How to compare logarithms with different bases. Basic properties of logarithms. What to do with logarithms

    Let's start with properties of the logarithm of one. Its formulation is as follows: the logarithm of unity is equal to zero, that is, log a 1=0 for any a>0, a≠1. The proof is not difficult: since a 0 =1 for any a satisfying the above conditions a>0 and a≠1, then the equality log a 1=0 to be proved follows immediately from the definition of the logarithm.

    Let us give examples of the application of the considered property: log 3 1=0, log1=0 and .

    Let's move on to the next property: the logarithm of a number equal to the base is equal to one, that is, log a a=1 for a>0, a≠1. Indeed, since a 1 =a for any a, then by definition of the logarithm log a a=1.

    Examples of using this property of logarithms are the equalities log 5 5=1, log 5.6 5.6 and lne=1.

    For example, log 2 2 7 =7, log10 -4 =-4 and .

    Logarithm of the product of two positive numbers x and y is equal to the product of the logarithms of these numbers: log a (x y)=log a x+log a y, a>0 , a≠1 . Let us prove the property of the logarithm of a product. Due to the properties of the degree a log a x+log a y =a log a x ·a log a y, and since by the main logarithmic identity a log a x =x and a log a y =y, then a log a x ·a log a y =x·y. Thus, a log a x+log a y =x·y, from which, by the definition of a logarithm, the equality being proved follows.

    Let's show examples of using the property of the logarithm of a product: log 5 (2 3)=log 5 2+log 5 3 and .

    The property of the logarithm of a product can be generalized to the product of a finite number n of positive numbers x 1 , x 2 , …, x n as log a (x 1 ·x 2 ·…·x n)= log a x 1 +log a x 2 +…+log a x n . This equality can be proven without problems.

    For example, the natural logarithm of the product can be replaced by the sum of three natural logarithms of the numbers 4, e, and.

    Logarithm of the quotient of two positive numbers x and y is equal to the difference between the logarithms of these numbers. The property of the logarithm of a quotient corresponds to a formula of the form , where a>0, a≠1, x and y are some positive numbers. The validity of this formula is proven as well as the formula for the logarithm of a product: since , then by definition of a logarithm.

    Here is an example of using this property of the logarithm: .

    Let's move on to property of the logarithm of the power. The logarithm of a degree is equal to the product of the exponent and the logarithm of the modulus of the base of this degree. Let us write this property of the logarithm of a power as a formula: log a b p =p·log a |b|, where a>0, a≠1, b and p are numbers such that the degree b p makes sense and b p >0.

    First we prove this property for positive b. The basic logarithmic identity allows us to represent the number b as a log a b , then b p =(a log a b) p , and the resulting expression, due to the property of power, is equal to a p·log a b . So we come to the equality b p =a p·log a b, from which, by the definition of a logarithm, we conclude that log a b p =p·log a b.

    It remains to prove this property for negative b. Here we note that the expression log a b p for negative b makes sense only for even exponents p (since the value of the degree b p must be greater than zero, otherwise the logarithm will not make sense), and in this case b p =|b| p. Then b p =|b| p =(a log a |b|) p =a p·log a |b|, from where log a b p =p·log a |b| .

    For example, and ln(-3) 4 =4·ln|-3|=4·ln3 .

    It follows from the previous property property of the logarithm from the root: the logarithm of the nth root is equal to the product of the fraction 1/n by the logarithm of the radical expression, that is, , where a>0, a≠1, n is a natural number greater than one, b>0.

    The proof is based on the equality (see), which is valid for any positive b, and the property of the logarithm of the power: .

    Here is an example of using this property: .

    Now let's prove formula for moving to a new logarithm base kind . To do this, it is enough to prove the validity of the equality log c b=log a b·log c a. The basic logarithmic identity allows us to represent the number b as a log a b , then log c b=log c a log a b . It remains to use the property of the logarithm of the degree: log c a log a b =log a b log c a. This proves the equality log c b=log a b·log c a, which means that the formula for transition to a new base of the logarithm has also been proven.

    Let's show a couple of examples of using this property of logarithms: and .

    The formula for moving to a new base allows you to move on to working with logarithms that have a “convenient” base. For example, it can be used to go to natural or decimal logarithms so that you can calculate the value of a logarithm from a table of logarithms. The formula for moving to a new logarithm base also allows, in some cases, to find the value of a given logarithm when the values ​​of some logarithms with other bases are known.

    A special case of the formula for transition to a new logarithm base for c=b of the form is often used . This shows that log a b and log b a – . Eg, .

