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 a 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 (log) 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 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 went through the inequalities several 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 always want to easily solve any inequality.
What is ODU? ODZ 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 example above. We will consider the DHS based on it, so that you understand the principle, and the solution of 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 must be positive by definition. Solve the inequality above. This can be done even 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 \u200b\u200binto the system. Thus,
This will be the range of admissible values \u200b\u200bfor 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 \u200b\u200bin 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 \u200b\u200bis 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 you won't have any problems solving such a simple equation. The 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 \u200b\u200bare positive, we put "+" there.
Answer: x cannot be more than -4 and less than -2.
We found the range of valid values \u200b\u200bonly for the left side, now we need to find the range of valid values \u200b\u200bfor 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.
The solution of logarithmic equations and inequalities with different bases involves the 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 beneficial for your educational process. We will understand 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 reduce the right-hand 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 method of rationalization, we pass to an equivalent system of inequalities. You will understand the rule itself when you substitute the corresponding values \u200b\u200band 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 the left), two expressions are multiplied and set under the original sign with respect to zero.
The 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 in solving a variety of problems within the exam and you will be able to get the highest score. Good luck in your difficult business!
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 range of 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 \u200b\u200bmust be written out and solved separately:
f (x)\u003e 0; g (x)\u003e 0; k (x)\u003e 0; k (x) ≠ 1.
These four inequalities constitute a system and must be fulfilled simultaneously. When the range of acceptable values \u200b\u200bis found, it remains to cross it with the solution of rational inequality - and the answer is ready.
Task. Solve the inequality:
First, 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 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 carry out the transition from a logarithmic inequality to a rational one. 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 \u003d 3; x \u003d −3; x \u003d 0. Moreover, x \u003d 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.
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 valid values. Since the original inequality may contain several logarithms, it is required to find the ODV for each of them. Thus, the general scheme for solving logarithmic inequalities is 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 (ODV) of the first logarithm:
We solve by the method of intervals. Find the zeros of the numerator:
3x - 2 \u003d 0;
x \u003d 2/3.
Then the zeros of the denominator:
x - 1 \u003d 0;
x \u003d 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. Do not believe it - you can check. 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 the same base. 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 we select the intervals filled in on both arrows. We get x ∈ (−1; 2/3) ∪ (1; 3) - all points are punctured.
Often, when solving logarithmic inequalities, problems with a variable base of the logarithm are encountered. So, an inequality of the form
is a standard school inequality. As a rule, to solve it, a transition to an equivalent set of systems is applied:
The disadvantage of this method is the need to solve seven inequalities, not counting two systems and one set. Already with given quadratic functions, solving a set can be time-consuming.
An alternative, less laborious way of solving this standard inequality can be proposed. For this we take into account the following theorem.
Theorem 1. Let a continuous increasing function on the set X. Then on this set the sign of the increment of the function will coincide with the sign of the increment of the argument, that is, where 
.
Note: if a continuous decreasing function on the set X, then.
Let's go back to inequality. Let's go to the decimal logarithm (you can go to any with a constant base greater than one).
Now you can use the theorem, noting in the numerator the increment of the functions 
and in the denominator. So it is true
As a result, the number of calculations leading to the answer is approximately halved, which not only saves time, but also allows you to potentially make fewer arithmetic and "carelessness" errors.
Example 1.
Comparing with (1) we find 
, 
, .
Passing to (2) we will have:
Example 2.
Comparing with (1) we find,,.
Passing to (2) we will have:
Example 3.
Since the left side of the inequality is an increasing function for and 
, then the answer is set.
The set of examples in which Theorem 1 can be applied can be easily extended if Theorem 2 is taken into account.
Let on the set X functions,,, and on this set the signs and coincide, i.e. , then it will be fair.
Example 4.
Example 5.
With the standard approach, the example is solved according to the scheme: the product is less than zero, when the factors are of opposite signs. Those. the set of two systems of inequalities is considered, in which, as was indicated at the beginning, each inequality splits into seven more.
If we take into account Theorem 2, then each of the factors, taking into account (2), can be replaced by another function that has the same sign in this example O.D.Z.
The method of replacing the increment of a function with an increment of the argument, taking into account Theorem 2, turns out to be very convenient when solving typical problems C3 of the exam.
Example 6.
Example 7.
... Let us denote. We get
... Note that the replacement implies:. Returning to the equation, we get
.
Example 8.
In the theorems we use, there is no restriction on the classes of functions. In this article, for example, the theorems have been applied to the solution of logarithmic inequalities. The next few examples will demonstrate the promise of the method for solving other types of inequalities.