Logarithmic inequalities solution examples online. Logarithmic inequalities

An inequality is called logarithmic if it contains a logarithmic function.

The methods for solving logarithmic inequalities are no different from, except for two things.

First, when passing from a logarithmic inequality to an inequality of sub-logarithmic functions, it follows that watch the sign of the resulting inequality... He obeys the following rule.

If the base of the logarithmic function is greater than $ 1 $, then when passing from the logarithmic inequality to the inequality of sub-logarithmic functions, the sign of the inequality is preserved, and if it is less than $ 1 $, then it changes to the opposite.

Secondly, the solution to any inequality is an interval, and, therefore, at the end of the solution to the inequality of sub-logarithmic functions, it is necessary to compose a system of two inequalities: the first inequality of this system will be the inequality of sub-logarithmic functions, and the second is the interval of the domain of definition of logarithmic functions included in the logarithmic inequality.

Practice.

Let's solve the inequalities:

1. $ \ log_ (2) ((x + 3)) \ geq 3. $

$ D (y): \ x + 3> 0. $

$ x \ in (-3; + \ infty) $

The base of the logarithm is $ 2> 1 $, so the sign does not change. Using the definition of the logarithm, we get:

$ x + 3 \ geq 2 ^ (3), $

$ x \ in \)

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

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

Solution:

\ (\ log \) \ (_ (\ frac (1) (3)) ⁡ (\ frac (3x-2) (2x-3)) \)\(≤-1\)	Let's write out ODZ.
ODZ: \ (\ frac (3x-2) (2x-3) \) \ (> 0 \)
\ (⁡ \ frac (3x-2-3 (2x-3)) (2x-3) \)\(≥\) \(0\)	We open the brackets, we give.
\ (⁡ \ frac (-3x + 7) (2x-3) \) \ (≥ \) \ (0 \)	We multiply the inequality by \ (- 1 \), not forgetting to reverse the comparison sign.
\ (⁡ \ frac (3x-7) (2x-3) \) \ (≤ \) \ (0 \)
\ (⁡ \ frac (3 (x- \ frac (7) (3))) (2 (x- \ frac (3) (2))) \)\(≤\) \(0\)	Let's construct a number axis and mark the points \ (\ frac (7) (3) \) and \ (\ frac (3) (2) \ on it. Note that the dot from the denominator is punctured, despite the fact that the inequality is not strict. The point is that this point will not be a solution, since when substituted into inequality, it will lead us to division by zero.
\ (x∈ (\) \ (\ frac (3) (2) \) \ (; \) \ (\ frac (7) (3)] \)	Now, on the same numerical axis, we plot the ODZ and write in response the interval that falls into the ODZ.
	We write down the final answer.

Answer: \ (x∈ (\) \ (\ frac (3) (2) \) \ (; \) \ (\ frac (7) (3)] \)

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

Solution:

\ (\ log ^ 2_3⁡x- \ log_3⁡x-2> 0 \)	Let's write out ODZ.
ODZ: \ (x> 0 \)	Let's get down to the solution.
Solution: \ (\ log ^ 2_3⁡x- \ log_3⁡x-2> 0 \)	We have before us a typical square-logarithmic inequality. We do it.
\ (t = \ log_3⁡x \) \ (t ^ 2-t-2> 0 \)	Expand the left side of the inequality into.
\ (D = 1 + 8 = 9 \) \ (t_1 = \ frac (1 + 3) (2) = 2 \) \ (t_2 = \ frac (1-3) (2) = - 1 \) \ ((t + 1) (t-2)> 0 \)
	Now you need to go back to the original variable - x. To do this, go to one that has the same solution and make the reverse replacement.
\ (\ left [\ begin (gathered) t> 2 \\ t<-1 \end{gathered} \right.\) \(\Leftrightarrow\) \(\left[ \begin{gathered} \log_3⁡x>2 \\ \ log_3⁡x<-1 \end{gathered} \right.\)	Convert \ (2 = \ log_3⁡9 \), \ (- 1 = \ log_3⁡ \ frac (1) (3) \).
\ (\ left [\ begin (gathered) \ log_3⁡x> \ log_39 \\ \ log_3⁡x<\log_3\frac{1}{3} \end{gathered} \right.\)	We make the transition to the comparison of arguments. The bases of logarithms are greater than \ (1 \), so the sign of the inequalities does not change.
\ (\ left [\ begin (gathered) x> 9 \\ x<\frac{1}{3} \end{gathered} \right.\)	Let us combine the solution of inequality and the DHS in one figure.
	Let's write down the answer.

