How to solve log equations. Logarithmic equations! you can get acquainted with functions and derivatives

Examples:

\(\log_(2)(⁡x) = 32\)
\(\log_3⁡x=\log_3⁡9\)
\(\log_3⁡((x^2-3))=\log_3⁡((2x))\)
\(\log_(x+1)((x^2+3x-7))=2\)
\(\lg^2⁡((x+1))+10=11 \lg⁡((x+1))\)

How to solve logarithmic equations:

When solving a logarithmic equation, you should strive to transform it to the form \(\log_a⁡(f(x))=\log_a⁡(g(x))\), and then make the transition to \(f(x)=g(x) \).

\(\log_a⁡(f(x))=\log_a⁡(g(x))\) \(⇒\) \(f(x)=g(x)\).

Example:\(\log_2⁡(x-2)=3\)

Solution: \(\log_2⁡(x-2)=\log_2⁡8\) \(x-2=8\) \(x=10\) Examination:\(10>2\) - suitable for DL Answer:\(x=10\)

ODZ: \(x-2>0\) \(x>2\)

Very important! This transition can only be made if:

You have written for the original equation, and at the end you will check whether those found are included in the DL. If this is not done, they may appear extra roots, which means it’s a wrong decision.

The number (or expression) on the left and right is the same;

The logarithms on the left and right are “pure”, that is, there should be no multiplications, divisions, etc. – only single logarithms on either side of the equal sign.

For example:

Note that Equations 3 and 4 can be easily solved by applying the necessary properties of logarithms.

Example . Solve the equation \(2\log_8⁡x=\log_8⁡2.5+\log_8⁡10\)

Solution :

		Let's write the ODZ: \(x>0\).
\(2\log_8⁡x=\log_8⁡2.5+\log_8⁡10\) ODZ: \(x>0\)		On the left in front of the logarithm is the coefficient, on the right is the sum of the logarithms. This bothers us. Let's move the two to the exponent \(x\) according to the property: \(n \log_b(⁡a)=\log_b⁡(a^n)\). Let us represent the sum of logarithms as one logarithm according to the property: \(\log_a⁡b+\log_a⁡c=\log_a(⁡bc)\)
\(\log_8⁡(x^2)=\log_8⁡25\)		We reduced the equation to the form \(\log_a⁡(f(x))=\log_a⁡(g(x))\) and wrote down the ODZ, which means we can move to the form \(f(x)=g(x)\ ).
		Happened . We solve it and get the roots.
\(x_1=5\) \(x_2=-5\)		We check whether the roots are suitable for ODZ. To do this, in \(x>0\) instead of \(x\) we substitute \(5\) and \(-5\). This operation can be performed orally.
\(5>0\), \(-5>0\)		The first inequality is true, the second is not. This means that \(5\) is the root of the equation, but \(-5\) is not. We write down the answer.

Answer : \(5\)

Example : Solve the equation \(\log^2_2⁡(x)-3 \log_2(⁡x)+2=0\)

Solution :

		Let's write the ODZ: \(x>0\).
\(\log^2_2⁡(x)-3 \log_2(⁡x)+2=0\) ODZ: \(x>0\)		A typical equation solved using . Replace \(\log_2⁡x\) with \(t\).
\(t=\log_2⁡x\)
		We received the usual one. We are looking for its roots.
\(t_1=2\) \(t_2=1\)		Making a reverse replacement
\(\log_2(⁡x)=2\) \(\log_2(⁡x)=1\)		We transform the right-hand sides, representing them as logarithms: \(2=2 \cdot 1=2 \log_2⁡2=\log_2⁡4\) and \(1=\log_2⁡2\)
\(\log_2(⁡x)=\log_2⁡4\) \(\log_2(⁡x)=\log_2⁡2 \)		Now our equations are \(\log_a⁡(f(x))=\log_a⁡(g(x))\), and we can transition to \(f(x)=g(x)\).
\(x_1=4\) \(x_2=2\)		Checking compliance ODZ roots. To do this, substitute \(4\) and \(2\) into the inequality \(x>0\) instead of \(x\).
\(4>0\) \(2>0\)		Both inequalities are true. This means that both \(4\) and \(2\) are roots of the equation.

Answer : \(4\); \(2\).

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Introduction

The increase in mental load in mathematics lessons makes us think about how to maintain students’ interest in the material being studied and their activity throughout the lesson. In this regard, a search is underway for new effective teaching methods and methodological techniques that would activate students’ thoughts and stimulate them to independently acquire knowledge.

