Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely taught in 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.
This way 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 acceptable values. If you have forgotten the ODZ of a 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 satisfied simultaneously. When the range of acceptable values is found, all that remains is to intersect it with the solution rational inequality- and the answer is ready.
Task. Solve the inequality:
First, let’s write out the logarithm’s ODZ:
The first two inequalities are satisfied automatically, but the last one will have to be written out. Since the square of a number is zero if and only if the number itself is zero, we have:
x 2 + 1 ≠ 1;
x 2 ≠ 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 make the transition from logarithmic inequality to rational one. The original inequality has a “less than” sign, which means the resulting inequality must also have a “less than” sign. We have:
(10 − (x 2 + 1)) · (x 2 + 1 − 1)< 0;
(9 − x 2) x 2< 0;
(3 − x ) (3 + x ) x 2< 0.
The zeros of this expression are: x = 3; x = −3; x = 0. Moreover, x = 0 is a 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 this is the answer.
Often the original inequality is different from the one above. This can be easily corrected using the standard rules for working with logarithms - see “Basic properties of logarithms”. Namely:
Separately, I would like 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 VA of each of them. Thus, general scheme solutions to logarithmic inequalities are as follows:
Task. Solve the inequality:
Let's find the domain of definition (DO) of the first logarithm:
We solve using 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 will have the same VA. If you don't believe me, you can check it. Now we transform the second logarithm so that the base is two:
As you can see, the threes at the base and in front of the logarithm have been reduced. We got two logarithms with the same base. Let's add them up:
log 2 (x − 1) 2< 2;
log 2 (x − 1) 2< log 2 2 2 .
We obtained the standard logarithmic inequality. We get rid of logarithms using the formula. Since the original inequality contains a “less than” sign, 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:
It remains to intersect these sets - we get the real answer:
We are interested in the intersection of sets, so we select intervals that are shaded on both arrows. We get x ∈ (−1; 2/3)∪(1; 3) - all points are punctured.
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Lesson objectives:
Didactic:
Educational: develop memory, attention, logical thinking, comparison skills, ability to generalize and draw conclusions
Educational: cultivate accuracy, responsibility for the task being performed, and mutual assistance.
Teaching methods: verbal , visual , practical , partial-search , self-government , control.
Forms of organization cognitive activity students: frontal , individual , work in pairs.
Equipment: kit test tasks, supporting notes, blank sheets for solutions.
Lesson type: learning new material.
Lesson progress
1. Organizational moment. The topic and goals of the lesson, the lesson plan are announced: each student is given an assessment sheet, which the student fills out during the lesson; for each pair of students - printed materials with tasks; tasks must be completed in pairs; blank solution sheets; support sheets: definition of logarithm; graph of a logarithmic function, its properties; properties of logarithms; algorithm for solving logarithmic inequalities.
All decisions after self-assessment are submitted to the teacher.
Student's score sheet
2. Updating knowledge.
Teacher's instructions. Recall the definition of a logarithm, the graph of a logarithmic function, and its properties. To do this, read the text on pp. 88–90, 98–101 of the textbook “Algebra and the beginnings of analysis 10–11” edited by Sh.A Alimov, Yu.M Kolyagin and others.
Students are given sheets on which are written: the definition of a logarithm; shows a graph of a logarithmic function and its properties; properties of logarithms; algorithm for solving logarithmic inequalities, an example of solving a logarithmic inequality that reduces to a quadratic one.
3. Studying new material.
Solving logarithmic inequalities is based on the monotonicity of the logarithmic function.
Algorithm for solving logarithmic inequalities:
A) Find the domain of definition of the inequality (the sublogarithmic expression is greater than zero).
B) Represent (if possible) the left and right sides of the inequality as logarithms to the same base.
C) Determine whether the logarithmic function is increasing or decreasing: if t>1, then increasing; if 0
D) Go to more simple inequality(sublogarithmic expressions), taking into account that the inequality sign will remain if the function increases, and will change if it decreases.
Learning element #1.
Goal: consolidate the solution to the simplest logarithmic inequalities
Form of organization of students' cognitive activity: individual work.
Tasks for independent work for 10 minutes. For each inequality there are several possible answers; you need to choose the correct one and check it using the key.
KEY: 13321, maximum number of points – 6 points.
Learning element #2.
Goal: consolidate the solution of logarithmic inequalities using the properties of logarithms.
Teacher's instructions. Remember the basic properties of logarithms. To do this, read the text of the textbook on pp. 92, 103–104.
Tasks for independent work for 10 minutes.
KEY: 2113, maximum number of points – 8 points.
Learning element #3.
Purpose: to study the solution of logarithmic inequalities by the method of reduction to quadratic.
Teacher's instructions: the method of reducing an inequality to a quadratic is to transform the inequality to such a form that a certain logarithmic function is denoted by a new variable, thereby obtaining a quadratic inequality with respect to this variable.
Let's use the interval method.
You have passed the first level of mastering the material. Now you have to choose your own solution method logarithmic equations using all your knowledge and capabilities.
Learning element #4.
Goal: consolidate the solution to logarithmic inequalities by independently choosing a rational solution method.
Tasks for independent work for 10 minutes
Learning element #5.
Teacher's instructions. Well done! You have mastered solving equations of the second level of complexity. The goal of your further work is to apply your knowledge and skills in more complex and non-standard situations.
Tasks for independent solution:
Teacher's instructions. It's great if you completed the whole task. Well done!
The grade for the entire lesson depends on the number of points scored for all educational elements:
Submit the assessment papers to the teacher.
5. Homework: if you scored no more than 15 points, work on your mistakes (solutions can be taken from the teacher), if you scored more than 15 points, complete a creative task on the topic “ Logarithmic inequalities”.
An inequality is called logarithmic if it contains a logarithmic function.
Methods for solving logarithmic inequalities are no different from, except for two things.
Firstly, when moving from the logarithmic inequality to the inequality of sublogarithmic functions, one should follow the sign of the resulting inequality. It obeys the following rule.
If the base of the logarithmic function is greater than $1$, then when moving from the logarithmic inequality to the inequality of sublogarithmic functions, the sign of the inequality is preserved, but if it is less than $1$, then it changes to the opposite.
Secondly, the solution to any inequality is an interval, and, therefore, at the end of solving the inequality of sublogarithmic functions it is necessary to create a system of two inequalities: the first inequality of this system will be the inequality of sublogarithmic functions, and the second will be the interval of the domain of definition of the logarithmic functions included in the logarithmic inequality.
Let's solve the inequalities:
1. $\log_(2)((x+3)) \geq 3.$
$D(y): \x+3>0.$
$x \in (-3;+\infty)$
The base of the logarithm is $2>1$, so the sign does not change. Using the definition of logarithm, we get:
$x+3 \geq 2^(3),$
$x \in )
