Presentation for the lesson of mathematics "solution of logarithmic equations". Presentation on the topic "logarithmic equations" bracketing the common factor
"Logarithmic equations."
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Why were logarithms invented? To speed up calculations. To simplify calculations. To solve astronomical problems.
In a modern school, the lesson is still the main form of teaching mathematics, the main link in the integration of various organizational forms of education. In the process of learning, 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 allotted in the curriculum, it is necessary to have a confident knowledge of the formulas for logarithms and the properties of the logarithmic function. The Logarithmic Equations topic in the curriculum comes after logarithmic functions and the properties of logarithms. The situation is somewhat more complicated in comparison with exponential equations by the presence of restrictions on the domain of definition of logarithmic functions. The use of formulas for the logarithm of the 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.
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“The invention of logarithms, having shortened the work of the astronomer, extended his life”
Topic: "Logarithmic equations." Objectives: Educational: 1. To introduce and consolidate the basic methods for solving logarithmic equations, to prevent the appearance of typical errors. 2. Provide each trainee with the opportunity to test their knowledge and improve their level. 3.Activate the work of the class through different forms of work. Developing: 1.Develop self-control skills. Educational: 1. To cultivate a responsible attitude to work. 2. To cultivate the will and perseverance to achieve the final results.
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Lesson number 1. Theme of the lesson: "Methods for solving logarithmic equations" Type of lesson: Lesson of familiarization with new material Equipment: Multimedia.
During the classes. 1 Organizational moment: 2. Actualization of basic knowledge; Simplify:
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Definition: An equation containing a variable under the logarithm sign is called a logarithmic equation. The simplest example of a logarithmic equation is the equation logax = b (a > 0, a≠ 1, b>0) Solutions Solving equations based on the definition of a logarithm, for example, the equation logax = b (a > 0, a≠ 1, b>0) has solution x = ab. potentiation method. Potentiation is understood as 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. The method of introducing a new variable. The method of taking the logarithm of both parts of the equation. Method for reducing logarithms to the same base. Functional - graphical method.
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1 method:
Based on the definition of the logarithm, equations are solved in which the logarithm is determined by the given bases and number, the number is determined by the given logarithm and base, and the base is determined by 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 \u003d 43, x \u003d 5/2. x = 1/27. x = 4.
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2 method:
Solve the equations: lg(x2-6x+9) - 2lg(x - 7) = lg9. The condition for verification is always compiled according to the original equation. (x2-6x+9) >0, x≠ 3, X-7 >0; x>7; x>7. From the beginning, you need to transform the equation to bring it to the form log ((x-3) / (x-7)) 2 = lg9 using the formula of the logarithm of the quotient. ((x-3)/(x-7))2 = 9, (x-3)/(x-7) = 3, (x-3)/(x-7)= - 3, x-3 = 3x -21, x -3 \u003d- 3x +21, x \u003d 9. x=6. foreign root. The check shows the 9 root of the equation. Answer: 9
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3 method:
Solve the equations: log62 x + log6 x +14 \u003d (√16 - x2) 2 + x2, 16 - x2 ≥0; - 4≤ x ≤ 4; x>0, x>0, O.D.Z. [ 0.4). log62 x + log6 x +14 \u003d 16 - x2 + x2, log62 x + log6 x -2 = 0 replace log6 x \u003d t t 2 + t -2 \u003d 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.
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4method:
Solve the equations = ZX, take the logarithm in base 3 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 get: 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 log3x = 1, x=3, log3x = -1/2, x= 1/√3. Answer: (3 ; 1/√3. ).
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5 method:
Solve equations: log9(37-12x) log7-2x 3 = 1, 37-12x >0, x0, x
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6 method
Solve the equations: log3 x = 12-x. Since the function y \u003d log3 x is increasing, and the function y \u003d 12 x is decreasing on (0; + ∞), then the given equation on this interval has one root. Which is easy to find. At x=10, the given equation turns into the correct numerical equality 1=1. The answer is x=10.
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Summary of the lesson. What methods of solving logarithmic equations did we meet in the lesson? 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.
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2 lesson. Lesson topic: "Application of various methods for solving logarithmic equations." Lesson type: Lesson to reinforce what has been learned Lesson progress. 1. Organizational moment: 2. "Test yourself" 1) log-3 ((x-1) / 5) =? 2) log5 (121 – x2), (121 – x2) ≥ 0, x
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3. Performing exercises: No. 1563 (b)
How can this equation be solved? (method introducing a new variable) log3 2x +3 log3x +9 = 37/log3 (x/27); х>0 Denote log3х = t ; t 2 -3 t +9 \u003d 37 / (t-3) ; t ≠ 3, (t-3) (t 2 -3 t +9) = 37, t3-27 = 37; t3= 64 ; t=4. log3x = 4; x \u003d 81. By checking we make sure that x \u003d 81 is the root of the equation.
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No. 1564 (a); (logarithm method)
log3 x X \u003d 81, take the logarithm in 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 \u003d -2, x \u003d 1/9. By checking we are convinced that x=9 and x=1/9 are the roots of the equation.
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4. Physical education minute (at desks, sitting).
1 The domain of definition of the logarithmic function y \u003d log3 X is the set of positive numbers. 2 The function y = log3 X is monotonically increasing. 3. Range of values of the logarithmic function from 0 to infinity. 4 logas / in = loga with - loga in. 5 It is true that log8 8-3 =1.
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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 is easy to find. At 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 \u003d 1- log3 (x - 2y), log3 (x 2 - 4y 2) \u003d log3 48, log1 / 4 (x -2y) \u003d -1; log1/4 (x -2y) = -1; x 2 - 4y 2 - 48 \u003d 0, x \u003d 4 + 2y, x \u003d 8, x -2y \u003d 4; 16y = 32; y=2. By checking, we make sure that the found values are the solutions of the system.
Slide 19
5. What a delight Logarithmic “comedy 2 > 3”
1/4 > 1/8 is undeniably correct. (1/2)2 > (1/2)3, which also does not inspire doubt. A larger number corresponds to a larger logarithm, which means that lg(1/2)2 > lg(1/2)3; 2lg(1/2) > 3lg(1/2). After reduction by lg(1/2) we have 2 > 3. - Where is the mistake?
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6.Perform the test:
1 Find the domain of definition: y \u003d log0.3 (6x -x2). 1(-∞ ;0) Ư(6 ; + ∞); 2. (-∞ ; -6) Ư(0 ; + ∞); 3.(-6; 0). 4.(0; 6). 2. Find the range: y \u003d 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="">
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Answer: 4; 3;2;1;2.
Lesson summary: To solve logarithmic equations well, you need to improve your skills in solving practical tasks, since they are the main content of the exam and life. Homework: No. 1563 (a, b), No. 1464 (b, c), No. 1567 (b).
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Lesson 3. Theme of the lesson: “Solution of logarithmic equations” Type of lesson: generalization lesson, systematization of knowledge. Course of the lesson.
№1 Which of the numbers -1; 0; one; 2; four; 8 are the roots of the equation log2 x=x-2? №2 Solve the equations: a) log16x= 2; c) log2 (2x-x2) -=0; d) log3 (х-1)=log3 (2х+1) №3 Solve inequalities: a) log3х> log3 5; b) log0.4x0. No. 4 Find the domain of the function: y \u003d log2 (x + 4) No. 5 Compare the numbers: log3 6/5 and log3 5/6; log0.2 5 i. Log0,2 17. №6 Determine the number of roots of the equation: log3 X==-2x+4.
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Slides captions:
Logarithms Solving logarithmic equations and inequalities
The concept of the logarithm For any and, a power 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 a base.
The logarithm of a positive number in a positive and unequal base: the exponent is called, when raised to which the number is obtained. or, then
PROPERTIES OF LOGARITHMS 1) If then. If then. 2) If then. If then.
In all equalities. 3) ; four) ; 5) ; 6); 7); eight) ; 9) ; ;
ten) , ; 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 natural logarithm is a logarithm if its base is equal to a number. Natural logarithm notation: .
Examples with logarithms Find the value of the expression: No. 1. ; No. 2.; Number 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.; № 26. Find the value of the expression if; № 27. Find the value of the expression if; № 28. Find the value of the expression if.
Solution of examples with logarithms No. 1. . Answer. . No. 2. . Answer. . Number 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 a. 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 the equation. B) The equation is equivalent to the system
2. Method of potentiation. A) If then the equation is equivalent to the system B) The equation is equivalent to the system
Solution of 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-graphical method. 3. Method of factorization. 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 the logarithm. Application of the properties of the logarithmic function.