The emergence of interest in mathematics among a significant number of students depends to a large extent on the methodology of its teaching, on how skillfully the educational work will be structured. Directing students' attention in a timely manner to what mathematics is studying. general properties objects and phenomena of the surrounding world, deals not with objects, but with abstract abstract concepts, one can achieve an understanding that mathematics does not violate the connection with reality, but, on the contrary, makes it possible to study it more deeply, to draw generalized theoretical conclusions that are widely used in practice.

Participating in the festival of pedagogical ideas "Open Lesson" 2004-2005 school year, I presented a lesson-lecture on the topic “Logarithmic function” (diploma No. 204044). I consider this method the most successful in this particular case. As a result of studying, students have a detailed outline and a brief outline of the topic, which will make it easier for them to prepare for the next lessons. In particular, on the topic "Solving logarithmic equations", which is entirely based on the study of the logarithmic function and its properties.

When forming fundamental mathematical concepts, it is important to create in students an idea of ​​the appropriateness of introducing each of them and the possibility of their application. To do this, it is necessary that when formulating the definition of a certain concept, working on its logical structure, questions about the history of its occurrence should be considered. this concept. This approach will help students realize that the new concept serves as a generalization of the facts of reality.

The history of the emergence of logarithms is presented in detail in last year's work.

Considering the importance of continuity in teaching mathematics in a secondary specialized educational institution and at a university and the need to comply uniform requirements To students, I consider it appropriate to use the following method to familiarize students with solving logarithmic equations.

Equations containing a variable under the logarithm sign (in particular, in the base of the logarithm) are called logarithmic. Let's consider logarithmic equations type:

The solution to these equations is based on the following theorem.

Theorem 1. The equation is equivalent to the system

(2)

To solve equation (1), it is enough to solve the equation

and substitute his solutions into the system of inequalities

defining the domain of definition of equation (1).

The roots of equation (1) will be only those solutions to equation (3) that satisfy system (4), i.e. belong to the domain of definition of equation (1).

When solving logarithmic equations, an expansion of the domain of definition (acquisition of extraneous roots) or narrowing (loss of roots) may occur. Therefore, substituting the roots of equation (3) into system (4), i.e. verification of the solution is required.

Example 1: Solve the equation

Solution:

Both meanings X satisfy the conditions of the system.

Answer:

Consider equations of the form:

Their solution is based on the following theorem

Theorem 2: Equation (5) is equivalent to the system

(6)

The roots of equation (5) will be only those roots of the equation that

belong to the domain of definition specified by the conditions .

A logarithmic equation of the form (5) can be solved in various ways. Let's look at the main ones.

1. POTENTIZATION (application of the properties of the logarithm).

Example 2: Solve the equation

Solution: By virtue of Theorem 2 given equation is equivalent to the system:

Let's solve the equation:

Only one root satisfies all the conditions of the system. Answer:

2. USING THE DEFINITION OF LOGARITHM .

Example 3: Find X, If

Solution:

Meaning X= 3 belongs to the domain of definition of the equation. Answer X = 3

3. REDUCTION TO A QUADRATE EQUATION.

Example 4: Solve the equation

Both meanings X are the roots of the equation.

Answer:

4. LOGARIFTHING.

Example 5: Solve the equation

Solution: Let's take the logarithm of both sides of the equation to base 10 and apply the "logarithm of power" property.

Both roots belong to the range of permissible values ​​of the logarithmic function.

Answer: X = 0,1; X = 100

5. REDUCTION TO ONE BASIS.

Example 6: Solve the equation

Let's use the formula and let’s go to the base 2 logarithm in all terms:

Then this equation will take the form:

Since , then this is the root of the equation.

Answer: X = 16

As you know, when multiplying expressions with powers, their exponents always add up (a b *a c = a b+c). This mathematical law was derived by Archimedes, and later, in the 8th century, the mathematician Virasen created a table of integer exponents. They were the ones who served for further opening logarithms. Examples of using this function can be found almost everywhere where you need to simplify cumbersome multiplication by simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. In simple and accessible language.

Definition in mathematics

A logarithm is an expression of the following form: log a b=c, that is, the logarithm of any non-negative number(that is, any positive) “b” by its base “a” is considered to be the power of “c” to which the base “a” must be raised in order to ultimately obtain the value “b”. Let's analyze the logarithm using examples, let's say there is an expression log 2 8. How to find the answer? It’s very simple, you need to find a power such that from 2 to the required power you get 8. After doing some calculations in your head, we get the number 3! And that’s true, because 2 to the power of 3 gives the answer as 8.

Types of logarithms

For many pupils and students, this topic seems complicated and incomprehensible, but in fact logarithms are not so scary, the main thing is to understand their general meaning and remember their properties and some rules. There are three individual species logarithmic expressions:

1. Natural logarithm ln a, where the base is the Euler number (e = 2.7).
2. Decimal a, where the base is 10.
3. Logarithm of any number b to base a>1.

Each of them is decided in a standard way, which includes simplification, reduction and subsequent reduction to one logarithm using logarithmic theorems. To obtain the correct values ​​of logarithms, you should remember their properties and the sequence of actions when solving them.

Rules and some restrictions

In mathematics, there are several rules-constraints that are accepted as an axiom, that is, they are not subject to discussion and are the truth. For example, it is impossible to divide numbers by zero, and it is also impossible to extract an even root from negative numbers. Logarithms also have their own rules, following which you can easily learn to work even with long and capacious logarithmic expressions:

• The base “a” must always be greater than zero, and not equal to 1, otherwise the expression will lose its meaning, because “1” and “0” to any degree are always equal to their values;
• if a > 0, then a b >0, it turns out that “c” must also be greater than zero.

How to solve logarithms?

For example, the task is given to find the answer to the equation 10 x = 100. This is very easy, you need to choose a power by raising the number ten to which we get 100. This, of course, is 10 2 = 100.

Now let's imagine this expression in logarithmic form. We get log 10 100 = 2. When solving logarithms, all actions practically converge to find the power to which it is necessary to enter the base of the logarithm in order to obtain a given number.

To accurately determine the value unknown degree you need to learn how to work with the table of degrees. It looks like this:

As you can see, some exponents can be guessed intuitively if you have a technical mind and knowledge of the multiplication table. However for large values you will need a table of degrees. It can be used even by those who know nothing at all about complex mathematical topics. The left column contains numbers (base a), the top row of numbers is the value of the power c to which the number a is raised. At the intersection, the cells contain the number values ​​that are the answer (a c =b). Let's take, for example, the very first cell with the number 10 and square it, we get the value 100, which is indicated at the intersection of our two cells. Everything is so simple and easy that even the most true humanist will understand!

Equations and inequalities

It turns out that under certain conditions the exponent is the logarithm. Therefore, any mathematical numerical expressions can be written as a logarithmic equation. For example, 3 4 =81 can be written as the base 3 logarithm of 81 equal to four (log 3 81 = 4). For negative powers the rules are the same: 2 -5 = 1/32 we write it as a logarithm, we get log 2 (1/32) = -5. One of the most fascinating sections of mathematics is the topic of “logarithms”. We will look at examples and solutions of equations below, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.

Given an expression of the following form: log 2 (x-1) > 3 - it is logarithmic inequality, since the unknown value "x" is under the sign of the logarithm. And also in the expression two quantities are compared: the logarithm of the desired number to base two is greater than the number three.

The most important difference between logarithmic equations and inequalities is that equations with logarithms (for example, the logarithm 2 x = √9) imply one or more specific answers. numerical values, while when solving the inequalities are defined as the region acceptable values, and the breakpoints of this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer to an equation, but rather continuous series or a set of numbers.

Basic theorems about logarithms

When solving primitive tasks of finding the values ​​of the logarithm, its properties may not be known. However, when it comes to logarithmic equations or inequalities, first of all, it is necessary to clearly understand and apply in practice all the basic properties of logarithms. We will look at examples of equations later; let's first look at each property in more detail.

1. The main identity looks like this: a logaB =B. It applies only when a is greater than 0, not equal to one, and B is greater than zero.
2. The logarithm of the product can be represented in the following formula: log d (s 1 *s 2) = log d s 1 + log d s 2. In this case prerequisite is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this logarithmic formula, with examples and solution. Let log a s 1 = f 1 and log a s 2 = f 2, then a f1 = s 1, a f2 = s 2. We obtain that s 1 * s 2 = a f1 *a f2 = a f1+f2 (properties of degrees ), and then by definition: log a (s 1 * s 2) = f 1 + f 2 = log a s1 + log a s 2, which is what needed to be proven.
3. The logarithm of the quotient looks like this: log a (s 1/ s 2) = log a s 1 - log a s 2.
4. The theorem in the form of a formula takes on next view: log a q b n = n/q log a b.

This formula is called the “property of the degree of logarithm.” It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics is based on natural postulates. Let's look at the proof.

Let log a b = t, it turns out a t =b. If we raise both parts to the power m: a tn = b n ;

but since a tn = (a q) nt/q = b n, therefore log a q b n = (n*t)/t, then log a q b n = n/q log a b. The theorem has been proven.

Examples of problems and inequalities

The most common types of problems on logarithms are examples of equations and inequalities. They are found in almost all problem books, and are also included in mandatory part mathematics exams. For admission to university or passing entrance examinations in mathematics you need to know how to solve such problems correctly.

Unfortunately, there is no single plan or scheme for solving and determining unknown value There is no such thing as a logarithm, but you can apply it to every mathematical inequality or logarithmic equation. certain rules. First of all, you should find out whether the expression can be simplified or lead to general appearance. Simplify long ones logarithmic expressions possible if you use their properties correctly. Let's get to know them quickly.

When solving logarithmic equations, we must determine what type of logarithm we have: an example expression may contain a natural logarithm or a decimal one.

Here are examples ln100, ln1026. Their solution boils down to the fact that they need to determine the power to which the base 10 will be equal to 100 and 1026, respectively. For solutions natural logarithms need to apply logarithmic identities or their properties. Let's look at the solution with examples logarithmic problems different types.

How to Use Logarithm Formulas: With Examples and Solutions

So, let's look at examples of using the basic theorems about logarithms.

1. The property of the logarithm of a product can be used in tasks where it is necessary to expand great importance numbers b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4*128) = log 2 512. The answer is 9.
2. log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, using the fourth property of the logarithm power, we managed to solve a seemingly complex and unsolvable expression. You just need to factor the base and then take the exponent values ​​out of the sign of the logarithm.

Assignments from the Unified State Exam

Logarithms are often found in entrance exams, especially a lot of logarithmic problems in the Unified State Exam ( State exam for all school leavers). Typically, these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most complex and voluminous tasks). The exam requires accurate and perfect knowledge of the topic “Natural logarithms”.

Examples and solutions to problems are taken from official Unified State Exam options. Let's see how such tasks are solved.

Given log 2 (2x-1) = 4. Solution:
let's rewrite the expression, simplifying it a little log 2 (2x-1) = 2 2, by the definition of the logarithm we get that 2x-1 = 2 4, therefore 2x = 17; x = 8.5.

• It is best to reduce all logarithms to the same base so that the solution is not cumbersome and confusing.
• All expressions under the logarithm sign are indicated as positive, therefore, when the exponent of an expression that is under the logarithm sign and as its base is taken out as a multiplier, the expression remaining under the logarithm must be positive.

Introduction

Logarithms were invented to speed up and simplify calculations. The idea of ​​a logarithm, that is, the idea of ​​expressing numbers as powers of the same base, belongs to Mikhail Stiefel. But in Stiefel’s time, mathematics was not so developed and the idea of ​​the logarithm was not developed. Logarithms were later invented simultaneously and independently of each other by the Scottish scientist John Napier (1550-1617) and the Swiss Jobst Burgi (1552-1632). Napier was the first to publish the work in 1614. entitled "Description of the amazing table of logarithms", Napier's theory of logarithms was given in sufficient in full, the method for calculating logarithms is given the simplest, therefore Napier’s merits in the invention of logarithms are greater than those of Bürgi. Bürgi worked on the tables at the same time as Napier, but for a long time kept them secret and published them only in 1620. Napier mastered the idea of ​​the logarithm around 1594. although the tables were published 20 years later. At first he called his logarithms “artificial numbers” and only then proposed to call these “artificial numbers” in one word “logarithm”, which translated from Greek means “correlated numbers”, taken one from an arithmetic progression, and the other from a geometric progression specially selected for it. progress. The first tables in Russian were published in 1703. with the participation of a wonderful teacher of the 18th century. L. F. Magnitsky. The works of St. Petersburg academician Leonhard Euler were of great importance in the development of the theory of logarithms. He was the first to consider logarithms as the inverse of raising to a power; he introduced the terms “logarithm base” and “mantissa.” Briggs compiled tables of logarithms with base 10. Decimal tables are more convenient for practical use, their theory is simpler than that of Napier’s logarithms . That's why decimal logarithms sometimes called brigs. The term "characterization" was introduced by Briggs.

In those distant times, when the sages first began to think about equalities containing unknown quantities, there were probably no coins or wallets. But there were heaps, as well as pots and baskets, which were perfect for the role of storage caches that could hold an unknown number of items. In the ancients mathematical problems Mesopotamia, India, China, Greece, unknown quantities expressed the number of peacocks in the garden, the number of bulls in the herd, the totality of things taken into account when dividing property. Scribes, officials and initiates well trained in the science of accounts secret knowledge The priests coped quite successfully with such tasks.

Sources that have reached us indicate that ancient scientists owned some general techniques solving problems with unknown quantities. However, not a single papyrus or clay tablet contains a description of these techniques. The authors only occasionally supplied their numerical calculations with skimpy comments such as: “Look!”, “Do this!”, “You found the right one.” In this sense, the exception is the “Arithmetic” of the Greek mathematician Diophantus of Alexandria (III century) - a collection of problems for composing equations with a systematic presentation of their solutions.

However, the first manual for solving problems that became widely known was the work of the Baghdad scientist of the 9th century. Muhammad bin Musa al-Khwarizmi. The word "al-jabr" from the Arabic name of this treatise - "Kitab al-jaber wal-mukabala" ("Book of restoration and opposition") - over time turned into the well-known word "algebra", and al-Khwarizmi's work itself served the starting point in the development of the science of solving equations.

Logarithmic equations and inequalities

1. Logarithmic equations

An equation containing an unknown under the logarithm sign or at its base is called a logarithmic equation.

The simplest logarithmic equation is an equation of the form

log a x = b . (1)

Statement 1. If a > 0, a≠ 1, equation (1) for any real b It has only decision x = a b .

Example 1. Solve the equations:

a)log 2 x= 3, b) log 3 x= -1, c)

Solution. Using Statement 1, we obtain a) x= 2 3 or x= 8; b) x= 3 -1 or x= 1 / 3 ; c)

or x = 1.

Let us present the basic properties of the logarithm.

P1. Basic logarithmic identity:

Where a > 0, a≠ 1 and b > 0.

P2. Logarithm of the product of positive factors equal to the sum logarithms of these factors:

log a N 1 · N 2 = log a N 1 + log a N 2 (a > 0, a ≠ 1, N 1 > 0, N 2 > 0).

Comment. If N 1 · N 2 > 0, then property P2 takes the form

log a N 1 · N 2 = log a |N 1 | + log a |N 2 | (a > 0, a ≠ 1, N 1 · N 2 > 0).

P3. The logarithm of the quotient of two positive numbers is equal to the difference between the logarithms of the dividend and the divisor

(a > 0, a ≠ 1, N 1 > 0, N 2 > 0).

Comment. If

, (which is equivalent N 1 N 2 > 0) then property P3 takes the form (a > 0, a ≠ 1, N 1 N 2 > 0).

P4. Logarithm of degree positive number equal to the product exponent per logarithm of this number:

log a N k = k log a N (a > 0, a ≠ 1, N > 0).

Comment. If k - even number (k = 2s), That

log a N 2s = 2s log a |N | (a > 0, a ≠ 1, N ≠ 0).

P5. Formula for moving to another base:

(a > 0, a ≠ 1, b > 0, b ≠ 1, N > 0),

in particular if N = b, we get

(a > 0, a ≠ 1, b > 0, b ≠ 1). (2)

Using properties P4 and P5, it is easy to obtain the following properties

(a > 0, a ≠ 1, b > 0, c ≠ 0), (3) (a > 0, a ≠ 1, b > 0, c ≠ 0), (4) (a > 0, a ≠ 1, b > 0, c ≠ 0), (5)

and, if in (5) c- even number ( c = 2n), occurs

(b > 0, a ≠ 0, |a | ≠ 1). (6)

Let us list the main properties of the logarithmic function f (x) = log a x :

1. The domain of definition of a logarithmic function is the set of positive numbers.

2. The range of values ​​of the logarithmic function is the set of real numbers.

3. When a > 1 logarithmic function strictly increasing (0< x 1 < x 2log a x 1 < loga x 2), and at 0< a < 1, - строго убывает (0 < x 1 < x 2log a x 1 > log a x 2).

4.log a 1 = 0 and log a a = 1 (a > 0, a ≠ 1).

5. If a> 1, then the logarithmic function is negative when x(0;1) and positive at x(1;+∞), and if 0< a < 1, то логарифмическая функция положительна при x (0;1) and negative at x (1;+∞).

6. If a> 1, then the logarithmic function is convex upward, and if a(0;1) - convex downwards.

The following statements (see, for example,) are used when solving logarithmic equations.