Examples of sum of logarithms with different bases. Solving Logarithmic Equations - Final Lesson

We are all familiar with equations from primary school. There we also learned to solve the simplest examples, and we must admit that they find their application even in higher mathematics. Everything is simple with equations, including quadratic equations. If you are having trouble with this topic, we highly recommend that you review it.

You've probably already gone through logarithms too. However, we consider it important to tell what it is for those who do not yet know. A logarithm is equated to the power to which the base must be raised to obtain the number to the right of the logarithm sign. Let's give an example based on which everything will become clear to you.

If you raise 3 to the fourth power, you get 81. Now substitute the numbers by analogy, and you will finally understand how logarithms are solved. Now all that remains is to combine the two concepts discussed. Initially, the situation seems extremely complicated, but upon closer examination the weight falls into place. We are sure that after this short article you will not have problems in this part of the Unified State Exam.

Today there are many ways to solve such structures. We will tell you about the simplest, most effective and most applicable in the case of Unified State Examination tasks. Solving logarithmic equations should start with the simplest example. The simplest logarithmic equations consist of a function and one variable in it.

It's important to note that x is inside the argument. A and b must be numbers. In this case, you can simply express the function in terms of a number to a power. It looks like this.

Of course, solving a logarithmic equation using this method will lead you to the correct answer. The problem for the vast majority of students in this case is that they do not understand what comes from where. As a result, you have to put up with mistakes and not get the desired points. The most offensive mistake will be if you mix up the letters. To solve the equation this way, you need to memorize this standard school formula because it is difficult to understand.

To make it easier, you can resort to another method - the canonical form. The idea is extremely simple. Turn your attention back to the problem. Remember that the letter a is a number, not a function or variable. A is not equal to one and greater than zero. There are no restrictions on b. Now, of all the formulas, let us remember one. B can be expressed as follows.

It follows from this that all original equations with logarithms can be represented in the form:

Now we can drop the logarithms. The result is a simple design, which we have already seen earlier.

The convenience of this formula lies in the fact that it can be used in a wide variety of cases, and not just for the simplest designs.

Don't worry about OOF!

Many experienced mathematicians will notice that we have not paid attention to the domain of definition. The rule boils down to the fact that F(x) is necessarily greater than 0. No, we did not miss this point. Now we are talking about another serious advantage of the canonical form.

There will be no extra roots here. If a variable will only appear in one place, then a scope is not necessary. It is done automatically. To verify this judgment, try solving several simple examples.

How to solve logarithmic equations with different bases

These are already complex logarithmic equations, and the approach to solving them must be special. Here it is rarely possible to limit ourselves to the notorious canonical form. Let's begin our detailed story. We have the following construction.

Pay attention to the fraction. It contains the logarithm. If you see this in a task, it’s worth remembering one interesting trick.

What does it mean? Each logarithm can be represented as the quotient of two logarithms with a convenient base. And this formula has a special case that is applicable with this example (we mean if c=b).

This is exactly the fraction we see in our example. Thus.

Essentially, we turned the fraction around and got a more convenient expression. Remember this algorithm!

Now it is necessary that the logarithmic equation does not contain different bases. Let's represent the base as a fraction.

In mathematics there is a rule based on which you can derive a degree from a base. The following construction results.

It would seem that what is preventing us from now turning our expression into the canonical form and simply solving it? Not so simple. There should be no fractions before the logarithm. Let's fix this situation! A fraction is allowed to be used as a degree.

Respectively.

If the bases are the same, we can remove the logarithms and equate the expressions themselves. This way the situation will become much simpler than it was. What will remain is an elementary equation that each of us knew how to solve back in 8th or even 7th grade. You can do the calculations yourself.

We have obtained the only true root of this logarithmic equation. Examples of solving a logarithmic equation are quite simple, aren't they? Now you will be able to independently deal with even the most complex tasks for preparing for and passing the Unified State Exam.

What's the result?

In the case of any logarithmic equations, we proceed from one very important rule. It is necessary to act in such a way as to reduce the expression to the simplest possible form. In this case, you will have a better chance of not only solving the task correctly, but also doing it in the simplest and most logical way possible. This is exactly how mathematicians always work.

We strongly do not recommend that you look for difficult paths, especially in this case. Remember a few simple rules that will allow you to transform any expression. For example, reduce two or three logarithms to the same base or derive a power from the base and win on this.

It is also worth remembering that solving logarithmic equations requires constant practice. Gradually you will move on to more and more complex structures, and this will lead you to confidently solving all variants of problems on the Unified State Exam. Prepare well in advance for your exams, and good luck!

Logarithm of the number b (b > 0) to base a (a > 0, a ≠ 1)– exponent to which the number a must be raised to obtain b.

