Among all the 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 told at 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.
So we get rid of logarithms and reduce the problem to rational inequality. The latter is much easier to solve, but when dropping logarithms, unnecessary roots may appear. To cut them off, it is enough to find the area acceptable values... If you have forgotten the ODZ of the logarithm, I strongly recommend repeating it - see "What is a logarithm".
Everything related to the range of permissible 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 fulfilled simultaneously. When the range of valid values is found, it remains to cross it with the solution rational inequality- and the answer is ready.
Task. Solve the inequality:
To begin with, let's write out the ODZ of the logarithm:
The first two inequalities are fulfilled automatically, and the last one will have to be described. Since the square of the 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 the rational. In the original inequality there is a "less" sign, which means that the resulting inequality must also be with a "less" 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: 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 that this is the answer.
Transforming Logarithmic Inequalities
Often the original inequality differs from the one above. It is easy to fix it according to the standard rules for 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 with one logarithm.
I would also like to remind you about the range of acceptable values. Since the original inequality may contain several logarithms, it is required to find the ODV for each of them. In this way, general scheme solutions of logarithmic inequalities are as follows:
- Find the ODV of each logarithm included in the inequality;
- Reduce inequality to the standard one according to the formulas for addition and subtraction of logarithms;
- Solve the resulting inequality according to the scheme given above.
Task. Solve the inequality:
Let's find the domain of definition (ODZ) of the first logarithm:
We solve by the method of intervals. Find the zeros of the numerator:
3x - 2 = 0;
x = 2/3.
Then - the zeros of the denominator:
x - 1 = 0;
x = 1.
We mark the zeros and signs on the coordinate arrow:
We get x ∈ (−∞ 2/3) ∪ (1; + ∞). The second logarithm of ODV will be the same. If you don’t believe it, you can check it. Now we transform the second logarithm so that there is a two at the base:
As you can see, the triplets at the base and in front of the logarithm have contracted. Received two logarithms with on the same basis... We add them:
log 2 (x - 1) 2< 2;
log 2 (x - 1) 2< log 2 2 2 .
Received the standard logarithmic inequality. We get rid of the logarithms by 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:
- ODZ: x ∈ (−∞ 2/3) ∪ (1; + ∞);
- Candidate answer: x ∈ (−1; 3).
It remains to cross these sets - we get the real answer:
We are interested in the intersection of sets, so select the intervals filled in on both arrows. We get x ∈ (−1; 2/3) ∪ (1; 3) - all points are punctured.
Logarithmic inequalities
In the previous lessons, we met with logarithmic equations and now we know what it is and how to solve them. And 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 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 the logarithmic equation, will be under the sign of the logarithm.
The simplest logarithmic inequalities are as follows:
where f (x) and g (x) are some expressions that depend on x.
Let's look at this with an example: f (x) = 1 + 2x + x2, g (x) = 3x − 1.
Solving logarithmic inequalities
Before solving logarithmic inequalities, it is worth noting that when solving them, they resemble exponential inequalities, namely:
First, when moving from logarithms to expressions under the sign of the logarithm, we also need to compare the base of the logarithm with one;
Second, solving the logarithmic inequality using a change of variables, we need to solve the inequality about the change until we get the simplest inequality.
But you and I have considered similar aspects of solving logarithmic inequalities. And 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, passing from logarithms to expressions under the sign of the logarithm, it is necessary to take into account the range of permissible values (ADV).
That is, it should be borne in mind that when deciding logarithmic equation you and I, we can first find the roots of the equation, and then check this solution. But solving the logarithmic inequality will not work that way, since passing from logarithms to expressions under the sign of the logarithm, it will be necessary to write down the ODZ of 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 must use the following record: 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.
Solving inequalities with a variable, you need to find all its solutions. If two inequalities have one variable x, then such inequalities are equivalent, provided that their solutions coincide.
When performing tasks for solving logarithmic inequalities, it is necessary to 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.
Ways to solve logarithmic inequalities
Now let's look at some of the ways that take place when solving logarithmic inequalities. For a better understanding and assimilation, we will try to understand them with specific examples.
You and I know that the simplest logarithmic inequality has the following form:
In this inequality, V - is one of such inequality signs as:<,>, ≤ or ≥.
When the base of this logarithm is greater than one (a> 1), making the transition from logarithms to expressions under the sign of the logarithm, then in this version the inequality sign is preserved, and the inequality will look like this:
which is equivalent to such a 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 see more examples of solving the simplest logarithmic inequalities shown in the picture below: Exercise. Let's try to solve this inequality: Solution of the range of valid values. Now let's try to multiply its right side by: Let's see what we get: Now, let's move on to the transformation of sub-logarithmic 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; And from this it follows that the interval that we have obtained is wholly and completely owned by the GDZ and is a solution to such an inequality. Here's our answer: Now let's try to analyze what we need to successfully solve logarithmic inequalities? First, focus 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 prevent the expansion and contraction of the ODZ 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 inequality, while being guided by its ODV. Thirdly, in order to successfully solve such inequalities, each of you must perfectly know all the properties of elementary functions and clearly understand their meaning. These functions include not only logarithmic, but also rational, power, trigonometric, etc., in a word, all those that you studied during schooling algebra. As you can see, having studied the topic of logarithmic inequalities, there is nothing difficult in solving these inequalities, provided you are attentive and persistent in achieving your goals. To avoid any problems in solving inequalities, you need to train as much as possible, solving various tasks and at the same time memorize the main ways of solving such inequalities and their systems. In case of unsuccessful solutions to logarithmic inequalities, you should carefully analyze your mistakes so as not to return to them again in the future. For a better understanding of the topic and consolidation of the passed material, solve the following inequalities: 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 read our privacy policy and let us know if you have any questions. Personal information refers to data that can be used to identify a specific person or contact him. 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 we collect: How we use your personal information: We do not disclose information received from you to third parties. Exceptions: We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and abuse, as well as from unauthorized access, disclosure, alteration and destruction. In order to make sure that your personal information is safe, we bring the rules of confidentiality and security to our employees, and strictly monitor the implementation of confidentiality measures. 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 under logarithmic functions should 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. 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)
Solution examples
3x> 24;
x> 8. 
What is needed to solve logarithmic inequalities?
Homework
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Practice.