The article is devoted to the analysis of tasks 15 of profile exam in mathematics for 2017. In this task, students are offered to solve inequalities, most often logarithmic ones. Although they can be indicative. This article provides an analysis of examples of logarithmic inequalities, including those containing a variable at the base of the logarithm. All examples are taken from open bank tasks of the USE in mathematics (profile), so such inequalities are very likely to come across to you on the exam as task 15. Ideal for those who want to learn how to solve task 15 from the second part of the profile USE in mathematics in a short period of time in order to get more scores on the exam.

Analysis of tasks 15 from the profile exam in mathematics

In tasks 15 of the Unified State Examination in mathematics (profile), logarithmic inequalities are often found. The solution of logarithmic inequalities begins with the definition of the range of acceptable values. In this case, there is no variable in the base of both logarithms, there is only the number 11, which greatly simplifies the task. Therefore, the only restriction we have here is that both expressions under the logarithm sign are positive:

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The first inequality in the system is square inequality. To solve it, we would really do well to factorize the left side. I think you know that any square trinomial kind It is factorized as follows:

where and are the roots of the equation . In this case, the coefficient is 1 (this is the numerical coefficient in front of ). The coefficient is also equal to 1, and the coefficient is a free term, it is equal to -20. The roots of a trinomial are easiest to determine using Vieta's theorem. Our equation is reduced, which means the sum of the roots and will be equal to the coefficient with the opposite sign, that is, -1, and the product of these roots will be equal to the coefficient, that is, -20. It is easy to guess that the roots will be -5 and 4.

Now the left side of the inequality can be factored: title="Rendered by QuickLaTeX.com" height="20" width="163" style="vertical-align: -5px;"> Решаем это неравенство. График соответствующей функции — это парабола, ветви которой направлены вверх. Эта парабола пересекает ось !} X at points -5 and 4. Hence, the desired solution to the inequality is the interval . For those who do not understand what is written here, you can see the details in the video, starting from now. There you will also find a detailed explanation of how the second inequality of the system is solved. It is being resolved. Moreover, the answer is exactly the same as for the first inequality of the system. That is, the set written above is the area of ​​​​admissible values ​​of inequality.

So, taking into account factorization, the original inequality takes the form:

Using the formula, let's add 11 to the power of the expression under the sign of the first logarithm, and transfer the second logarithm to left side inequality, while changing its sign to the opposite:

After reduction we get:

The last inequality, due to the increase in the function , is equivalent to the inequality , whose solution is the interval . It remains to cross it with the area of ​​​​admissible values ​​of inequality, and this will be the answer to the entire task.

So, the desired answer to the task has the form:

We figured out this task, now we move on to the next example of task 15 of the Unified State Examination in mathematics (profile).

We begin the solution by determining the range of admissible values ​​of this inequality. The base of each logarithm must be positive number, which is not equal to 1. All expressions under the sign of the logarithm must be positive. The denominator of a fraction must not be zero. The last condition is equivalent to , since only otherwise both logarithms in the denominator vanish. All these conditions determine the range of admissible values ​​of this inequality, which is given by the following system of inequalities:

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In the range of acceptable values, we can use logarithm transformation formulas in order to simplify the left side of the inequality. Using the formula get rid of the denominator:

Now we have only base logarithms. It's already more convenient. Next, we use the formula, and also the formula in order to bring the expression worth glory to the following form:

In the calculations, we used what is in the range of acceptable values. Using the substitution, we arrive at the expression:

Let's use one more substitution: . As a result, we arrive at the following result:

So, gradually return to the original variables. First to the variable:

USE in mathematics profile level

The work consists of 19 tasks.
Part 1:
8 tasks with a short answer of the basic level of complexity.
Part 2:
4 tasks with a short answer
7 tasks with a detailed answer high level difficulties.

Run time - 3 hours 55 minutes.

Examples of USE assignments

Solving USE tasks in mathematics.

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USE 2020 in mathematics task 15 with a solution

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USE 2020 in mathematics task 15

USE 2020 in mathematics profile level task 15 with a solution

USE in mathematics task 15

Condition:

Solve the inequality:
log 2 ((7 -x 2 - 3) (7 -x 2 +16 -1)) + log 2 ((7 -x 2 -3)/(7 -x 2 +16 - 1)) > log 2 ( 7 7-x 2 - 2) 2

Solution:

Dealing with ODZ:
1. The expression under the first sign of the logarithm must be greater than zero:
(7 (-(x 2))-3) (7 (-(x 2) + 16) -1) > 0

X 2 is always less than or equal to zero, therefore,
7 (-x2)< = 1, следовательно,
7 (-x 2) - 3< = -2 < 0

This means that in order for the first condition on the ODZ to be satisfied, it is necessary that
7 (-(x 2)+16) - 1< 0
7 (-(x2)+16)< 1 = 7 0
-(x2)+16< 0
x2 > 16
x belongs to (-infinity; -4) U (4, +infinity)

2. The expression under the second sign of the logarithm must be greater than zero. But there the result will be the same as in the first paragraph, since the same expressions are in brackets.

3. The expression under the third sign of the logarithm must be greater than zero.
(7 (7-x 2) -2) 2 > 0
This inequality is always true except when
7(7-x2)-2=0
7 (7-x 2) = 7 (log_7(2))
7-x 2 = log_7(2)
x 2 = 7 - log_7(2)
x = (+-)sqrt(7-log_7(x))

Let's estimate what is approximately equal to sqrt(7-log_7(x)).
1/3 = log_8(2)< log_7(2) < log_4(2) = 1/2
2 = sqrt(4)< sqrt(7-1/2) < sqrt(7-log_7(2)) < sqrt(7-1/3) < sqrt(9) = 3

That is, the condition x is not equal to (+-)sqrt(7-log_7(x)) is already superfluous, since in paragraph (1) we have already thrown out the interval including these points from the DPV.

So, again ODZ:
x belongs to (- infinity; -4) U (4, + infinity)

4. Now, using the properties of the logarithm, the original inequality can be transformed like this:
log_2((7 (-x 2) - 3) 2) > log_2((7 (7 - x 2) - 2) 2)

Log_2(x) is an increasing function, so we get rid of the logarithm without changing the sign:
(7 (-x 2) -3) 2 > (7 (7-x 2) -2) 2

Let us estimate from above and below the expressions (7 (-x 2) -3) 2 And (7(7-x2)-2)2, taking into account the ODZ:

x2< -16
0 < 7 (-x 2) < 1
-3 < 7 (-x 2) -3 < -2
4 < (7 (-x 2) -3) 2 < 9

x2< -16
0 < 7 (7-x 2) < 1
-2 < 7 (-x 2) -2 < -1
1 < (7 (-x 2) -3) 2 < 4

Hence, the inequality holds for any x belonging to the ODZ.