Examples of solving logarithmic inequalities of increased complexity. All about logarithmic inequalities. Analysis of examples

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 area 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 present 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 with a solution to this simple equation you won't have any problems. 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 for different reasons presupposes an initial reduction to one base. Next, use the method described above. But there is more difficult case. Let's consider one of the most complex species 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 benefit your educational process. Let's understand 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.

The 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 the most different tasks within the exam and you will be able to get the highest score. Good luck to you in your difficult task!

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 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 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 ​​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;
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 will have the same VA. 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.

Definition of logarithm The easiest way to write it mathematically is:

The definition of logarithm can be written in another way:

Pay attention to the restrictions that are imposed on the base of the logarithm ( a) and to the sublogarithmic expression ( x). In the future, these conditions will turn into important restrictions for OD, which will need to be taken into account when solving any equation with logarithms. So, now, in addition to the standard conditions leading to restrictions on ODZ (positivity of expressions under the roots of even powers, non-equal denominator to zero, etc.), the following conditions must also be taken into account:

• Sublogarithmic expression can only be positive.
• The base of the logarithm can only be positive and not equal to one.

Note that neither the base of the logarithm nor the sublogarithmic expression can be equal to zero. Please also note that the logarithm value itself can take on all possible values, i.e. The logarithm can be positive, negative or zero. Logarithms have a lot various properties, which follow from the properties of powers and the definition of the logarithm. Let's list them. So, the properties of logarithms:

Logarithm of the product:

Logarithm of a fraction:

Taking the degree out of the sign of the logarithm:

Pay especially close attention to those of the last listed properties in which the modulus sign appears after the degree is taken. Do not forget that when placing an even power outside the logarithm sign, under the logarithm or at the base, you must leave the modulus sign.

Other beneficial features logarithms:

The last property is very often used in complex logarithmic equations and inequalities. He should be remembered as well as everyone else, although he is often forgotten.

The simplest logarithmic equations look like:

And their solution is given by a formula that directly follows from the definition of the logarithm:

Other simplest logarithmic equations are those that, using algebraic transformations and the above formulas and properties of logarithms, can be reduced to the form:

The solution to such equations taking into account the ODZ is as follows:

Some others logarithmic equations with a variable at the base can be reduced to the form:

In such logarithmic equations general form the solution also follows directly from the definition of the logarithm. Only in this case there are additional restrictions for DZ that need to be taken into account. As a result, to solve a logarithmic equation with a variable in the base, you need to solve the following system:

When solving more complex logarithmic equations that cannot be reduced to one of the equations presented above, it is also actively used variable replacement method. As usual, when using this method, you need to remember that after introducing the replacement, the equation should simplify and no longer contain the old unknown. You also need to remember to perform reverse substitution of variables.

Sometimes when solving logarithmic equations you also have to use graphic method . This method consists in constructing graphs of functions that are on the left and right sides of the equation as accurately as possible on one coordinate plane, and then finding the coordinates of their intersection points from the drawing. The roots obtained in this way must be checked by substitution into the original equation.

When solving logarithmic equations it is often also useful grouping method. When using this method, the main thing to remember is that: in order for the product of several factors to be equal to zero, it is necessary that at least one of them is equal to zero, and the rest existed. When the factors are logarithms or parentheses with logarithms, and not just parentheses with variables as in rational equations, then many errors may occur. Since logarithms have many restrictions on the region where they exist.

When deciding systems of logarithmic equations most often you have to use either the substitution method or the variable replacement method. If there is such a possibility, then when solving systems of logarithmic equations, one must strive to ensure that each of the equations of the system is individually brought to a form in which it will be possible to make the transition from a logarithmic equation to a rational one.

