Logarithmic inequalities negative value. Logarithmic inequalities - Knowledge Hypermarket

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

The definition of the logarithm can be written in another way:

Pay attention to the restrictions that are imposed on the base of the logarithm ( a) and on the sublogarithmic expression ( x). In the future, these conditions will turn into important restrictions for the ODZ, 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 the ODZ (positiveness of expressions under roots of even degrees, non-equality of the denominator to zero, etc.), the following conditions must also be taken into account:

The 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. Also note that the value of the logarithm itself can take on all possible values, i.e. logarithm can be positive, negative or zero. Logarithms have so many different properties that follow from the properties of powers and the definition of a logarithm. Let's list them. So, the properties of logarithms:

The logarithm of the product:

Fraction logarithm:

Taking the degree out of the sign of the logarithm:

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

Other useful properties of logarithms:

The last property is very often used in complex logarithmic equations and inequalities. It must be remembered as well as everyone else, although it is often forgotten.

The simplest logarithmic equations are:

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 of such equations, taking into account the ODZ, is as follows:

Some others logarithmic equations with a variable in the base can be summarized as:

In such logarithmic equations, the general form of the solution also follows directly from the definition of the logarithm. Only in this case, there are additional restrictions for DHS 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 above equations, it is also actively used variable change method. As usual, when applying this method, one must remember that after the introduction of the replacement, the equation should be simplified and no longer contain the old unknown. You also need to remember to reverse-replace variables.

Sometimes, when solving logarithmic equations, one also has to use graphic method. This method consists in constructing as accurately as possible on the same coordinate plane the graphs of functions that are on the left and right sides of the equation, and then finding the coordinates of their intersection points according to the drawing. The roots obtained in this way must be verified 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 be equal to zero, and the rest existed. When the factors are logarithms or brackets with logarithms, and not just brackets with variables as in rational equations, then many errors can occur. Since logarithms have many restrictions on the area where they exist.

When deciding systems of logarithmic equations most often you have to use either the substitution method or the variable substitution method. If there is such a possibility, then when solving systems of logarithmic equations, one should strive to ensure that each of the equations of the system is individually reduced to such 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 much the same way as similar equations. First, with the help of algebraic transformations and the properties of logarithms, one should 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 that, you need to go to a rational inequality, given that this transition should be performed as follows: if the base of the logarithm is greater than one, then the inequality sign does not need to be changed, and if the base of the logarithm is less than one, then you need to change the inequality sign to the opposite (this means changing "less" to "greater" or vice versa). At the same time, the minus signs to plus, bypassing the previously studied rules, do not need to be changed anywhere. Let's write down mathematically what we get as a result of such a transition. If the base is greater than one, we get:

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

As we can see, when solving logarithmic inequalities, as usual, 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 it is sufficient to require the positivity of only the smaller of them.

When deciding logarithmic inequalities with a variable in the base logarithm, it is necessary to independently consider both options (when the base is less than one, and more than one) and combine the solutions of these cases in the aggregate. At the same time, one should not forget about the ODZ, 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 get the following set of systems:

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

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

Determine the ODZ;
Transform the inequality so that there is zero on the right side (on the left side, if possible, bring to a common denominator, factorize, etc.);
Find all the roots of the numerator and denominator and put them on the number line, 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 dots;
Find the sign of the whole expression on each of the intervals, substituting a number from the given interval into the transformed inequality. At the same time, it is no longer possible to alternate signs in any way by passing through points on the axis. It is necessary to determine the sign of the expression on each interval by substituting the value from the interval into this expression, and so on for each interval. There is no other way (this is, by and large, the difference between the generalized method of intervals and the usual one);
Find the intersection of the ODZ and the intervals that satisfy the inequality, while not losing individual points that satisfy the inequality (numerator roots in non-strict inequalities), and do not forget to exclude all denominator roots in all inequalities from the answer.

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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 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.

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 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.

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 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!

With them are inside logarithms.

Examples:

\(\log_3⁡x≥\log_3⁡9\)
\(\log_3⁡ ((x^2-3))< \log_3⁡{(2x)}\)
\(\log_(x+1)⁡((x^2+3x-7))>2\)
\(\lg^2⁡((x+1))+10≤11 \lg⁡((x+1))\)

How to solve logarithmic inequalities:

Any logarithmic inequality should be reduced to the form \(\log_a⁡(f(x)) ˅ \log_a(⁡g(x))\) (symbol \(˅\) means any of ). This form allows us to get rid of logarithms and their bases by passing to the inequality of expressions under logarithms, that is, to the form \(f(x) ˅ g(x)\).

But when making this transition, there is one very important subtlety:
\(-\) if - a number and it is greater than 1 - the inequality sign remains the same during the transition,
\(-\) if the base is a number greater than 0 but less than 1 (between zero and one), then the inequality sign must be reversed, i.e.

Examples:

\(\log_2⁡((8-x))<1\)
ODZ: \(8-x>0\)
\(-x>-8\)
\(x<8\)

Solution:
\(\log\)\(_2\) \((8-x)<\log\)\(_2\) \({2}\)
\(8-x\)\(<\) \(2\)
\(8-2\(x>6\)
Answer: \((6;8)\)

\(\log\)\(_(0.5⁡)\) \((2x-4)\)≥\(\log\)\(_(0.5)\) ⁡\(((x+ one))\)
ODZ: \(\begin(cases)2x-4>0\\x+1 > 0\end(cases)\)
\(\begin(cases)2x>4\\x > -1\end(cases)\) \(\Leftrightarrow\) \(\begin(cases)x>2\\x > -1\end(cases) \) \(\Leftrightarrow\) \(x\in(2;\infty)\)

Solution:
\(2x-4\)\(≤\)\(x+1\)
\(2x-x≤4+1\)
\(x≤5\)
Answer: \((2;5]\)

Very important! In any inequality, the transition from the form \(\log_a(⁡f(x)) ˅ \log_a⁡(g(x))\) to comparing expressions under logarithms can only be done if:

Example . Solve the inequality: \(\log\)\(≤-1\)

Solution:

\(\log\) \(_(\frac(1)(3))⁡(\frac(3x-2)(2x-3))\)\(≤-1\)	Let's write out the ODZ.
ODZ: \(\frac(3x-2)(2x-3)\) \(>0\)
\(⁡\frac(3x-2-3(2x-3))(2x-3)\)\(≥\) \(0\)	We open the brackets, give .
\(⁡\frac(-3x+7)(2x-3)\) \(≥\) \(0\)	We multiply the inequality by \(-1\), remembering to reverse the comparison sign.
\(⁡\frac(3x-7)(2x-3)\) \(≤\) \(0\)
\(⁡\frac(3(x-\frac(7)(3)))(2(x-\frac(3)(2)))\)\(≤\) \(0\)	Let's build a number line and mark the points \(\frac(7)(3)\) and \(\frac(3)(2)\) on it. Note that the point from the denominator is punctured, despite the fact that the inequality is not strict. The fact is that this point will not be a solution, since when substituting into an inequality, it will lead us to division by zero.
\(x∈(\)\(\frac(3)(2)\) \(;\)\(\frac(7)(3)]\)	Now we plot the ODZ on the same numerical axis and write down in response the interval that falls into the ODZ.
	Write down the final answer.

Answer: \(x∈(\)\(\frac(3)(2)\) \(;\)\(\frac(7)(3)]\)

Example . Solve the inequality: \(\log^2_3⁡x-\log_3⁡x-2>0\)