Now we will look at the transformation of expressions containing logarithms from a general point of view. Here we will analyze not only the transformation of expressions using the properties of logarithms, but we will consider the transformation of expressions with logarithms general view, which contain not only logarithms, but also powers, fractions, roots, etc. As usual, we will supply all material with characteristic examples with detailed descriptions solutions.
Page navigation.
Expressions with logarithms and logarithmic expressions
Performing actions with fractions
In the previous paragraph, we analyzed the main transformations that are carried out with individual fractions containing logarithms. These transformations, of course, can be carried out with each individual fraction that is part of a more complex expression, for example, representing the sum, difference, product, and quotient of similar fractions. But in addition to working with individual fractions, the transformation of expressions of this kind often involves performing appropriate actions with fractions. Next, we will consider the rules by which these actions are carried out.
From grades 5-6, we know the rules by which . In the article general view of operations with fractions we have circulated these rules with ordinary fractions into fractions of general form A/B , where A and B are some numeric, literal or variable expressions, and B is identically non-zero. It is clear that fractions with logarithms are special cases of general fractions. And in this regard, it is clear that actions with fractions that contain logarithms in their records are carried out according to the same rules. Namely:
- To add or subtract two fractions with the same denominators, add or subtract the numerators accordingly, and leave the denominator the same.
- To add or subtract two fractions with different denominators, you need to bring them to a common denominator and perform the appropriate actions according to the previous rule.
- To multiply two fractions, you need to write a fraction whose numerator is the product of the numerators of the original fractions, and the denominator is the product of the denominators.
- To divide a fraction by a fraction, it is necessary to multiply the divisible fraction by the reciprocal of the divisor, that is, by the fraction with the numerator and denominator rearranged.
Here are some examples for performing operations with fractions containing logarithms.
Example.
Perform actions with fractions containing logarithms: a), b) 
, in) 
, G) 
.
Solution.
a) The denominators of the added fractions are obviously the same. Therefore, according to the rule for adding fractions with the same denominators, we add the numerators, and leave the denominator the same: 
.
b) Here the denominators are different. Therefore, first you need bring fractions to the same denominator. In our case, the denominators are already presented as products, and it remains for us to take the denominator of the first fraction and add to it the missing factors from the denominator of the second fraction. So we get a common denominator of the form 
. In this case, the subtracted fractions are reduced to a common denominator using additional factors in the form of a logarithm and the expression x 2 ·(x+1), respectively. After that, it remains to subtract fractions with the same denominators, which is not difficult.
So the solution is: 
c) It is known that the result of multiplying fractions is a fraction, the numerator of which is the product of the numerators, and the denominator is the product of the denominators, therefore 
It is easy to see that it is possible to fraction reduction by two and by the decimal logarithm, as a result we have 
.
d) We pass from division of fractions to multiplication, replacing the fraction-divisor with its reciprocal. So 
The numerator of the resulting fraction can be represented as 
, from which the common factor of the numerator and denominator is clearly visible - the factor x, you can reduce the fraction by it: 
Answer:
a), b) 
, in) 
, G) 
.
It should be remembered that actions with fractions are carried out taking into account the order in which actions are performed: first multiplication and division, then addition and subtraction, and if there are brackets, then actions in brackets are performed first.
Example.
Do actions with fractions 
.
Solution.
First, we perform the addition of fractions in brackets, after which we will carry out the multiplication: 
Answer:
At this point, it remains to say out loud three rather obvious, but at the same time important points:
Converting expressions using the properties of logarithms
Most often, the transformation of expressions with logarithms involves the use of identities expressing the definition of the logarithm and
Maths. Thematic tests. Part II. Preparation for the Unified State Exam-2010. 10-11 grades. Ed. Lysenko F.F. - Rostov n / D .: Legion, 2009. - 176s.
Maths. USE-2009. Thematic tests. Part II (B4-B8, C1-C2) Ed. Lysenko F.F. - Rostov n / D: Legion, 2008 - 160 p.
The manual consists of tests on individual topics, which are traditional in the course of mathematics and therefore, as a rule, are included in the exam. They fully cover the groups of tasks of an increased and high level of complexity of the USE, except for text tasks and problems in geometry. One or more test suites are offered for each topic. Each set contains 10 tests, each test contains 8 tasks.
The purpose of this book is to work out tasks with a short and detailed answer of the USE tests. It is necessary primarily for graduates who expect to receive USE good assessment, as well as for students of the 10th grade, who can consolidate the topics covered from the point of view of the USE. The proposed manual can be useful to all graduates preparing for the USE in mathematics, as well as teachers preparing students for the USE.
Format: djvu/zip (2009 , 176s.)
The size: 2.5 MB
Download / Download file 14
Format: pdf (2009 , 176s.)
The size: 8.6 MB
Download: 14 .12.2018, links removed at the request of the Legion publishing house (see note)
Format: djvu/zip (2008 , 160s.)
The size: 3 MB
Download / Download file 14 .12.2018, links removed at the request of the Legion publishing house (see note)
Format: pdf (2008 , 160s.)
The size: 9.9 MB
Download: 14 .12.2018, links removed at the request of the Legion publishing house (see note)
Educational-methodical complex "Mathematics. Unified State Exam-2010" ed. Lysenko F.F. and Kulabukhov S.Yu. includes tutorials:
1.
Maths. Preparation for the Unified State Exam-2010.
2. Reshebnik. Maths. Preparation for the Unified State Exam-2010.
3. Maths. Thematic tests. Part I (basic level). Preparation for the Unified State Exam-2010. 10-11 grades.
4. Maths. Thematic tests. Part II. Preparation for the Unified State Exam-2010. 10-11 grades.
5.
Maths. Thematic tests: geometry, word problems. Preparation for the Unified State Exam-2010. 10-11 grades.
6.
Maths. Collection of USE tests 2001 - 2010.
7. Maths. Preparation for the Unified State Exam-2010. Training tests.
8. Pocket guide to mathematics.
Table of contents
From authors 11
§ 1. Identity transformations of logarithmic expressions 13
Option number 1 13
Option number 2 13
Option number 3 14
Option number 4 14
Option number 5 15
Option number 6 15
Option number 7 16
Option number 8 16
Option number 9 17
Option number 10 17
§ 2. Identity transformations of expressions containing degree 18
Option number 1 18
Option number 2 19
Option number 3 19
Option number 4 20
Option number 5 21
Option number 6 21
Option number 7 22
Option number 8 23
Option number 9 23
Option number 10 24
§ 3. Identity transformations of irrational expressions 25
Option number 1 25
Option number 2 25
Option number 3 26
Option number 4 26
Option number 5 27
Option number 6 28
Option number 7 28
Option number 8 29
Option number 9 30
Option number 10 30
§ 4. Systems of equations 31
Option number 1 31
Option number 2 32
Option number 3 33
Option number 4 33
Option number 5 34
Option number 6 35
Option number 7 36
Option number 8 37
Option number 9 38
Option number 10 39
§ 5. geometric sense derivative 39
Option number 1 39
Option number 2 41
Option number 3 43
Option number 4 44
Option number 5 46
Option number 6 48
Option number 7 50
Option number 8 52
Option number 9 54
Option number 10 55
§ 6. Inequalities 56
Option number 1 g 56
Option number 2 57
Option number 3 58
Option number 4 58
Option number 5 59
Option number 6 60
Option number 7 60
Option number 8 61
Option number 9 62
Option number 10 63
§ 7. Irrational Equations 63
Option number 1 63
Option number 2 64
Option number 3 65
Option number 4 65
Option number 5 66
Option number 6 66
Option number 7 67
Option number 8 67
Option number 9 68
Option No. Yu 68
§ 8. Trigonometric Equations 69
Option number 1 69
Option number 2 69
Option number 3 70
Option number 4 70
Option number 5 71
Option number 6 72
Option number 7 72
Option number 8 73
Option number 9 74
Option number 10 74
§ 9. Logarithmic equations 75
Option number 1 75
Option number 2 75
Option number 3 76
Option number 4 76
Option number 5 77
Option number 6 77
Option number 7 78
Option No. 8 * 78
Option number 9 79
Option number 10 79
§ 10. Exponential Equations 80
Option number 1 80
Option number 2 80
Option number 3 81
Option number 4 81
Option number 5 82
Option number 6 82
Option number 7 83
Option number 8 83
Option number 9 84
Option number 10 84
§eleven. Periodicity, even and odd functions 85
Option number 1 85
Option number 2 86
Option number 3 87
Option number 4 89
Option number 5 90
Option number 6 91
Option number 7 92
Option number 8 93
Option number 9 94
Option number 10 95
§ 12. Zeros of a complex function. Limited function 97
Option number 1 97
Option number 2 97
Option number 3 98
Option number 4 98
Option number 5 99
Option number 6 99
Option number 7 100
Option number 8 100
Option number 9 101
Option number 10 101
§ 13. Domain of definition, set of values, monotonicity of functions 102
Option number 1 102
Option number 2 102
Option number 3 103
Option number 4 103
Option number 5 104
Option number 6 104
Option number 7 105
Option number 8 105
Option number 9 106
Option number 10 107
§ 14. Extrema of a function. The largest and smallest values of the function 107
Option number 1 107
Option number 2 108
Option number 3 108
Option number 4 109
Option number 5 109
Option number 6 110
Option number 7 110
Option number 8 111
Option number 9 111
Option number 10 112
§ 15. Various techniques for solving logarithmic equations 113
Option number 1 113
Option number 2 113
Option number 3 114
Option number 4 114
Option number 5 115
Option number 6 115
Option number 7 116
Option number 8 116
Option number 9 117
Option number 10 117
§ 16. Various techniques for solving trigonometric equations 118
Option number 1 118
Option number 2 118
Option number 3 118
Option number 4 119
Option number 5 119
Option number 6 120
Option number 7 120
Option number 8 121
Option number 9 121
Option number 10 122
§ 17. Various techniques for solving irrational equations 123
Option number 1 123
Option number 2 123
Option number 3 124
Option number 4 124
Option number 5 125
Option number 6 125
Option number 7 125
Option number 8 126
Option number 9 126
Option No. 10 127
§ 18. Equations containing a variable under the modulo sign 127
Option number 1 127
Option number 2 128
Option number 3 128
Option number 4 129
Option number 5 129
Option number 6 130
Option number 7 130
Option number 8 131
Option number 9 131
Option number 10 131
§ 19. Various methods for solving exponential equations.132
Option number 1 132
Option number 2 133
Option number 3 133
Option number 4 134
Option number 5 134
Option number 6 135
Option number 7 135
Option number 8 135
Option number 9 136
Option number 10 136
§ 20. Various techniques for solving combined equations 137
Option number 1 137
Option number 2 137
Option number 3 138
Option number 4 138
Option number 5 139
Option number 6 139
Option number 7 140
Option number 8 140
Option number 9 141
Option number 10 141
§ 21. Equations with a parameter containing module 142
Option number 1 142
Option number 2 142
Option number 3 143
Option number 4 144
Option number 5 144
Option number 6 145
Option number 7 146
Option number 8 146
Option number 9 147
Option number 10 148
Answers 149
§ 1. Identity transformations of logarithmic expressions 149
§ 2. Identity transformations of expressions containing the degree 150
§ 3. Identity transformations of irrational expressions 150
§ 4. Systems of equations 151
§ 5. The geometric meaning of the derivative 151
§ 6. Inequalities 152
§ 7. Irrational equations 152
§ 8. Trigonometric equations 153
§ 9. Logarithmic equations 153
§ 10. Exponential Equations 154
§eleven. Periodicity, even and odd functions 154
§ 12. Zeros of a complex function. Limited function 155
§ 13. Domain of definition, set of values, monotonicity of functions 156
§ 14. Extrema of a function. The largest and smallest values of the function 158
§ fifteen. Various tricks when solving logarithmic equations 159
§ 16. Various techniques for solving trigonometric equations 160
§ 17. Various tricks for solving irrational equations 164
§ 18. Equations containing a variable under the modulo sign 165
§ 19. Various techniques for solving exponential equations.166
§ 20. Various techniques for solving combined equations 167
§ 21. Equations with a parameter containing module 169
Literature 170
EGOROVA VICTORIA VALEREVNA
Mathematic teacher
highest qualification category
TOPIC: “IDENTITY TRANSFORMATION
LOGARITHMIC EXPRESSIONS"
Knowledge and skills that students should master after studying this lesson:
know the definition of the logarithm of a number, the basic logarithmic identity, the properties of logarithms;
be able to perform transformations of expressions containing logarithms, calculate logarithms.
Literature:
1. Alimov Sh.A., Kolyagin Yu.M., Sidorov Yu.V. and others. Algebra and the beginning of analysis: a textbook for grades 10-11 educational institutions. - M .: Education, 2001.
2. Kochagin V.V., Kochagina M.V., Intensive course of preparation for the exam. – M.: Eksmo, 2009.
3. A. G. Merzlyak, V. B. Polonsky, and M. S. Yakir, Algebraic simulator: A guide for schoolchildren and university entrants. – M.: Ileksa, 2005.
4. Gusev V.A., Mordkovich A.G. Mathematics: Reference materials: A book for students. - M .: Education, 2001.
Lesson plan:
During the classes:
1) The logarithm is Greek word, which consists of 2 words: “logos” - relation, “arithmos” - number. So the logarithm is the number that measures the ratio. A publication in 1614 reported that Napier had invented logarithms. Later, he compiled logarithmic tables, which are now known to us as the tables of Bradys. In less than a century, tables have spread throughout the world and have become an indispensable computing tool. In the future, they were, as it were, built into a convenient device that extremely speeds up the calculation process - a slide rule, which was used until the seventies of the twentieth century.
Attachment 1.
2) logarithm positive numberb by reason a, moreover greater than zero and not equal to one,is the exponent to which a number must be raiseda to get the numberb.
This equality expressing the definition of the logarithm is calledbasic logarithmic identity .
C 
OR 1
P 
The base of the degree and the base of the logarithm are seventeen, which means that according to the basic logarithmic identity, the value of the expression is three.
We will work verbally:
SCH 
ELCHOK
O the bottom second is zero point five tenths, so the expression is equal to the arithmetic square root of five.
P 
appendix 2.
Equality means that
From the definition of the logarithm, the following important equalities are obtained:
For example:
P 
appendix 3.
Let's move on to USE assignments:
Appendix 4
3
)
There is a special notation and name for the base ten logarithm.decimal logarithm
.
L 
arithm to basee
callednatural logarithm
.
H 
for example,
4) The following properties follow from the definition of the logarithm. All properties are formulated and proved only for positive values of variables contained under logarithm signs.
Base logarithm of the product of two positive numbers a is equal to the sum logarithms of these numbers with the same base.
DER 2
For example,
W 
assignment 1.
Task 2. Simplify the expression
AT 
Let's use the previous example. Let's replace
Note that the logarithm is squared, so the sum must also be squared. Using the formula for the square of the sum, open the brackets. We present similar terms.
5) The logarithm of the quotient is equal to the difference between the logarithms of the dividend and the divisor.
C
Pay attention to the base of the degree and the base of the logarithm - they are the same.
OR 3R 
Let's look at the application of this formula with an example:
W 
assignment 1. Find the value of the expression if
Task 2. Find the value b by its logarithm
6) Logarithm of the degree to the basea , is equal to the product exponent per logarithm in the same base.
DER 4
For example,
W 
assignment 1. Calculate if
Let's simplify the expression
Formula
called formula for transition to a new basis.
W 
assignment 1. Express in terms of a logarithm with base 2.
Task 2. Calculate
CER 5
CER 6
For example,
W 
assignment 1. Calculate
W 
assignment 2. Calculate
9) You can start logarithmic transformations only if if you remember all the properties of logarithms. Having repeated them, we will consider tasks for transforming logarithmic expressions from the other side.
To convert the sum or difference of logarithmic expressions, sometimes it is enough to use the definition of the logarithm, and most often the properties of the logarithm of the product or quotient.
W 
assignment 1. Calculate
Let's solve it in two ways.
1 way, using the definition of the logarithm:
Method 2 based on quotient logarithm property:
Task 2. Find the value of an expression
Let us first apply the formula the logarithm of the product, then the definition of the logarithm.
The basic logarithmic identity is used when converting expressions containing a logarithm in the exponent. The idea of such operations is to get the base of the exponent and the base of the logarithm equal.
Sometimes it is necessary to transform the expression by the properties of the logarithm and by the properties of the degree, also one can easily move from one base to another using the transition formula. In other cases, multiple properties should be applied.
W 
assignment 3. Calculate
W 
assignment 4. Find the value of an expression
Task 5. Find the value of an expression
W 
task 6. Express as a difference of logarithms
H 
The greatest difficulty is the transformation of logarithmic expressions under the radical. In the process of transformations, one has to consider the modules of logarithmic expressions, for the disclosure of which one needs to compare irrational numbers or rational and irrational numbers. We will act consistently. Consider the expression under the inner radical.
Substitute in the original expression.
It should be noted that the transformation of logarithmic expressions can also be encountered when solving equations and inequalities or studying functions, therefore, in an implicit form, they can also be present in the assignments of groups B and C.
10) Summing up. Questions:
The base 10 logarithm is called
basic logarithm
main logarithm
natural logarithm
decimal logarithm
2) What values canx
in expression 
Value is not defined
5) Indicate the ratio that is true for allx ≠ 0 .
6) Specify the correct ratio for the formula for the transition to a new base.
7) Indicate the correct equality for
11)
Control testing.
Transnistrian State University
them. T.G. Shevchenko
Faculty of Physics and Mathematics
Department of Mathematical Analysis
and methods of teaching mathematics
COURSE WORK
"Identity Transformations
exponential and logarithmic
expressions"
Work completed:
student of ______ group
Faculty of Physics and Mathematics
_________________________
Checked work:
_________________________
Tiraspol, 2003
Introduction………………………………………………………………………2
Chapter 1
§one. Formation of skills for applying specific types of transformations………………………………………………………………………………….4
§2. Features of the organization of a knowledge system in the study of identical transformations.…….………………………….………..………….5
§3. Mathematics Program ……………………………………….11
Chapter 2
§one. Generalization of the concept of degree……………………………………..13
§2. The exponential function…………………………………………..15
§3. Logarithmic function………………………………………….16
Chapter 3. Identical transformations of exponential and logarithmic expressions in practice ............................................ ...................................19
Conclusion………………………………………………………………..24
List of used literature…………………………………….25
Introduction
This course work will consider the identical transformations of the exponential and logarithmic functions, consider the methodology for teaching them in the school course of algebra and the beginning of analysis.
The first chapter of this work describes the methodology for teaching identical transformations in the school course of mathematics, it also includes a program in mathematics in the course "Algebra and the beginning of analysis" with the study of exponential and logarithmic functions.
The second chapter deals directly with the exponential and logarithmic functions themselves, their main properties used in identical transformations.
The third chapter is the solution of examples and problems using the identical transformations of the exponential and logarithmic functions.
The study of various transformations of expressions and formulas takes up a significant part of the study time in the course of school mathematics. The simplest transformations based on the properties of arithmetic operations are already performed in primary school and in grades IV-V. But the main burden on the formation of skills and abilities to perform transformations is borne by the course of school algebra. This is connected both with a sharp increase in the number and variety of transformations performed, and with the complication of activities to substantiate them and clarify the conditions of applicability, with the identification and study of generalized concepts of identity, identical transformation, equivalent transformation, logical consequence.
The culture of performing identical transformations develops in the same way as the culture of computing, based on a solid knowledge of the properties of operations on objects (numbers, vectors, polynomials, etc.) and algorithms for their implementation. It manifests itself not only in the ability to correctly justify transformations, but also in the ability to find the shortest path to the transition from the original analytical expression to the expression that best suits the purpose of the transformation, in the ability to track changes in the domain of definition of analytical expressions in a chain of identical transformations, in the speed and error-free execution of transformations. .
Ensuring a high culture of computation and identical transformations is an important problem in teaching mathematics. However, this problem is still far from being solved satisfactorily. Proof of this is the statistical data of the public education authorities, in which errors and irrational methods of calculations and transformations made by students are annually stated. various classes while doing control works. This is confirmed by the opinions of higher educational institutions about the quality of mathematical knowledge and skills of applicants. One cannot but agree with the conclusions of the public education authorities and universities that it is not enough high level culture of computing and identical transformations in high school is a consequence of formalism in the knowledge of students, the separation of theory from practice.
Identity transformations and teaching methods
in the school course of algebra and the beginning of analysis.
§one. Formation of application skills
specific types of transformations.
The system of techniques and rules for carrying out transformations, used at the stage of the beginnings of algebra, has a very wide range of applications: it is used in the study of the entire course of mathematics. However, precisely because of its low specificity, this system needs additional transformations that take into account the peculiarities of the structure of the transformed expressions and the properties of newly introduced operations and functions. The development of the corresponding types of transformations begins with the introduction of abbreviated multiplication formulas. Then we consider the transformations associated with the exponentiation operation, with different classes elementary functions- exponential, power, logarithmic, trigonometric. Each of these types of transformations goes through a stage of study, in which attention is focused on the assimilation of their characteristic features.
With the accumulation of material, it becomes possible to single out the common features of all the transformations under consideration and, on this basis, introduce the concepts of identical and equivalent transformations.
It should be noted that the concept of an identical transformation is given in the school course of algebra not in full generality, but only in application to expressions. Transformations are divided into two classes: identical transformations are transformations of expressions, and equivalent transformations are transformations of formulas. In the case when there is a need to simplify one part of the formula, an expression is highlighted in this formula, which serves as an argument for the applied identical transformation. The corresponding predicate is considered unchanged.
As for the organization of an integral system of transformations (synthesis), its main goal is to form a flexible and powerful; apparatus suitable for use in solving a variety of educational tasks.
In the course of algebra and the beginning of analysis, an integral system of transformations, already formed in its main features, continues to be gradually improved. Some new types of transformations are also added to it, but they only enrich it, expand its capabilities, but do not change its structure. The methodology for studying these new transformations practically does not differ from that used in the course of algebra.
§2. Features of the organization of the task system
in the study of identical transformations.
The basic principle of organizing any system of tasks is to present them from simple to complex, taking into account the need for students to overcome feasible difficulties and create problem situations. The specified basic principle requires concretization in relation to the features of this educational material. To describe various systems of tasks in the methodology of mathematics, the concept of a cycle of exercises is used. The cycle of exercises is characterized by the combination in the sequence of exercises of several aspects of the study and methods of arranging the material. In relation to identical transformations, the idea of a cycle can be given as follows.
The cycle of exercises is connected with the study of one identity, around which other identities are grouped, which are in a natural connection with it. The composition of the cycle, along with executive tasks, includes tasks that require recognition of the applicability of the considered identity. The identity under study is used to perform calculations on various numerical domains. The specificity of the identity is taken into account; in particular, turns of speech associated with it are organized.
The tasks in each cycle are divided into two groups. The first includes tasks performed during the initial acquaintance with identity. They serve educational material for several consecutive lessons, united by one topic. The second group of exercises relates the identity under study to various applications. This group does not form a compositional unity - the exercises here are scattered over various topics.
The described structure of the cycle refers to the stage of formation of skills for applying specific types of transformations. At the final stage - the stage of synthesis, the cycles are modified. First, both groups of tasks are combined, forming an "unfolded" cycle, and the simplest ones in terms of wording or the complexity of the task are excluded from the first group. The remaining types of tasks become more difficult. Secondly, there is a merging of cycles related to different identities, due to which the role of actions to recognize the applicability of one or another identity increases.
We note the features of task cycles related to identities for elementary functions. These features are due to the fact that, firstly, the corresponding identities are studied in connection with the study of functional material and, secondly, they appear later than the identities of the first group and are studied using already formed skills for carrying out identical transformations.
Each newly introduced elementary function sharply expands the range of numbers that can be designated and named individually. Therefore, the first group of tasks of the cycles should include tasks to establish a connection between these new numerical regions and the original region of rational numbers. We give examples of such tasks.
Example 1 Calculate:
Next to each expression, there is an identity, in the cycles for which the proposed tasks may be present. The purpose of such tasks is to master the features of records, including symbols of new operations and functions, and to develop mathematical speech skills.
A significant part of the use of identity transformations associated with elementary functions falls on the solution of irrational and transcendental equations. The cycles related to the assimilation of identities include only the simplest equations, but already here it is advisable to carry out work on mastering the method of solving such equations: reducing it by replacing the unknown to an algebraic equation.
The sequence of steps for this solution is as follows:
a) find a function for which this equation can be represented as ;
b) make a substitution and solve the equation;
c) solve each of the equations , where is the set of roots of the equation .
When using the described method, step b) is often performed implicitly, without introducing a notation for . In addition, students often choose from the various paths leading to finding an answer to choose the one that leads to the algebraic equation faster and easier.
Example 2. Solve the equation.
First way:
Second way:
a) 
b) 
It can be seen here that step a) is more difficult in the first method than in the second. The first way is “harder to start”, although the further course of the solution is much easier. On the other hand, the second method has advantages, consisting in greater ease, greater sophistication in teaching reduction to an algebraic equation.
For a school course in algebra, tasks are typical in which the transition to an algebraic equation is even easier than in this example. The main load of such tasks relates to the selection of step c) as an independent part of the solution process associated with the use of the properties of the elementary function under study.
Example 3. Solve the equation:
a) 
; b) 
.
These equations are reduced to the equations: a) or ; b) or. To solve these equations, knowledge of only the simplest facts about the exponential function is required: its monotonicity, range of values. Like the previous example, equations a) and b) can be attributed to the first group of a cycle of exercises for solving quadratic exponential equations.
Thus, we come to the classification of tasks in cycles related to the solution of transcendental equations, including an exponential function:
1) equations that are reduced to equations of the form and have a simple answer, general in form: ;
2) equations that reduce to equations , where is an integer, or , where ;
3) equations that reduce to equations and require an explicit analysis of the form in which the number is written.
Similar tasks can be classified for other elementary functions.
A significant part of the identities studied in the courses on algebra and algebra and the beginnings of analysis are proved in them or at least explained. This side of the study of identities has great importance for both courses, since demonstrative reasoning in them is carried out with the greatest clarity and rigor precisely in relation to identities. Outside of this material, the evidence is usually less complete, it is not always distinguished from the composition of the applied means of justification.
The properties of arithmetic operations are used as a support on which the proofs of identities are built.
The educational impact of calculations and identical transformations can be directed to the development logical thinking, if only students are systematically required to justify calculations and identical transformations, to develop functional thinking, which is achieved in various ways. The importance of calculations and identical transformations in the development of will, memory, ingenuity, self-control, and creative initiative is quite obvious.
The demands of everyday, industrial computing practice require the formation of strong, automated skills of rational calculations and identical transformations in students. These skills are developed in the process of any computational work, however, special skills are needed. training exercises in fast calculations and transformations.
So, if the lesson involves solving logarithmic equations using the basic logarithmic identity, then it is useful to include oral exercises in the lesson plan to simplify or calculate the values of expressions: , , 
. The purpose of the exercises is always communicated to the students. During the exercise, it may be necessary to require students to justify individual transformations, actions, or solve the entire problem, even if this was not planned. Where different ways of solving a problem are possible, it is always desirable to ask questions: “How was the problem solved?”, “Who solved the problem in a different way?”
The concepts of identity and identical transformation, they are explicitly introduced in the class VI algebra course. The very definition of identical expressions cannot be practically used to prove the identity of two expressions, and to understand that the essence of identical transformations consists in applying to the expression the definitions and properties of those actions that are indicated in the expression, or in adding to it an expression that is identically equal to 0, or in multiplying it by an expression identically equal to one. But, even having mastered these provisions, students often do not understand why these transformations allow us to assert that the original and resulting expressions are identical, i.e. take the same values for any systems (sets) of variable values.
It is also important to ensure that students understand well that such conclusions of identical transformations are consequences of the definitions and properties of the corresponding actions.
The apparatus of identical transformations, accumulated in previous years, is being expanded in the 6th grade. This extension begins with the introduction of an identity expressing the property of the product of powers with the same bases: , where , are integers.
§3. Mathematics Program. In the school course "Algebra and the Beginnings of Analysis", students systematically study exponential and logarithmic functions and their properties, identical transformations of logarithmic and exponential expressions and their application to solving the corresponding equations and inequalities, get acquainted with the basic concepts, statements. In the 11th grade, algebra lessons take 3 hours a week, for a total of 102 hours a year. It takes 36 hours to study the exponential, logarithmic and power functions according to the program. The program includes consideration and study of the following issues: The concept of a degree with a rational indicator. Solution of irrational equations. An exponential function, its properties and graph. identical transformations of exponential expressions. Solution of exponential equations and inequalities. The logarithm of a number. Basic properties of logarithms. Logarithmic function, its properties and graph. Solution of logarithmic equations and inequalities. Derivative of exponential function. Number and natural logarithm. Derivative of a power function. The main purpose of the section on the study of exponential and logarithmic functions is to familiarize students with exponential, logarithmic and power functions; teach students to solve exponential and logarithmic equations and inequalities. The concepts of the root of the th degree and the degree with a rational exponent are a generalization of the concepts of the square root and the degree with an integer exponent. Students should pay attention to the fact that the properties of roots and degrees with a rational exponent considered here are similar to those properties that have been studied earlier. square roots and degrees with integer exponents. It is necessary to devote enough time to working out the properties of degrees and the formation of skills for identical transformations. The concept of degree with irrational indicator introduced on a visual-intuitive basis. This material plays an auxiliary role and is used when introducing the exponential function. The study of the properties of the exponential, logarithmic and power functions is built in accordance with the accepted general scheme function research. In this case, an overview of the properties is given depending on the parameter values. The exponential and logarithmic inequalities are solved based on the studied properties of functions. A characteristic feature of the course is the systematization and generalization of students' knowledge, the consolidation and development of the skills and abilities acquired in the course of algebra, which is carried out both when studying new material and when conducting a generalizing repetition.
Chapter 2
§one. Generalization of the concept of degree.
Definition: The root of the th degree of pure is such a number, the th degree of which is equal to.
According to this definition, the root of the th degree of a number is a solution to the equation. The number of roots of this equation depends on and . Let's consider a function. As is known, on the interval this function increases for any and takes all values from the interval . By the root theorem, the equation for any has a non-negative root, and moreover, only one. It is called the arithmetic root of the th degree of a number and denoted; the number is called the index of the root, and the number itself is called the radical expression. The sign is also called a radical.
Definition: The arithmetic root of the th degree of a number is a non-negative number whose th degree is .
For even, the function is even. It follows that if , then the equation , in addition to the root , also has a root . If , then there is only one root: ; if , then this equation has no roots, since the even power of any number is non-negative.
For odd values, the function increases along the entire number line; its range is the set of all real numbers. Applying the root theorem, we find that the equation has one root for any and, in particular, for . This root for any value is denoted by .
For roots of odd degree, equality is true. Indeed, , i.e. number is the th root of . But such a root for odd is unique. Consequently, .
Remark 1: For any real
Recall the well-known properties of arithmetic roots of the th degree.
For any natural , integer and any non-negative integers and equalities are true:
1. 
2. 
3. 
4. 
Degree with a rational exponent.
The expression is defined for all and , except for the case when . Recall the properties of such powers.
For any numbers , and any integers and equalities are true:
We also note that if , then for and for .. and
For students studying for the Unified State Examination, mathematics teachers at secondary school No. 26 in Yakutsk use a list of content questions (codifier) of the school mathematics course, the assimilation of which is checked when passing the unified state exam in 2007. elective course in preparation for the Unified State Exam is based on repetition, systematization and deepening of knowledge acquired earlier. Classes are held in the form of free...
The listed equalities when converting expressions with logarithms are used both from right to left and from left to right.
It is worth noting that it is not necessary to memorize the consequences of the properties: when carrying out transformations, you can get by with the basic properties of logarithms and other facts (for example, those for b≥0), from which the corresponding consequences follow. The "side effect" of this approach is only that the solution will be a little longer. For example, in order to do without the consequence, which is expressed by the formula 
, and starting only from the basic properties of logarithms, you will have to carry out a chain of transformations of the following form: 
.
The same can be said about the last property from the above list, which corresponds to the formula 
, since it also follows from the basic properties of logarithms. The main thing to understand is that it is always possible for the degree of a positive number with a logarithm in the exponent to swap the base of the degree and the number under the sign of the logarithm. In fairness, we note that examples involving the implementation of transformations of this kind are rare in practice. We will give a few examples below.
Converting numeric expressions with logarithms
We remembered the properties of logarithms, now it's time to learn how to put them into practice to transform expressions. It is natural to start with the transformation of numeric expressions, and not expressions with variables, since it is more convenient and easier to learn the basics on them. So we will do, and we will start with a very simple examples to learn how to choose the desired property of the logarithm, but we will gradually complicate the examples, up to the moment when several properties will need to be applied in a row to obtain the final result.
Selecting the desired property of logarithms
There are not so few properties of logarithms, and it is clear that you need to be able to choose the appropriate one from them, which in this particular case will lead to the desired result. Usually this is not difficult to do by comparing the form of the logarithm or expression being converted with the types of the left and right parts of the formulas expressing the properties of logarithms. If left or right part one of the formulas coincides with the given logarithm or expression, then, most likely, it is this property that should be used during the transformation. The following examples clearly demonstrate this.
Let's start with examples of transforming expressions using the definition of the logarithm, which corresponds to the formula a log a b =b , a>0 , a≠1 , b>0 .
Example.
Calculate, if possible: a) 5 log 5 4 , b) 10 log(1+2 π) , c) 
, d) 2 log 2 (−7) , e) .
Solution.
In the example, letter a) clearly shows the structure a log a b , where a=5 , b=4 . These numbers satisfy the conditions a>0 , a≠1 , b>0 , so you can safely use the equality a log a b =b . We have 5 log 5 4=4 .
b) Here a=10 , b=1+2 π , conditions a>0 , a≠1 , b>0 are fulfilled. In this case, the equality 10 lg(1+2 π) =1+2 π takes place.
c) And in this example we are dealing with a degree of the form a log a b , where and b=ln15 . So 
.
Despite belonging to the same form a log a b (here a=2 , b=−7 ), the expression under the letter d) cannot be converted by the formula a log a b =b . The reason is that it doesn't make sense because it contains a negative number under the logarithm sign. Moreover, the number b=−7 does not satisfy the condition b>0 , which makes it impossible to resort to the formula a log a b =b , since it requires the conditions a>0 , a≠1 , b>0 . So, we can't talk about computing the value 2 log 2 (−7) . In this case, writing 2 log 2 (−7) = −7 would be an error.
Similarly, in the example under the letter e) it is impossible to give a solution of the form 
, since the original expression does not make sense.
Answer:
a) 5 log 5 4 =4 , b) 10 log(1+2 π) =1+2 π , c) , d), e) expressions do not make sense.
It is often useful to transform positive number is represented as a power of some positive and non-one number with a logarithm in the exponent. It is based on the same definition of the logarithm a log a b =b , a>0 , a≠1 , b>0 , but the formula is applied from right to left, that is, in the form b=a log a b . For example, 3=e ln3 or 5=5 log 5 5 .
Let's move on to using the properties of logarithms to transform expressions.
Example.
Find the value of the expression: a) log −2 1, b) log 1 1, c) log 0 1, d) log 7 1, e) ln1, f) lg1, g) log 3.75 1, h) log 5 π 7 1 .
Solution.
In the examples under the letters a), b) and c), the expressions log −2 1 , log 1 1 , log 0 1 are given, which do not make sense, since the base of the logarithm should not contain a negative number, zero or one, because we have defined logarithm only for a positive and non-unit base. Therefore, in examples a) - c) there can be no question of finding the value of the expression.
In all other tasks, obviously, the bases of the logarithms contain positive and non-unit numbers 7 , e , 10 , 3.75 and 5 π 7 respectively, and units are everywhere under the signs of the logarithms. And we know the property of the logarithm of unity: log a 1=0 for any a>0 , a≠1 . Thus, the values of expressions b) - f) are equal to zero.
Answer:
a), b), c) the expressions do not make sense, d) log 7 1=0, e) ln1=0, f) log1=0, g) log 3.75 1=0, h) log 5 e 7 1=0 .
Example.
Calculate: a) , b) lne , c) lg10 , d) log 5 π 3 −2 (5 π 3 −2), e) log −3 (−3) , f) log 1 1 .
Solution.
It is clear that we have to use the property of the logarithm of the base, which corresponds to the formula log a a=1 for a>0 , a≠1 . Indeed, in tasks under all letters, the number under the sign of the logarithm coincides with its base. Thus, I want to say right away that the value of each of the given expressions is 1 . However, do not rush to conclusions: in tasks under the letters a) - d) the values of the expressions are really equal to one, and in tasks e) and f) the original expressions do not make sense, so it cannot be said that the values of these expressions are equal to 1.
Answer:
a) , b) lne=1 , c) lg10=1 , d) log 5 π 3 −2 (5 π 3 −2)=1, e), f) expressions do not make sense.
Example.
Find the value: a) log 3 3 11 , b) 
, c) , d) log −10 (−10) 6 .
Solution.
Obviously, under the signs of the logarithms are some degrees of base. Based on this, we understand that the property of the degree of the base is useful here: log a a p =p, where a>0, a≠1 and p is any real number. Considering this, we have the following results: a) log 3 3 11 =11 , b) 
, in) 
. Is it possible to write a similar equality for the example under the letter d) of the form log −10 (−10) 6 =6? No, you can't, because log −10 (−10) 6 doesn't make sense.
Answer:
a) log 3 3 11 =11, b) , in) d) the expression does not make sense.
Example.
Express the expression as the sum or difference of logarithms in the same base: a) 
, b) , c) log((−5) (−12)) .
Solution.
a) The product is under the sign of the logarithm, and we know the property of the logarithm of the product log a (x y)=log a x+log a y , a>0 , a≠1 , x>0 , y>0 . In our case, the number in the base of the logarithm and the numbers in the product are positive, that is, they satisfy the conditions of the selected property, therefore, we can safely apply it: 
.
b) Here we use the property of the logarithm of the quotient , where a>0 , a≠1 , x>0 , y>0 . In our case, the base of the logarithm is a positive number e, the numerator and denominator π are positive, which means they satisfy the conditions of the property, so we have the right to use the chosen formula: 
.
c) First, note that the expression lg((−5) (−12)) makes sense. But at the same time, we do not have the right to apply the formula for the logarithm of the product log a (x y)=log a x+log a y , a>0 , a≠1 , x>0 , y>0 , since the numbers −5 and −12 are negative and do not satisfy the conditions x>0 , y>0 . That is, it is impossible to carry out such a transformation: log((−5)(−12))=log(−5)+log(−12). But what to do? In such cases, the original expression needs to be pre-transformed to avoid negative numbers. About similar cases of transformation of expressions with negative numbers under the sign of the logarithm, we will talk in detail in one of, but for now we will give a solution to this example, which is clear in advance and without explanation: lg((−5)(−12))=lg(5 12)=lg5+lg12.
Answer:
a) , b) , c) lg((−5) (−12))=lg5+lg12 .
Example.
Simplify the expression: a) log 3 0.25 + log 3 16 + log 3 0.5, b) .
Solution.
Here, all the same properties of the logarithm of the product and the logarithm of the quotient that we used in the previous examples will help us, only now we will apply them from right to left. That is, we convert the sum of logarithms to the logarithm of the product, and the difference of the logarithms to the logarithm of the quotient. We have
a) log 3 0.25+log 3 16+log 3 0.5=log 3 (0.25 16 0.5)=log 3 2.
b) 
.
Answer:
a) log 3 0.25+log 3 16+log 3 0.5=log 3 2, b) 
.
Example.
Get rid of the degree under the sign of the logarithm: a) log 0.7 5 11, b) 
, c) log 3 (−5) 6 .
Solution.
It is easy to see that we are dealing with expressions like log a b p . The corresponding property of the logarithm is log a b p =p log a b , where a>0 , a≠1 , b>0 , p is any real number. That is, under the conditions a>0 , a≠1 , b>0 from the logarithm of the degree log a b p we can go to the product p·log a b . Let's carry out this transformation with the given expressions.
a) In this case a=0.7 , b=5 and p=11 . So log 0.7 5 11 =11 log 0.7 5 .
b) Here , the conditions a>0 , a≠1 , b>0 are fulfilled. That's why 
c) The expression log 3 (−5) 6 has the same structure log a b p , a=3 , b=−5 , p=6 . But for b, the condition b>0 is not satisfied, which makes it impossible to apply the formula log a b p =p log a b . So why can't you get the job done? It is possible, but a preliminary transformation of the expression is required, which we will discuss in detail below in the paragraph under the heading . The solution will be like this: log 3 (−5) 6 =log 3 5 6 =6 log 3 5.
Answer:
a) log 0.7 5 11 =11 log 0.7 5 ,
b)
c) log 3 (−5) 6 =6 log 3 5 .
Quite often, the formula for the logarithm of the degree when carrying out transformations has to be applied from right to left in the form p log a b \u003d log a b p (this requires the same conditions for a, b and p). For example, 3 ln5=ln5 3 and lg2 log 2 3=log 2 3 lg2 .
Example.
a) Calculate the value of log 2 5 if it is known that lg2≈0.3010 and lg5≈0.6990. b) Write the fraction as a logarithm to base 3.
Solution.
a) The formula for the transition to a new base of the logarithm allows us to represent this logarithm as a ratio of decimal logarithms, the values of which are known to us: . It remains only to carry out the calculations, we have 
.
b) Here it is enough to use the formula for the transition to a new base, and apply it from right to left, that is, in the form 
. We get 
.
Answer:
a) log 2 5≈2.3223, b) .
At this stage, we have rather scrupulously considered the transformation of the simplest expressions using the basic properties of logarithms and the definition of a logarithm. In these examples, we had to use one property and nothing else. Now, with a clear conscience, you can move on to examples whose transformation requires the use of several properties of logarithms and other additional transformations. We will deal with them in the next paragraph. But before that, let us briefly dwell on examples of the application of consequences from the basic properties of logarithms.
Example.
a) Get rid of the root under the sign of the logarithm. b) Convert the fraction to a base 5 logarithm. c) Get rid of the powers under the sign of the logarithm and at its base. d) Calculate the value of the expression 
. e) Replace the expression with a power with base 3.
Solution.
a) If we recall the corollary from the property of the logarithm of the degree 
, then you can immediately answer: 
.
b) Here we use the formula 
from right to left, we have 
.
c) B this case formula leads to the result . We get 
.
d) And here it suffices to apply the corollary to which the formula corresponds 
. So 
.
e) The property of the logarithm allows us to achieve the desired result: 
.
Answer:
a) . b) . in) 
. G) 
. e) .
Consistently Applying Multiple Properties
Real tasks for transforming expressions using the properties of logarithms are usually more complicated than those that we dealt with in the previous paragraph. In them, as a rule, the result is not obtained in one step, but the solution already consists in the sequential application of one property after another, together with additional identical transformations, such as opening brackets, reducing like terms, reducing fractions, etc. So let's get closer to such examples. There is nothing complicated about this, the main thing is to act carefully and consistently, observing the order in which the actions are performed.
Example.
Calculate the value of an expression (log 3 15−log 3 5) 7 log 7 5.
Solution.
The difference of logarithms in brackets by the property of the logarithm of the quotient can be replaced by the logarithm log 3 (15:5) , and then calculate its value log 3 (15:5)=log 3 3=1 . And the value of the expression 7 log 7 5 by the definition of the logarithm is 5 . Substituting these results into the original expression, we get (log 3 15−log 3 5) 7 log 7 5 =1 5=5.