    The formula is also often used , which is convenient for finding logarithm values. To confirm our words, we will show how it can be used to calculate the value of a logarithm of the form . We have . To prove the formula it is enough to use the formula for transition to a new base of the logarithm a: .

    It remains to prove the properties of comparison of logarithms.

    Let us prove that for any positive numbers b 1 and b 2, b 1 log a b 2 , and for a>1 – the inequality log a b 1

    Finally, it remains to prove the last of the listed properties of logarithms. Let us limit ourselves to the proof of its first part, that is, we will prove that if a 1 >1, a 2 >1 and a 1 1 is true log a 1 b>log a 2 b . The remaining statements of this property of logarithms are proved according to a similar principle.

    Let's use the opposite method. Suppose that for a 1 >1, a 2 >1 and a 1 1 is true log a 1 b≤log a 2 b . Based on the properties of logarithms, these inequalities can be rewritten as And respectively, and from them it follows that log b a 1 ≤log b a 2 and log b a 1 ≥log b a 2, respectively. Then, according to the properties of powers with the same bases, the equalities b log b a 1 ≥b log b a 2 and b log b a 1 ≥b log b a 2 must hold, that is, a 1 ≥a 2 . So we came to a contradiction to the condition a 1

Bibliography.

  • Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
  • Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).

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.

Comparing the values ​​of logarithms or the value of a logarithm with a certain number occurs in school problem solving practice not only as an independent task. You have to compare logarithms, for example, when solving equations and inequalities. The materials of the article (problems and their solutions) are arranged according to the principle “from simple to complex” and can be used to prepare and conduct a lesson (lessons) on this topic, as well as in elective classes. The number of tasks considered in a lesson depends on the level of the class and its specialized area. In advanced mathematics classes, this material can be used for a two-hour lecture lesson.

1. (Orally.) Which of the functions are increasing and which are decreasing:

Comment. This exercise is a preparatory exercise.

2. (Orally.)Compare with zero:

Comment. When solving exercise No. 2, you can use both the properties of the logarithmic function using the graph of the logarithmic function, and the following useful property:

if positive numbers a and b lie on the number line to the right of 1 or to the left of 1 (that is, a>1 and b>1 or 0 0 ;
if positive numbers a and b lie on the number line on opposite sides of 1 (that is, 0 .

Let's show the use of this property in decision No. 2(a).

Since the function y = log 7 t increases by R+, 10 > 7, then log 7 10 > log 7 7, that is, log 7 10 > 1. Thus, positive numbers sin3 and log 7 10 lie on opposite sides of 1. Therefore, log sin3 log 7 10< 0.

3. (Orally.) Find the error in reasoning:

Function y = lgt increases by R + , then ,

Let us divide both sides of the last inequality by . We get that 2 > 3.

Solution.

Positive numbers and 10 (the base of the logarithm) lie on opposite sides of 1. This means that< 0. При делении обеих частей неравенства на число знак неравенства следует изменить на противоположный.

4. (Orally.) Compare the numbers:

Comment. When solving exercises No. 4(a–c), we use the property of monotonicity of the logarithmic function. For solution No. 4(d), we use the property:

if c > a >1, then for b>1 the inequality log a b > log c b is true.

Solution 4(d).

Since 1< 5 < 7 и 13 >1, then log 5 13 > log 7 13.

5. Compare numbers log 2 6 and 2.

Solution.

First way (using the monotonicity of the logarithmic function).

Function y = log 2 t increases by R+, 6 > 4. So, log 2 6 > log 2 4 And log 2 5 > 2.

The second method (composing the difference).

Let's make up the difference.

6. Compare numbers And -1.

Function y = decreases by R+ , 3 < 5. Значит, >And > -1 .

7. Compare numbers And 3log 8 26 .

Function y = log 2 t increases by R+, 25 < 26. Значит, log 2 25 < log 2 26 и.

First way.

Let's multiply both sides of the inequality by 3:

Function y = log 5 t increases by R+ , 27 > 25. So,

Second way.

Let's make up the difference
. From here.

9. Compare the numbers log 4 26 And log 6 17.

Let's estimate the logarithms, taking into account that the functions y = log 4 t and y = log 6 t increasing by R+:

Considering that the functions decreasing by R+, we have:

Means,

Comment. The proposed comparison method is called “insertion” method or “separation” method(we found the number 4 separating these two numbers).

11. Compare the numbers log 2 3 And log 3 5.

Note that both logarithms are greater than 1 but less than 2.