The base 10 logarithm of b can be written as log(b), and the logarithm to base e (natural logarithm) is ln(b).

Often used when solving problems with logarithms:

Properties of logarithms

There are four main properties of logarithms.

Let a > 0, a ≠ 1, x > 0 and y > 0.

Property 1. Logarithm of the product

Logarithm of the product equal to the sum of logarithms:

log a (x ⋅ y) = log a x + log a y

Property 2. Logarithm of the quotient

Logarithm of the quotient equal to the difference of logarithms:

log a (x / y) = log a x – log a y

Property 3. Logarithm of power

Logarithm of degree equal to the product of the power and the logarithm:

If the base of the logarithm is in the degree, then another formula applies:

Property 4. Logarithm of the root

This property can be obtained from the property of the logarithm of a power, since the nth root of the power is equal to the power of 1/n:

Formula for converting from a logarithm in one base to a logarithm in another base

This formula is also often used when solving various tasks on logarithms:

Special case:

Comparing logarithms (inequalities)

Let us have 2 functions f(x) and g(x) under logarithms with the same bases and between them there is an inequality sign:

To compare them, you need to first look at the base of the logarithms a:

• If a > 0, then f(x) > g(x) > 0
• If 0< a < 1, то 0 < f(x) < g(x)

How to solve problems with logarithms: examples

Problems with logarithms included in the Unified State Exam in mathematics for grade 11 in task 5 and task 7, you can find tasks with solutions on our website in the appropriate sections. Also, tasks with logarithms are found in the math task bank. You can find all examples by searching the site.

What is a logarithm

Logarithms have always been considered a difficult topic in school mathematics courses. There are many different definitions of logarithm, but for some reason most textbooks use the most complex and unsuccessful of them.

We will define the logarithm simply and clearly. To do this, let's create a table:

So, we have powers of two.

Logarithms - properties, formulas, how to solve

If you take the number from the bottom line, you can easily find the power to which you will have to raise two to get this number. For example, to get 16, you need to raise two to the fourth power. And to get 64, you need to raise two to the sixth power. This can be seen from the table.

And now - actually, the definition of the logarithm:

the base a of the argument x is the power to which the number a must be raised to obtain the number x.

Designation: log a x = b, where a is the base, x is the argument, b is what the logarithm is actually equal to.

For example, 2 3 = 8 ⇒log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). With the same success, log 2 64 = 6, since 2 6 = 64.

The operation of finding the logarithm of a number to a given base is called. So, let's add a new line to our table:

Unfortunately, not all logarithms are calculated so easily. For example, try to find log 2 5. The number 5 is not in the table, but logic dictates that the logarithm will lie somewhere on the interval. Because 2 2< 5 < 2 3 , а чем больше степень двойки, тем больше получится число.

Such numbers are called irrational: the numbers after the decimal point can be written ad infinitum, and they are never repeated. If the logarithm turns out to be irrational, it is better to leave it that way: log 2 5, log 3 8, log 5 100.

It is important to understand that a logarithm is an expression with two variables (the base and the argument). At first, many people confuse where the basis is and where the argument is. To avoid annoying misunderstandings, just look at the picture:

Before us is nothing more than the definition of a logarithm. Remember: logarithm is a power, into which the base must be built in order to obtain an argument. It is the base that is raised to a power - it is highlighted in red in the picture. It turns out that the base is always at the bottom! I tell my students this wonderful rule at the very first lesson - and no confusion arises.

How to count logarithms

We've figured out the definition - all that remains is to learn how to count logarithms, i.e. get rid of the "log" sign. To begin with, we note that two important facts follow from the definition:

1. The argument and the base must always be greater than zero. This follows from the definition of a degree by a rational exponent, to which the definition of a logarithm is reduced.
2. The base must be different from one, since one to any degree still remains one. Because of this, the question “to what power must one be raised to get two” is meaningless. There is no such degree!

Such restrictions are called range of acceptable values(ODZ). It turns out that the ODZ of the logarithm looks like this: log a x = b ⇒x > 0, a > 0, a ≠ 1.

Note that there are no restrictions on the number b (the value of the logarithm). For example, the logarithm may well be negative: log 2 0.5 = −1, because 0.5 = 2 −1.

However, now we are considering only numerical expressions, where it is not required to know the VA of the logarithm. All restrictions have already been taken into account by the authors of the problems. But when logarithmic equations and inequalities come into play, DL requirements will become mandatory. After all, the basis and argument may contain very strong constructions that do not necessarily correspond to the above restrictions.

Now let's look at the general scheme for calculating logarithms. It consists of three steps:

1. Express the base a and the argument x as a power with the minimum possible base greater than one. Along the way, it’s better to get rid of decimals;
2. Solve the equation for variable b: x = a b ;
3. The resulting number b will be the answer.

That's all! If the logarithm turns out to be irrational, this will be visible already in the first step. The requirement that the base be greater than one is very important: this reduces the likelihood of error and greatly simplifies the calculations. It’s the same with decimal fractions: if you immediately convert them into ordinary ones, there will be many fewer errors.

Let's see how this scheme works using specific examples:

Task. Calculate the logarithm: log 5 25

1. Let's imagine the base and argument as a power of five: 5 = 5 1 ; 25 = 5 2 ;

Let's create and solve the equation:
log 5 25 = b ⇒(5 1) b = 5 2 ⇒5 b = 5 2 ⇒ b = 2;

2. We received the answer: 2.

Task. Calculate the logarithm:

Task. Calculate the logarithm: log 4 64

1. Let's imagine the base and argument as a power of two: 4 = 2 2 ; 64 = 2 6 ;
2. Let's create and solve the equation:
   log 4 64 = b ⇒(2 2) b = 2 6 ⇒2 2b = 2 6 ⇒2b = 6 ⇒ b = 3;
3. We received the answer: 3.

Task. Calculate the logarithm: log 16 1

1. Let's imagine the base and argument as a power of two: 16 = 2 4 ; 1 = 2 0 ;
2. Let's create and solve the equation:
   log 16 1 = b ⇒(2 4) b = 2 0 ⇒2 4b = 2 0 ⇒4b = 0 ⇒ b = 0;
3. We received the answer: 0.

Task. Calculate the logarithm: log 7 14

1. Let's imagine the base and argument as a power of seven: 7 = 7 1 ; 14 cannot be represented as a power of seven, since 7 1< 14 < 7 2 ;
2. From the previous paragraph it follows that the logarithm does not count;
3. The answer is no change: log 7 14.

A small note on the last example. How can you be sure that a number is not an exact power of another number? It’s very simple - just factor it into prime factors. If the expansion has at least two different factors, the number is not an exact power.

Task. Find out whether the numbers are exact powers: 8; 48; 81; 35; 14.

8 = 2 · 2 · 2 = 2 3 - exact degree, because there is only one multiplier;
48 = 6 · 8 = 3 · 2 · 2 · 2 · 2 = 3 · 2 4 - is not an exact power, since there are two factors: 3 and 2;
81 = 9 · 9 = 3 · 3 · 3 · 3 = 3 4 - exact degree;
35 = 7 · 5 - again not an exact power;
14 = 7 · 2 - again not an exact degree;

Note also that the prime numbers themselves are always exact powers of themselves.

Decimal logarithm

Some logarithms are so common that they have a special name and symbol.

of the argument x is the logarithm to base 10, i.e. The power to which the number 10 must be raised to obtain the number x. Designation: lg x.

For example, log 10 = 1; lg 100 = 2; lg 1000 = 3 - etc.

From now on, when a phrase like “Find lg 0.01” appears in a textbook, know that this is not a typo. This is a decimal logarithm. However, if you are unfamiliar with this notation, you can always rewrite it:
log x = log 10 x

Everything that is true for ordinary logarithms is also true for decimal logarithms.

Natural logarithm

There is another logarithm that has its own designation. In some ways, it's even more important than decimal. We are talking about the natural logarithm.

of the argument x is the logarithm to base e, i.e. the power to which the number e must be raised to obtain the number x. Designation: ln x.

Many people will ask: what is the number e? This is an irrational number; its exact value cannot be found and written down. I will give only the first figures:
e = 2.718281828459…

We will not go into detail about what this number is and why it is needed. Just remember that e is the base of the natural logarithm:
ln x = log e x

Thus ln e = 1; ln e 2 = 2; ln e 16 = 16 - etc. On the other hand, ln 2 is an irrational number. In general, the natural logarithm of any rational number is irrational. Except, of course, for one: ln 1 = 0.

For natural logarithms, all the rules that are true for ordinary logarithms are valid.

See also:

Logarithm. Properties of the logarithm (power of the logarithm).

How to represent a number as a logarithm?

We use the definition of logarithm.

A logarithm is an exponent to which the base must be raised to obtain the number under the logarithm sign.

Thus, in order to represent a certain number c as a logarithm to base a, you need to put a power with the same base as the base of the logarithm under the sign of the logarithm, and write this number c as the exponent:

Absolutely any number can be represented as a logarithm - positive, negative, integer, fractional, rational, irrational:

In order not to confuse a and c under stressful conditions of a test or exam, you can use the following memorization rule:

what is below goes down, what is above goes up.

For example, you need to represent the number 2 as a logarithm to base 3.

We have two numbers - 2 and 3. These numbers are the base and the exponent, which we will write under the sign of the logarithm. It remains to determine which of these numbers should be written down, to the base of the degree, and which – up, to the exponent.

