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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 dedicate 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. 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 went through the inequalities a few years ago. And since then, you constantly meet them in mathematics. If you're having trouble solving inequalities, check out the appropriate section.
Now, when we have got acquainted with concepts separately, we will pass to their consideration 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 give a more applicable example, still quite simple, we leave complex logarithmic inequalities for later.
How to solve it? It all starts with ODZ. You should know more about it if you want to always easily solve any inequality.
What is ODZ? DPV for logarithmic inequalities
The abbreviation stands for area allowed values. In assignments for the exam, this wording often pops up. DPV is 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 the solution of logarithmic inequalities does not raise questions. It follows from the definition of the logarithm 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 presented above. This can even be done orally, here it is clear that X cannot be less than 2. The solution of 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 parts of the inequality. What is left for us as a result? simple inequality.
It's easy to solve. X must be greater than -0.5. Now we combine the two obtained values into the system. In this way,
This will be the region of admissible values for the considered logarithmic inequality.
Why is ODZ needed 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 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 steps. First, it is necessary to find the range of acceptable values. There will be two values in the ODZ, we considered this above. The next step is to solve the inequality itself. The solution methods are as follows:
- multiplier replacement method;
- decomposition;
- rationalization method.
Depending on the situation, one of the above methods should be used. Let's go straight to the solution. We will reveal the most popular method that is suitable for solving USE tasks in almost all cases. Next, we will consider the decomposition method. It can help if you come across a particularly "tricky" inequality. So, the algorithm for solving the logarithmic inequality.
Solution examples :
It is not in vain that we took precisely such an inequality! Pay attention to the base. Remember: if it is greater than one, the sign remains the same when finding the range of valid values; 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, zero. Instead of the “less than” sign, we put “equal”, we solve the equation. Thus, we will find the ODZ. We hope that with the solution of such simple equation you won't have a problem. The answers are -4 and -2. That's not all. You need to display these points on the chart, place "+" and "-". What needs to be done for this? Substitute 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 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 by no means easier. Answer: -2. We intersect both received areas.
And only now we begin to solve the inequality itself.
Let's simplify it as much as possible to make it easier to decide.
We again use the interval method in the solution. Let's skip 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 bases.
Solution logarithmic equations and inequalities with different grounds presupposes an initial reduction to one basis. Then use the above method. But there is more difficult case. Consider one of the most complex types logarithmic inequalities.
Logarithmic inequalities with variable base
How to solve inequalities with such characteristics? Yes, and such can be found in the exam. Solving inequalities in the following way will also have a beneficial effect on your educational process. Let's understand the issue in detail. Let's put theory aside and go straight to practice. To solve logarithmic inequalities, it is enough to once familiarize yourself with the example.
To solve the logarithmic inequality of the presented form, it is necessary to bring the 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 follow their changes. The system will have the following inequalities.
Using the rationalization method, when solving inequalities, you need to remember the following: you need to subtract one from the base, x, by definition of the logarithm, is subtracted from both parts of the inequality (the right from the left), the two expressions are multiplied and set under the original sign relative to zero.
The further solution is carried out by the interval method, everything is simple here. It is important for you to understand the differences in the 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 it so that to 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 various problems within the exam and you will be able to get the highest score. Good luck in your difficult work!
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.
Collection and use of personal information
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.
The following are some examples of the types of personal information we may collect and how we may use such information.
What personal information we collect:
- When you submit an application on the site, we may collect various information, including your name, phone number, email address, etc.
How we use your personal information:
- The personal information we collect allows us to contact you and inform you about unique offers, promotions and other events and upcoming events.
- From time to time, we may use your personal information to send you important notices and messages.
- We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
- If you enter a prize draw, contest or similar incentive, we may use the information you provide to administer such programs.
Disclosure to third parties
We do not disclose information received from you to third parties.
Exceptions:
- In the event that it is necessary - in accordance with the law, judicial order, in legal proceedings, and / or based on public requests or requests from state bodies in the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public interest purposes.
- In the event of a reorganization, merger or sale, we may transfer the personal information we collect to the relevant third party successor.
Protection of personal information
We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as from unauthorized access, disclosure, alteration and destruction.
Maintaining your privacy at the company level
To ensure that your personal information is secure, we communicate privacy and security practices to our employees and strictly enforce privacy practices.
Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved according to a special formula, which for some reason is rarely taught at school:
log k (x ) f (x ) ∨ log k (x ) g (x ) ⇒ (f (x ) − g (x )) (k (x ) − 1) ∨ 0
Instead of a jackdaw "∨", 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 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 admissible values. If you forgot the ODZ of the 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 fulfilled simultaneously. When the range of acceptable values is found, it remains to cross it with the solution rational inequality- and the answer is ready.
A task. Solve the inequality:
First, let's write the ODZ of the logarithm:
The first two inequalities are performed automatically, and the last one will have to be written. 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 perform the transition from the logarithmic inequality to the rational one. The original inequality has a “less than” sign, which means that the resulting inequality should also have a “less than” sign. We have:
(10 − (x 2 + 1)) (x 2 + 1 − 1)< 0;
(9 − x2) x2< 0;
(3 − x) (3 + x) x 2< 0.
Zeros of this expression: x = 3; x = -3; x = 0. Moreover, x = 0 is the 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.
Transformation of logarithmic inequalities
Often the original inequality differs from the one above. This is easy to fix 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 base can be replaced by a single logarithm.
Separately, I want 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 DPV of each of them. In this way, general scheme solution of logarithmic inequalities is the following:
- Find the ODZ of each logarithm included in the inequality;
- Reduce the inequality to the standard one using the formulas for adding and subtracting logarithms;
- Solve the resulting inequality according to the scheme above.
A task. Solve the inequality:
Find the domain of definition (ODZ) of the first logarithm:
We solve by 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 of the ODZ will be the same. If you don't believe me, you can check. Now we transform the second logarithm so that the base is two:
As you can see, the triples at the base and before the logarithm have shrunk. We got two logarithms from the same base. Let's put them together:
log 2 (x − 1) 2< 2;
log 2 (x − 1) 2< log 2 2 2 .
We have obtained the standard logarithmic inequality. We get rid of the logarithms by the formula. Since there is a less than sign in the original inequality, 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; +∞);
- Answer candidate: x ∈ (−1; 3).
It remains to cross these sets - we get the real answer:
We are interested in the intersection of sets, so we choose the intervals shaded on both arrows. We get x ∈ (−1; 2/3)∪(1; 3) - all points are punctured.