LOGARITHMIC INEQUALITIES IN THE USE
Sechin Mikhail Alexandrovich
Small Academy of Sciences for students of the Republic of Kazakhstan "Seeker"
MBOU "Sovetskaya secondary school №1", grade 11, town. Sovetsky Sovetsky district
Gunko Lyudmila Dmitrievna, teacher of MBOU "Soviet school №1"
Soviet district
Objective: investigation of the mechanism for solving logarithmic inequalities C3 using non-standard methods, identifying interesting facts logarithm.
Subject of study:
3) Learn to solve specific logarithmic inequalities C3 using non-standard methods.
Results:
Content
Introduction ………………………………………………………………………… .4
Chapter 1. Background ………………………………………………… ... 5
Chapter 2. Collection of logarithmic inequalities ………………………… 7
2.1. Equivalent transitions and the generalized method of intervals …………… 7
2.2. Rationalization method ………………………………………………… 15
2.3. Non-standard substitution ……………… .......................................... ..... 22
2.4. Trap Missions ………………………………………………… 27
Conclusion ………………………………………………………………… 30
Literature……………………………………………………………………. 31
Introduction
I am in 11th grade and plan to enter a university, where profile subject is math. Therefore, I work a lot with the problems of part C. In task C3, you need to solve a non-standard inequality or a system of inequalities, usually associated with logarithms. While preparing for the exam, I faced the problem of a shortage of methods and techniques for solving the exam logarithmic inequalities offered in C3. The methods that are studied in the school curriculum on this topic do not provide a basis for solving C3 tasks. The math teacher invited me to work with the C3 tasks on my own under her guidance. In addition, I was interested in the question: are there logarithms in our life?
With this in mind, the topic was chosen:
"Logarithmic inequalities in the exam"
Objective: investigation of the mechanism for solving C3 problems using non-standard methods, revealing interesting facts of the logarithm.
Subject of study:
1) Find the necessary information about non-standard methods for solving logarithmic inequalities.
2) Find more information about logarithms.
3) Learn to solve specific C3 problems using non-standard methods.
Results:
The practical significance lies in the expansion of the apparatus for solving C3 problems. This material can be used in some lessons, for circles, extracurricular activities in mathematics.
The project product will be the collection "Logarithmic C3 inequalities with solutions".
Chapter 1. Background
Throughout the 16th century, the number of approximate calculations increased rapidly, primarily in astronomy. Improving instruments, studying planetary movements and other work required colossal, sometimes many years, calculations. Astronomy was in real danger of drowning in unfulfilled calculations. Difficulties arose in other areas, for example, in the insurance business, tables of compound interest were needed for various values \u200b\u200bof interest. The main difficulty was represented by multiplication, division of multidigit numbers, especially trigonometric quantities.
The discovery of logarithms was based on the well-known properties of progressions by the end of the 16th century. Archimedes spoke about the connection between the members of the geometric progression q, q2, q3, ... and the arithmetic progression of their exponents 1, 2, 3, ... in the Psalm. Another premise was the extension of the concept of degree to negative and fractional indicators. Many authors have pointed out that multiplication, division, exponentiation, and extraction of a root exponentially correspond in arithmetic - in the same order - addition, subtraction, multiplication and division.
This was the idea behind the logarithm as an exponent.
Several stages have passed in the history of the development of the doctrine of logarithms.
Stage 1
Logarithms were invented no later than 1594 by the Scottish baron Napier (1550-1617) and ten years later by the Swiss mechanic Burghi (1552-1632). Both wanted to give a new convenient means of arithmetic calculations, although they approached this task in different ways. Neper expressed kinematically the logarithmic function and thus entered a new field of function theory. Burghi remained on the basis of considering discrete progressions. However, the definition of the logarithm for both does not resemble the modern one. The term "logarithm" (logarithmus) belongs to Napier. It arose from a combination of Greek words: logos - "relationship" and ariqmo - "number", which meant "number of relationships." Initially, Napier used a different term: numeri artificiales - "artificial numbers", as opposed to numeri naturalts - "natural numbers".
