Methods for solving logarithmic equations. Logarithms: examples and solutions Solving logarithm equations at the root

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. It was they who served for the further discovery of 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” to its base “a” is considered to be the power “c” to which the base “a” must be raised in order to ultimately get 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 separate types of 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 solved in a standard way, including simplification, reduction and subsequent reduction to a single 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 the even root of 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 represent 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 of an unknown degree, you need to learn how to work with a 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 larger values ​​you will need a power table. 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 equality. 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.

The following expression is given: log 2 (x-1) > 3 - it is a logarithmic inequality, since the unknown value “x” is under the logarithmic sign. 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 numerical values ​​in the answer, while when solving an inequality, both the range of acceptable values ​​​​and the points are determined breaking this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer to an equation, but a continuous series or 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, the mandatory condition 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 the following form: 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 a required part of mathematics exams. To enter a university or pass entrance examinations in mathematics, you need to know how to correctly solve such tasks.

Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, but certain rules can be applied to each mathematical inequality or logarithmic equation. First of all, you should find out whether the expression can be simplified or reduced to a general form. You can simplify long logarithmic expressions 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. To solve natural logarithms, you need to apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various 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 decompose a large value of the number 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 many logarithmic problems in the Unified State Exam (state exam for all school graduates). 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 the official versions of the Unified State Exam. 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.

Preparation for the final test in mathematics includes an important section - “Logarithms”. Tasks from this topic are necessarily contained in the Unified State Examination. Experience from past years shows that logarithmic equations caused difficulties for many schoolchildren. Therefore, students with different levels of training must understand how to find the correct answer and quickly cope with them.

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The Shkolkovo educational portal allows you to prepare for the Unified State Exam anywhere at any time. Our website offers the most convenient approach to repeating and assimilating a large amount of information on logarithms, as well as with one and several unknowns. Start with easy equations. If you cope with them without difficulty, move on to more complex ones. If you have trouble solving a particular inequality, you can add it to your Favorites so you can return to it later.

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In this lesson we will review the basic theoretical facts about logarithms and consider solving the simplest logarithmic equations.

Let us recall the central definition - the definition of a logarithm. It involves solving an exponential equation. This equation has a single root, it is called the logarithm of b to base a:

Definition:

The logarithm of b to base a is the exponent to which base a must be raised to get b.

Let us remind you basic logarithmic identity.

The expression (expression 1) is the root of the equation (expression 2). Substitute the value x from expression 1 instead of x into expression 2 and get the main logarithmic identity:

So we see that each value is associated with a value. We denote b by x(), c by y, and thus obtain a logarithmic function:

For example:

Let us recall the basic properties of the logarithmic function.

Let us pay attention once again, here, since under the logarithm there can be a strictly positive expression, as the base of the logarithm.

Rice. 1. Graph of a logarithmic function with different bases

The graph of the function at is shown in black. Rice. 1. If the argument increases from zero to infinity, the function increases from minus to plus infinity.

The graph of the function at is shown in red. Rice. 1.

Properties of this function:

Domain: ;

Range of values: ;

The function is monotonic throughout its entire domain of definition. When monotonically (strictly) increases, a larger value of the argument corresponds to a larger value of the function. When monotonically (strictly) decreases, a larger value of the argument corresponds to a smaller value of the function.

The properties of the logarithmic function are the key to solving a variety of logarithmic equations.

Let's consider the simplest logarithmic equation; all other logarithmic equations, as a rule, are reduced to this form.

Since the bases of logarithms and the logarithms themselves are equal, the functions under the logarithm are also equal, but we must not miss the domain of definition. Only a positive number can appear under the logarithm, we have:

We found out that the functions f and g are equal, so it is enough to choose any one inequality to comply with the ODZ.

Thus, we have a mixed system in which there is an equation and an inequality:

As a rule, it is not necessary to solve an inequality; it is enough to solve the equation and substitute the found roots into the inequality, thus performing a check.

Let us formulate a method for solving the simplest logarithmic equations:

Equalize the bases of logarithms;

Equate sublogarithmic functions;

Perform check.

Let's look at specific examples.