No. 1 . Solve the equation. Solution. We 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. . We use the definition of the logarithm, we get. Let's check, substituting the found values of the variable into the square trinomial, we get, therefore, the values are the roots of this equation. Answer. .
#3 Solve the equation. Solution. Find the domain of the equation: . We 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: . We solve by changing the variable. Let the equation then take the form:
Considering that, we get the equation Reverse replacement: Answer.
#5 Solve the equation. Solution. You can guess the root of this equation:. We check: ; ; . True equality, therefore, is the root of this equation. And now: DIFFICULT LOGARIFM! Let's take the logarithm of both sides of the equation to the base. We get an equivalent equation: .
We got a quadratic equation, which has one root. According to the Vieta theorem, we find the sum of the roots: therefore, we find the second root:. Answer. .
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Slides captions:
Logarithmic inequalities Logarithmic inequalities are inequalities of the form, where are expressions containing. If in 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.
Monotonicity properties 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 a variable, then
Methods for solving logarithmic inequalities 1. Potentiation method. 2. Application of the simplest properties of logarithms. 3 . Factoring 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) We transform this inequality, therefore, .
3) Given 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 figure it out. Consider the inequality. The condition must be met: . If, then, then.
2) We transform this inequality, therefore, We solve the equation. The sum of the coefficients, hence one of the roots. We divide the quadrilateral 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) We transform this equation. 2) Inequality is equivalent to a system of inequalities:
3) We solve the inequality. 4) We consider the system and solve it. 5) We solve the inequality. a) If, then, therefore,
Solution of inequality. b) If, then, therefore, . Considering what we have considered, we obtain a solution to the inequality. 6) We receive. Answer. .
No. 5 . Solve the inequality. Solution. 1) We transform this inequality 2) The inequality is equivalent to the system of inequalities:
Answer. . No. 6 . Solve the inequality. Solution. 1) We 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) We transform this inequality. 3) We apply the variable replacement method. Let, then the inequality can be represented as: . 4) Let's perform the reverse replacement:
5) We solve the inequality.
6) Solve the inequality
7) We get 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 is “Logarithms. Solution of logarithmic equations and inequalities”. This work is presented in the form of a presentation for the lessons.
USED RESOURCES AND LITERATURE 1. Algebra and the beginnings of mathematical analysis. 10 11 classes. At 2 hours. Part 1. A textbook for students of educational institutions (basic level) / A.G. Mordkovich. Moscow: Mnemosyne, 2012. 2. Algebra and the beginnings of analysis. 10 11 classes. Modular triactive course / A.R. Ryazanovsky, S.A. Shestakov, I.V. Yashchenko. Moscow: National Education Publishing House, 2014. 3. USE. Mathematics: typical examination options: 36 options / ed. I.V.Yashchenko. Moscow: National Education Publishing House, 2015.
4. USE 2015. Mathematics. 30 variants of typical test tasks and 800 tasks of part 2 / I.R. Vysotsky, P.I. Zakharov, V.S. Panferov, S.E. Positselsky, A.V. Semyonov, M.A. Semyonova, I.N. Sergeev, V.A. Smirnov, S.A. Shestakov, D.E. Shnol, I.V. Yaschenko; ed. I.V. Yashchenko. M.: Exam Publishing House, MTsNMO Publishing House, 2015. 5. Unified State Examination-2016: Mathematics: 30 options for examination papers to prepare for the unified state exam: profile level / ed. I.V. Yashchenko. M.: AST: Astrel, 2016. 6. mathege.ru. Open bank of tasks in mathematics.
1. Introductory part.
Grade 11 is a crucial stage in your life's journey, the year of graduation, and, of course, the year when the results of the most important topics you study in algebra lessons are summed up. We will devote our lesson to repetition.Lesson objective : systematize methods for solving exponential and logarithmic equations. And the epigraph to our lesson will be the wordscontemporary Polish mathematician Stanisław Koval: "Equations are the golden key that unlocks all mathematical sesame." (SLIDE 2)
2. Oral account.
The English philosopher Herbert Spencer said: “The roads are not the knowledge that is stored in the brain like fat, the roads are those that turn into mental muscles.”(SLIDE 3)
(Work is being done with cards for 2 options, followed by verification.)
SOLVE AND WRITE ANSWERS. (1 option)370 + 230 3 0.3 7 - 2.1 -23 - 29 -19 + 100
: 50 + 4,1: 7: (-13) : (-3)
30: 100 1.4 (-17) - 13
340 20 + 0.02 - 32 + 40
________ __________ __________ _________ _________
? ? ? ? ?
SOLVE AND WRITE ANSWERS. (Option 2)280 + 440 2 0.4 8 - 3.2 -35 - 33 -64 + 100
: 60 +1,2: 8: (-17) : (-2)
40: 100 1.6 (-13) - 12
220 50 +0.04 – 48 + 30
_________ ________ _________ _________ _________
? ? ? ? ?
The time has expired. Exchange a card with a neighbor.
Check the correctness of the solution and the answers.(SLIDE 4)
And rate according to the following criteria. (SLIDE 5)
3. Repetition of material.
a) Graphs and properties of exponential and logarithmic functions. (SLIDE 6-9)
b) Orally complete the tasks written on the board. (From the bank of USE assignments)
c) Let us recall the solution of the simplest exponential and logarithmic equations.
4 x - 1 = 1 27 x = 2 4 X = 64 5 X = 8 X
log 6 x = 3log 7 (x+3) = 2log 11 (2x - 5) =log 11 (x+6)log 5 X 2 = 0
4. Work in groups.
Ancient Greek poet Nivei argued that "mathematics cannot be learned by watching your neighbor do it." Therefore, we will now work independently.
A group of weak students solves the equations of the 1st part of the exam.
1.logarithmic
.
.
If the equation has more than one root, indicate the smaller one in your answer.
2.Demonstration
A group of stronger students continue to repeat methods for solving equations.
Suggest a method for solving equations.
1. 4. log 6x (X 2 – 8x) =log 6x (2x - 9)
2. 5 lg 2 x 4 –lgx 14 = 2
3. 6 log 3 x + log 9 x + log 81 x=7
5. Homework:
№ 163-165(a), 171(a), 194(a), 195(a)
6. The results of the lesson.
Let's return to the epigraph of our lesson "Solving equations is the golden key that opens all sesame."
I would like to wish you that each of you will find in life your own golden key, with the help of which any doors will open in front of you.
Evaluation of the work of the class and each student individually, checking the evaluation sheets and grading.
7. Reflection.
The teacher needs to know how independently and with what confidence the student performed the task. To do this, the students will answer the test questions (questionnaire), and then the teacher will process the results.
I worked actively / passively in the lessonI am satisfied/dissatisfied with my work at the lesson
The lesson seemed short / long for me
For the lesson I'm not tired / tired
My mood got better / got worse
The material of the lesson was clear / not clear to me
useful / useless
interesting / boring
Counting and calculation - 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
Mutual check
True equalities
Calculate
-2
-2
22
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 sign of the logarithm or at the base of the logarithm is called logarithmic
For example, or
- If the equation contains a variable that is not under the sign of the logarithm, then it will not be logarithmic.
For example,
Are not logarithmic
Are logarithmic
1. By definition of the logarithm
The solution of the simplest logarithmic equation is based on applying the definition of the logarithm and solving the equivalent equation
Example 1
2. Potentiation
By potentiation is meant the transition from an equality containing logarithms to an equality that does not contain them:
Having solved the resulting equality, you should check the roots,
since the use of potentiation formulas expands
domain of the equation
Example 2
Solve the Equation
Potentiating, we get:
Examination:
If a
Answer
Example 2
Solve the Equation
Potentiating, we get:
is the root of the original equation.
REMEMBER!
Logarithm and ODZ
together
are toiling
everywhere!
Sweet couple!
Two of a Kind!
HE
- LOGARIFM !
SHE IS
-
ODZ!
Two in one!
Two banks on one river!
We don't live
friend without
friend!
Close and inseparable!
3. Application of the properties of logarithms
Example 3
Solve the Equation
0 Passing to the variable x, we get: ; x \u003d 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
Passing to the variable x, we get:
; X = 4 satisfy the condition x 0, so
roots of the original equation.
Determine the method for solving equations:
Applying
holy logarithms
By definition
Introduction
new variable
Potentiation
The nut of knowledge is very hard,
But don't you dare back down.
Orbit will help to gnaw it,
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]
3)
2) [ - 2;1]
4) }