The base 3 in the notation of a logarithm is at the bottom, which means that when we represent two as a logarithm to the base 3, we will also write 3 down to the base.

2 is higher than three. And in notation of the degree two we write above the three, that is, as an exponent:

Logarithms. First level.

Logarithms

Logarithm positive number b based on a, Where a > 0, a ≠ 1, is called the exponent to which the number must be raised a, To obtain b.

Definition of logarithm can be briefly written like this:

This equality is valid for b > 0, a > 0, a ≠ 1. It is usually called logarithmic identity.
The action of finding the logarithm of a number is called by logarithm.

Properties of logarithms:

Logarithm of the product:

Logarithm of the quotient:

Replacing the logarithm base:

Logarithm of degree:

Logarithm of the root:

Logarithm with power base:

Decimal and natural logarithms.

Decimal logarithm numbers call the logarithm of this number to base 10 and write   lg b
Natural logarithm numbers are called the logarithm of that number to the base e, Where e- an irrational number approximately equal to 2.7. At the same time they write ln b.

Other notes on algebra and geometry

Basic properties of logarithms

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: 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 the 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.

How to solve logarithms

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:

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

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:

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 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:

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. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
2. log a 1 = 0 is. The 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.

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:

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:

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:

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 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:

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.

With this video I begin a long series of lessons about logarithmic equations. Now you have three examples on the basis of which we will learn to solve the simplest problems, which are called - protozoa.

log 0.5 (3x − 1) = −3

log (x + 3) = 3 + 2 log 5

Let me remind you that the simplest logarithmic equation is the following:

log a f(x) = b

In this case, it is important that the variable x is present only inside the argument, that is, only in the function f (x). And the numbers a and b are just numbers, and in no case are functions containing the variable x.

Basic solution methods

There are many ways to solve such structures. For example, most teachers at school offer this method: Immediately express the function f (x) using the formula f ( x) = a b . That is, when you come across the simplest construction, you can immediately move on to the solution without additional actions and constructions.

Yes, of course, the decision will be correct. However, the problem with this formula is that most students do not understand, where it comes from and why we raise the letter a to the letter b.

As a result, I often see very annoying mistakes when, for example, these letters are swapped. This formula must either be understood or crammed, and the second method leads to mistakes at the most inopportune and most crucial moments: during exams, tests, etc.

That is why I suggest to all my students to abandon the standard school formula and use the second approach to solve logarithmic equations, which, as you probably guessed from the name, is called canonical form.

The idea of ​​the canonical form is simple. Let's look at our problem again: on the left we have log a, and by the letter a we mean a number, and in no case a function containing the variable x. Consequently, this letter is subject to all the restrictions that are imposed on the base of the logarithm. namely:

1 ≠ a > 0

On the other hand, from the same equation we see that the logarithm must be equal to the number b, and no restrictions are imposed on this letter, because it can take any value - both positive and negative. It all depends on what values ​​the function f(x) takes.

And here we remember our wonderful rule that any number b can be represented as a logarithm to the base a of a to the power of b:

b = log a a b

How to remember this formula? Yes, very simple. Let's write the following construction:

b = b 1 = b log a a

Of course, in this case all the restrictions that we wrote down at the beginning arise. Now let's use the basic property of the logarithm and introduce the multiplier b as the power of a. We get:

b = b 1 = b log a a = log a a b

As a result, the original equation will be rewritten as follows:

log a f (x) = log a a b → f (x) = a b

That's all. The new function no longer contains a logarithm and can be solved using standard algebraic techniques.

Of course, someone will now object: why was it necessary to come up with some kind of canonical formula at all, why perform two additional unnecessary steps if it was possible to immediately move from the original design to the final formula? Yes, if only because most students do not understand where this formula comes from and, as a result, regularly make mistakes when applying it.

But this sequence of actions, consisting of three steps, allows you to solve the original logarithmic equation, even if you do not understand where the final formula comes from. By the way, this entry is called the canonical formula:

log a f (x) = log a a b

The convenience of the canonical form also lies in the fact that it can be used to solve a very wide class of logarithmic equations, and not just the simplest ones that we are considering today.

Examples of solutions

Now let's look at real examples. So, let's decide:

log 0.5 (3x − 1) = −3

Let's rewrite it like this:

log 0.5 (3x − 1) = log 0.5 0.5 −3

Many students are in a hurry and try to immediately raise the number 0.5 to the power that came to us from the original problem. Indeed, when you are already well trained in solving such problems, you can immediately perform this step.

However, if you are now just starting to study this topic, it is better not to rush anywhere in order to avoid making offensive mistakes. So, we have the canonical form. We have:

3x − 1 = 0.5 −3

This is no longer a logarithmic equation, but linear with respect to the variable x. To solve it, let's first look at the number 0.5 to the power of −3. Note that 0.5 is 1/2.

(1/2) −3 = (2/1) 3 = 8

Convert all decimal fractions to common fractions when solving a logarithmic equation.

We rewrite and get:

3x − 1 = 8
3x = 9
x = 3

That's it, we got the answer. The first problem has been solved.

Second task

Let's move on to the second task:

As we see, this equation is no longer the simplest. If only because there is a difference on the left, and not a single logarithm to one base.

Therefore, we need to somehow get rid of this difference. In this case, everything is very simple. Let's take a closer look at the bases: on the left is the number under the root:

General recommendation: in all logarithmic equations, try to get rid of radicals, i.e., from entries with roots and move on to power functions, simply because the exponents of these powers are easily taken out of the sign of the logarithm and, ultimately, such an entry significantly simplifies and speeds up calculations. Let's write it down like this:

Now let us remember the remarkable property of the logarithm: powers can be derived from the argument, as well as from the base. In the case of grounds, the following happens:

log a k b = 1/k loga b

In other words, the number that was in the base power is brought forward and at the same time inverted, that is, it becomes a reciprocal number. In our case, the base degree was 1/2. Therefore, we can take it out as 2/1. We get:

5 2 log 5 x − log 5 x = 18
10 log 5 x − log 5 x = 18

Please note: under no circumstances should you get rid of logarithms at this step. Remember 4th-5th grade math and the order of operations: multiplication is performed first, and only then addition and subtraction. In this case, we subtract one of the same elements from 10 elements:

9 log 5 x = 18
log 5 x = 2

Now our equation looks as it should. This is the simplest construction, and we solve it using the canonical form:

log 5 x = log 5 5 2
x = 5 2
x = 25

That's all. The second problem has been solved.

Third example

Let's move on to the third task:

log (x + 3) = 3 + 2 log 5

Let me remind you of the following formula:

log b = log 10 b

If for some reason you are confused by the notation log b , then when performing all the calculations you can simply write log 10 b . You can work with decimal logarithms in the same way as with others: take powers, add and represent any numbers in the form lg 10.

It is these properties that we will now use to solve the problem, since it is not the simplest one that we wrote down at the very beginning of our lesson.

First, note that the factor 2 in front of lg 5 can be added and becomes a power of base 5. In addition, the free term 3 can also be represented as a logarithm - this is very easy to observe from our notation.

Judge for yourself: any number can be represented as log to base 10:

3 = log 10 10 3 = log 10 3

Let's rewrite the original problem taking into account the obtained changes:

log (x − 3) = log 1000 + log 25
log (x − 3) = log 1000 25
log (x − 3) = log 25,000

We have before us the canonical form again, and we got it without going through the transformation stage, i.e. the simplest logarithmic equation did not appear anywhere.

This is exactly what I talked about at the very beginning of the lesson. The canonical form allows you to solve a wider class of problems than the standard school formula that most school teachers give.

Well, that’s it, we get rid of the sign of the decimal logarithm, and we get a simple linear construction:

x + 3 = 25,000
x = 24,997

All! The problem is solved.

A note on scope

Here I would like to make an important remark regarding the scope of definition. Surely now there will be students and teachers who will say: “When we solve expressions with logarithms, we must remember that the argument f (x) must be greater than zero!” In this regard, a logical question arises: why did we not require this inequality to be satisfied in any of the problems considered?

Do not worry. In these cases, no extra roots will appear. And this is another great trick that allows you to speed up the solution. Just know that if in the problem the variable x occurs only in one place (or rather, in one single argument of a single logarithm), and nowhere else in our case does the variable x appear, then write down the domain of definition no need, because it will be executed automatically.

Judge for yourself: in the first equation we got that 3x − 1, i.e. the argument should be equal to 8. This automatically means that 3x − 1 will be greater than zero.

With the same success, we can write that in the second case x should be equal to 5 2, i.e. it is certainly greater than zero. And in the third case, where x + 3 = 25,000, i.e., again, obviously greater than zero. In other words, the scope is satisfied automatically, but only if x occurs only in the argument of only one logarithm.

That's all you need to know to solve the simplest problems. This rule alone, together with the transformation rules, will allow you to solve a very wide class of problems.

But let's be honest: in order to finally understand this technique, to learn how to apply the canonical form of the logarithmic equation, it is not enough to just watch one video lesson. Therefore, right now, download the options for independent solutions that are attached to this video lesson and start solving at least one of these two independent works.

It will take you literally a few minutes. But the effect of such training will be much higher than if you simply watched this video lesson.

I hope this lesson will help you understand logarithmic equations. Use the canonical form, simplify expressions using the rules for working with logarithms - and you won’t be afraid of any problems. That's all I have for today.

Taking into account the domain of definition

Now let's talk about the domain of definition of the logarithmic function, and how this affects the solution of logarithmic equations. Consider a construction of the form

log a f (x) = b

Such an expression is called the simplest - it contains only one function, and the numbers a and b are just numbers, and in no case a function that depends on the variable x. It can be solved very simply. You just need to use the formula:

b = log a a b

This formula is one of the key properties of the logarithm, and when substituting into our original expression we get the following:

log a f (x) = log a a b

f (x) = a b

This is a familiar formula from school textbooks. Many students will probably have a question: since in the original expression the function f (x) is under the log sign, the following restrictions are imposed on it:

f(x) > 0

This limitation applies because the logarithm of negative numbers does not exist. So, perhaps, as a result of this limitation, a check on answers should be introduced? Perhaps they need to be inserted into the source?

No, in the simplest logarithmic equations additional checking is unnecessary. And that's why. Take a look at our final formula:

f (x) = a b

The fact is that the number a is in any case greater than 0 - this requirement is also imposed by the logarithm. The number a is the base. In this case, no restrictions are imposed on the number b. But this doesn’t matter, because no matter what power we raise a positive number to, we will still get a positive number at the output. Thus, the requirement f (x) > 0 is satisfied automatically.

What's really worth checking is the domain of the function under the log sign. There may be quite complex structures, and you definitely need to keep an eye on them during the solution process. Let's get a look.

First task:

First step: convert the fraction on the right. We get:

We get rid of the logarithm sign and get the usual irrational equation:

Of the obtained roots, only the first one suits us, since the second root is less than zero. The only answer will be the number 9. That's it, the problem is solved. No additional checks are required to ensure that the expression under the logarithm sign is greater than 0, because it is not just greater than 0, but according to the condition of the equation it is equal to 2. Therefore, the requirement “greater than zero” is satisfied automatically.

Let's move on to the second task:

Everything is the same here. We rewrite the construction, replacing the triple:

We get rid of the logarithm signs and get an irrational equation:

We square both sides taking into account the restrictions and get:

4 − 6x − x 2 = (x − 4) 2

4 − 6x − x 2 = x 2 + 8x + 16

x 2 + 8x + 16 −4 + ​​6x + x 2 = 0

2x 2 + 14x + 12 = 0 |:2

x 2 + 7x + 6 = 0

We solve the resulting equation through the discriminant:

D = 49 − 24 = 25

x 1 = −1

x 2 = −6

But x = −6 does not suit us, because if we substitute this number into our inequality, we get:

−6 + 4 = −2 < 0

In our case, it is required that it be greater than 0 or, in extreme cases, equal. But x = −1 suits us:

−1 + 4 = 3 > 0

The only answer in our case will be x = −1. That's the solution. Let's go back to the very beginning of our calculations.

The main takeaway from this lesson is that you don't need to check constraints on a function in simple logarithmic equations. Because during the solution process all constraints are satisfied automatically.

However, this in no way means that you can forget about checking altogether. In the process of working on a logarithmic equation, it may well turn into an irrational one, which will have its own restrictions and requirements for the right side, which we have seen today in two different examples.

Feel free to solve such problems and be especially careful if there is a root in the argument.

Logarithmic equations with different bases

We continue to study logarithmic equations and look at two more quite interesting techniques with which it is fashionable to solve more complex constructions. But first, let’s remember how the simplest problems are solved:

log a f (x) = b

In this entry, a and b are numbers, and in the function f (x) the variable x must be present, and only there, that is, x must only be in the argument. We will transform such logarithmic equations using the canonical form. To do this, note that

b = log a a b

Moreover, a b is precisely an argument. Let's rewrite this expression as follows:

log a f (x) = log a a b

This is exactly what we are trying to achieve, so that there is a logarithm to base a on both the left and the right. In this case, we can, figuratively speaking, cross out the log signs, and from a mathematical point of view we can say that we are simply equating the arguments:

f (x) = a b

As a result, we will get a new expression that will be much easier to solve. Let's apply this rule to our problems today.

So, the first design:

First of all, I note that on the right is a fraction whose denominator is log. When you see an expression like this, it’s a good idea to remember a wonderful property of logarithms:

Translated into Russian, this means that any logarithm can be represented as the quotient of two logarithms with any base c. Of course 0< с ≠ 1.

So: this formula has one wonderful special case, when the variable c is equal to the variable b. In this case we get a construction like:

This is exactly the construction we see from the sign on the right in our equation. Let's replace this construction with log a b , we get:

In other words, in comparison with the original task, we swapped the argument and the base of the logarithm. Instead, we had to reverse the fraction.

We recall that any degree can be derived from the base according to the following rule:

In other words, the coefficient k, which is the power of the base, is expressed as an inverted fraction. Let's render it as an inverted fraction:

The fractional factor cannot be left in front, because in this case we will not be able to represent this notation as a canonical form (after all, in the canonical form there is no additional factor before the second logarithm). Therefore, let's add the fraction 1/4 to the argument as a power:

Now we equate arguments whose bases are the same (and our bases are really the same), and write:

x + 5 = 1

x = −4

That's all. We got the answer to the first logarithmic equation. Please note: in the original problem, the variable x appears in only one log, and it appears in its argument. Therefore, there is no need to check the domain, and our number x = −4 is indeed the answer.

Now let's move on to the second expression:

log 56 = log 2 log 2 7 − 3log (x + 4)

Here, in addition to the usual logarithms, we will have to work with log f (x). How to solve such an equation? To an unprepared student it may seem like this is some kind of tough task, but in fact everything can be solved in an elementary way.

Take a close look at the term lg 2 log 2 7. What can we say about it? The bases and arguments of log and lg are the same, and this should give some ideas. Let's remember once again how powers are taken out from under the sign of the logarithm:

log a b n = nlog a b

In other words, what was a power of b in the argument becomes a factor in front of log itself. Let's apply this formula to the expression lg 2 log 2 7. Don't be scared by lg 2 - this is the most common expression. You can rewrite it as follows:

All the rules that apply to any other logarithm are valid for it. In particular, the factor in front can be added to the degree of the argument. Let's write it down:

Very often, students do not see this action directly, because it is not good to enter one log under the sign of another. In fact, there is nothing criminal about this. Moreover, we get a formula that is easy to calculate if you remember an important rule:

This formula can be considered both as a definition and as one of its properties. In any case, if you are converting a logarithmic equation, you should know this formula just like you would know the log representation of any number.

Let's return to our task. We rewrite it taking into account the fact that the first term to the right of the equal sign will be simply equal to lg 7. We have:

lg 56 = lg 7 − 3lg (x + 4)

Let's move lg 7 to the left, we get:

lg 56 − lg 7 = −3lg (x + 4)

We subtract the expressions on the left because they have the same base:

lg (56/7) = −3lg (x + 4)

Now let's take a closer look at the equation we got. It is practically the canonical form, but there is a factor −3 on the right. Let's add it to the right lg argument:

log 8 = log (x + 4) −3

Before us is the canonical form of the logarithmic equation, so we cross out the lg signs and equate the arguments:

(x + 4) −3 = 8

x + 4 = 0.5

That's all! We solved the second logarithmic equation. In this case, no additional checks are required, because in the original problem x was present in only one argument.

Let me list the key points of this lesson again.

The main formula that is taught in all the lessons on this page dedicated to solving logarithmic equations is the canonical form. And don’t be scared by the fact that most school textbooks teach you to solve such problems differently. This tool works very effectively and allows you to solve a much wider class of problems than the simplest ones that we studied at the very beginning of our lesson.

In addition, to solve logarithmic equations it will be useful to know the basic properties. Namely:

1. The formula for moving to one base and the special case when we reverse log (this was very useful to us in the first problem);
2. Formula for adding and subtracting powers from the logarithm sign. Here, many students get stuck and do not see that the degree taken out and introduced can itself contain log f (x). Nothing wrong with that. We can introduce one log according to the sign of the other and at the same time significantly simplify the solution of the problem, which is what we observe in the second case.

In conclusion, I would like to add that it is not necessary to check the domain of definition in each of these cases, because everywhere the variable x is present in only one sign of log, and at the same time is in its argument. As a consequence, all requirements of the scope are fulfilled automatically.

Problems with variable base

Today we will look at logarithmic equations, which for many students seem non-standard, if not completely unsolvable. We are talking about expressions based not on numbers, but on variables and even functions. We will solve such constructions using our standard technique, namely through the canonical form.

First, let's remember how the simplest problems are solved, based on ordinary numbers. So, the simplest construction is called

log a f (x) = b

To solve such problems we can use the following formula:

b = log a a b

We rewrite our original expression and get:

log a f (x) = log a a b

Then we equate the arguments, i.e. we write:

f (x) = a b

Thus, we get rid of the log sign and solve the usual problem. In this case, the roots obtained from the solution will be the roots of the original logarithmic equation. In addition, a record when both the left and the right are in the same logarithm with the same base is called the canonical form. It is to such a record that we will try to reduce today's designs. So, let's go.

First task:

log x − 2 (2x 2 − 13x + 18) = 1

Replace 1 with log x − 2 (x − 2) 1 . The degree that we observe in the argument is actually the number b that stood to the right of the equal sign. Thus, let's rewrite our expression. We get:

log x − 2 (2x 2 − 13x + 18) = log x − 2 (x − 2)

What do we see? Before us is the canonical form of the logarithmic equation, so we can safely equate the arguments. We get:

2x 2 − 13x + 18 = x − 2

But the solution does not end there, because this equation is not equivalent to the original one. After all, the resulting construction consists of functions that are defined on the entire number line, and our original logarithms are not defined everywhere and not always.

Therefore, we must write down the domain of definition separately. Let's not split hairs and first write down all the requirements:

First, the argument of each of the logarithms must be greater than 0:

2x 2 − 13x + 18 > 0

x − 2 > 0

Secondly, the base must not only be greater than 0, but also different from 1:

x − 2 ≠ 1

As a result, we get the system:

But don’t be alarmed: when processing logarithmic equations, such a system can be significantly simplified.

Judge for yourself: on the one hand, we are required that the quadratic function be greater than zero, and on the other hand, this quadratic function is equated to a certain linear expression, which is also required that it be greater than zero.

In this case, if we require that x − 2 > 0, then the requirement 2x 2 − 13x + 18 > 0 will automatically be satisfied. Therefore, we can safely cross out the inequality containing the quadratic function. Thus, the number of expressions contained in our system will be reduced to three.

Of course, we could just as easily cross out the linear inequality, that is, cross out x − 2 > 0 and require that 2x 2 − 13x + 18 > 0. But you must agree that solving the simplest linear inequality is much faster and simpler, than quadratic, even under the condition that as a result of solving this entire system we get the same roots.

In general, try to optimize calculations whenever possible. And in the case of logarithmic equations, cross out the most difficult inequalities.

Let's rewrite our system:

Here is a system of three expressions, two of which we, in fact, have already dealt with. Let's write out the quadratic equation separately and solve it:

2x 2 − 14x + 20 = 0

x 2 − 7x + 10 = 0

Before us is a reduced quadratic trinomial and, therefore, we can use Vieta’s formulas. We get:

(x − 5)(x − 2) = 0

x 1 = 5

x 2 = 2

Now we return to our system and find that x = 2 does not suit us, because we are required that x be strictly greater than 2.

But x = 5 suits us perfectly: the number 5 is greater than 2, and at the same time 5 is not equal to 3. Therefore, the only solution to this system will be x = 5.

That's it, the problem is solved, including taking into account the ODZ. Let's move on to the second equation. More interesting and informative calculations await us here:

The first step: like last time, we bring this whole matter to canonical form. To do this, we can write the number 9 as follows:

You don’t have to touch the base with the root, but it’s better to transform the argument. Let's move from the root to the power with a rational exponent. Let's write down:

Let me not rewrite our entire large logarithmic equation, but just immediately equate the arguments:

x 3 + 10x 2 + 31x + 30 = x 3 + 9x 2 + 27x + 27

x 2 + 4x + 3 = 0

Before us is a newly reduced quadratic trinomial, let’s use Vieta’s formulas and write:

(x + 3)(x + 1) = 0

x 1 = −3

x 2 = −1

So, we got the roots, but no one guaranteed us that they would fit the original logarithmic equation. After all, the log signs impose additional restrictions (here we should have written down the system, but due to the cumbersome nature of the whole structure, I decided to calculate the domain of definition separately).

First of all, remember that the arguments must be greater than 0, namely:

These are the requirements imposed by the scope of definition.

Let us immediately note that since we equate the first two expressions of the system to each other, we can cross out any of them. Let's cross out the first one because it looks more threatening than the second one.

In addition, note that the solution to the second and third inequalities will be the same sets (the cube of some number is greater than zero, if this number itself is greater than zero; similarly, with a root of the third degree - these inequalities are completely analogous, so we can cross it out).

But with the third inequality this will not work. Let's get rid of the radical sign on the left by raising both parts to a cube. We get:

So we get the following requirements:

− 2 ≠ x > −3

Which of our roots: x 1 = −3 or x 2 = −1 meets these requirements? Obviously, only x = −1, because x = −3 does not satisfy the first inequality (since our inequality is strict). So, returning to our problem, we get one root: x = −1. That's it, problem solved.

Once again, the key points of this task:

1. Feel free to apply and solve logarithmic equations using canonical form. Students who make such a notation, rather than moving directly from the original problem to a construction like log a f (x) = b, make much fewer errors than those who rush somewhere, skipping intermediate steps of calculations;
2. As soon as a variable base appears in a logarithm, the problem ceases to be the simplest. Therefore, when solving it, it is necessary to take into account the domain of definition: the arguments must be greater than zero, and the bases must not only be greater than 0, but they also must not be equal to 1.

The final requirements can be applied to the final answers in different ways. For example, you can solve an entire system containing all the requirements for the domain of definition. On the other hand, you can first solve the problem itself, and then remember the domain of definition, separately work it out in the form of a system and apply it to the resulting roots.

Which method to choose when solving a particular logarithmic equation is up to you. In any case, the answer will be the same.

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

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