Example 1 - solve the equation:

The bases of logarithms are initially equal, we have the right to equate sublogarithmic expressions, do not forget about the ODZ, we choose the first logarithm to compose the inequality:

Example 2 - solve the equation:

This equation differs from the previous one in that the bases of the logarithms are less than one, but this does not affect the solution in any way:

Let's find the root and substitute it into the inequality:

We received an incorrect inequality, which means that the found root does not satisfy the ODZ.

Example 3 - solve the equation:

The bases of logarithms are initially equal, we have the right to equate sublogarithmic expressions, do not forget about the ODZ, we choose the second logarithm to compose the inequality:

Let's find the root and substitute it into the inequality:

Obviously, only the first root satisfies the ODZ.

Logarithmic equations and inequalities in the Unified State Examination in mathematics it is devoted to problem C3 . Every student must learn to solve C3 tasks from the Unified State Exam in mathematics if he wants to pass the upcoming exam with “good” or “excellent”. This article provides a brief overview of commonly encountered logarithmic equations and inequalities, as well as basic methods for solving them.

So, let's look at a few examples today. logarithmic equations and inequalities, which were offered to students in the Unified State Examination in mathematics of previous years. But it will begin with a brief summary of the main theoretical points that we will need to solve them.

Logarithmic function

Definition

Function of the form

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called logarithmic function.

Basic properties

Basic properties of the logarithmic function y=log a x:

The graph of a logarithmic function is logarithmic curve:

Properties of logarithms

Logarithm of the product two positive numbers is equal to the sum of the logarithms of these numbers:

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Logarithm of the quotient two positive numbers is equal to the difference between the logarithms of these numbers:

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If a And b a≠ 1, then for any number r equality is true:

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Equality log a t=log a s, Where a > 0, a ≠ 1, t > 0, s> 0, valid if and only if t = s.

If a, b, c are positive numbers, and a And c are different from unity, then the equality ( formula for moving to a new logarithm base):

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Theorem 1. If f(x) > 0 and g(x) > 0, then the logarithmic equation log a f(x) = log a g(x) (Where a > 0, a≠ 1) is equivalent to the equation f(x) = g(x).

Solving logarithmic equations and inequalities

Example 1. Solve the equation:

Solution. The range of acceptable values ​​includes only those x, for which the expression under the logarithm sign is greater than zero. These values ​​are determined by the following system of inequalities:

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Considering that

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we obtain the interval that defines the range of permissible values ​​of this logarithmic equation:

Based on Theorem 1, all conditions of which are satisfied here, we proceed to the following equivalent quadratic equation:

The range of acceptable values ​​includes only the first root.

Answer: x = 7.

Example 2. Solve the equation:

Solution. The range of acceptable values ​​of the equation is determined by the system of inequalities:

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Solution. The range of acceptable values ​​of the equation is determined here easily: x > 0.

We use substitution:

The equation becomes:

Reverse substitution:

Both answer are within the range of acceptable values ​​of the equation because they are positive numbers.

Example 4. Solve the equation:

Solution. Let's start the solution again by determining the range of acceptable values ​​of the equation. It is determined by the following system of inequalities:

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The bases of the logarithms are the same, so in the range of acceptable values ​​we can proceed to the following quadratic equation:

The first root is not within the range of acceptable values ​​of the equation, but the second is.

Answer: x = -1.

Example 5. Solve the equation:

Solution. We will look for solutions in between x > 0, x≠1. Let's transform the equation to an equivalent one:

Both answer are within the range of acceptable values ​​of the equation.

Example 6. Solve the equation:

Solution. The system of inequalities defining the range of permissible values ​​of the equation this time has the form:

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Using the properties of the logarithm, we transform the equation to an equation that is equivalent in the range of acceptable values:

Using the formula for moving to a new logarithm base, we get:

The range of acceptable values ​​includes only one answer: x = 4.