The simplest logarithmic inequalities are solved in approximately the same way as similar equations. First, using algebraic transformations and the properties of logarithms, we must try to bring them to a form where the logarithms on the left and right sides of the inequality will have the same bases, i.e. get an inequality of the form:

After which you need to move to a rational inequality, taking into account that this transition should be performed as follows: if the base of the logarithm is greater than one, then the sign of the inequality does not need to be changed, and if the base of the logarithm is less than one, then you need to change the sign of the inequality to the opposite (this means changing "less" to "more" or vice versa). In this case, there is no need to change the minus signs to plus ones, bypassing the previously learned rules. Let's write down mathematically what we get as a result of performing such a transition. If the base is greater than one we get:

If the base of the logarithm is less than one, we change the sign of the inequality and get the following system:

As we see, when solving logarithmic inequalities, as usual, the ODZ is also taken into account (this is the third condition in the systems above). Moreover, in this case it is possible not to require the positivity of both sublogarithmic expressions, but rather to require only the positivity of the smaller of them.

When deciding logarithmic inequalities with a variable at the base logarithm, it is necessary to independently consider both options (when the base is less than one and greater than one) and combine the solutions of these cases into a set. At the same time, we must not forget about DL, i.e. about the fact that both the base and all sublogarithmic expressions must be positive. Thus, when solving an inequality of the form:

We obtain the following set of systems:

More complex logarithmic inequalities can also be solved using changes of variables. Some other logarithmic inequalities (as well as logarithmic equations) require the procedure of taking the logarithm of both sides of the inequality or equation to be solved. same basis. So, when carrying out such a procedure with logarithmic inequalities, there is a subtlety. Please note that when taking logarithms to a base greater than one, the inequality sign does not change, but if the base is less than one, then the inequality sign is reversed.

If a logarithmic inequality cannot be reduced to a rational one or solved by substitution, then in this case one must use generalized interval method, which is as follows:

• Define DL;
• Transform the inequality so that there is a zero on the right side (on the left side, if possible, reduce to common denominator, factorize, etc.);
• Find all the roots of the numerator and denominator and plot them on the number axis, and if the inequality is not strict, paint over the roots of the numerator, but in any case leave the roots of the denominator as dotted out;
• Find the sign of the entire expression on each of the intervals by substituting a number from a given interval into the transformed inequality. In this case, it is no longer possible to alternate signs in any way when passing through points on the axis. It is necessary to determine the sign of an expression on each interval by substituting the value from the interval into this expression, and so on for each interval. There’s no way else (that’s what it’s all about, by and large, the difference between the generalized interval method and the usual one);
• Find the intersection of the ODZ and intervals that satisfy the inequality, but do not lose individual points that satisfy the inequality (the roots of the numerator in non-strict inequalities), and do not forget to exclude from the answer all the roots of the denominator in all inequalities.

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How to successfully prepare for the CT in physics and mathematics?

In order to successfully prepare for the CT in physics and mathematics, among other things, it is necessary to fulfill three most important conditions:

1. Study all topics and complete all tests and assignments given in the educational materials on this site. To do this, you need nothing at all, namely: devote three to four hours every day to preparing for the CT in physics and mathematics, studying theory and solving problems. The fact is that CT is an exam where it is not enough just to know physics or mathematics, you also need to be able to solve it quickly and without failures a large number of tasks for different topics And of varying complexity. The latter can only be learned by solving thousands of problems.
2. Learn all the formulas and laws in physics, and formulas and methods in mathematics. In fact, this is also very simple to do, necessary formulas in physics there are only about 200 pieces, and in mathematics even a little less. Each of these subjects has about a dozen standard methods for solving problems basic level difficulties that can also be learned, and thus solved completely automatically and without difficulty right moment most of the DH. After this, you will only have to think about the most difficult tasks.
3. Attend all three stages of rehearsal testing in physics and mathematics. Each RT can be visited twice to decide on both options. Again, on the CT, in addition to the ability to quickly and efficiently solve problems, and knowledge of formulas and methods, you must also be able to properly plan time, distribute forces, and most importantly, correctly fill out the answer form, without confusing the numbers of answers and problems, or your own last name. Also, during RT, it is important to get used to the style of asking questions in problems, which may seem very unusual to an unprepared person at the DT.

Successful, diligent and responsible implementation of these three points will allow you to show up on the CT excellent result, the maximum of what you are capable of.

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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 logarithmic function has limited area definitions, therefore, moving from logarithms to expressions under the logarithm sign, you need to take into account the range of permissible values ​​(APV).

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 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 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: