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 of a 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. In the original inequality there is a “less than” sign, so the resulting inequality should also be with 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.
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.
Protozoa 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.
Solution logarithmic equations and inequalities with different bases implies an 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 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 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!
The solution of the simplest logarithmic inequalities and inequalities, where the base of the logarithm is fixed, we considered in the last lesson.
But what if the base of the logarithm is a variable?
Then we will come to the rescue rationalization of inequalities. To understand how this works, let's consider, for example, the inequality:
$$\log_(2x) x^2 > \log_(2x) x.$$
As expected, let's start with the ODZ.
ODZ
$$\left[ \begin(array)(l)x>0,\\ 2x ≠ 1. \end(array)\right.$$
Solving the inequality
Let's reason as if we were solving an inequality with a fixed base. If the base is greater than one, we get rid of the logarithms, and the inequality sign does not change, if it is less than one, it changes.
Let's write it as a system:
$$\left[ \begin(array)(l) \left\( \begin(array)(l)2x>1,\\ x^2 > x; \end(array)\right. \\ \left\ ( \begin(array)(l)2x<1,\\ x^2 < x; \end{array}\right. \end{array} \right.$$
For further reasoning, we transfer all the right-hand sides of the inequalities to the left.
$$\left[ \begin(array)(l) \left\( \begin(array)(l)2x-1>0,\\ x^2 -x>0; \end(array)\right. \ \ \left\( \begin(array)(l)2x-1<0,\\ x^2 -x<0; \end{array}\right. \end{array} \right.$$
What did we get? It turned out that we need the expressions `2x-1` and `x^2 - x` to be either positive or negative at the same time. The same result will be obtained if we solve the inequality:
$$(2x-1)(x^2 - x) >0.$$
This inequality, like the original system, is true if both factors are either positive or negative. It turns out that it is possible to move from the logarithmic inequality to the rational one (taking into account the ODZ).
Let's formulate rationalization method for logarithmic inequalities$$\log_(f(x)) g(x) \vee \log_(f(x)) h(x) \Leftrightarrow (f(x) - 1)(g(x)-h(x)) \ vee 0,$$ where `\vee` is any inequality sign. (For the `>` sign, we just checked the validity of the formula. For the rest, I suggest checking it yourself - this way it will be remembered better).
Let's return to the solution of our inequality. Expanding into brackets (to better see the zeros of the function), we get
$$(2x-1)x(x - 1) >0.$$
The interval method will give the following picture:
(Since the inequality is strict and the ends of the intervals are of no interest to us, they are not filled in.) As can be seen, the resulting intervals satisfy the ODZ. Got the answer: `(0,\frac(1)(2)) \cup (1,∞)`.
Second example. Solution of logarithmic inequality with variable base
$$\log_(2-x) 3 \leqslant \log_(2-x) x.$$
ODZ
$$\left\(\begin(array)(l)2-x > 0,\\ 2-x ≠ 1, \\ x > 0. \end(array)\right.$$
$$\left\(\begin(array)(l)x< 2,\\ x ≠ 1, \\ x >0. \end(array)\right.$$
Solving the inequality
According to the rule we have just obtained rationalization of logarithmic inequalities, we obtain that this inequality is identical (taking into account the ODZ) to the following:
$$(2-x -1) (3-x) \leqslant 0.$$
$$(1-x) (3-x) \leqslant 0.$$
Combining this solution with the ODZ, we get the answer: `(1,2)`.
Third example. Logarithm of a fraction
$$\log_x\frac(4x+5)(6-5x) \leqslant -1.$$
ODZ
$$\left\(\begin(array)(l) \dfrac(4x+5)(6-5x)>0, \\ x>0,\\ x≠ 1.\end(array) \right.$ $
Since the system is relatively complex, let's immediately plot the solution of inequalities on the number line:
Thus, ODZ: `(0,1)\cup \left(1,\frac(6)(5)\right)`.
Solving the inequality
Let's represent `-1` as a logarithm with base `x`.
$$\log_x\frac(4x+5)(6-5x) \leqslant \log_x x^(-1).$$
By using rationalization of the logarithmic inequality we get a rational inequality:
$$(x-1)\left(\frac(4x+5)(6-5x) -\frac(1)(x)\right)\leqslant0,$$
$$(x-1)\left(\frac(4x^2+5x - 6+5x)(x(6-5x))\right)\leqslant0,$$
$$(x-1)\left(\frac(2x^2+5x - 3)(x(6-5x))\right)\leqslant0.$$
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. The presentation presents solutions to tasks C3 USE - 2014 in mathematics.
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Solving logarithmic inequalities containing a variable at the base of the logarithm: methods, techniques, equivalent transitions teacher of mathematics MBOU secondary school No. 143 Knyazkina T.V.
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 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 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. Don't forget the ODZ of the 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 of a rational inequality - and the answer is ready.
Solve the inequality: Solution To begin with, let's write out the ODZ of the logarithm. The first two inequalities are performed automatically, and the last one will have to be painted. Since the square of a number is equal to zero if and only if the number itself is equal to 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. In the original inequality there is a “less than” sign, so the resulting inequality should also be with a “less than” sign.
We have: (10 − (x 2 + 1)) (x 2 + 1 − 1)
Converting logarithmic inequalities Often the original inequality differs from the one above. This is easy to fix using the standard rules for working with 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. Thus, the general scheme for solving logarithmic inequalities is as follows: Find the ODZ for 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.
Solve the inequality: Solution 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: 3 x − 2 = 0; x = 2/3. Then - denominator zeros: x − 1 = 0; x = 1. We mark zeros and signs on the coordinate line:
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 let's transform the second logarithm so that there is a 2 at the base: As you can see, the 3s at the base and in front of the logarithm have shrunk. Get two logarithms with the same base. Add them up: log 2 (x − 1) 2
(f (x) − g (x)) (k (x) − 1)
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. Answer: x ∈ (−1; 2/3)∪(1; 3)
Solving tasks of the Unified State Exam-2014 type C3
Solve the system of inequalities Solution. ODZ: 1) 2)
Solve the system of inequalities 3) -7 -3 - 5 x -1 + + + − − (continued)
Solve the system of inequalities 4) General solution: and -7 -3 - 5 x -1 -8 7 log 2 129 (continued)
Solve the inequality (continued) -3 3 -1 + - + - x 17 + -3 3 -1 x 17 -4
Solve the inequality Solution. ODZ:
Solve the inequality (continued)
Solve the inequality Solution. ODZ: -2 1 -1 + - + - x + 2 -2 1 -1 x 2