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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.
The abbreviation stands for the range of valid 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.
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:
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 equal to zero. Instead of the “less than” sign, we put “equal”, we solve the equation. Thus, we will find the ODZ. We hope that you will have no problems with 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 "+" 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.
Solving logarithmic equations and inequalities with different bases involves initial reduction to one base. Then use the above method. But there is also a more complicated case. Consider one of the most complex types of 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 have a beneficial effect on your educational process. Let's look at 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 reduce 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!
In previous lessons, we got acquainted with logarithmic equations and now we know what they are 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 inequalities?
Logarithmic inequalities are inequalities that have a variable under the sign of the logarithm or at its base.
Or, one can also say that a logarithmic inequality is such an inequality in which its unknown value, as in the logarithmic equation, will be under the sign of the logarithm.
The simplest logarithmic inequalities look like this:
where f(x) and g(x) are some expressions that depend on x.
Let's look at this using the following example: f(x)=1+2x+x2, g(x)=3x−1.
Before solving logarithmic inequalities, it is worth noting that when they are solved, they are similar to 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;
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 it was we who considered the similar moments of solving logarithmic inequalities. Now let's look at a rather significant difference. You and I know that the logarithmic function has a limited domain of definition, so when moving from logarithms to expressions under the sign of the logarithm, you need to take into account the range of acceptable values (ODV).
That is, it should be borne in mind that when solving a logarithmic equation, we can first find the roots of the equation, and then check this solution. But solving the logarithmic inequality will not work this way, since moving from logarithms to expressions under the sign of the logarithm, 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 the following notation must be used: a > 0. In this case, both the sum and the product of such these numbers will also be positive.
The basic principle of solving an inequality is to replace it with a simpler inequality, but the main thing is that it be equivalent to the given one. Further, we also obtained an inequality and again replaced it with one that has a simpler form, and so on.
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 are the same.
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.
Now let's look at some of the methods that take place when solving logarithmic inequalities. For a better understanding and assimilation, we will try to understand them using specific examples.
We 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 the following system: