Presentation on the topic "logarithmic equations". Presentation for a mathematics lesson "solving logarithmic equations" roots of the original equation

"Logarithmic equations."

Slide 2

Why were logarithms invented? To speed up calculations. To simplify calculations. To solve astronomical problems.

In a modern school, the main form of teaching mathematics, the main link in the integration of various organizational forms of teaching, is still the lesson. During the learning process, mathematical material is realized and assimilated mainly in the process of solving problems, therefore, in mathematics lessons, theory is not studied in isolation from practice. In order to successfully solve logarithmic equations, for which only 3 hours are allocated in the curriculum, you must have confident knowledge of formulas for logarithms and the properties of the logarithmic function. The topic “Logarithmic Equations” in the curriculum follows logarithmic functions and properties of logarithms. The situation is somewhat complicated compared to exponential equations by the presence of restrictions on the domain of definition of logarithmic functions. Using formulas for the logarithm of a product, quotient, and others without additional reservations can lead to both the acquisition of extraneous roots and the loss of roots. Therefore, it is necessary to carefully monitor the equivalence of the transformations being made.

Slide 3

“The invention of logarithms, while reducing the astronomer’s work, extended his life.”

Topic: “Logarithmic equations.” Objectives: Educational: 1. To familiarize and consolidate the basic methods for solving logarithmic equations, to prevent the occurrence of typical errors. 2. Provide each teacher with the opportunity to test their knowledge and improve their level. 3. Activate the work of the class through different forms of work. Developmental: 1.Develop self-control skills. Educational: 1. Foster a responsible attitude to work. 2. Cultivate the will and perseverance to achieve final results.

Slide 4

Lesson No. 1. Lesson topic: “Methods for solving logarithmic equations” Lesson type: Lesson on introducing new material Equipment: Multimedia.

During the classes. 1Organizational point: 2.Updating basic knowledge; Simplify:

Slide 5

Definition: An equation containing a variable under the logarithmic sign is called logarithmic. The simplest example of a logarithmic equation is the equation logax = b (a > 0, a≠ 1, b>0) Methods of solution Solving equations based on the definition of the logarithm, for example, the equation logax = b (a > 0, a≠ 1, b>0) has solution x = ab. Potentiation method. By potentiation we mean the transition from an equality containing logarithms to an equality that does not contain them: if logaf(x) = logag(x), then f(x) = g(x), f(x)>0, g(x )>0, a>0, a≠ 1. Method of introducing a new variable. Method of taking logarithms of both sides of an equation. A method for reducing logarithms to the same base. Functional - graphic method.

Slide 6

1method:

Based on the definition of the logarithm, equations are solved in which the logarithm is determined from the given bases and number, the number is determined from the given logarithm and base, and the base is determined from the given number and logarithm. Log2 4√2= x, log3√3 x = - 2, logx 64= 3, 2x= 4√2, x =3√3 – 2, x3 =64, 2x = 25/2, x =3- 3, x3 = 43, x =5/2. x = 1/27. x =4.

Slide 7

2method:

Solve the equations: lg(x2-6x+9) - 2lg(x - 7) = log9. The condition for verification is always made using the original equation. (x2-6x+9) >0, x≠ 3, X-7 >0; x >7; x >7. First, you need to transform the equation to the form log ((x-3)/(x-7))2 = log9 using the logarithm of the quotient formula. ((x-3)/(x-7))2 = 9, (x-3)/(x-7) = 3, (x-3)/(x-7)= - 3, x- 3 = 3x -21, x -3 = - 3x +21, x =9. x=6. extraneous root. Checking shows the 9th root of the equation. Answer: 9

Slide 8

Method 3:

Solve the equations: log62 x + log6 x +14 = (√16 – x2)2 + x2, 16 – x2 ≥0 ; - 4≤ x ≤ 4; x >0, x >0, O.D.Z. [ 0.4). log62 x + log6 x +14 = 16 – x2 + x2, log62 x + log6 x -2 = 0 replace log6 x = t t 2 + t -2 =0 ; D = 9; t1 =1, t2 = -2. log6 x = 1, x = 6 extraneous root. log6 x = -2, x = 1/36, check shows 1/36 is the root. Answer: 1/36.

Slide 9

4method:

Solve the equation = ZX, take the base 3 logarithm from both sides of the equation Question: 1. Is this an equivalent transformation? 2.If so, why? We get log3=log3(3x) . Taking into account Theorem 3, we obtain: log3 x2 log3x = log3 3x, 2log3x log3x = log3 3+ log3x, 2 log32x = log3x +1, 2 log32x - log3x -1=0, replace log3x = t, x >0 2 t2 + t - 2 =0 ; D = 9; t1 =1, t2 = -1/2 log3х = 1, x=3, log3х = -1/2, x= 1/√3. Answer: (3; 1/√3. ).

Slide 10

Method 5:

Solve the equations: log9(37-12x) log7-2x 3 = 1, 37-12x >0, x0, x

Slide 11

6 method

Solve the equations: log3 x = 12's. Since the function y = log3 x is increasing, and the function y = 12 is decreasing on (0; + ∞), then the given equation on this interval has one root. Which can be easily found. When x=10, the given equation turns into the correct numerical equality 1=1. The answer is x=10.

Slide 12

Lesson summary. What methods of solving logarithmic equations did we learn in class? Homework: Determine the solution method and solve No. 1547 (a, b), No. 1549 (a, b), No. 1554 (a, b). Work through all the theoretical material and analyze examples §52.

Slide 13

Lesson 2. Lesson topic: “Application of various methods in solving logarithmic equations.” Lesson type: Lesson to consolidate what has been learned. Progress of the lesson. 1. Organizational point: 2. “Test yourself” 1)log-3 ((x-1)/5)=? 2) log5 (121 – x2), (121 – x2) ≥ 0, x

Slide 14

3. Performing exercises: No. 1563 (b)

How can you solve this equation? (method of introducing a new variable) log3 2x +3 log3x +9 = 37/ log3 (x/27); x>0 Let us denote log3x = t ; t 2 -3 t +9 =37/(t-3) ; t ≠ 3, (t-3) (t 2 -3 t +9) = 37, t3-27 = 37; t3= 64 ; t=4. log3x = 4; x=81. By checking we are convinced that x=81 is the root of the equation.

Slide 15

No. 1564 (a); (logarithm method)

log3 x X = 81, take the logarithm to base 3 from both sides of the equation; log3 x log3 X = log3 81; log3x log3x = log381; log3 2x =4; log3x =2, x=9 ; log3 x = -2, x = 1/9. By checking we are convinced that x=9 and x=1/9 are the roots of the equation.

Slide 16

4. Physical education minute (at desks, sitting).

1 The domain of definition of the logarithmic function y = log3 X is the set of positive numbers. 2The function y = log3 X increases monotonically. 3. The range of values of the logarithmic function is from 0 to infinity. 4 logас/в = logа с - logа в. 5 It is true that log8 8-3 =1.

Slide 17

No. 1704.(a)

1-√x =In x Since the function y=In x is increasing, and the function y =1-√x is decreasing on (0; + ∞), then the given equation on this interval has one root. Which can be easily found. When x=1, the given equation turns into the correct numerical equality 1=1. Answer: x=1.

Slide 18

No. 1574(b)

log3 (x + 2y) -2log3 4 =1- log3 (x - 2y), log3 (x 2 - 4y 2) = log3 48, log1/4 (x -2y) = -1; log1/4 (x -2y) = -1; x 2 - 4y 2 – 48 =0, x =4 +2y, x =8, x -2y = 4; 16у = 32; y =2. By checking we make sure that the found values are solutions of the system.

Slide 19

5. What a delight Logarithmic “comedy 2 > 3”

1/4 > 1/8 is undoubtedly correct. (1/2)2 > (1/2)3, which also does not inspire doubt. A larger number corresponds to a larger logarithm, which means log(1/2)2 > log(1/2)3; 2lg(1/2) > 3lg(1/2). After reduction by lg(1/2) we have 2 > 3. - Where is the error?

Slide 20

6.Run the test:

1Find the domain of definition: y = log0.3 (6x –x2). 1(-∞ ;0) Ư(6 ; + ∞); 2. (-∞ ; -6) Ư(0 ; + ∞); 3.(-6; 0). 4.(0; 6). 2. Find the range of values: y = 2.5 + log1.7 x. 1(2.5 ; + ∞); 2. (-∞; 2.5); 3 (- ∞ ; + ∞); 4. (0 ; + ∞). 3.Compare: log0.5 7 and log0.5 5. 1.>. 2.<. :="" log5x="х" .="" log4="">

Slide 21

Answer: 4; 3;2;1;2.

Lesson summary: To solve logarithmic equations well, you need to improve your skills in solving practical problems, since they are the main content of the exam and life. Homework: No. 1563 (a, b), No. 1464 (b, c), No. 1567 (b).

Slide 22

Lesson 3. Lesson topic: “Solving logarithmic equations” Lesson type: generalization lesson, systematization of knowledge. Lesson progress. 1. Updating background knowledge:

No. 1 Which of the numbers are -1; 0; 1; 2; 4; 8 are the roots of the equation log2 x=x-2? No. 2 Solve the equations: a) log16x= 2; c) log2 (2x-x2) -=0; d) log3 (x-1)=log3 (2x+1) No. 3 Solve the inequalities: a) log3x> log3 5; b) log0.4x0. No. 4 Find the domain of definition of the function: y = log2 (x + 4) No. 5 Compare the numbers: log3 6/5 and log3 5/6; log0.2 5 and. Log0.2 17. No. 6 Determine the number of roots of the equation: log3 X= =-2x+4.

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Slide captions:

Logarithms Solving logarithmic equations and inequalities

The concept of a logarithm For any and degree with an arbitrary real exponent is defined and equal to some positive real number: The exponent 𝑝 of the degree is called the logarithm of this degree with the base.

The logarithm of a positive number to a positive and unequal base: is the exponent that, when raised to which the number is obtained. or, then

PROPERTIES OF LOGARITHMES 1) If then. If then. 2) If then. If then.

In all equalities. 3) ; 4) ; 5) ; 6) ; 7) ; 8) ; 9) ; ;

10) , ; eleven) , ; 12) if; 13), if is an even number, if is an odd number.

Decimal logarithm and natural logarithm A decimal logarithm is a logarithm if its base is 10. Decimal logarithm notation: . A logarithm is called a natural logarithm if its base is equal to a number. Notation for natural logarithm: .

Examples with logarithms Find the meaning of the expression: No. 1. ; No. 2. ; No. 3. ; No. 4. ; No. 5. ; No. 6. ; No. 7. ; No. 8. ; No. 9. ;

№ 10. ; № 11. ; № 12. ; № 13. ; № 14. ; № 15. ; № 16. ; № 17. ; № 18. ; № 19. ; № 20. ; № 21. ;

No. 22. ; No. 23. ; No. 24. ; No. 25. ; No. 26. Find the value of the expression if; No. 27. Find the value of the expression if; No. 28. Find the value of the expression if.

Solving examples with logarithms No. 1. . Answer. . No. 2. . Answer. . No. 3. . Answer. . No. 4. . Answer. . No. 5. . Answer. .

No. 6. . Answer. . No. 7. . Answer. . No. 8. . Answer. . No. 9. . Answer. . No. 10. . Answer. .

No. 11. Answer. . No. 12. . Answer. . No. 13. . Answer. No. 14. . Answer. .

No. 15. . Answer. No. 16. . Answer. No. 17. . Answer. . No. 18. . Answer. . No. 19. . Answer. .

No. 20. . Answer. . No. 21. . Answer. . No. 22. . Answer. . No. 23. . No. 24. . Answer. . No. 25. . Answer. .

No. 26. . E if, then. Answer. . No. 27. . E if, then. Answer. . No. 28. . If. Answer. .

The simplest logarithmic equations The simplest logarithmic equation is an equation of the form: ; , where and are real numbers, are expressions containing.

Methods for solving the simplest logarithmic equations 1. By definition of the logarithm. A) If, then the equation is equivalent to Eq. B) The equation is equivalent to the system

2. Potentiation method. A) If that equation is equivalent to the system B) The equation is equivalent to the system

Solving the simplest logarithmic equations No. 1. Solve the equation. Solution. ; ; ; ; . Answer. . #2: Solve the equation. Solution. ; ; ; . Answer. .

#3: Solve the equation. Solution. . Answer. .

#4: Solve the equation. Solution. . Answer. .

Methods for solving logarithmic equations 1. Potentiation method. 2. Functional-graphic method. 3. Factorization method. 4. Variable replacement method. 5. Logarithm method.

Features of solving logarithmic equations Apply the simplest properties of logarithms. Distribute terms containing unknowns, using the simplest properties of logarithms, in such a way that logarithms of ratios do not arise. Apply chains of logarithms: the chain is expanded based on the definition of a logarithm. Applying the properties of the logarithmic function.

No. 1. Solve the equation. Solution. Let's transform this equation using the properties of the logarithm. This equation is equivalent to the system:

Let's solve the first equation of the system: . Considering that and, we get. Answer. .

#2: Solve the equation. Solution. . Using the definition of a logarithm, we get: Let's check by substituting the found values of the variable into the quadratic trinomial, we obtain, therefore, the values are the roots of this equation. Answer. .

#3: Solve the equation. Solution. We find the domain of definition of the equation: . Let's transform this equation

Taking into account the domain of definition of the equation, we obtain. Answer. .

#4: Solve the equation. Solution. Equation domain: . Let's transform this equation: . Solve using the variable replacement method. Let then the equation take the form:

Considering that, we get the equation Reverse substitution: Answer.

#5: Solve the equation. Solution. You can guess the root of this equation: . We check: ; ; . The true equality is therefore the root of this equation. And now: LOGARIFTH HARD! Let's take the logarithm of both sides of the equation to the base. We obtain an equivalent equation: .

We have obtained a quadratic equation for which one root is known. Using Vieta's theorem, we find the sum of the roots: , therefore, we find the second root: . Answer. .

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Slide captions:

Logarithmic inequalities Logarithmic inequalities are inequalities of the form, where are expressions containing. If in the inequalities the unknown is under the sign of the logarithm, then the inequalities are classified as logarithmic inequalities.

Properties of logarithms expressed by inequalities 1. Comparison of logarithms: A) If, then; B) If, then. 2. Comparison of a logarithm with a number: A) If, then; B) If, then.

Properties of monotonicity of logarithms 1) If, then and. 2) If, then and 3) If, then. 4) If, then 5) If, then and

6) If, then and 7) If the base of the logarithm is variable, then

Methods for solving logarithmic inequalities 1. Potentiation method. 2. Application of the simplest properties of logarithms. 3. Factorization method. 4. Variable replacement method. 5. Application of the properties of the logarithmic function.

Solving Logarithmic Inequalities #1: Solve the inequality. Solution. 1) Find the domain of definition of this inequality. 2) Let us transform this inequality, therefore, .

3) Considering that, we get. Answer. . #2: Solve the inequality. Solution. 1) Find the domain of definition of this inequality

From the first two inequalities: . Let's estimate. Let's consider inequality. The following condition must be met: . If, then, then.

2) Let's transform this inequality, therefore, Solve the equation. The sum of the coefficients is therefore one of the roots. Divide the fournomial by the binomial, we get.

Then, therefore, solving this inequality by the method of intervals, we determine. Considering that, we find the values of the unknown quantity. Answer. .

#3: Solve the inequality. Solution. 1) Let's transform. 2) This inequality takes the form: and

Answer. . No. 4. Solve the inequality. Solution. 1) Transform this equation. 2) Inequality is equivalent to a system of inequalities:

3) Solve the inequality. 4) Consider the system and solve it. 5) Solving inequality. a) If, then, therefore,

Solution of inequality. b) If, then, therefore, . Taking into account what we have considered, we obtain a solution to the inequality. 6) We get it. Answer. .

No. 5. Solve the inequality. Solution. 1) Transform this inequality 2) The inequality is equivalent to a system of inequalities:

Answer. . No. 6. Solve the inequality. Solution. 1) Transform this inequality. 2) Taking into account the transformations of the inequality, this inequality is equivalent to the system of inequalities:

No. 7. Solve the inequality. Solution. 1) Find the domain of definition of this inequality: .

2) Transform this inequality. 3) We use the variable replacement method. Let, then the inequality can be represented as: . 4) Let's perform the reverse replacement:

5) Solving inequality.

6) Solving inequality

7) We obtain a system of inequalities. Answer. .

The topic of my methodological work in the 2013–2014 academic year, and later in the 2015–2016 academic year “Logarithms. Solving logarithmic equations and inequalities.” This work is presented in the form of a presentation for lessons.

RESOURCES AND LITERATURE USED 1. Algebra and principles of mathematical analysis. 10 11 grades. In 2 hours. Part 1. Textbook for students of general education institutions (basic level) / A.G. Mordkovich. M.: Mnemosyne, 2012. 2. Algebra and the beginnings of analysis. 10 11 grades. Modular triactive course / A.R. Ryazanovsky, S.A. Shestakov, I.V. Yashchenko. M.: Publishing house “National Education”, 2014. 3. Unified State Examination. Mathematics: standard exam options: 36 options / ed. I.V. Yashchenko. M.: Publishing house “National Education”, 2015.

4. Unified State Exam 2015. Mathematics. 30 variants of standard test tasks and 800 tasks of part 2 / I.R. Vysotsky, P.I. Zakharov, V.S. Panferov, S.E. Positselsky, A.V. Semenov, M.A. Semyonova, I.N. Sergeev, V.A. Smirnov, S.A. Shestakov, D.E. Shnol, I.V. Yashchenko; edited by I.V. Yashchenko. M.: Publishing house “Examination”, publishing house MTsNMO, 2015. 5. Unified State Exam-2016: Mathematics: 30 options of examination papers for preparing for the unified state exam: profile level / ed. I.V. Yashchenko. M.: AST: Astrel, 2016. 6. mathege.ru. Open bank of tasks in mathematics.

Counting and calculations are the basis of order in the head

Johann Heinrich Pestalozzi

Find errors:

log 3 24 – log 3 8 = 16
log 3 15 + log 3 3 = log 3 5
log 5 5 3 = 2
log 2 16 2 = 8
3log 2 4 = log 2 (4*3)
3log 2 3 = log 2 27
log 3 27 = 4
log 2 2 3 = 8

Calculate:

log 2 11 – log 2 44
log 1/6 4 + log 1/6 9
2log 5 25 +3log 2 64

Find x:

log 3 x = 4
log 3 (7x-9) = log 3 x

Peer review

True equalities

Calculate

-2

Find x

Results of oral work:

“5” - 12-13 correct answers

“4” - 10-11 correct answers

“3” - 8-9 correct answers

“2” - 7 or less

Find x:

log 3 x = 4
log 3 (7x-9) = log 3 x

Definition

An equation containing a variable under the logarithm sign or in the base of the logarithm is called logarithmic

For example, or

If an equation contains a variable that is not under the logarithmic sign, then it will not be logarithmic.

For example,

Are not logarithmic

Are logarithmic

1. By definition of logarithm

The solution to the simplest logarithmic equation is based on applying the definition of logarithm and solving the equivalent equation

Example 1

2. Potentization

By potentiation we mean the transition from an equality containing logarithms to an equality not containing them:

Having solved the resulting equality, you should check the roots,

because the use of potentiation formulas expands

domain of equation

Example 2

Solve the equation

Potentiating, we get:

Examination:

Answer

Example 2

Solve the equation

Potentiating, we get:

is the root of the original equation.

REMEMBER!

Logarithm and ODZ

together

are working

everywhere!

Sweet couple!

Two of a Kind!

- LOGARITHM !

SHE

ODZ!

Two in one!

Two banks of one river!

We can't live

friend without

friend!

Close and inseparable!

3. Application of the properties of logarithms

Example 3

Solve the equation

0 Moving on to the variable x, we get: ; x = 4 satisfy the condition x 0, therefore, the roots of the original equation. "width="640"

4. Introduction of a new variable

Example 4

Solve the equation

Moving on to the variable x, we get:

; X = 4 satisfy the condition x 0 therefore

roots of the original equation.

Determine the method for solving the equations:

Applying

holy of logarithms

A-priory

Introduction

new variable

Potentiation

The nut of knowledge is very hard,

But don't you dare back down.

“Orbit” will help you crack it,

And pass the knowledge exam.

№ 1 Find the product of the roots of the equation

4) 1,21

3) 0 , 81

2) - 0,9

1) - 1,21

№ 2 Specify the interval to which the root of the equation

1) (- ∞;-2]

2) [ - 2;1]

4) }