First way. Let's try to use the “separation” method. Let's compare logarithms with the number.

Second method ( multiplication by a natural number).

Note 1. The essence methodmultiplying by a natural number” is that we are looking for a natural number k, when multiplied by which the compared numbers a And b get these numbers ka And kb that there is at least one integer between them.

Note 2. The implementation of the above method can be very labor-intensive if the numbers being compared are very close to each other.
In this case, you can try comparison method of “subtracting one”" Let's demonstrate it with the following example.

12. Compare the numbers log 7 8 And log 6 7.

First way (subtract one).

Subtract 1 from the numbers being compared.

In the first inequality we used the fact that

if c > a > 1, then for b > 1 the inequality log a b > log c b is true.

In the second inequality – the monotonicity of the function y = log a x.

Second way (application of Cauchy's inequality).

13. Compare the numbers log 24 72 And log 12 18.

14. Compare the numbers log 20 80 And log 80 640.

Let log 2 5 = x. notice, that x > 0.

We get inequality.

Let us find many solutions to the inequality, satisfying the condition x > 0.

Let us construct both sides of the inequality squared (at x> 0 both sides of the inequality are positive). We have 9x 2< 9x + 28.

The set of solutions to the last inequality is the interval.

Considering that x> 0, we get: .

Answer: The inequality is true.

Problem solving workshop.

1. Compare the numbers:

2. Arrange the numbers in ascending order:

3. Solve the inequality 4 4 – 2 2 4+1 – 3< 0 . Is the number √2 solution to this inequality? (Answer:(–∞; log 2 3) ; number √2 is a solution to this inequality.)

Conclusion.

There are many methods for comparing logarithms. The purpose of the lessons on this topic is to teach one to navigate the variety of methods, to choose and apply the most rational method of solution in each specific situation.

In classes with in-depth study of mathematics, material on this topic can be presented in the form of a lecture. This form of educational activity presupposes that the lecture material must be carefully selected, worked out, and arranged in a certain logical sequence. The notes the teacher makes on the board must be thoughtful and mathematically accurate.

It is advisable to consolidate lecture material and practice problem-solving skills in practical lessons. The purpose of the workshop is not only to consolidate and test the acquired knowledge, but also to expand it. Therefore, tasks should contain tasks of different levels, from the simplest tasks to tasks of increased complexity. The teacher at such workshops acts as a consultant.

Literature.

  1. Galitsky M.L. and others. In-depth study of the course of algebra and mathematical analysis: Method. recommendations and teaching materials: A manual for teachers. – M.: Education, 1986.
  2. Ziv B.G., Goldich V.A. Didactic materials on algebra and basic analysis for grade 10. – St. Petersburg: “CheRo-on-Neva”, 2003.
  3. Litvinenko V.N., Mordkovich A.G. Workshop on elementary mathematics. Algebra. Trigonometry: Educational publication. – M.: Education, 1990.
  4. Ryazanovsky A.R. Algebra and the beginnings of analysis: 500 ways and methods of solving problems in mathematics for schoolchildren and those entering universities. – M.: Bustard, 2001.
  5. Sadovnichy Yu.V. Mathematics. Competition problems in algebra with solutions. Part 4. Logarithmic equations, inequalities, systems. Textbook.-3rd ed., ster.-M.: Publishing department of UNTsDO, 2003.
  6. Sharygin I.F., Golubev V.I. Optional course in mathematics: Problem solving: Proc. allowance for 11th grade. secondary school – M.: Prosveshchenie, 1991.

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Slide captions:

Properties of monotonicity of a logarithm. Comparison of logarithms. Algebra 11th grade. Completed by mathematics teacher: Liliya Anasovna Kinzyabulatova, Noyabrsk, 2014.

y= log a x , where a>0; a≠1. a) If a> 1, then y= log a x – increasing b) If 0

Methods for comparing logarithms. ① Monotonicity property Compare log a b log a c bases are a If a> 1, then y= log a t is increasing, then from b> c = > log a b > log a c ; If 0 c => log a b log 1/3 8;

Methods for comparing logarithms. ② Graphical method Compare log a b log with b bases are different, numbers are equal to b 1) If a> 1; с > 1, then y=log a t, y=log с t – age. a) If a> c, b>1, then log a b log c b

Methods for comparing logarithms. ② Graphical method Compare log a b log with b bases are different, numbers are equal to b 2) If 0 c, b>1, then log a b > log c b b) If a