Here is a solution without explanation:
(log 3 15−log 3 5) 7 log 7 5 =log 3 (15:5) 5=
= log 3 3 5=1 5=5 .
Answer:
(log 3 15−log 3 5) 7 log 7 5 =5.
Example.
What is the value of the numerical expression log 3 log 2 2 3 −1 ?
Solution.
Let's first transform the logarithm, which is under the sign of the logarithm, according to the formula of the logarithm of the degree: log 2 2 3 =3. So log 3 log 2 2 3 =log 3 3 and then log 3 3=1 . So log 3 log 2 2 3 −1=1−1=0 .
Answer:
log 3 log 2 2 3 −1=0 .
Example.
Simplify the expression.
Solution.
The formula for converting to a new base of the logarithm allows the ratio of logarithms to one base to be represented as log 3 5 . In this case, the original expression will take the form . By definition of the logarithm 3 log 3 5 =5 , that is 
, and the value of the resulting expression, by virtue of the same definition of the logarithm, is equal to two.
Here short version solution, which is usually given: 
.
Answer:
.
For a smooth transition to the information of the next paragraph, let's take a look at the expressions 5 2+log 5 3 , and lg0.01 . Their structure does not fit any of the properties of logarithms. So what happens if they cannot be converted using the properties of logarithms? It is possible if you carry out preliminary transformations that prepare these expressions for applying the properties of logarithms. So 5 2+log 5 3 =5 2 5 log 5 3 =25 3=75, 
and lg0,01=lg10 −2 = −2 . Further we will understand in detail how such preparation of expressions is carried out.
Preparing expressions to apply the properties of logarithms
Logarithms in the converted expression very often differ in the structure of the notation from the left and right parts of formulas that correspond to the properties of logarithms. But just as often, the transformation of these expressions involves the use of the properties of logarithms: their use only requires preliminary preparation. And this preparation consists in carrying out certain identical transformations that bring logarithms to a form convenient for applying properties.
In fairness, we note that almost any transformation of expressions can act as preliminary transformations, from the banal reduction of similar terms to the application trigonometric formulas. This is understandable, since the converted expressions can contain any mathematical objects: brackets, modules, fractions, roots, degrees, etc. Thus, one must be prepared to perform any required transformation in order to further benefit from the properties of logarithms.
Let's say right away that in this section we do not set ourselves the task of classifying and analyzing all conceivable preliminary transformations that allow us to apply the properties of logarithms or the definition of a logarithm in the future. Here we will focus on only four of them, which are the most characteristic and most often encountered in practice.
And now in detail about each of them, after which, within the framework of our topic, it remains only to deal with the transformation of expressions with variables under the signs of logarithms.
Selection of powers under the sign of the logarithm and in its base
Let's start right away with an example. Let us have a logarithm. Obviously, in this form, its structure is not conducive to the use of the properties of logarithms. Is it possible to somehow transform this expression in order to simplify it, or even better calculate its value? To answer this question, let's take a closer look at the numbers 81 and 1/9 in the context of our example. It is easy to see here that these numbers can be represented as a power of 3 , indeed, 81=3 4 and 1/9=3 −2 . In this case, the original logarithm is presented in the form and it becomes possible to apply the formula . So, 
.
The analysis of the analyzed example gives rise to the following idea: if possible, you can try to highlight the degree under the sign of the logarithm and at its base in order to apply the property of the logarithm of the degree or its consequence. It remains only to figure out how to single out these degrees. We will give some recommendations on this issue.
Sometimes it is quite obvious that the number under the sign of the logarithm and / or in its base represents some integer power, as in the example discussed above. Almost constantly you have to deal with powers of two, which are well familiar: 4=2 2 , 8=2 3 , 16=2 4 , 32=2 5 , 64=2 6 , 128=2 7 , 256=2 8 , 512= 2 9 , 1024=2 10 . The same can be said about the degrees of the triple: 9=3 2 , 27=3 3 , 81=3 4 , 243=3 5 , ... In general, it does not hurt if there is degree table natural numbers within ten. It is also not difficult to work with integer powers of ten, hundred, thousand, etc.
Example.
Calculate the value or simplify the expression: a) log 6 216 , b) , c) log 0.000001 0.001 .
Solution.
a) Obviously, 216=6 3 , so log 6 216=log 6 6 3 =3 .
b) The table of powers of natural numbers allows us to represent the numbers 343 and 1/243 as powers of 7 3 and 3 −4, respectively. Therefore it is possible next transformation given logarithm: 
c) Since 0.000001=10 −6 and 0.001=10 −3, then log 0.000001 0.001=log 10 −6 10 −3 =(−3)/(−6)=1/2.
Answer:
a) log 6 216=3, b) 
, c) log 0.000001 0.001=1/2 .
In more complex cases, to highlight the powers of numbers, you have to resort to.
Example.
Change the expression to the simpler form log 3 648 log 2 3 .
Solution.
Let's see what the expansion of the number 648 into prime factors:
That is, 648=2 3 3 4 . In this way, log 3 648 log 2 3=log 3 (2 3 3 4) log 2 3.
Now we convert the logarithm of the product to the sum of logarithms, after which we apply the properties of the logarithm of the degree:
log 3 (2 3 3 4) log 2 3=(log 3 2 3 + log 3 3 4) log 2 3=
=(3 log 3 2+4) log 2 3 .
By virtue of the corollary of the property of the logarithm of the degree, which corresponds to the formula , the product log32 log23 is the product , and it is known to be equal to one. Considering this, we get 3 log 3 2 log 2 3+4 log 2 3=3 1+4 log 2 3=3+4 log 2 3.
Answer:
log 3 648 log 2 3=3+4 log 2 3.
Quite often, expressions under the sign of the logarithm and in its base are products or ratios of the roots and / or powers of some numbers, for example, , . Similar expressions can be represented as a degree. To do this, the transition from roots to degrees is carried out, and and are applied. These transformations allow you to select the degrees under the sign of the logarithm and in its base, and then apply the properties of logarithms.
Example.
Calculate: a) 
, b).
Solution.
a) The expression in the base of the logarithm is the product of powers with the same bases, by the corresponding property of powers we have 5 2 5 −0.5 5 −1 =5 2−0.5−1 =5 0.5.
Now let's convert the fraction under the sign of the logarithm: let's move from the root to the degree, after which we will use the property of the ratio of degrees with the same bases: 
.
It remains to substitute the results obtained into the original expression, use the formula and finish the transformation: 
b) Since 729=3 6 , and 1/9=3 −2 , the original expression can be rewritten as .
Next, apply the property of the root of the exponent, move from the root to the exponent, and use the ratio property of the powers to convert the base of the logarithm to a power: 
.
Taking into account the last result, we have 
.
Answer:
a) 
, b).
It is clear that in the general case, to obtain powers under the sign of the logarithm and in its base, various transformations of various expressions may be required. Let's give a couple of examples.
Example.
What is the value of the expression: a) 
, b) 
.
Solution.
Further, we note that the given expression has the form log A B p , where A=2 , B=x+1 and p=4 . We transformed numerical expressions of this kind according to the property of the logarithm of the degree log a b p \u003d p log a b, therefore, with a given expression, I want to do the same, and go from log 2 (x + 1) 4 to 4 log 2 (x + 1) . And now let's calculate the value of the original expression and the expression obtained after the transformation, for example, with x=−2 . We have log 2 (−2+1) 4 =log 2 1=0 , and 4 log 2 (−2+1)=4 log 2 (−1)- meaningless expression. This raises a legitimate question: “What did we do wrong”?
And the reason is as follows: we performed the transformation log 2 (x+1) 4 =4 log 2 (x+1) , based on the formula log a b p =p log a b , but we have the right to apply this formula only if the conditions a >0 , a≠1 , b>0 , p - any real number. That is, the transformation we have done takes place if x+1>0 , which is the same x>−1 (for A and p, the conditions are met). However, in our case, the ODZ of the variable x for the original expression consists not only of the interval x> −1 , but also of the interval x<−1 . Но для x<−1 мы не имели права осуществлять преобразование по выбранной формуле.
The need to take into account ODZ
Let's continue to analyze the transformation of the expression log 2 (x+1) 4 we have chosen, and now let's see what happens to the ODZ when passing to the expression 4 log 2 (x+1) . In the previous paragraph, we found the ODZ of the original expression - this is the set (−∞, −1)∪(−1, +∞) . Now let's find the area of acceptable values of the variable x for the expression 4 log 2 (x+1) . It is determined by the condition x+1>0 , which corresponds to the set (−1, +∞) . It is obvious that when going from log 2 (x+1) 4 to 4·log 2 (x+1), the range of admissible values narrows. And we agreed to avoid reforms that lead to a narrowing of the ODZ, as this can lead to various negative consequences.
Here it is worth noting for yourself that it is useful to control the ODZ at each step of the transformation and not allow it to narrow. And if suddenly at some stage of the transformation there was a narrowing of the ODZ, then it is worth looking very carefully at whether this transformation is permissible and whether we had the right to carry it out.
In fairness, we say that in practice we usually have to work with expressions in which the ODZ of variables is such that it allows us to use the properties of logarithms without restrictions in the form already known to us, both from left to right and from right to left, when carrying out transformations. You quickly get used to this, and you begin to carry out the transformations mechanically, without thinking about whether it was possible to carry them out. And at such moments, as luck would have it, more complex examples slip through, in which the inaccurate application of the properties of logarithms leads to errors. So you need to always be on the alert, and make sure that there is no narrowing of the ODZ.
It does not hurt to separately highlight the main transformations based on the properties of logarithms, which must be carried out very carefully, which can lead to a narrowing of the ODZ, and as a result, to errors:
Some transformations of expressions according to the properties of logarithms can also lead to the opposite - the expansion of the ODZ. For example, going from 4 log 2 (x+1) to log 2 (x+1) 4 extends the ODZ from the set (−1, +∞) to (−∞, −1)∪(−1, +∞) . Such transformations take place if you remain within the ODZ for the original expression. So the transformation just mentioned 4 log 2 (x+1)=log 2 (x+1) 4 takes place on the ODZ variable x for the original expression 4 log 2 (x+1) , that is, when x+1> 0 , which is the same as (−1, +∞) .
Now that we have discussed the nuances that you need to pay attention to when converting expressions with variables using the properties of logarithms, it remains to figure out how these conversions should be carried out correctly.
X+2>0 . Does it work in our case? To answer this question, let's take a look at the DPV of the x variable. It is determined by the system of inequalities 
, which is equivalent to the condition x+2>0 (if necessary, see the article solution of systems of inequalities). Thus, we can safely apply the property of the logarithm of the degree.
We have
3 log(x+2) 7 −log(x+2)−5 log(x+2) 4 =
=3 7 log(x+2)−log(x+2)−5 4 log(x+2)=
=21 log(x+2)−log(x+2)−20 log(x+2)=
=(21−1−20)lg(x+2)=0 .
You can act differently, since the ODZ allows you to do this, for example like this: 
Answer:
3 log(x+2) 7 −log(x+2)−5 log(x+2) 4 =0.
And what to do when the conditions associated with the properties of logarithms are not met on the ODZ? We will deal with this with examples.
Let us be required to simplify the expression lg(x+2) 4 −lg(x+2) 2 . The transformation of this expression, unlike the expression from the previous example, does not allow the free use of the property of the logarithm of the degree. Why? The ODZ of the variable x in this case is the union of two intervals x>−2 and x<−2 . При x>−2 we can safely apply the property of the logarithm of the degree and proceed as in the example above: log(x+2) 4 −log(x+2) 2 =4 log(x+2)−2 log(x+2)=2 log(x+2). But the ODZ contains another interval x+2<0 , для которого последнее преобразование будет некорректно. Что же делать при x+2<0 ? В подобных случаях на помощь приходит . Определение модуля позволяет выражение x+2 при x+2<0 представить как −|x+2| . Тогда при x+2<0 от lg(x+2) 4 −lg(x+2) 2 переходим к log(−|x+2|) 4 −log(−|x+2|) 2 and further, due to the power properties of lg|x+2| 4−lg|x+2| 2. The resulting expression can be transformed according to the property of the logarithm of the degree, since |x+2|>0 for any values of the variable. We have log|x+2| 4−lg|x+2| 2 =4 log|x+2|−2 log|x+2|=2 log|x+2|. Now you can get rid of the module, since it has done its job. Since we are transforming at x+2<0 , то 2·lg|x+2|=2·lg(−(x+2)) . Итак, можно считать, что мы справились с поставленной задачей. Ответ: . Полученный результат можно записать компактно с использованием модуля как .
Let's consider one more example to make working with modules familiar. Let us conceive from the expression 
pass to the sum and difference of the logarithms of the linear binomials x−1 , x−2 and x−3 . First we find the ODZ: 
On the interval (3, +∞), the values of the expressions x−1 , x−2 and x−3 are positive, so we can safely apply the properties of the logarithm of the sum and difference: 
And on the interval (1, 2), the values of the expression x−1 are positive, and the values of the expressions x−2 and x−3 are negative. Therefore, on the interval under consideration, we represent x−2 and x−3 using the modulo as −|x−2| and −|x−3| respectively. Wherein 
Now we can apply the properties of the logarithm of the product and the quotient, since on the considered interval (1, 2) the values of the expressions x−1 , |x−2| and |x−3| - positive.
We have 
The results obtained can be combined:
In general, similar reasoning allows, based on the formulas for the logarithm of the product, ratio and degree, to obtain three practically useful results that are quite convenient to use:
- The logarithm of the product of two arbitrary expressions X and Y of the form log a (X·Y) can be replaced by the sum of logarithms log a |X|+log a |Y| , a>0 , a≠1 .
- The special logarithm log a (X:Y) can be replaced by the difference of the logarithms log a |X|−log a |Y| , a>0 , a≠1 , X and Y are arbitrary expressions.
- From the logarithm of some expression B to an even power p of the form log a B p, one can pass to the expression p log a |B| , where a>0 , a≠1 , p is an even number and B is an arbitrary expression.
Similar results are given, for example, in instructions for solving exponential and logarithmic equations in the collection of problems in mathematics for applicants to universities, edited by M. I. Skanavi.
Example.
Simplify the expression 
.
Solution.
It would be good to apply the properties of the logarithm of the degree, sum and difference. But can we do it here? To answer this question, we need to know the ODZ.
Let's define it: 
It is quite obvious that the expressions x+4 , x−2 and (x+4) 13 on the range of possible values of the variable x can take both positive and negative values. Therefore, we will have to work through modules. 
Module properties allow you to rewrite as , so 
Also, nothing prevents you from using the property of the logarithm of the degree, and then bring like terms: 
Another sequence of transformations leads to the same result: 
and since the expression x−2 can take both positive and negative values on the ODZ, when taking an even exponent 14
