How to compare logarithms with different bases. Comparison of numbers. The Comprehensive Guide (2019). Logarithms. First level

main properties.

1. logax + logay = loga(x y);
2. logax − logay = loga (x: y).

identical grounds

Log6 4 + log6 9.

Now let's complicate the task a little.

Examples of solving logarithms

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:

Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x >

Task. Find the meaning of the expression:

Transition to a new foundation

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

Task. Find the meaning of the expression:

See also:

Basic properties of the logarithm

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is equal to 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy.

Basic properties of logarithms

Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.

Examples for logarithms

Logarithm expressions

Example 1.
A). x=10ac^2 (a>0,c>0).

Using properties 3.5 we calculate

2.

3.

4. Where .

Example 2. Find x if

Example 3. Let the value of logarithms be given

Calculate log(x) if

Basic properties of logarithms

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 problem can be solved. logarithmic problem. 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: logax and logay. Then they can be added and subtracted, and:

1. logax + logay = loga(x y);
2. logax − logay = loga (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 the lesson “What is a logarithm”). Take a look at the examples and see:

Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

Task. Find the value of the expression: log2 48 − log2 3.

The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 − log3 5.

Again the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 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 are built on this fact test papers. 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

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: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

Task. Find the meaning of the expression:

Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 24; 49 = 72. We have:

I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator.

Logarithm formulas. Logarithms examples solutions.

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: log2 7. Since log2 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 logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

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

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: log5 16 log2 25.

Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let’s “reverse” the second logarithm:

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: log9 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:

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

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 the exponent in the argument. The number n can be absolutely anything, because it is just a logarithm value.

The second formula is actually a paraphrased definition. That's what it's called: .

In fact, what happens if the number b is raised to such a power that the number b to this power gives the number a? That's right: the result is the 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:

Note that log25 64 = log5 8 - simply took the square from the base and argument of the logarithm. Considering the rules for multiplying powers with the same basis, we get:

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. logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
2. loga 1 = 0 is. The base a can be anything, but if the argument contains one - logarithm equal to zero! Because a0 = 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.

See also:

The logarithm of b to base a denotes the expression. To calculate the logarithm means to find a power x () at which the equality is satisfied

Basic properties of the logarithm

It is necessary to know the above properties, since almost all problems and examples related to logarithms are solved on their basis. The rest of the exotic properties can be derived through mathematical manipulations with these formulas

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

When calculating the formula for the sum and difference of logarithms (3.4) you come across quite often. The rest are somewhat complex, but in a number of tasks they are indispensable for simplifying complex expressions and calculating their values.

Common cases of logarithms

Some of the common logarithms are those in which the base is even ten, exponential or two.
The logarithm to base ten is usually called the decimal logarithm and is simply denoted by lg(x).

It is clear from the recording that the basics are not written in the recording. For example

A natural logarithm is a logarithm whose base is an exponent (denoted by ln(x)).

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is equal to 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy. Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.

And another important logarithm to base two is denoted by

The derivative of the logarithm of a function is equal to one divided by the variable

The integral or antiderivative logarithm is determined by the relationship

The given material is enough for you to solve a wide class of problems related to logarithms and logarithms. To help you understand the material, I will give just a few common examples from school curriculum and universities.

Examples for logarithms

Logarithm expressions

Example 1.
A). x=10ac^2 (a>0,c>0).

Using properties 3.5 we calculate

2.
By the property of difference of logarithms we have

3.
Using properties 3.5 we find

4. Where .

By the look complex expression using a number of rules is simplified to form

Finding logarithm values

Example 2. Find x if

Solution. For calculation, we apply to the last term 5 and 13 properties

We put it on record and mourn

Since the bases are equal, we equate the expressions

Logarithms. First level.

Let the value of logarithms be given

Calculate log(x) if

Solution: Let's take a logarithm of the variable to write the logarithm through the sum of its terms

This is just the beginning of our acquaintance with logarithms and their properties. Practice calculations, enrich your practical skills - you will soon need the knowledge you gain to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge for another no less important topic- logarithmic inequalities...

Basic properties of logarithms

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: logax and logay. Then they can be added and subtracted, and:

1. logax + logay = loga(x y);
2. logax − logay = loga (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 the lesson “What is a logarithm”). Take a look at the examples and see:

Task. Find the value of the expression: log6 4 + log6 9.

Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

Task. Find the value of the expression: log2 48 − log2 3.

The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 − log3 5.

Again the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 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.

How to solve logarithms

This is what is most often required.

Task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

Task. Find the meaning of the expression:

Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 24; 49 = 72. We have:

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: log2 7. Since log2 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 logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

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

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: log5 16 log2 25.

Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let’s “reverse” the second logarithm:

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: log9 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:

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

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 the exponent in the argument. The number n can be absolutely anything, because it is just a logarithm value.

The second formula is actually a paraphrased definition. That's what it's called: .

In fact, what happens if the number b is raised to such a power that the number b to this power gives the number a? That's right: the result is the 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:

Note that log25 64 = log5 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:

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. logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
2. loga 1 = 0 is. The base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a0 = 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.

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 classes with in-depth study mathematics, this material can be used for a two-hour lesson-lecture.

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 method “multiplying 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 show it in 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.

To use presentation previews, create a Google account and log in to it: https://accounts.google.com

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 Graphic method Compare log a b log with b different bases, numbers 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:

[Caption for the picture]

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:

[Caption for the picture]

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.