Methods for comparing logarithms. ② Graphical method Compare log a b log with b bases are different, numbers are equal to b Examples log 2 3 > log 4 3 2 1 Log 3 1/4 0.25; 3>1 Log 0.3 0.6

Methods for comparing logarithms. ③ Functions of different monotonicity a>1 y=log a x – increases 0 1, then log a c > log b d b) If 0 1) Log 0.5 1/3 > log 5 1/2

Methods for comparing logarithms. ⑤ Evaluation method log 3 5 log 4 17 1 > > > >

Methods for comparing logarithms. ⑦ Comparison with the middle of the segment log 2 3 log 5 8 1 3/2 log 5 8 2* 3/2 2*log 5 8 2 log 5 64 log 2 8 log 5 64

Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called main properties.

You definitely need to know these rules - without them, not a single serious logarithmic problem can be solved. In addition, there are very few of them - you can learn everything in one day. So let's get started.

Adding and subtracting logarithms

Consider two logarithms with the same bases: log a x and log a y. Then they can be added and subtracted, and:

  1. log a x+ log a y=log a (x · y);
  2. log a x− log a y=log a (x : y).

So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to the logarithm of the quotient. Please note: the key point here is identical grounds. If the reasons are different, these rules do not work!

These formulas will help you calculate a logarithmic expression even when its individual parts are not considered (see lesson “What is a logarithm”). Take a look at the examples and see:

Log 6 4 + log 6 9.

Since logarithms have the same bases, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.

Task. Find the value of the expression: log 2 48 − log 2 3.

The bases are the same, we use the difference formula:
log 2 48 − log 2 3 = log 2 (48: 3) = log 2 16 = 4.

Task. Find the value of the expression: log 3 135 − log 3 5.

Again the bases are the same, so we have:
log 3 135 − log 3 5 = log 3 (135: 5) = log 3 27 = 3.

As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many tests are based on this fact. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.

Extracting the exponent from the logarithm

Now let's complicate the task a little. What if the base or argument of a logarithm is a power? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:

It is easy to see that the last rule follows the first two. But it’s better to remember it anyway - in some cases it will significantly reduce the amount of calculations.

Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x> 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. You can enter the numbers before the logarithm sign into the logarithm itself. This is what is most often required.

Task. Find the value of the expression: log 7 49 6 .

Let's get rid of the degree in the argument using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12

Task. Find the meaning of the expression:

[Caption for the picture]

Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2. We have:

[Caption for the picture]

I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator. We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator contain the same number: log 2 7. Since log 2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which is what was done. The result was the answer: 2.

Transition to a new foundation

Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the reasons are different? What if they are not exact powers of the same number?

Formulas for transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:

Let the logarithm log be given a x. Then for any number c such that c> 0 and c≠ 1, the equality is true:

[Caption for the picture]

In particular, if we put c = x, we get:

[Caption for the picture]

From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears in the denominator.

These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.

However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:

Task. Find the value of the expression: log 5 16 log 2 25.

Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2log 2 5;

Now let’s “reverse” the second logarithm:

[Caption for the picture]

Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.

Task. Find the value of the expression: log 9 100 lg 3.

The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:

[Caption for the picture]

Now let's get rid of the decimal logarithm by moving to a new base:

[Caption for the picture]

Basic logarithmic identity

Often in the solution process it is necessary to represent a number as a logarithm to a given base. In this case, the following formulas will help us:

In the first case, the number n becomes an indicator of the degree standing in the argument. Number n can be absolutely anything, because it’s just a logarithm value.

The second formula is actually a paraphrased definition. That’s what it’s called: the basic logarithmic identity.

In fact, what will happen if the number b raise to such a power that the number b to this power gives the number a? That's right: you get this same number a. Read this paragraph carefully again - many people get stuck on it.

Like formulas for moving to a new base, the basic logarithmic identity is sometimes the only possible solution.

Task. Find the meaning of the expression:

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Note that log 25 64 = log 5 8 - simply took the square from the base and argument of the logarithm. Taking into account the rules for multiplying powers with the same base, we get:

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If anyone doesn't know, this was a real task from the Unified State Exam :)

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that can hardly be called properties - rather, they are consequences of the definition of the logarithm. They constantly appear in problems and, surprisingly, create problems even for “advanced” students.

  1. log a a= 1 is a logarithmic unit. Remember once and for all: logarithm to any base a from this very base is equal to one.
  2. log a 1 = 0 is logarithmic zero. Base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a 0 = 1 is a direct consequence of the definition.

That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.