Elementary logarithmic equations. Equations that are square relative to the logarithm and other non-standard tricks

Logarithmic equation is called an equation in which the unknown (x) and expressions with it are under the sign of a logarithmic function. Solving logarithmic equations assumes that you are already familiar with and.
How to solve logarithmic equations?

The simplest equation is log a x \u003d b, where a and b are some numbers, x is unknown.
By solving the logarithmic equation is x \u003d a b provided: a\u003e 0, a 1.

It should be noted that if x is somewhere outside the logarithm, for example log 2 x \u003d x-2, then such an equation is already called mixed and a special approach is needed to solve it.

The ideal case is a situation when you come across an equation in which only numbers are under the sign of the logarithm, for example x + 2 \u003d log 2 2. Here it is enough to know the properties of logarithms to solve it. But this kind of luck doesn't happen often, so get ready for more difficult things.

But first, after all, let's start with simple equations. To solve them, it is desirable to have the most general understanding of the logarithm.

Solving the simplest logarithmic equations

These include equations like log 2 x \u003d log 2 16. The naked eye can see that dropping the sign of the logarithm, we get x \u003d 16.

In order to solve a more complex logarithmic equation, it is usually reduced to solving an ordinary algebraic equation or to solving the simplest logarithmic equation log a x \u003d b. In the simplest equations, this happens in one motion, which is why they are called the simplest ones.

The above method of lowering the logarithms is one of the main ways to solve logarithmic equations and inequalities. In mathematics, this operation is called potentiation. There are certain rules or restrictions for this kind of operations:

• identical numerical bases for logarithms
• logarithms in both sides of the equation are found freely, i.e. without any coefficients and other various kinds of expressions.

Let's say in the equation log 2 x \u003d 2log 2 (1-x) potentiation is not applicable - the coefficient 2 on the right does not allow. In the following example, log 2 x + log 2 (1 - x) \u003d log 2 (1 + x) also fails one of the constraints - on the left there are two logarithms. That would be one - a completely different matter!

In general, you can remove logarithms only if the equation looks like:

log a (...) \u003d log a (...)

Absolutely any expressions can be found in brackets; this has absolutely no effect on the potentiation operation. And after the elimination of logarithms, a simpler equation will remain - linear, quadratic, exponential, etc., which I hope you already know how to solve.

Let's take another example:

log 3 (2x-5) \u003d log 3 x

We apply potentiation, we get:

log 3 (2x-1) \u003d 2

Based on the definition of the logarithm, namely that the logarithm is the number to which the base must be raised in order to obtain an expression that is under the sign of the logarithm, i.e. (4x-1), we get:

We got a nice answer again. Here we have dispensed with the elimination of logarithms, but potentiation is applicable here, because a logarithm can be made from any number, and exactly the one that we need. This method is very helpful in solving logarithmic equations and especially inequalities.

Let's solve our logarithmic equation log 3 (2x-1) \u003d 2 using potentiation:

Let's represent the number 2 as a logarithm, for example, such log 3 9, because 3 2 \u003d 9.

Then log 3 (2x-1) \u003d log 3 9 and again we get the same equation 2x-1 \u003d 9. I hope everything is clear.

So we examined how to solve the simplest logarithmic equations, which are actually very important, because solving logarithmic equations, even the most terrible and twisted, in the end always comes down to solving the simplest equations.

In everything that we did above, we lost sight of one very important point, which in the future will have a decisive role. The fact is that the solution of any logarithmic equation, even the most elementary one, consists of two equivalent parts. The first is the solution of the equation itself, the second is the work with the range of permissible values \u200b\u200b(ADV). That's just the first part we have mastered. In the above examples, the DHS does not affect the answer in any way, so we did not consider it.

Let's take another example:

log 3 (x 2 -3) \u003d log 3 (2x)

Outwardly, this equation is no different from the elementary one, which is very successfully solved. But it is not so. No, we will, of course, solve it, but most likely it will be wrong, because there is a small ambush in it, into which both C and A students immediately come across. Let's take a closer look at it.

Let's say you need to find the root of an equation or the sum of roots, if there are several:

log 3 (x 2 -3) \u003d log 3 (2x)

We use potentiation, here it is permissible. As a result, we get the usual quadratic equation.

Find the roots of the equation:

It turned out two roots.

Answer: 3 and -1

At first glance, everything is correct. But let's check the result and plug it into the original equation.