Let's now move on to logarithmic inequalities . This is exactly what you will have to deal with on the Unified State Exam in mathematics. To solve further examples we need the following theorem:

Theorem 2. If f(x) > 0 and g(x) > 0, then:
at a> 1 logarithmic inequality log a f(x) > log a g(x) is equivalent to an inequality of the same meaning: f(x) > g(x);
at 0< a < 1 логарифмическое неравенство log a f(x) > log a g(x) is equivalent to an inequality with the opposite meaning: f(x) < g(x).

Example 7. Solve the inequality:

Solution. Let's start by defining the range of acceptable values ​​of the inequality. The expression under the sign of the logarithmic function must take only positive values. This means that the required range of acceptable values ​​is determined by the following system of inequalities:

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Since the base of the logarithm is a number less than one, the corresponding logarithmic function will be decreasing, and therefore, according to Theorem 2, the transition to the following quadratic inequality will be equivalent:

Finally, taking into account the range of acceptable values, we obtain answer:

Example 8. Solve the inequality:

Solution. Let's start again by defining the range of acceptable values:

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On the set of admissible values ​​of the inequality we carry out equivalent transformations:

After reduction and transition to the inequality equivalent by Theorem 2, we obtain:

Taking into account the range of acceptable values, we obtain the final answer:

Example 9. Solve logarithmic inequality:

Solution. The range of acceptable values ​​of inequality is determined by the following system:

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It can be seen that in the range of acceptable values, the expression at the base of the logarithm is always greater than one, and therefore, according to Theorem 2, the transition to the following inequality will be equivalent:

Taking into account the range of acceptable values, we obtain the final answer:

Example 10. Solve the inequality:

Solution.

The range of acceptable values ​​of inequality is determined by the system of inequalities:

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Method I Let us use the formula for transition to a new base of the logarithm and move on to an inequality that is equivalent in the range of acceptable values.

Mathematics is more than science, this is the language of science.

Danish physicist and public figure Niels Bohr

Logarithmic equations

Among the typical tasks, offered at entrance (competitive) tests, are the tasks, related to solving logarithmic equations. To successfully solve such problems, you must have a good knowledge of the properties of logarithms and have the skills to use them.

This article first introduces the basic concepts and properties of logarithms., and then examples of solving logarithmic equations are considered.

Basic concepts and properties

First, we present the basic properties of logarithms, the use of which allows one to successfully solve relatively complex logarithmic equations.

The main logarithmic identity is written as

, (1)

Among the most well-known properties of logarithms are the following equalities:

1. If , , and , then , ,

2. If , , , and , then .

3. If , , and , then .

4. If , , and natural number, That

5. If , , and natural number, That

6. If , , and , then .

7. If , , and , then .

More complex properties of logarithms are formulated through the following statements:

8. If , , , and , then

9. If , , and , then

10. If , , , and , then

The proof of the last two properties of logarithms is given in the author’s textbook “Mathematics for high school students: additional sections of school mathematics” (M.: Lenand / URSS, 2014).

Also worth noting what is the function is increasing, if , and decreasing , if .

Let's look at examples of problems for solving logarithmic equations, arranged in order of increasing difficulty.

Examples of problem solving

Example 1. Solve the equation

. (2)

Solution. From equation (2) we have . Let's transform the equation as follows: , or .

Because , then the root of equation (2) is.

Answer: .

Example 2. Solve the equation

Solution. Equation (3) is equivalent to the equations

Or .

From here we get .

Answer: .

Example 3. Solve the equation

Solution. From equation (4) it follows, What . Using the basic logarithmic identity (1), we can write

or .

If you put then from here we get a quadratic equation, which has two roots And . However, therefore and a suitable root of the equation is only . Since , then or .

Answer: .

Example 4. Solve the equation

Solution.Range of permissible values ​​of the variablein equation (5) are.

Let it be . Since the functionon the domain of definition is decreasing, and the function increases along the entire number line, then the equation cannot have more than one root.

By selection we find the only root.

Answer: .

Example 5. Solve the equation.

Solution. If both sides of the equation are taken logarithmically to base 10, then

Or .

Solving the quadratic equation for , we obtain and . Therefore, here we have and .

Answer: , .

Example 6. Solve the equation

. (6)

Solution.Let us use identity (1) and transform equation (6) as follows:

Or .

Answer: , .

Example 7. Solve the equation

. (7)

Solution. Taking into account property 9, we have . In this regard, equation (7) takes the form

From here we get or .

Answer: .

Example 8. Solve the equation

. (8)

Solution.Let us use property 9 and rewrite equation (8) in the equivalent form.

If we then designate, then we get a quadratic equation, Where . Since the equationhas only one positive root, then or . This implies .

Answer: .

Example 9. Solve the equation

. (9)

Solution. Since from equation (9) it follows then here. According to property 10, can be written down.

In this regard, equation (9) will be equivalent to the equations

Or .

From here we obtain the root of equation (9).

Example 10. Solve the equation

. (10)

Solution. The range of permissible values ​​of the variable in equation (10) is . According to property 4, here we have

. (11)

Since , then equation (11) takes the form of a quadratic equation, where . The roots of a quadratic equation are and .

Since , then and . From here we get and .

Answer: , .

Example 11. Solve the equation

. (12)

Solution. Let us denote then and equation (12) takes the form

Or

. (13)

It is easy to see that the root of equation (13) is . Let us show that this equation has no other roots. To do this, divide both sides by and obtain the equivalent equation

. (14)

Since the function is decreasing, and the function is increasing on the entire numerical axis, then equation (14) cannot have more than one root. Since equations (13) and (14) are equivalent, equation (13) has a single root.

Since , then and .

Answer: .

Example 12. Solve the equation

. (15)

Solution. Let's denote and . Since the function decreases on the domain of definition, and the function is increasing for any values, the equation cannot have the same root. By direct selection we establish that the desired root of equation (15) is .

Answer: .

Example 13. Solve the equation

. (16)

Solution. Using the properties of logarithms, we get

Since then and we have inequality

The resulting inequality coincides with equation (16) only in the case when or .

By value substitutioninto equation (16) we are convinced that, What is its root.

Answer: .

Example 14. Solve the equation

. (17)

Solution. Since here , then equation (17) takes the form .

If we put , then we get the equation

, (18)

Where . From equation (18) it follows: or . Since, the equation has one suitable root. However, that's why.

Example 15. Solve the equation

. (19)

Solution. Let us denote , then equation (19) takes the form . If we take this equation to base 3, we get

Or

It follows that and . Since , then and . In this regard, and.

Answer: , .

Example 16. Solve the equation

. (20)

Solution. Let's enter the parameterand rewrite equation (20) in the form of a quadratic equation with respect to the parameter, i.e.

. (21)

The roots of equation (21) are

or , . Since , we have equations and . From here we get and .

Answer: , .

Example 17. Solve the equation

. (22)

Solution. To establish the domain of definition of the variable in equation (22), it is necessary to consider a set of three inequalities: , and .

Applying property 2, from equation (22) we obtain

Or

. (23)

If in equation (23) we put, then we get the equation

. (24)

Equation (24) will be solved as follows:

Or

It follows that and , i.e. equation (24) has two roots: and .

Since , then , or , .

Answer: , .

Example 18. Solve the equation

. (25)

Solution. Using the properties of logarithms, we transform equation (25) as follows:

, , .

From here we get .

Example 19. Solve the equation

. (26)

Solution. Since, then.

Next, we have. Hence , equality (26) is satisfied only if, when both sides of the equation are equal to 2 at the same time.

Thus , equation (26) is equivalent to the system of equations

From the second equation of the system we obtain

Or .

It's easy to see what's the meaning also satisfies the first equation of the system.

Answer: .

For a more in-depth study of methods for solving logarithmic equations, you can refer to the textbooks from the list of recommended literature.

1. Kushnir A.I. Masterpieces of school mathematics (problems and solutions in two books). – Kyiv: Astarte, book 1, 1995. – 576 p.

2. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace and Education, 2013. – 608 p.

3. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. – 216 p.

4. Suprun V.P. Mathematics for high school students: tasks of increased complexity. – M.: CD “Librocom” / URSS, 2017. – 200 p.

5. Suprun V.P. Mathematics for high school students: non-standard methods for solving problems. – M.: CD “Librocom” / URSS, 2017. – 296 p.

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