Logarithmic inequalities with different variable bases. Logarithmic inequalities

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:

We should strive to reduce any logarithmic inequality to the form \(\log_a⁡(f(x)) ˅ \log_a(⁡g(x))\) (the symbol \(˅\) means any of ). This type allows you to get rid of logarithms and their bases, making the transition 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 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 inequality sign should change to the opposite, 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: \((6;8)\)

\(\log\)\(_(0.5⁡)\) \((2x-4)\)≥\(\log\)\(_(0.5)\) ⁡\(((x+ 1))\) 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 be done only if:

Example . Solve 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 and bring .
\(⁡\frac(-3x+7)(2x-3)\) \(≥\) \(0\)	We multiply the inequality by \(-1\), not forgetting 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 construct a number line and mark the points \(\frac(7)(3)\) and \(\frac(3)(2)\) on it. Please note that the dot is removed from the denominator, despite the fact that the inequality is not strict. The fact is that this point will not be a solution, since when substituted into 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.
	We write down the final answer.

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

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

Solution:

\(\log^2_3⁡x-\log_3⁡x-2>0\)	Let's write out the ODZ.
ODZ: \(x>0\)	Let's get to the solution.
Solution: \(\log^2_3⁡x-\log_3⁡x-2>0\)	Here we have a typical square-logarithmic inequality. Let's do it.
\(t=\log_3⁡x\) \(t^2-t-2>0\)	We expand the left side of the inequality into .
\(D=1+8=9\) \(t_1= \frac(1+3)(2)=2\) \(t_2=\frac(1-3)(2)=-1\) \((t+1)(t-2)>0\)
	Now we need to return to the original variable - x. To do this, let's go to , which has the same solution, and make the reverse substitution.
\(\left[ \begin(gathered) t>2 \\ t<-1 \end{gathered} \right.\) \(\Leftrightarrow\) \(\left[ \begin{gathered} \log_3⁡x>2\\\log_3⁡x<-1 \end{gathered} \right.\)	Transform \(2=\log_3⁡9\), \(-1=\log_3⁡\frac(1)(3)\).
\(\left[ \begin(gathered) \log_3⁡x>\log_39 \\ \log_3⁡x<\log_3\frac{1}{3} \end{gathered} \right.\)	Let's move on to comparing arguments. The bases of logarithms are greater than \(1\), so the sign of the inequalities does not change.
\(\left[ \begin(gathered) x>9 \\ x<\frac{1}{3} \end{gathered} \right.\)	Let us combine the solution to the inequality and the ODZ in one figure.
	Let's write down the answer.

Answer: \((0; \frac(1)(3))∪(9;∞)\)

LOGARITHMIC INEQUALITIES IN THE USE

Sechin Mikhail Alexandrovich

Small Academy of Sciences for Students of the Republic of Kazakhstan “Iskatel”

MBOU "Sovetskaya Secondary School No. 1", 11th grade, town. Sovetsky Sovetsky district

Gunko Lyudmila Dmitrievna, teacher of the Municipal Budgetary Educational Institution “Sovetskaya Secondary School No. 1”

Sovetsky district

Goal of the work: study of the solution mechanism logarithmic inequalities C3 using non-standard methods, identifying interesting facts about the logarithm.

Subject of study:

3) Learn to solve specific logarithmic inequalities C3 using non-standard methods.

Results:

Content

Introduction………………………………………………………………………………….4

Chapter 1. History of the issue……………………………………………………...5

Chapter 2. Collection of logarithmic inequalities ………………………… 7

2.1. Equivalent transitions and generalized interval method…………… 7

2.2. Rationalization method……………………………………………………………… 15

2.3. Non-standard substitution……………….................................................... ..... 22

2.4. Tasks with traps……………………………………………………27

Conclusion……………………………………………………………………………… 30

Literature……………………………………………………………………. 31

Introduction

I am in 11th grade and plan to enter a university where the core subject is mathematics. That’s why I work a lot with problems in part C. In task C3, I need to solve a non-standard inequality or system of inequalities, usually related to logarithms. When preparing for the exam, I was faced with the problem of a shortage of methods and techniques for solving 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 suggested that I work on C3 assignments independently under her guidance. In addition, I was interested in the question: do we encounter logarithms in our lives?

With this in mind, the topic was chosen:

“Logarithmic inequalities in the Unified State Exam”

Goal of the work: study of the mechanism for solving C3 problems using non-standard methods, identifying interesting facts about the logarithm.

Subject of study:

1) Find the necessary information about non-standard methods for solving logarithmic inequalities.

2) Find additional 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 clubs, and elective classes in mathematics.

The project product will be the collection “C3 Logarithmic 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 multi-year, calculations. Astronomy was in real danger of drowning in unfulfilled calculations. Difficulties arose in other areas, for example, in the insurance business, compound interest tables were needed for various interest rates. The main difficulty was multiplication and division of multi-digit numbers, especially trigonometric quantities.

The discovery of logarithms was based on the properties of progressions that were well known by the end of the 16th century. Archimedes spoke about the connection between the terms of the geometric progression q, q2, q3, ... and the arithmetic progression of their exponents 1, 2, 3,... in the Psalm. Another prerequisite was the extension of the concept of degree to negative and fractional exponents. Many authors have pointed out that multiplication, division, exponentiation and root extraction in geometric progression correspond in arithmetic - in the same order - addition, subtraction, multiplication and division.

Here was the idea of ​​the logarithm as an exponent.

In the history of the development of the doctrine of logarithms, several stages have passed.

Stage 1

Logarithms were invented no later than 1594 independently by the Scottish Baron Napier (1550-1617) and ten years later by the Swiss mechanic Bürgi (1552-1632). Both wanted to provide a new, convenient means of arithmetic calculations, although they approached this problem in different ways. Napier kinematically expressed the logarithmic function and thereby entered a new field of function theory. Bürgi remained on the basis of considering discrete progressions. However, the definition of the logarithm for both is not similar to the modern one. The term "logarithm" (logarithmus) belongs to Napier. It arose from a combination of Greek words: logos - “relation” and ariqmo - “number”, which meant “number of relations”. 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), a professor of mathematics at Gresh College in London, Napier suggested taking zero as the logarithm of one, and 100 as the logarithm of ten, or, what amounts to the same thing, just 1. This is how decimal logarithms and The first logarithmic tables were printed. Later, Briggs' tables were supplemented by the Dutch bookseller and mathematics enthusiast Adrian Flaccus (1600-1667). Napier and Briggs, although they came to logarithms earlier than everyone else, published their tables later than the others - in 1620. The signs log and Log were introduced in 1624 by I. Kepler. The term “natural logarithm” was introduced by Mengoli in 1659 and 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 name “New Logarithms”.

The first logarithmic tables were published in Russian in 1703. But in all logarithmic tables there were calculation errors. 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 analytical geometry and infinitesimal calculus. By that time, the connection between the quadrature of an equilateral hyperbola and the natural logarithm had been established. 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 an essay

"Logarithmotechnics" (1668) gives a series giving the expansion of ln(x+1) in

powers of x:

This expression exactly corresponds to his train of thought, although, of course, he did not use the signs d, ..., but more cumbersome symbolism. 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 a Higher Point of View,” given in 1907-1908, F. Klein proposed using the formula as the starting point for constructing the theory of logarithms.

Stage 3

Definition of a logarithmic function as an inverse function

exponential, logarithm as an exponent of a given base

was not formulated immediately. Essay by Leonhard Euler (1707-1783)

"An Introduction to the Analysis of Infinitesimals" (1748) served to further

development of the theory of logarithmic functions. Thus,

134 years have passed since logarithms were first introduced

(counting from 1614), before mathematicians came to the definition

the concept of logarithm, which is now the basis of the school course.

Chapter 2. Collection of logarithmic inequalities

2.1. Equivalent transitions and the generalized method of intervals.

Equivalent transitions

, if a > 1

, if 0 < а < 1

Generalized interval method

This method most universal when solving inequalities of almost any type. The solution diagram looks like this:

1. Bring the inequality to the form where the function on the left side is
, and on the right 0.

2. Find the domain of the function
.

3. Find the zeros of the function
, that is, solve the equation
(and solving an equation is usually easier than solving an inequality).

4. Draw the domain of definition and zeros of the function on the number line.

5. Determine the signs of the function
on the obtained intervals.

6. Select intervals where the function takes the required values ​​and write down the answer.

Example 1.

Solution:

Let's apply the interval method

where

For these values, all expressions under the logarithmic signs are positive.

Answer:

Example 2.

Solution:

1st way . ADL is determined by inequality x> 3. Taking logarithms for such x in base 10, we get

The last inequality could be solved by applying expansion rules, i.e. comparing factors to zero. However, in in this case easy to determine intervals of constant sign of a function

therefore, the interval method can be applied.

Function f(x) = 2x(x- 3.5)lgǀ x- 3ǀ is continuous at x> 3 and vanishes at points x 1 = 0, x 2 = 3,5, x 3 = 2, x 4 = 4. Thus, we determine the intervals of constant sign of the function f(x):

Answer:

2nd method . Let us directly apply the ideas of the interval method 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 at x> 3 is equivalent to inequality

or

The last inequality is solved using the interval method

Answer:

Example 3.

Solution:

Let's apply the interval method

Answer:

Example 4.

Solution:

Since 2 x 2 - 3x+ 3 > 0 for all real x, That

To solve the second inequality we use the interval method

In the first inequality we make the replacement

then we come to the inequality 2y 2 - y - 1 < 0 и, применив метод интервалов, получаем, что решениями будут те y, which satisfy the inequality -0.5< y < 1.

From where, because

we get the inequality

which is carried out when x, for which 2 x 2 - 3x - 5 < 0. Вновь применим метод интервалов

Now, taking into account the solution to the second inequality of the system, we finally obtain

Answer:

Example 5.

Solution:

Inequality is equivalent to a collection of systems

or

Let's use the interval method or

Answer:

Example 6.