In 1615, in a conversation with Henry Briggs (1561-1631), professor of mathematics at Gresch College in London, Napier proposed to take zero for the logarithm of unity, and 100 for the logarithm of ten, or, which comes down to the same thing, simply 1. This is how decimal logarithms appeared and the first logarithmic tables were printed. Later, the Dutch bookseller and lover of mathematics Andrian Flakk (1600-1667) supplemented the Briggs tables. Napier and Briggs, although they came to logarithms earlier than anyone else, published their tables later than others - in 1620. The log and Log signs were introduced in 1624 by I. Kepler. The term "natural logarithm" was introduced by Mengoli in 1659, followed by N. Mercator in 1668, and the London teacher John Speidel published tables of natural logarithms of numbers from 1 to 1000 under the title "New Logarithms".
The first logarithmic tables in Russian were published in 1703. But in all logarithmic tables, errors were made in the calculation. The first error-free tables were published in 1857 in Berlin, processed by the German mathematician K. Bremiker (1804-1877).
Stage 2
Further development of the theory of logarithms is associated with a wider application of analytic geometry and infinitesimal calculus. The establishment of a connection between the quadrature of an equilateral hyperbola and the natural logarithm dates back to that time. The theory of logarithms of this period is associated with the names of a number of mathematicians.
German mathematician, astronomer and engineer Nikolaus Mercator in the composition
"Logarithmtechnics" (1668) gives a series that decomposes ln (x + 1) in
powers of x:
This expression exactly corresponds to the line of his thought, although, of course, he used not the signs d, ..., but more cumbersome symbols. With the discovery of the logarithmic series, the technique for calculating logarithms changed: they began to be determined using infinite series. In his lectures "Elementary mathematics from the highest point of view", read in 1907-1908, F. Klein proposed using the formula as a starting point for constructing the theory of logarithms.
Stage 3
Definition logarithmic function as a function of the inverse
exponential, logarithm as an indicator of the degree of a given base
was not immediately formulated. Composition by Leonard Euler (1707-1783)
An Introduction to the Analysis of the Infinitesimal (1748) served as a further
development of the theory of the logarithmic function. In this way,
134 years have passed since logarithms were first introduced
(counting from 1614) before mathematicians came to the definition
the concept of the logarithm, which is now the basis of the school course.
Chapter 2. Collection of logarithmic inequalities
2.1. Equivalent transitions and the generalized interval method.
Equivalent transitions
if a\u003e 1
if 0 <
а <
1
Generalized interval method
This method is the most versatile for solving inequalities of almost any type. The solution scheme looks like this:
1. Bring the inequality to the form where the function 
, and on the right 0.
2. Find the domain of the function .
3. Find the zeros of the function , that is, to solve the equation 
(and solving an equation is usually easier than solving an inequality).
4. Draw on the number line the domain and zeros of the function.
5. Determine the signs of the function at the intervals obtained.
6. Select intervals where the function takes the required values \u200b\u200band write down the answer.
Example 1.
Decision:
Let's apply the spacing method
from where
For these values, all expressions under the sign of the logarithms are positive.
Answer:
Example 2.
Decision:
1st way . ODZ is determined by the inequality x \u003e 3. Taking the logarithm for such x base 10, we get
The last inequality could be solved by applying the decomposition rules, i.e. comparing the factors to zero. However, in this case it is easy to determine the intervals of constancy of the function
therefore, you can apply the method of intervals.
Function f(x) = 2x(x- 3,5) lgǀ x- 3ǀ is continuous at x \u003e 3 and vanishes at points x 1 = 0, x 2 = 3,5, x 3 = 2, x 4 \u003d 4. Thus, we define the intervals of constancy of the function f(x):
Answer:
2nd way . Let us apply the ideas of the method of intervals directly to the original inequality.
To do this, recall that the expressions a b - a c and ( a - 1)(b - 1) have one sign. Then our inequality for x \u003e 3 is equivalent to the inequality
or
The last inequality is solved by the method of intervals
Answer:
Example 3.
Decision:
Let's apply the spacing method
Answer:
Example 4.
Decision:
Since 2 x 2 - 3x + 3\u003e 0 for all real xthen
To solve the second inequality, we use the method of intervals
In the first inequality, we make the replacement
then we arrive at the inequality 2y 2 - y - 1 < 0 и, применив метод интервалов, получаем, что решениями будут те ythat satisfy the inequality -0.5< y < 1.
Where, since
we obtain the inequality
which is carried out with those xfor which 2 x 2 - 3x - 5 < 0. Вновь применим метод интервалов
Now, taking into account the solution of the second inequality of the system, we finally obtain
Answer:
Example 5.