Answer: \ ((0; \ frac (1) (3)) ∪ (9; ∞) \)

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Introduction

Logarithms were invented to speed up and simplify calculations. The idea of ​​the logarithm, that is, the idea of ​​expressing numbers as a power of the same base, belongs to Mikhail Shtifel. But at the time of Stiefel, mathematics was not so developed and the idea of ​​the logarithm did not find its development. 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 his work in 1614. under the title "Description of the amazing table of logarithms", Napier's theory of logarithms was given in a fairly complete volume, the method for calculating logarithms was given the simplest, therefore Napier's contribution to the invention of logarithms was greater than that of Burghi. Burghi worked on the tables at the same time as Napier, but for a long time kept them secret and published only in 1620. Napier mastered the idea of ​​the logarithm around 1594. although the tables were published after 20 years. At first, he called his logarithms "artificial numbers" and only then suggested that these "artificial numbers" be called in one word "logarithm", which is translated from Greek as "related numbers" 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, the works of the St. Petersburg academician Leonard Euler were of great importance. He was the first to consider logarithm as the inverse of raising to a power, he introduced the terms "base of the logarithm" and "mantissa" Briggs compiled tables of logarithms with base 10. Decimal tables are more convenient for practical use, their theory is simpler than Napier's logarithms ... Therefore, decimal logarithms are sometimes called brigs logarithms. The term "characteristic" was coined by Briggs.

In those distant times, when sages first began to think about equalities containing unknown quantities, there probably were no coins or wallets yet. But on the other hand, there were heaps, as well as pots, baskets, which perfectly suited the role of caches-storage, containing 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, the totality of things taken into account when dividing property. Scribes, officials well trained in the science of counting, and priests initiated into secret knowledge were quite successful in coping with such tasks.

Sources that have come down to us testify that ancient scientists possessed some general methods of solving problems with unknown quantities. However, not a single papyrus or a single clay tablet contains a description of these techniques. The authors only occasionally supplied their numerical calculations with scanty comments such as: "Look!", "Do this!", "You found it right." In this sense, an exception is the "Arithmetic" of the Greek mathematician Diophantus of Alexandria (III century) - a collection of problems for the compilation of equations with a systematic presentation of their solutions.

However, the first widely known guide to solving problems was the work of a Baghdad scholar of the 9th century. Muhammad bin Musa al-Khwarizmi. The word "al-jabr" from the Arabic name of this treatise - "Kitab al-jerber wal-muqabala" ("Book of restoration and opposition") - over time turned into the well-known word "algebra", and al-Khwarizmi's work itself served starting point in the formation of the science of solving equations.

Logarithmic equations and inequalities

1. Logarithmic equations

An equation containing an unknown under the sign of the logarithm 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 It has only decision 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.

Here are the main 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 to 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 exponent per 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), then

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

P5. The formula for the transition 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 the 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), occurs

(b > 0, a ≠ 0, |a | ≠ 1). (6)

We also list the main properties of the logarithmic function f (x) = log a x :

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

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

3. When a> 1 the logarithmic function is strictly increasing (0< x 1 < x 2 log a x 1 < loga x 2), and at 0< a < 1, - строго убывает (0 < x 1 < x 2 log 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 for x(0; 1) and is positive for x(1; + ∞), and if 0< a < 1, то логарифмическая функция положительна при x (0; 1) and is negative for x (1;+∞).

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

The following statements (see, for example) are used to solve logarithmic equations.

Do you think that there is still time before the exam, and you will have time to prepare? Perhaps this is so. But in any case, the earlier the student begins training, the more successfully he passes the exams. Today we decided to devote an article to logarithmic inequalities. This is one of the tasks, which means an opportunity to get an extra point.

Do you already know what a logarithm is? We really hope so. But even if you don’t have an answer to this question, it’s not a problem. It is very easy to understand what a logarithm is.