Solution:

Inequality equals system

Let

Then y > 0,

and the first inequality

system takes the form

or, unfolding

quadratic trinomial factored,

Applying the interval method to the last inequality,

we see that its solutions satisfying the condition y> 0 will be all y > 4.

Thus, the original inequality is equivalent to the system:

So, the solutions to the inequality are all

2.2. Rationalization method.

Previously method rationalization of inequality was not solved, it was not known. This is the "new modern" effective method solutions to exponential and logarithmic inequalities" (quote from the book by S.I. Kolesnikova)
And even if the teacher knew him, there was a fear - does the Unified State Exam expert know him, and why don’t they give him at school? There were situations when the teacher said to the student: “Where did you get it? Sit down - 2.”
Now the method is being promoted everywhere. And for experts there is guidelines, associated with this method, and in the "Most Complete Editions of Model Options..." solution C3 uses this method.
WONDERFUL METHOD!

"Magic Table"

In other sources

If a >1 and b >1, then log a b >0 and (a -1)(b -1)>0;

If a >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)>0.

The reasoning carried out is simple, but significantly simplifies the solution of logarithmic inequalities.

Example 4.

log x (x 2 -3)<0

Solution:

Example 5.

log 2 x (2x 2 -4x +6)≤log 2 x (x 2 +x )

Solution:

Answer. (0; 0.5)U.

Example 6.

To solve this inequality, instead of the denominator, we write (x-1-1)(x-1), and instead of the numerator, we write the product (x-1)(x-3-9 + x).

Answer : (3;6)

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=3 x -1; then this inequality will take the form

Log 4 log 0.25
.

Because log 0.25 = -log 4 = -(log 4 y -log 4 16)=2-log 4 y , then we rewrite the last inequality as 2log 4 y -log 4 2 y ≤.

Let us make the replacement t =log 4 y and obtain the inequality t 2 -2t +≥0, the solution of which is the intervals - .

Thus, to find the values ​​of y we have a set of two simple inequalities
The solution to this set is the intervals 0<у≤2 и 8≤у<+.

Therefore, the original inequality is equivalent to the set of two exponential inequalities,
that is, aggregates

The solution to the first inequality of this set is the interval 0<х≤1, решением второго – промежуток 2≤х<+. Thus, the original inequality is satisfied for all values ​​of x from the intervals 0<х≤1 и 2≤х<+.

Example 8.

Solution:

Inequality equals system

The solution to the second inequality defining the ODZ will be the set of those x,

for which x > 0.

To solve the first inequality we make the substitution

Then we get 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

Lots of those x, which satisfy the last inequality

belongs to ODZ ( x> 0), therefore, is a solution to the system,

and hence the original inequality.

Answer:

2.4. Tasks with traps.

Example 1.

.

Solution. The ODZ of the inequality is all x satisfying the condition 0 . Therefore, all x are from the interval 0

Example 2.

log 2 (2 x +1-x 2)>log 2 (2 x-1 +1-x)+1.. ? The point is that the second number is obviously greater than

Conclusion

It was not easy to find specific methods for solving C3 problems from a 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 ODZ. These methods are not included in the school curriculum.

Using different methods, I solved 27 inequalities proposed on the Unified State Exam in part C, namely C3. These inequalities with solutions by methods formed the basis of the collection “C3 Logarithmic Inequalities with Solutions,” which became a project product of my activity. The hypothesis I posed at the beginning of the project was confirmed: C3 problems can be effectively solved if you know these methods.

In addition, I discovered interesting facts about logarithms. It was interesting for me to do this. My project products will be useful for both students and teachers.

Conclusions:

Thus, the project goal has been achieved and the problem has been solved. And I received the most complete and varied experience of project activities at all stages of work. While 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, and activity.

A guarantee of success when creating a research project for I gained: significant school experience, the ability to obtain information from various sources, check its reliability, and rank it by importance.

In addition to direct subject knowledge in mathematics, I expanded my 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. During the project activities, organizational, intellectual and communicative general educational skills were developed.

Literature

1. Koryanov A. G., Prokofiev A. A. Systems of inequalities with one variable (standard tasks C3).

2. Malkova A. G. Preparation for the Unified State Exam in Mathematics.

3. Samarova S. S. Solving logarithmic inequalities.

4. Mathematics. Collection of training works edited by A.L. Semenov and I.V. Yashchenko. -M.: MTsNMO, 2009. - 72 p.-

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 ​​has been found, all that remains is to intersect it with the solution of the 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.

Converting logarithmic inequalities

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:

1. Any number can be represented as a logarithm with a given base;
2. The sum and difference of logarithms with the same bases can be replaced by one logarithm.

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, the general scheme for solving logarithmic inequalities is as follows:

1. Find the VA of each logarithm included in the inequality;
2. Reduce the inequality to a standard one using the formulas for adding and subtracting logarithms;
3. Solve the resulting inequality using the scheme given above.

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 it, 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:

1. ODZ: x ∈ (−∞ 2/3)∪(1; +∞);
2. Candidate answer: x ∈ (−1; 3).

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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