Decision:
Inequality is equivalent to a set of systems
or
Let's apply the method of intervals or
Answer:
Example 6.
Decision:
Inequality is equivalent to the system
Let be
then y > 0,
and the first inequality
system takes the form
or by expanding
square trinomial by factors,
Applying the method of intervals to the last inequality,
we see that its solutions satisfying the condition y \u003e 0 will be all y > 4.
Thus, the original inequality is equivalent to the system:
So, solutions to inequality are all
2.2. Method of rationalization.
Previously, the method of rationalizing inequality was not solved, it was not known. This is "a new modern effective method for solving exponential and logarithmic inequalities" (quote from the book of S. I. Kolesnikova)
And even if the teacher knew him, there was apprehension - does the examiner know him, and why is he not given at school? There were situations when the teacher told the student: "Where did you get it? Sit down - 2."
Now the method is widely promoted. And for experts there is guidelinesrelated to this method, and in "The most complete editions standard options... "solution C3 uses this method.
WONDERFUL METHOD!
"Magic table"
In other sources
if a a\u003e 1 and b\u003e 1, then log a b\u003e 0 and (a -1) (b -1)\u003e 0;
if a a\u003e 1 and 0 if 0<a<1 и b
>1, then log a b<0 и (a
-1)(b
-1)<0;
if 0<a<1 и 00 and (a -1) (b -1)\u003e 0. The above reasoning is simple, but it considerably simplifies the solution of logarithmic inequalities. Example 4.
log x (x 2 -3)<0
Decision:
Example 5.
log 2 x (2x 2 -4x +6) ≤log 2 x (x 2 + x) Decision: Example 6.
To solve this inequality, instead of the denominator, we write down (x-1-1) (x-1), and instead of the numerator - the product (x-1) (x-3-9 + x). Example 7.
Example 8.
2.3. Non-standard substitution. Example 1.
Example 2.
Example 3.
Example 4.
Example 5.
Example 6.
Example 7.
log 4 (3 x -1) log 0.25 Let's make the replacement y \u003d 3 x -1; then this inequality takes the form Log 4 log 0.25 Because log 0.25 We make the change t \u003d log 4 y and obtain the inequality t 2 -2t + ≥0, the solution of which is the intervals - Thus, to find the values \u200b\u200bof y, we have a set of two simplest inequalities Therefore, the original inequality is equivalent to the collection of two exponential inequalities, The solution to the first inequality of this set is the interval 0<х≤1, решением второго – промежуток 2≤х<+ Example 8.
Decision:
Inequality is equivalent to the system The solution to the second inequality, which determines the DHS, will be the set of those x,
for which x > 0.
To solve the first inequality, we make the change Then we obtain the inequality or The set of solutions to the last inequality is found by the method intervals: -1< t < 2. Откуда, возвращаясь к переменной x, we get or Many of those xthat satisfy the last inequality belongs to ODZ ( x \u003e 0), therefore, is a solution to the system and hence the original inequality. Answer: 2.4. Trap quests. Example 1.
Decision. ODZ inequalities are all x satisfying the condition 0 Example 2.
log 2 (2 x + 1-x 2)\u003e log 2 (2 x-1 + 1-x) +1.
Answer... (0; 0.5) U.
Answer :
(3;6)
.
\u003d -log 4 \u003d - (log 4 y -log 4 16) \u003d 2-log 4 y, then rewrite the last inequality as 2log 4 y -log 4 2 y ≤.
The solution to this set is the intervals 0<у≤2 и 8≤у<+.
that is, the totality 
... Thus, the original inequality holds for all values \u200b\u200bof x from the intervals 0<х≤1 и 2≤х<+.
.
... Therefore, all x from the interval 0
Conclusion
It was not easy to find special methods for solving C3 problems from the large abundance of different educational sources. In the course of the work done, I was able to study non-standard methods for solving complex logarithmic inequalities. These are: equivalent transitions and the generalized method of intervals, the method of rationalization , non-standard substitution , tasks with traps on the ODZ. These methods are absent in the school curriculum.
Using different methods, I solved the 27 inequalities proposed in the exam in part C, namely C3. These inequalities with solutions by methods formed the basis of the collection "Logarithmic C3 inequalities with solutions", which became a project product of my work. The hypothesis that I posed at the beginning of the project was confirmed: the C3 tasks can be effectively solved, knowing these methods.
In addition, I found interesting facts about logarithms. It was interesting for me to do it. My design products will be useful for both students and teachers.
Conclusions:
Thus, the set goal of the project has been achieved, the problem has been resolved. And I got the most complete and versatile experience in project activities at all stages of work. In the course of working on the project, my main developmental impact was on mental competence, activities related to logical mental operations, the development of creative competence, personal initiative, responsibility, perseverance, activity.
A guarantee of success when creating a research project for I became: significant school experience, the ability to extract information from various sources, check its reliability, rank it by importance.
In addition to direct subject knowledge in mathematics, he expanded his practical skills in the field of computer science, gained new knowledge and experience in the field of psychology, established contacts with classmates, and learned to cooperate with adults. In the course of the project activities, organizational, intellectual and communicative general educational skills and abilities were developed.
Literature
1. Koryanov A. G., Prokofiev A. A. Systems of inequalities with one variable (typical tasks C3).
2. Malkova A. G. Preparation for the exam in mathematics.
3. Samarova SS Solution of logarithmic inequalities.
4. Mathematics. Collection of training works edited by A.L. Semyonov and I.V. Yashchenko. -M .: MTsNMO, 2009 .-- 72 p. -
We considered the solution of the simplest logarithmic inequalities and inequalities where the base of the logarithm is fixed in the last lesson.
But what if there is a variable at the base of the logarithm?
Then will come to our aid rationalization of inequalities.To understand how this works, let's consider, for example, inequality:
$$ \\ log_ (2x) x ^ 2\u003e \\ log_ (2x) x. $$
As expected, let's start with the ODZ.
ODZ
$$ \\ left [\\ begin (array) (l) x\u003e 0, \\\\ 2x ≠ 1. \\ end (array) \\ right. $$
Inequality solution
Let's think 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 down as a system:
$$ \\ left [\\ begin (array) (l) \\ left \\ (\\ begin (array) (l) 2x\u003e 1, \\\\ x ^ 2\u003e x; \\ end (array) \\ right. \\\\ \\ left \\ For further reasoning, we transfer all the right-hand sides of the inequalities to the left.<1,\\ x^2 < x; \end{array}\right. \end{array} \right.$$
$$ \\ left [\\ begin (array) (l) \\ left \\ (\\ begin (array) (l) 2x-1\u003e 0, \\\\ x ^ 2 -x\u003e 0; \\ end (array) \\ right. \\ What did we do? It turned out that we need the expressions `2x-1` and` x ^ 2 - x` to be simultaneously either positive or negative. The same result will be obtained if we solve the inequality:
$$ (2x-1) (x ^ 2 - x)\u003e 0. $$<0,\\ x^2 -x<0; \end{array}\right. \end{array} \right.$$
This inequality, just like the original system, is true if both factors are either positive or negative. It turns out that it is possible to go from a logarithmic inequality to a rational one (taking into account the ODZ).
Let's formulate
method of rationalizing logarithmic inequalities
$$ \\ log_ (f (x)) g (x) \\ vee \\ log_ (f (x)) h (x) \\ Leftrightarrow (f (x) - 1) (g (x) -h (x)) \\ (For the `\u003e` sign, we have just checked the validity of the formula. For the rest, I propose to check it yourself - it will be remembered better this way). let's go back to solving our inequality. Expanding into brackets (to make the zeros of the function better visible), we get $$ (2x-1) x (x - 1)\u003e 0. $$
The spacing method will give the following picture:
(Since the inequality is strict and the ends of the intervals are not of interest to us, they are not shaded.) As can be seen, the obtained intervals satisfy the ODZ. Received the answer: `(0, \\ frac (1) (2)) \\ cup (1, ∞)`.
Example two. Variable base logarithmic inequality solution
$$ \\ log_ (2-x) 3 \\ leqslant \\ log_ (2-x) x. $$
$$ \\ left \\ (\\ begin (array) (l) 2-x\u003e 0, \\\\ 2-x ≠ 1, \\\\ x\u003e 0. \\ end (array) \\ right. $$
$$ \\ left \\ (\\ begin (array) (l) x
ODZ
0. \\ end (array) \\ right. $$
By the rule we just got< 2,\\ x ≠ 1, \\ x > rationalizing logarithmic inequalities,
Inequality solution
{!LANG-08f6a1058f94f6b0d0e7061045ec3879!} {!LANG-c6ae2a46b8fce31cbb86800d68d8e5d6!} we obtain that this inequality is identical (taking into account the ODD) to the following:
$$ (2-x -1) (3-x) \\ leqslant 0. $$
$$ (1-x) (3-x) \\ leqslant 0. $$
Combining this solution with ODZ, we get the answer: `(1,2)`.