Let's start with x 1 \u003d 3:

log 3 6 \u003d log 3 6

The check was successful, now the queue x 2 \u003d -1:

log 3 (-2) \u003d log 3 (-2)

So stop! Outwardly, everything is perfect. One point - there are no logarithms of negative numbers! This means that the root x \u003d -1 is not suitable for solving our equation. And therefore the correct answer is 3, not 2, as we wrote.

It was here that the ODZ played its fatal role, which we forgot about.

Let me remind you that under the range of valid values, such values \u200b\u200bof x are accepted that are allowed or make sense for the original example.

Without ODZ, any solution, even absolutely correct, of any equation turns into a lottery - 50/50.

How could we get caught while solving a seemingly elementary example? But exactly at the moment of potentiation. Logarithms disappeared, and with them all restrictions.

What, then, to do? Refuse to eliminate logarithms? And completely refuse to solve this equation?

No, we just, like real heroes from one famous song, will go around!

Before proceeding with the solution of any logarithmic equation, we will write down the ODZ. But after that you can do whatever your heart desires with our equation. Having received the answer, we simply throw away those roots that are not included in our LDZ, and write down the final version.

Now let's decide how to write the ODZ. To do this, we carefully examine the original equation and look for suspicious places in it, such as division by x, an even root, etc. Until we solve the equation, we do not know what x equals, but we firmly know that such x, which, when substituted, will give division by 0 or taking the square root of a negative number, will certainly not work in the answer. Therefore, such x are unacceptable, while the rest will constitute the ODZ.

Let's use the same equation again:

log 3 (x 2 -3) \u003d log 3 (2x)

log 3 (x 2 -3) \u003d log 3 (2x)

As you can see, there is no division by 0, there are no square roots either, but there are expressions with x in the body of the logarithm. We immediately recall that the expression inside the logarithm must always be\u003e 0. We write this condition in the form of an ODZ:

Those. we haven’t decided anything yet, but we have already written down a prerequisite for the entire sub-logarithmic expression. The curly brace means that these conditions must be met simultaneously.

ODZ is written down, but it is also necessary to solve the resulting system of inequalities, which is what we will do. We get the answer x\u003e v3. Now we know for sure which x won't work for us. And then we are already starting to solve the logarithmic equation itself, which we did above.

Having received the answers x 1 \u003d 3 and x 2 \u003d -1, it is easy to see that only x1 \u003d 3 is suitable for us, and we write it down as the final answer.

For the future, it is very important to remember the following: we solve any logarithmic equation in 2 stages. The first one - we solve the equation itself, the second one - we solve the condition of ODD. Both stages are performed independently of each other and are compared only when writing an answer, i.e. we discard all unnecessary and write down the correct answer.

To consolidate the material, we strongly recommend watching the video:

On the video, there are other examples of the log. equations and working out the method of intervals in practice.

On this question, how to solve logarithmic equations, for now. If something is decided by the log. equations remained unclear or incomprehensible, write your questions in the comments.

Note: The Academy of Social Education (KSUI) is ready to accept new students.

Examples:

\\ (\\ log_ (2) (\u2061x) \u003d 32 \\)
\\ (\\ log_3\u2061x \u003d \\ log_3\u20619 \\)
\\ (\\ log_3\u2061 ((x ^ 2-3)) \u003d \\ log_3\u2061 ((2x)) \\)
\\ (\\ log_ (x + 1) ((x ^ 2 + 3x-7)) \u003d 2 \\)
\\ (\\ lg ^ 2\u2061 ((x + 1)) + 10 \u003d 11 \\ lg\u2061 ((x + 1)) \\)

How to solve logarithmic equations:

When solving a logarithmic equation, you need to strive to transform it to the form \\ (\\ log_a\u2061 (f (x)) \u003d \\ log_a\u2061 (g (x)) \\), then make the transition to \\ (f (x) \u003d g (x) \\).

\\ (\\ log_a\u2061 (f (x)) \u003d \\ log_a\u2061 (g (x)) \\) \\ (⇒ \\) \\ (f (x) \u003d g (x) \\).

Example: \\ (\\ log_2\u2061 (x-2) \u003d 3 \\)

Decision: \\ (\\ log_2\u2061 (x-2) \u003d \\ log_2\u20618 \\) \\ (x-2 \u003d 8 \\) \\ (x \u003d 10 \\) Verification: \\ (10\u003e 2 \\) - suitable for ODZ Answer: \\ (x \u003d 10 \\)

ODZ: \\ (x-2\u003e 0 \\) \\ (x\u003e 2 \\)

Very important! This transition can be done only if:

You wrote for the original equation, and at the end check if the ones found are included in the ODU. If this is not done, unnecessary roots may appear, which means the wrong decision.

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

The logarithms on the left and right are "pure", that is, there should be no multiplications, divisions, etc. - only lone logarithms on either side of the equal sign.

For example:

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

Example ... Solve the equation \\ (2 \\ log_8\u2061x \u003d \\ log_8\u20612,5 + \\ log_8\u206110 \\)

Decision :

		We write ODZ: \\ (x\u003e 0 \\).
\\ (2 \\ log_8\u2061x \u003d \\ log_8\u20612,5 + \\ log_8\u206110 \\) ODZ: \\ (x\u003e 0 \\)		On the left in front of the logarithm is the coefficient, on the right is the sum of the logarithms. This disturbs us. We transfer two to the exponent \\ (x \\) by the property: \\ (n \\ log_b (\u2061a) \u003d \\ log_b\u2061 (a ^ n) \\). We represent the sum of the logarithms as one logarithm by the property: \\ (\\ log_a\u2061b + \\ log_a\u2061c \u003d \\ log_a (\u2061bc) \\)
\\ (\\ log_8\u2061 (x ^ 2) \u003d \\ log_8\u206125 \\)		We brought the equation to the form \\ (\\ log_a\u2061 (f (x)) \u003d \\ log_a\u2061 (g (x)) \\) and wrote down the ODZ, so you can go to the form \\ (f (x) \u003d g (x) \\ Happened . We solve it and get the roots.
		\\ (x_1 \u003d 5 \\) \\ (x_2 \u003d -5 \\)
We check if the roots are suitable for ODZ. To do this, in \\ (x\u003e 0 \\) instead of \\ (x \\) we substitute \\ (5 \\) and \\ (- 5 \\). This operation can be performed orally.		The first inequality is true, the second is not. So \\ (5 \\) is the root of the equation, but \\ (- 5 \\) is not. We write down the answer.
\(5>0\), \(-5>0\)		Answer

: Solve the equation \\ (\\ log ^ 2_2\u2061 (x) -3 \\ log_2 (\u2061x) + 2 \u003d 0 \\) : \(5\)

Example \\ (\\ log ^ 2_2\u2061 (x) -3 \\ log_2 (\u2061x) + 2 \u003d 0 \\) ODZ: \\ (x\u003e 0 \\)

Decision :

		We write ODZ: \\ (x\u003e 0 \\).
A typical equation solved with. Replace \\ (\\ log_2\u2061x \\) with \\ (t \\).		\\ (t \u003d \\ log_2\u2061x \\)
We got the usual. We are looking for its roots.
		\\ (t_1 \u003d 2 \\) \\ (t_2 \u003d 1 \\)
We do the reverse replacement		\\ (\\ log_2 (\u2061x) \u003d 2 \\) \\ (\\ log_2 (\u2061x) \u003d 1 \\)
Transform the right-hand sides, representing them as logarithms: \\ (2 \u003d 2 \\ cdot 1 \u003d 2 \\ log_2\u20612 \u003d \\ log_2\u20614 \\) and \\ (1 \u003d \\ log_2\u20612 \\)		\\ (\\ log_2 (\u2061x) \u003d \\ log_2\u20614 \\) \\ (\\ log_2 (\u2061x) \u003d \\ log_2\u20612 \\)
Now our equations are of the form \\ (\\ log_a\u2061 (f (x)) \u003d \\ log_a\u2061 (g (x)) \\) and we can jump to \\ (f (x) \u003d g (x) \\).		\\ (x_1 \u003d 4 \\) \\ (x_2 \u003d 2 \\)
We check the correspondence of the roots of the ODZ. To do this, we substitute \\ (4 \\) and \\ (2 \\) in the inequality \\ (x\u003e 0 \\) instead of \\ (x \\).		Both inequalities are true. Hence both \\ (4 \\) and \\ (2 \\) are roots of the equation.
\(4>0\) \(2>0\)		We are all familiar with equations from elementary school. There we also learned to solve the simplest examples, and we must admit that they find their application even in higher mathematics. It's simple with equations, including square ones. If you have problems with this theme, we strongly recommend that you repeat it.

: Solve the equation \\ (\\ log ^ 2_2\u2061 (x) -3 \\ log_2 (\u2061x) + 2 \u003d 0 \\) : \(4\); \(2\).

You probably already passed the logarithms. Nevertheless, we consider it important to tell what it is for those who do not know yet. The logarithm is equated to the degree to which the base needs to be raised to get the number to the right of the logarithm sign. Let's give an example, based on which, everything will become clear to you.

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

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

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

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

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

We can now drop the logarithms. The result is a simple construction that we saw earlier.

The convenience of this formula lies in the fact that it can be used in a variety of cases, and not only 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 is reduced to the fact that F (x) is necessarily greater than 0. No, we did not miss this moment. Now we are talking about another major advantage of the canonical form.

No extra roots will arise here. If the variable will only appear in one place, then the scope is not necessary. It runs automatically. To verify this judgment, consider solving a few simple examples.

How to solve logarithmic equations with different bases

These are already complex logarithmic equations, and the approach to their solution should be special. It rarely turns out to be limited to the notorious canonical form. Let's start our detailed story. We have the following design.

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

What does it mean? Each logarithm can be represented as a quotient of two logarithms with a convenient base. And this formula has a special case that is applicable with this example (meaning, if c \u003d b).

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

In fact, they turned the fraction over and got a more convenient expression. Remember this algorithm!

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

In mathematics, there is a rule based on which you can take a degree from the base. The following construction turns out.

It would seem, what prevents now from turning our expression into a canonical form and solving it in an elementary way? Not so simple. There should be no fractions before the logarithm. We fix this situation! The fraction is allowed to be carried out as a degree.

Respectively.

If the bases are the same, we can remove the logarithms and equate the expressions themselves. So the situation will become much easier than it was. There will remain an elementary equation, which each of us was able to solve in 8th or even 7th grade. You can make the calculations yourself.

We got the only true root of this logarithmic equation. Examples of solving a logarithmic equation are pretty simple, aren't they? Now you will be able to independently figure out even the most difficult tasks for preparing and passing the exam.

What's the bottom line?

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 bring the expression to the most simple form. In this case, you will have more chances not only to solve the task correctly, but also to make it as simple and logical as possible. This is how mathematicians always do.

We strongly discourage you from looking for difficult paths, especially in this case. Remember a few simple rules that will allow you to transform any expression. For example, bring two or three logarithms to one base, or derive a degree from the base and win on that.

It is also worth remembering that you need to constantly train in solving logarithmic equations. Gradually, you will move on to more and more complex designs, and this will lead you to confidently solving all variants of problems on the exam. Prepare for your exams well in advance, and good luck!

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

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

Definition:

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

Recall basic logarithmic identity.

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

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

For example:

Let's recall the main properties of the logarithmic function.

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

Figure: 1. Graph of the logarithmic function at various bases

The function graph for is shown in black. Figure: 1. If the argument increases from zero to infinity, the function increases from minus to plus infinity.

The function graph for is shown in red. Figure: 1.

Properties of this function:

Domain: ;

Range of values:;

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

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

Consider the simplest logarithmic equation, all other logarithmic equations, as a rule, are reduced to this form.

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

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

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

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

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

Equalize the bases of logarithms;

Equate sub-logarithmic functions;

Check.

Let's look at specific examples.

Example 1 - Solve the equation:

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

Example 2 - Solve the equation:

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

Find the root and substitute it into the inequality:

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

Example 3 - Solve the equation:

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

Find the root and substitute it into the inequality:

Obviously, only the first root satisfies the ODV.

Introduction

Logarithms were invented to speed up and simplify calculations. The idea of \u200b\u200bthe logarithm, that is, the idea of \u200b\u200bexpressing numbers as a power of the same base, belongs to Mikhail Stifel. But at the time of Stiefel, mathematics was not so developed and the idea of \u200b\u200bthe logarithm did not find its development. Logarithms were later invented simultaneously and independently of each other by the Scottish scientist John Napier (1550-1617) and the Swiss Jobst Burgi (1552-1632). Napier was the first to publish his work in 1614. under the title "Description of the amazing table of logarithms", Napier's theory of logarithms was given in a fairly complete volume, the method for calculating logarithms was given the simplest, therefore Napier's contribution to the invention of logarithms is greater than that of Burghi. Burghi worked on the tables at the same time as Napier, but kept them secret for a long time and published them only in 1620. Napier mastered the idea of \u200b\u200bthe logarithm around 1594. although the tables were published after 20 years. At first, he called his logarithms "artificial numbers" and only then suggested that these "artificial numbers" be called in one word "logarithm", which translated from Greek means "related numbers", taken one from an arithmetic progression, and the other from a specially selected geometric progress. The first tables in Russian were published in 1703. with the participation of a wonderful teacher of the 18th century. L. F Magnitsky. In the development of the theory of logarithms, the works of St. Petersburg academician Leonard Euler were of great importance. He was the first to consider logarithm as the inverse of raising to a power, he introduced the terms "base of the logarithm" and "mantissa" Briggs compiled tables of logarithms with base 10. Decimal tables are more convenient for practical use, their theory is simpler than that of Napier's logarithms ... Therefore, decimal logarithms are sometimes called brigs logarithms. The term "characteristic" was introduced by Briggs.

In those distant times, when sages first began to think about equalities containing unknown quantities, there probably were no coins or wallets yet. But on the other hand, there were heaps, as well as pots, baskets, which perfectly suited the role of caches-storage, containing an unknown number of items. In the ancient mathematical problems of Mesopotamia, India, China, Greece, unknown values \u200b\u200bexpressed the number of peacocks in the garden, the number of bulls in the herd, the totality of things taken into account when dividing property. Scribes, officials well trained in the science of counting, and priests initiated into secret knowledge were quite successful in coping with such tasks.

Sources that have come down to us testify that ancient scientists possessed some general methods of solving problems with unknown quantities. However, not a single papyrus or a single clay tablet contains a description of these techniques. The authors only occasionally supplied their numerical calculations with scanty comments like: "Look!", "Do this!", "You found it right." In this sense, the exception is the "Arithmetic" of the Greek mathematician Diophantus of Alexandria (III century) - a collection of problems for drawing up equations with a systematic presentation of their solutions.

However, the first widely known guide to solving problems was the work of a Baghdad scholar of the 9th century. Mohammed bin Musa al-Khwarizmi. The word "al-jabr" from the Arabic name of this treatise - "Kitab al-jerber wal-muqabala" ("The Book of Restoration and Opposition") - eventually turned into the well-known word "algebra", and al-Khwarizmi's work itself served starting point in the formation of the science of solving equations.

Logarithmic equations and inequalities

1. Logarithmic equations

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

The simplest logarithmic equation is an equation of the form

log a x = b . (1)

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

Example 1. Solve the equations:

a) log 2 x \u003d 3, b) log 3 x \u003d -1, c)

Decision. Using Statement 1, we obtain a) x \u003d 2 3 or x \u003d 8; b) x \u003d 3 -1 or x \u003d 1/3; c)

or x = 1.

Here are the main properties of the logarithm.

P1. Basic logarithmic identity:

where a > 0, a ≠ 1 and b > 0.

P2. The logarithm of the product of positive factors is equal to the sum of the logarithms of these factors:

log a N 1 · N 2 \u003d log a N 1 + log a N 2 (a > 0, a ≠ 1, N 1 > 0, N 2 > 0).

Comment. If a N 1 · N 2\u003e 0, then property P2 takes the form

log a N 1 · N 2 \u003d log a |N 1 | + log a |N 2 | (a > 0, a ≠ 1, N 1 · N 2 > 0).

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

(a > 0, a ≠ 1, N 1 > 0, N 2 > 0).

Comment. If a

, (which is equivalent to N 1 N 2\u003e 0) then property P3 takes the form (a > 0, a ≠ 1, N 1 N 2 > 0).

P4. The logarithm of the power of a positive number is equal to the product of the exponent by the logarithm of this number:

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

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

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

P5. The formula for the transition to another base:

(a > 0, a ≠ 1, b > 0, b ≠ 1, N > 0),

in particular if N = b , we get

(a > 0, a ≠ 1, b > 0, b ≠ 1). (2)

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

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

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

(b > 0, a ≠ 0, |a | ≠ 1). (6)

We also list the main properties of the logarithmic function f (x) \u003d log a x :

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

2. The range of values \u200b\u200bof a logarithmic function is a set of real numbers.

3. When a \u003e 1 the logarithmic function is strictly increasing (0< x 1 < x 2 log a x 1 < log a x 2), and at 0< a < 1, - строго убывает (0 < x 1 < x 2 log a x 1\u003e log a x 2).

4.log a 1 \u003d 0 and log a a = 1 (a > 0, a ≠ 1).

5. If a \u003e 1, then the logarithmic function is negative for x (0; 1) and is positive for x (1; + ∞), and if 0< a < 1, то логарифмическая функция положительна при x  (0; 1) and is negative for x (1;+∞).

6. If a \u003e 1, then the logarithmic function is convex upward, and if a (0; 1) - convex down.

The following statements (see, for example,) are used to solve logarithmic equations.