Why exactly 4? You need to raise the number 3 to such a power to get 81. When you understand the principle, you can proceed to more complex calculations.

You passed the inequalities a few years ago. And since then they are constantly encountered in mathematics. If you have problems solving inequalities, see the corresponding section.
Now that we have gotten to know the concepts separately, 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 needed? To better understand how to solve inequality with logarithms. Now we will give a more applicable example, it is still quite simple, we will leave complex logarithmic inequalities for later.

How to solve this? It all starts with ODZ. It is worth knowing more about it if you want to always easily solve any inequality.

What is ODU? ODV for logarithmic inequalities

The abbreviation stands for range of valid values. In tasks for the exam, this wording often pops up. ODZ is useful to you not only in the case of logarithmic inequalities.

Take another look at the above example. We will consider the DHS based on it, so that you understand the principle, and the solution of logarithmic inequalities does not raise any questions. From the definition of the 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 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 admissible values.
Now let's move on to solving the simplest logarithmic inequality.

We discard the logarithms themselves from both sides of the inequality. What do we have left as a result? Simple inequality.

It is not difficult to solve it. X must be greater than -0.5. Now we combine the two obtained values ​​into the system. Thus,

This will be the range of admissible values ​​for the considered logarithmic inequality.

Why do you 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 exam often there is 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 valid values. There will be two values ​​in the ODZ, we discussed this above. Next, you need to solve the inequality itself. Solution methods are as follows:

• multiplier replacement method;
• decomposition;
• method of rationalization.

Depending on the situation, you should use one of the above methods. Let's go directly to the solution. We will reveal the most popular method that is suitable for solving USE tasks in almost all cases. Next, we'll look at the decomposition method. It can help if you come across particularly tricky inequalities. So, the algorithm for solving the logarithmic inequality.

Solution examples :

We have not taken just such an inequality for nothing! Pay attention to the base. Remember: if it is greater than one, the sign remains the same when the range of acceptable values ​​is found; otherwise, the inequality sign must be changed.

As a result, we get the inequality:

Now we bring the left side to the form of the equation, equal to zero... Instead of the sign “less” we put “equal”, solve the equation. Thus, we will find the ODZ. We hope that with a solution to this simple equation you won't have a problem. Answers are -4 and -2. That's not all. You need to display these points on the chart, place the "+" and "-". What needs to be done for this? Substitute numbers from intervals into the expression. Where the values ​​are positive, we put "+" there.

Answer: x cannot be more than -4 and less than -2.

We found the range of valid values ​​only for the left side, now we need to find the range of valid values ​​for the right side. This is much easier. Answer: -2. We intersect both obtained areas.

And only now are we beginning to address inequality itself.

Let's simplify it as much as possible to make it easier to solve.

Apply the spacing method again in the solution. Let's omit the calculations, with him everything is already clear from the previous example. Answer.

But this method is suitable if the logarithmic inequality has the same basis.

Solving logarithmic equations and inequalities with different bases assumes initial reduction to one base. Then follow the above method. But there is also a more complicated case. Consider one of the most difficult types of logarithmic inequalities.

Variable base logarithmic inequalities

How to solve inequalities with such characteristics? Yes, and such can be found in the exam. Solving inequalities in the following way will also be useful for your educational process... Let's look at the issue in detail. Let's discard the theory, let's go straight to practice. To solve logarithmic inequalities, it is enough to read the example once.

To solve the logarithmic inequality of the presented form, it is necessary to bring right side to the logarithm with the same base. The principle resembles equivalent transitions. As a result, the inequality will look like this.

Actually, it remains to create a system of inequalities without logarithms. Using the rationalization method, we pass to an equivalent system of inequalities. You will understand the rule itself when you substitute the appropriate values ​​and track their changes. The system will have the following inequalities.

Using the rationalization method when solving inequalities, you need to remember the following: it is necessary to subtract one from the base, x, by the 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 with respect to zero.

Further solution is carried out by the method of intervals, 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 easy enough to solve. How to make sure that you can 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. Practice consistently solving the most different tasks within the exam and you will be able to get the highest score. Good luck in your difficult business!