Third example. Logarithm of fraction
$$ \\ log_x \\ frac (4x + 5) (6-5x) \\ leqslant -1. $$
ODZ
$$ \\ left \\ (\\ begin (array) (l) \\ dfrac (4x + 5) (6-5x)\u003e 0, \\\\ x\u003e 0, \\\\ x ≠ 1. \\ end (array) \\ right. $ $
Since the system is relatively complex, let's immediately plot the solution to the inequalities on the number axis:
Thus, ODZ: `(0,1) \\ cup \\ left (1, \\ frac (6) (5) \\ right)`.
Inequality solution
Let's represent `-1` as a logarithm with base` x`.
$$ \\ log_x \\ frac (4x + 5) (6-5x) \\ leqslant \\ log_x x ^ (- 1). $$
Through rationalizing 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 all the 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 told at school. The presentation presents solutions to tasks C3 of the exam - 2014 in mathematics.
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Solution of logarithmic inequalities containing a variable at the base of the logarithm: methods, techniques, equivalent transitions mathematics teacher MBOU Secondary School No. 143 Knyazkina TV
Among all the 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 told 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. So we get rid of logarithms and reduce the problem to rational inequality. The latter is much easier to solve, but when dropping logarithms, extra roots may appear. To cut them off, it is enough to find the range of acceptable values. Don't forget the ODZ of the logarithm! Everything related to the range of permissible values \u200b\u200bmust be written out and solved separately: f (x)\u003e 0; g (x)\u003e 0; k (x)\u003e 0; k (x) ≠ 1. These four inequalities constitute a system and must be satisfied simultaneously. When the range of acceptable values \u200b\u200bis found, it remains to cross it with the solution of rational inequality - and the answer is ready.
Solve the inequality: Solution First, let's write the ODZ of the logarithm The first two inequalities are fulfilled automatically, and the last one will have to be written down. 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; 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 carry out the transition from the logarithmic inequality to the rational one. In the original inequality there is a “less” sign, which means that the resulting inequality must also be with a “less” sign.
We have: (10 - (x 2 + 1)) (x 2 + 1 - 1)
Transforming Logarithmic Inequalities Often the original inequality differs from the one above. It is easy to fix this by following 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 bases can be replaced with one logarithm. I would also like to remind you about the range of valid values. Since the original inequality may contain several logarithms, it is required to find the ODV for each of them. Thus, the general scheme for solving logarithmic inequalities is as follows: Find the ODV of each logarithm included in the inequality; Reduce inequality to the standard one according to the formulas for addition and subtraction of logarithms; Solve the resulting inequality according to the scheme given above.
Solve the inequality: Solution Let us find the domain of definition (ODD) of the first logarithm: Solve by the method of intervals. Find the zeros of the numerator: 3 x - 2 \u003d 0; x \u003d 2/3. Then - zeros of the denominator: x - 1 \u003d 0; x \u003d 1. Mark the zeros and signs on the coordinate line:
We get x ∈ (−∞ 2/3) ∪ (1; + ∞). The second logarithm of ODV will be the same. Do not believe it - you can check. Now we transform the second logarithm so that there is a 2 at the base: As you can see, the triples at the base and in front of the logarithm have canceled. Received 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 select the intervals filled in on both arrows. We get: x ∈ (−1; 2/3) ∪ (1; 3) -all points are punctured. Answer: x ∈ (−1; 2/3) ∪ (1; 3)
Solving tasks for the 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 Inequality Solution. ODZ:
Solve Inequality (continued)
Solve Inequality Solution. ODZ: -2 1 -1 + - + - x + 2 -2 1 -1 x 2
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 devote 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, this is 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 passed the inequalities several years ago. And since then, they are constantly encountered in mathematics. If you have problems solving inequalities, see the corresponding section.
Now that we have become acquainted with the concepts separately, let's move on to considering them 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 will give a more applicable example, it is still quite simple, we will leave complex logarithmic inequalities for later.
How to solve this? It all starts with ODZ. It is worth knowing more about it if you always want to easily solve any inequality.
What is ODU? ODZ for logarithmic inequalities
The abbreviation stands for range of valid values. In tasks for the exam, this wording often pops up. ODZ is useful to you not only in the case of logarithmic inequalities.
Take another look at the example above. We will consider the DHS based on it, so that you understand the principle, and the solution of logarithmic inequalities does not raise questions. From the definition of the logarithm, it follows 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 above. This can be done even orally, here it is clear that X cannot be less than 2. The solution to the inequality will be the definition of the range of admissible values.
Now let's move on to solving the simplest logarithmic inequality.
We discard the logarithms themselves from both sides of the inequality. What do we have left as a result? Simple inequality.
It is not difficult to solve it. X must be greater than -0.5. Now we combine the two obtained values \u200b\u200binto the system. In this way,
This will be the range of admissible values \u200b\u200bfor the considered logarithmic inequality.
Why do you need ODZ at all? This is an opportunity to weed out incorrect and impossible answers. If the answer is not within the range of valid values, then the answer simply does not make sense. This is worth remembering for a long time, since in the USE 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 stages. First, you need to find the range of valid values. There will be two values \u200b\u200bin the ODZ, we discussed this above. Next, you need to solve the inequality itself. Solution methods are as follows:
- multiplier replacement method;
- decomposition;
- method of rationalization.
Depending on the situation, you should use one of the above methods. Let's go directly to the solution. We will reveal the most popular method that is suitable for solving USE tasks in almost all cases. Next, we'll look at the decomposition method. It can help if you come across particularly tricky inequalities. So, the algorithm for solving the logarithmic inequality.
Solution examples :
We have not taken just such an inequality for nothing! Pay attention to the base. Remember: if it is greater than one, the sign remains the same when the range of acceptable values \u200b\u200bis found; 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 sign "less" we put "equal", we solve the equation. Thus, we will find the ODZ. We hope you won't have any problems solving such a simple equation. Answers are -4 and -2. That's not all. You need to display these points on the chart, place the "+" and "-". What needs to be done for this? Substitute numbers from intervals into the expression. Where the values \u200b\u200bare positive, we put "+" there.
Answer: x cannot be more than -4 and less than -2.
We found the range of valid values \u200b\u200bonly for the left side, now we need to find the range of valid values \u200b\u200bfor the right side. This is much easier. Answer: -2. We cross both obtained areas.
And only now are we beginning to address inequality itself.
Let's simplify it as much as possible to make it easier to solve.
Apply the spacing method again in the solution. Let's omit 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 basis.
The solution of logarithmic equations and inequalities with different bases involves the initial reduction to one base. Then follow the above method. But there is also a more complicated case. Consider one of the most difficult types of logarithmic inequalities.
Variable base 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 be beneficial for your educational process. We will understand the issue in detail. Let's discard the theory, let's go straight to practice. To solve logarithmic inequalities, it is enough to read the example once.
To solve the logarithmic inequality of the presented form, it is necessary to reduce the right-hand 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 method of rationalization, we pass to an equivalent system of inequalities. You will understand the rule itself when you substitute the corresponding values \u200b\u200band track their changes. The system will have the following inequalities.
Using the rationalization method when solving inequalities, you need to remember the following: it is necessary to subtract one from the base, x, by the definition of the logarithm, is subtracted from both sides of the inequality (right from left), two expressions are multiplied and set under the original sign with respect to zero.
The further solution is carried out by the method of intervals, everything is simple here. It is important for you to understand the differences in 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 sure that you can 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. Practice consistently in solving a variety of problems within the exam and you will be able to get the highest score. Good luck in your difficult business!
They are inside the logarithms.
Examples:
\\ (\\ log_3\u2061x≥ \\ log_3\u20619 \\)
\\ (\\ log_3\u2061 ((x ^ 2-3))< \log_3{(2x)}\)
\\ (\\ log_ (x + 1) \u2061 ((x ^ 2 + 3x-7))\u003e 2 \\)
\\ (\\ lg ^ 2\u2061 ((x + 1)) + 10≤11 \\ lg\u2061 ((x + 1)) \\)
How to solve logarithmic inequalities:
Any logarithmic inequality must be reduced to the form \\ (\\ log_a\u2061 (f (x)) ˅ \\ log_a (\u2061g (x)) \\) (the symbol \\ (˅ \\) means any of). This form allows you to get rid of the logarithms and their bases by making the transition to the inequality of expressions under the logarithms, that is, to the form \\ (f (x) ˅ g (x) \\).
But there is one very important subtlety when performing this transition:
\\ (- \\) if is 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 (lies between zero and one), then the sign of the inequality should be reversed, i.e.
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\\ (\\ log_2\u2061 ((8-x))<1\) Decision: |
\\ (\\ log \\) \\ (_ (0,5\u2061) \\) \\ ((2x-4) \\) ≥ \\ (\\ log \\) \\ (_ (0,5) \\) \u2061 \\ (((x + 1))\\) Decision: |
Very important! In any inequality, the transition from the form \\ (\\ log_a (\u2061f (x)) ˅ \\ log_a\u2061 (g (x)) \\) to the comparison of expressions under logarithms can be done only if:
Example ... Solve inequality: \\ (\\ log \\) \\ (≤-1 \\)
Decision:
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\\ (\\ log \\) \\ (_ (\\ frac (1) (3)) \u2061 (\\ frac (3x-2) (2x-3)) \\)\(≤-1\) |
Let's write out the ODZ. |
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ODZ: \\ (\\ frac (3x-2) (2x-3) \\) \\ (\u003e 0 \\) |
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\\ (\u2061 \\ frac (3x-2-3 (2x-3)) (2x-3) \\)\(≥\) \(0\) |
We open the brackets, we give. |
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\\ (\u2061 \\ frac (-3x + 7) (2x-3) \\) \\ (≥ \\) \\ (0 \\) |
We multiply the inequality by \\ (- 1 \\), not forgetting to reverse the comparison sign. |
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\\ (\u2061 \\ frac (3x-7) (2x-3) \\) \\ (≤ \\) \\ (0 \\) |
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\\ (\u2061 \\ frac (3 (x- \\ frac (7) (3))) (2 (x- \\ frac (3) (2))) \\)\(≤\) \(0\) |
Let's construct a number axis and mark the points \\ (\\ frac (7) (3) \\) and \\ (\\ frac (3) (2) \\) on it. Note that the dot from the denominator is punctured, despite the fact that the inequality is not strict. The point is that this point will not be a solution, since when substituted into an inequality, it will lead us to division by zero. |
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Now, on the same numerical axis, we put the ODZ and write in response the interval that falls into the ODZ. |
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We write down the final answer. |
Example ... Solve the inequality: \\ (\\ log ^ 2_3\u2061x- \\ log_3\u2061x-2\u003e 0 \\)
Decision:
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\\ (\\ log ^ 2_3\u2061x- \\ log_3\u2061x-2\u003e 0 \\) |
Let's write out the ODZ. |
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ODZ: \\ (x\u003e 0 \\) |
Let's get down to the solution. |
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Solution: \\ (\\ log ^ 2_3\u2061x- \\ log_3\u2061x-2\u003e 0 \\) |
Before us is a typical square-logarithmic inequality. We do it. |
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\\ (t \u003d \\ log_3\u2061x \\) |
Expand the left side of the inequality into. |
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\\ (D \u003d 1 + 8 \u003d 9 \\) |
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Now you need to go back to the original variable - x. To do this, go to one that has the same solution and make the reverse replacement. |
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\\ (\\ left [\\ begin (gathered) t\u003e 2 \\\\ t<-1 \end{gathered} \right.\) \(\Leftrightarrow\) \(\left[ \begin{gathered} \log_3x>2 \\\\ \\ log_3\u2061x<-1 \end{gathered} \right.\) |
Convert \\ (2 \u003d \\ log_3\u20619 \\), \\ (- 1 \u003d \\ log_3\u2061 \\ frac (1) (3) \\). |
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\\ (\\ left [\\ begin (gathered) \\ log_3\u2061x\u003e \\ log_39 \\\\ \\ log_3\u2061x<\log_3\frac{1}{3} \end{gathered} \right.\) |
We make the transition to the comparison of arguments. Logarithms have bases greater than \\ (1 \\), so the sign of the inequalities does not change. |
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\\ (\\ left [\\ begin (gathered) x\u003e 9 \\\\ x<\frac{1}{3} \end{gathered} \right.\) |
Let us combine the solution to inequality and the DHS in one figure. |
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Let's write down the answer. |
