers involved in preparing high school students for passing the Unified State Exam in mathematics. It is very dangerous to prepare schoolchildren for an exam strictly on given topics, because in this case there is a risk of completely “failing” it even with minor change previously stated task format. Mathematics education must be complete. Dear colleagues, please do not liken your students to robots by so-called “training” to solve a certain type of problem. After all, there is nothing worse than the formalization of human thinking.

Good luck and creative success to everyone!

Sergey Valerievich

If you try, there are two options: it will work or it won’t work. If you don’t try, there’s only one.
© Folk wisdom

Introduction

Logarithms were invented to speed up and simplify calculations. The idea of a logarithm, that is, the idea of expressing numbers as powers of the same base, belongs to Mikhail Stiefel. But in Stiefel’s time, mathematics was not so developed and the idea of the logarithm was not developed. Logarithms were later invented simultaneously and independently of each other by the Scottish scientist John Napier (1550-1617) and the Swiss Jobst Burgi (1552-1632). Napier was the first to publish the work in 1614. under the title “Description of an amazing table of logarithms”, Napier’s theory of logarithms was given in a fairly complete volume, the method of calculating logarithms was given the simplest, therefore Napier’s merits in the invention of logarithms were greater than those of Bürgi. Bürgi worked on the tables at the same time as Napier, but for a long time kept them secret and published them only in 1620. Napier mastered the idea of the logarithm around 1594. although the tables were published 20 years later. At first he called his logarithms “artificial numbers” and only then proposed to call these “artificial numbers” in one word “logarithm”, which translated from Greek means “correlated numbers”, taken one from an arithmetic progression, and the other from a geometric progression specially selected for it. progress. The first tables in Russian were published in 1703. with the participation of a wonderful teacher of the 18th century. L. F. Magnitsky. In the development of the theory of logarithms great value had the works of St. Petersburg academician Leonhard Euler. He was the first to consider logarithms as the inverse of raising to a power; he introduced the terms “logarithm base” and “mantissa.” Briggs compiled tables of logarithms with base 10. Decimal tables are more convenient for practical use, their theory is simpler than that of Napier logarithms. Therefore, decimal logarithms are sometimes called Briggs logarithms. The term "characterization" was introduced by Briggs.

In those distant times, when the sages first began to think about equalities containing unknown quantities, there were probably no coins or wallets. But there were heaps, as well as pots and baskets, which were perfect for the role of storage caches that could hold an unknown number of items. In the ancient mathematical problems of Mesopotamia, India, China, Greece, unknown quantities expressed the number of peacocks in the garden, the number of bulls in the herd, and the totality of things taken into account when dividing property. Scribes, officials and priests initiated into secret knowledge, well trained in the science of accounts, coped with such tasks quite successfully.

Sources that have reached us indicate that ancient scientists had some general techniques for solving problems with unknown quantities. However, not a single papyrus or clay tablet contains a description of these techniques. The authors only occasionally provided their numerical calculations with skimpy comments such as: “Look!”, “Do this!”, “You found the right one.” In this sense, the exception is the “Arithmetic” of the Greek mathematician Diophantus of Alexandria (III century) - a collection of problems for composing equations with a systematic presentation of their solutions.

However, the first manual for solving problems that became widely known was the work of the Baghdad scientist of the 9th century. Muhammad bin Musa al-Khwarizmi. The word "al-jabr" from the Arabic name of this treatise - "Kitab al-jaber wal-mukabala" ("Book of restoration and opposition") - over time turned into the well-known word "algebra", and the work of al-Khwarizmi itself served the starting point in the development of the science of solving equations.

Logarithmic equations and inequalities

1. Logarithmic equations

An equation containing an unknown under the logarithm sign or at its base is called a logarithmic equation.

The simplest logarithmic equation is an equation of the form

log a x = b . (1)

Statement 1. If a > 0, a≠ 1, equation (1) for any real b has the only solution x = a b .

Example 1. Solve the equations:

a)log 2 x= 3, b) log 3 x= -1, c)

Solution. Using Statement 1, we obtain a) x= 2 3 or x= 8; b) x= 3 -1 or x= 1 / 3 ; c)

or x = 1.

Let us present the basic properties of the logarithm.

P1. Basic logarithmic identity:

Where a > 0, a≠ 1 and b > 0.

P2. The logarithm of the product of positive factors is equal to the sum of the logarithms of these factors:

log a N 1 · N 2 = log a N 1 + log a N 2 (a > 0, a ≠ 1, N 1 > 0, N 2 > 0).

Comment. If N 1 · N 2 > 0, then property P2 takes the form

log a N 1 · N 2 = log a |N 1 | + log a |N 2 | (a > 0, a ≠ 1, N 1 · N 2 > 0).

P3. The logarithm of the quotient of two positive numbers is equal to the difference between the logarithms of the dividend and the divisor

(a > 0, a ≠ 1, N 1 > 0, N 2 > 0).

Comment. If

, (which is equivalent N 1 N 2 > 0) then property P3 takes the form

(a > 0, a ≠ 1, N 1 N 2 > 0).

P4. Logarithm of degree positive number is equal to the product of the exponent and the logarithm of this number:

log a N k = k log a N (a > 0, a ≠ 1, N > 0).

Comment. If k- even number ( k = 2s), That

log a N 2s = 2s log a |N | (a > 0, a ≠ 1, N ≠ 0).

P5. Formula for moving to another base:

(a > 0, a ≠ 1, b > 0, b ≠ 1, N > 0),

in particular if N = b, we get

(a > 0, a ≠ 1, b > 0, b ≠ 1). (2)

Using properties P4 and P5, it is easy to obtain following properties

(a > 0, a ≠ 1, b > 0, c ≠ 0), (3) (a > 0, a ≠ 1, b > 0, c ≠ 0), (4)

(a > 0, a ≠ 1, b > 0, c ≠ 0), (5)

and, if in (5) c- even number ( c = 2n), holds

(b > 0, a ≠ 0, |a | ≠ 1). (6)

Let us list the main properties of the logarithmic function f (x) = log a x :

1. The domain of definition of a logarithmic function is the set of positive numbers.

2. The range of values of the logarithmic function is the set of real numbers.

3. When a> 1 logarithmic function is strictly increasing (0< x 1 < x 2log a x 1 < loga x 2), and at 0< a < 1, - строго убывает (0 < x 1 < x 2log a x 1 > log a x 2).

4.log a 1 = 0 and log a a = 1 (a > 0, a ≠ 1).

5. If a> 1, then the logarithmic function is negative when x(0;1) and positive at x(1;+∞), and if 0< a < 1, то логарифмическая функция положительна при x (0;1) and negative at x (1;+∞).

6. If a> 1, then the logarithmic function is convex upward, and if a(0;1) - convex downwards.

The following statements (see, for example,) are used when solving logarithmic equations.

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- these are inequalities that have a variable under the sign of the logarithm 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 deciding 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 DZ 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 successful solution 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 ODZ of inequality, 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 studied during school 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: