What is a logarithmic equation. Solving logarithmic equations. The Complete Guide (2019)

Logarithmic equations. From simple to complex.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very ..."
And for those who "very much ...")

What is a logarithmic equation?

This is an equation with logarithms. I was surprised, right?) Then I'll clarify. This is an equation in which unknowns (x) and expressions with them are found inside logarithms. And only there! It is important.

Here are some examples logarithmic equations:

log 3 x = log 3 9

log 3 (x 2 -3) = log 3 (2x)

log x + 1 (x 2 + 3x-7) = 2

lg 2 (x + 1) +10 = 11lg (x + 1)

Well, you get the idea ... )

Note! A wide variety of expressions with x are located exclusively inside the logarithms. If, suddenly, an x is found in the equation somewhere outside, for example:

log 2 x = 3 + x,

this will already be a mixed-type equation. Such equations do not have clear rules for solving. We will not consider them yet. By the way, there are equations where inside the logarithms only numbers... For example:

What can I say? Lucky you if you come across this! Logarithm with numbers is some number. And that's all. It is enough to know the properties of logarithms to solve such an equation. Knowledge of special rules, techniques adapted specifically for solving logarithmic equations, not required here.

So, what is logarithmic equation- figured it out.

How to solve logarithmic equations?

Solution logarithmic equations- the thing, in fact, is not very simple. So the section we have - for four ... Requires a decent stock of knowledge on all sorts of related topics. In addition, there is a special feature in these equations. And this feature is so important that it can be safely called the main problem in solving logarithmic equations. We will deal with this problem in detail in the next lesson.

For now, don't worry. We'll go the right way from simple to complex. On specific examples... The main thing is to delve into simple things and do not be lazy to follow the links, I did not put them just like that ... And you will succeed. Necessarily.

Let's start with the most elementary, simplest equations. To solve them, it is desirable to have an idea of the logarithm, but nothing more. Just no idea logarithm, tackle a solution logarithmic equations - somehow embarrassing even ... Very boldly, I would say).

The simplest logarithmic equations.

These are equations of the form:

1.log 3 x = log 3 9

2.log 7 (2x-3) = log 7x

3.log 7 (50x-1) = 2

Decision process any logarithmic equation consists in the transition from an equation with logarithms to an equation without them. In the simplest equations, this transition is carried out in one step. Therefore, the simplest.)

And solving such logarithmic equations is surprisingly simple. See for yourself.

Solving the first example:

log 3 x = log 3 9

To solve this example, you don't need to know almost anything, yes ... Purely intuition!) especially don't like this example? What-what ... Logarithms are not pleasant! Right. So let's get rid of them. We look closely at an example, and we have a natural desire ... Downright irresistible! Take and throw out logarithms altogether. And what pleases me is can do! Mathematics allows. Logarithms disappear the answer is:

Great, isn't it? You can (and should) always do this. Eliminating logarithms in this way is one of the main ways to solve logarithmic equations and inequalities. In mathematics, this operation is called potentiation. There are, of course, their own rules for such liquidation, but they are few. Remember:

You can eliminate logarithms without any fear if they have:

a) identical numerical bases

c) left-right logarithms are pure (without any coefficients) and are in splendid isolation.

Let me explain the last point. In an equation, say

log 3 x = 2log 3 (3x-1)

you cannot remove logarithms. The deuce on the right does not allow. Coefficient, you know ... In the example

log 3 x + log 3 (x + 1) = log 3 (3 + x)

it is also impossible to potentiate the equation. There is no lone logarithm on the left. There are two of them.

In short, you can remove the logarithms if the equation looks like this and only like this:

log a (.....) = log a (.....)

In parentheses, where ellipsis can be any expressions. Simple, super complex, all sorts. Anything. The important thing is that after the elimination of logarithms, we still have a simpler equation. It is assumed, of course, that you already know how to solve linear, quadratic, fractional, exponential and other equations without logarithms.)

Now you can easily solve the second example:

log 7 (2x-3) = log 7 x

Actually, it is decided in the mind. Potentiating, we get:

Well, is it very difficult?) As you can see, logarithmic part of the solution to the equation is only in the elimination of logarithms ... And then the solution of the remaining equation goes without them. Trivial business.

Let's solve the third example:

log 7 (50x-1) = 2

We see that the logarithm is on the left:

We recall that this logarithm is some number to which the base (i.e., seven) must be raised in order to obtain a sub-logarithm expression, i.e. (50x-1).

But that number is two! According to the equation. That is:

That, in essence, is all. Logarithm disappeared, there is a harmless equation left:

We have solved this logarithmic equation based only on the meaning of the logarithm. Is it easier to eliminate the logarithms?) I agree. By the way, if you make a logarithm of two, you can solve this example through liquidation. From any number, you can make a logarithm. Moreover, the way we need it. Highly useful trick in solving logarithmic equations and (especially!) inequalities.

Do not know how to make a logarithm from a number !? It's OK. Section 555 describes this technique in detail. You can master and apply it to its fullest! It greatly reduces the number of errors.

The fourth equation is solved in exactly the same way (by definition):

That's all there is to it.

Let's summarize this lesson. We have considered by examples the solution of the simplest logarithmic equations. It is very important. And not only because there are such equations in control exams. The fact is that even the most evil and confused equations are necessarily reduced to the simplest ones!

Actually, the simplest equations are the finishing part of the solution. any equations. And this finishing part must be understood as a matter of course! And further. Be sure to read this page to the end. There is a surprise there ...)

Now we decide on our own. We fill our hand, so to speak ...)

Find the root (or the sum of the roots, if there are several) of the equations:

ln (7x + 2) = ln (5x + 20)

log 2 (x 2 +32) = log 2 (12x)

log 16 (0.5x-1.5) = 0.25

log 0.2 (3x-1) = -3

ln (e 2 + 2x-3) = 2

log 2 (14x) = log 2 7 + 2

Answers (in disarray, of course): 42; 12; nine; 25; 7; 1.5; 2; 16.

What, not everything is working out? It happens. Do not grieve! Section 555 describes the solution to all these examples in a clear and detailed manner. You will certainly figure it out there. Moreover, master useful practical techniques.

Everything worked out!? All examples are "one left"?) Congratulations!

The time has come to reveal to you the bitter truth. Successful solution of these examples does not at all guarantee success in solving all other logarithmic equations. Even the simplest ones like these. Alas.

The fact is that the solution to any logarithmic equation (even the most elementary one!) Consists of two equal parts. Solving the equation, and working with the ODZ. One part - solving the equation itself - we have mastered. It's not that hard right?

For this lesson, I have specially selected such examples in which the LDO does not affect the answer in any way. But not everyone is as kind as me, right? ...)

Therefore, it is imperative to master the other part. ODZ. This is the main problem in solving logarithmic equations. And not because it's difficult - this part is even easier than the first. But because the ODZ is simply forgotten. Or they don’t know. Or both). And fall out of the blue ...

In the next lesson, we will deal with this problem. Then you can confidently decide any simple logarithmic equations and get to quite solid tasks.

If you like this site ...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Instant validation testing. Learning - with interest!)

you can get acquainted with functions and derivatives.

Preparation for the final test in mathematics includes an important section - "Logarithms". Tasks from this topic are necessarily contained in the exam. The experience of the past years shows that logarithmic equations have caused difficulties for many schoolchildren. Therefore, students with different levels of training should understand how to find the correct answer, and quickly cope with them.

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

log 0.5 (3x - 1) = −3

lg (x + 3) = 3 + 2 lg 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 exactly numbers, and in no case are functions containing the variable x.

Basic solution methods

There are many ways to solve such designs. For example, most of the teachers in the school suggest this way: Immediately express the function f (x) by the formula f ( x) = a b. That is, when you meet the simplest construction, you can go straight to the solution without additional actions and constructions.

Yes, of course, the decision will turn out to 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 offensive 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 inappropriate and most crucial moments: at exams, tests, etc.

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

The idea behind the canonical form is simple. Let's take another look at our problem: on the left we have log a, while the letter a means exactly a number, and in no case a function containing a variable x. Therefore, this letter is subject to all 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 should be equal to the number b, and no restrictions are imposed on this letter, because it can take any values - 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 from a to the power of b:

b = log a a b

How do you remember this formula? It's very simple. Let's write the following construction:

b = b 1 = b log a a

Of course, all the restrictions that we wrote down at the beginning arise. Now let's use the basic property of the logarithm, and introduce the factor 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 the logarithm and is solved using standard algebraic techniques.

Of course, someone will now object: why bother to come up with some kind of canonical formula, why perform two additional unnecessary steps, if you could immediately go from the initial construction to the final formula? Yes, even then, that the majority of 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 record 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.

Solution examples

Now let's consider real examples... So, we 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 follow this step.

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

3x - 1 = 0.5 −3

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

(1/2) −3 = (2/1) 3 = 8

Everything decimals convert to normal when you solve a logarithmic equation.

We rewrite and get:

3x - 1 = 8
3x = 9
x = 3

That's it, we got an answer. The first task has been solved.

Second task

Let's move on to the second task:

As you can see, this equation is no longer the simplest one. If only because the difference is on the left, and not one single logarithm in one base.

Therefore, you need to somehow get rid of this difference. In this case, everything is very simple. Let's take a close 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, that is, from entries with roots and go to power functions, simply because the exponents of these degrees are easily taken out of the sign of the logarithm, and ultimately such a notation greatly simplifies and speeds up the calculations. So let's write it this way:

Now we recall the remarkable property of the logarithm: from the argument, as well as from the base, you can derive degrees. In the case of grounds, the following occurs:

log a k b = 1 / k loga b

In other words, the number that stood in the degree of the base is carried forward and at the same time turns over, that is, it becomes backward... In our case, there was a degree of foundation with an exponent of 1/2. Therefore, we can render it as 2/1. We get:

5 2 log 5 x - log 5 x = 18
10 log 5 x - log 5 x = 18

Please note: in no case should you get rid of the logarithms at this step. Remember the mathematics of grades 4-5 and the procedure: first, multiplication is performed, and only then addition and subtraction. In this case, we subtract one of the same from 10 elements:

9 log 5 x = 18
log 5 x = 2

Now our equation looks like it should. it simplest design and we solve it with the canonical form:

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

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

Third example

Let's move on to the third task:

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

Let me remind you the following formula:

lg b = log 10 b

If for some reason you are confused by the log b, then when performing all the calculations, you can simply log 10 b. You can work with decimal logarithms in the same way as with others: take out degrees, 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.

To begin with, note that the factor 2 before lg 5 can be introduced and becomes a power of the 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 base 10:

3 = log 10 10 3 = log 10 3

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

lg (x - 3) = lg 1000 + lg 25
lg (x - 3) = lg 1000 25
lg (x - 3) = lg 25,000

We have before us the canonical form again, and we got it bypassing the stage of transformations, that is, the simplest logarithmic equation did not appear anywhere in our country.

This is exactly what I talked about at the very beginning of the lesson. The canonical form allows solving a wider class of problems than the standard school formula given by most school teachers.

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

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

Everything! The problem has been solved.

A note on scope

Here I would like to make an important remark about the scope of definition. Surely now there are students and teachers who will say: "When we solve expressions with logarithms, it is imperative to remember that the argument f (x) must be greater than zero!" In this regard, a logical question arises: why in none of the considered problems did we require this inequality to be fulfilled?

Do not worry. No extra roots will arise in these cases. And this is another great trick that allows you to speed up the solution. Just know that if in a problem the variable x occurs only in one place (or rather, in a single argument of a single logarithm), and nowhere else in our case is there a variable x, then write the domain not necessary because it will run automatically.

Judge for yourself: in the first equation we got that 3x - 1, that is, 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 must be equal to 5 2, that is, it is certainly greater than zero. And in the third case, where x + 3 = 25,000, that is, again obviously greater than zero. In other words, the domain is automatically satisfied, 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 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, in order to learn how to apply the canonical form of the logarithmic equation, it is not enough just to watch one video tutorial. Therefore, download the options for independent decision that are attached to this video tutorial and start solving at least one of these two independent works.

It will take you just a few minutes. But the effect of such training will be much higher compared to if you just watched this video tutorial.

I hope this tutorial will help you understand logarithmic equations. Use the canonical form, simplify expressions using rules for working with logarithms - and no problem will be scary for you. And I have everything for today.

Consideration of the scope

Now let's talk about the domain of the logarithmic function, as well as 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 - there is only one function in it, and the numbers a and b are exactly numbers, and in no case is it a function that depends on the variable x. It is 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 substituted 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 is in effect because the logarithm of negative numbers does not exist. So, maybe because of this limitation, you should introduce a check for answers? Perhaps they need to be substituted in the source?

No, in the simplest logarithmic equations an additional check 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 does not matter, because no matter what degree we raise a positive number, at the output we will still get a positive number. Thus, the requirement f (x)> 0 is fulfilled automatically.

What is really worth checking is the scope of the function under the log sign. There may be rather complicated structures, and in the process of solving them, you must definitely follow them. Let's see.

First task:

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

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

Of the resulting 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 that the expression under the sign of the logarithm is greater than 0 is not required, because it is not just greater than 0, but by the condition of the equation it is equal to 2. Therefore, the requirement “greater than zero” is fulfilled automatically.

Let's move on to the second task:

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

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

We square both sides, taking into account the restrictions, and we 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 is x = −1. That's the whole solution. Let's go back to the very beginning of our calculations.

The main takeaway from this lesson is that you do not need to check the constraints for a function in the simplest logarithmic equations. Because in the process of solving all the constraints are met 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 limitations and requirements for the right-hand side, as we have seen today on 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 analyze two more rather interesting tricks with which it is fashionable to solve more complex structures... But first, let's remember how the simplest tasks are solved:

log a f (x) = b

In this notation, a and b are exactly numbers, and in the function f (x) the variable x must be present, and only there, that is, x must be only 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 exactly the 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 both the left and the right are the logarithm to the base a. In this case, we can, figuratively speaking, strike out the signs of log, and from the point of view of mathematics, we can say that we are simply equating the arguments:

f (x) = a b

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

So the first construct:

First of all, note that there is a fraction on the right with log in the denominator. When you see such an expression, it will not be superfluous to remember the wonderful property of logarithms:

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

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

It is this construction that we observe from the sign to the right in our equation. Let's replace this construction with log a b, we get:

In other words, compared to the original problem, we have swapped the argument and the base of the logarithm. Instead, we had to flip 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 degree of the base, is taken out as an inverted fraction. Let's take it out as an inverted fraction:

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

Now we equate arguments whose bases are the same (and we really have the same bases), 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 occurs only in one log, and it is 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:

lg 56 = lg 2 log 2 7 - 3lg (x + 4)

Here, in addition to the usual logarithms, we will have to work with lg f (x). How to solve such an equation? It may seem to an untrained student that this is some kind of toughness, but in fact, everything is solved in an elementary way.

Take a close look at the term lg 2 log 2 7. What can we say about it? The reasons and arguments for log and lg are the same, and that should be suggestive. Let's remember again how the degrees are taken out from under the sign of the logarithm:

log a b n = nlog a b

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

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

Very often students do not see this action point blank, 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 can be easily calculated 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 transform a logarithmic equation, you should know this formula in the same way as representing any number in the form of log.

We return to our task. We rewrite it taking into account the fact that the first term to the right of the equal sign will simply be 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)

Subtract the expressions on the left because they have the same base:

lg (56/7) = −3lg (x + 4)

Now let's take a close look at the equation we got. It is practically the canonical form, but there is a factor of −3 on the right. Let's put it in 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 signs of lg and equate the arguments:

(x + 4) −3 = 8

x + 4 = 0.5

That's all! We have 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.

I will list again key points of this tutorial.

The main formula that is studied in all the lessons on this page dedicated to solving logarithmic equations is the canonical form. And don't be intimidated by the fact that most school textbooks teach you to solve such problems in a different way. 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, it will be useful to know the basic properties for solving logarithmic equations. Namely:

The formula for the transition to one base and the special case when we flip log (this was very useful to us in the first problem);
The formula for adding and removing degrees from the sign of the logarithm. Here, many students freeze and do not see at close range that the exponential and inserted degree itself can contain log f (x). Nothing wrong with that. We can introduce one log by the sign of the other and at the same time significantly simplify the solution of the problem, which we observe in the second case.

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

Variable radix problems

Today we will look at logarithmic equations, which for many students seem to be non-standard, if not completely unsolvable. It is 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.

To begin with, let's remember how the simplest problems are solved, which are based on ordinary numbers. So, the simplest is a construction of the form

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, that is, we write:

f (x) = a b

Thus, we get rid of the log sign and solve the already common problem. In this case, the roots obtained in the solution will be the roots of the original logarithmic equation. In addition, the record, when both the left and the right are on 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 constructions. 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, in fact, the number b that stood to the right of the equal sign. Thus, we will 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 initial logarithms are not defined everywhere and not always.

Therefore, we must write down the scope separately. Let's not be smart 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 do not 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 to be greater than zero.

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

Of course, we could just as well cross out and linear inequality, i.e., delete x - 2> 0 and require that 2x 2 - 13x + 18> 0. But you must admit that solving the simplest linear inequality is much faster and easier than the quadratic one, even if the condition is that as a result of solving the entire of this system, we get the same roots.

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

Let's rewrite our system:

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

2x 2 - 14x + 20 = 0

x 2 - 7x + 10 = 0

Before us is the given square trinomial and, therefore, we can use Vieta's formulas. We get:

(x - 5) (x - 2) = 0

x 1 = 5

x 2 = 2

And 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 is fine for us: 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 has been solved, including taking into account the ODZ. Let's move on to the second equation. Here we will find more interesting and informative calculations:

The first step: just like last time, we bring the whole thing to the canonical form. For this, we can write the number 9 as follows:

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

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

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

x 2 + 4x + 3 = 0

Before us is the newly given square trinomial, we 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 would have to write the system, but due to the cumbersomeness of the whole construction, I decided to calculate the domain separately).

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

These are the requirements imposed by the domain of definition.

Immediately, we note that since we equate the first two expressions of the system to each other, then we can delete any of them. Let's delete 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 one of them 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, for which we will build both parts into 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 all, the problem is solved.

Once again, the key points of this task:

Feel free to apply and solve logarithmic equations using the canonical form. Students who make such a record, and do not go directly from the original problem to a construction like log a f (x) = b, make much fewer mistakes than those who rush somewhere, skipping intermediate steps of calculations;
As soon as a variable base appears in the logarithm, the problem ceases to be the simplest one. 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 also they must not be equal to 1.

There are different ways to impose the final requirements on the final answers. For example, you can solve the whole system containing all the requirements for the domain of definition. On the other hand, you can first solve the problem itself, and then remember about the domain of definition, work it out separately in the form of a system and superimpose it on the resulting roots.

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

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 related to the decision 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 of 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 us recall the basic 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.

Rice. 1. Graph of the logarithmic function at various bases

The function graph for is shown in black. Rice. 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. Rice. 1.

Properties of this function:

Domain: ;

Range of values:;

The function is monotonic throughout its entire domain of definition. When monotonically (strictly) increases, more meaning the argument matches the 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.

So we got 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 consider 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 got the wrong 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.

Logarithmic equations. We continue to consider the problems from part B of the exam in mathematics. We have already considered the solutions of some equations in the articles "", "". In this article, we'll look at logarithmic equations. I must say right away that there will be no complex transformations when solving such equations on the exam. They are simple.

It is enough to know and understand the basic logarithmic identity, to know the properties of the logarithm. Pay attention to the fact that after the solution, you MUST do a check - substitute the resulting value into the original equation and calculate, in the end you should get the correct equality.

Definition:

The logarithm of the number a to base b is the exponent,to which you need to raise b to get a.

For example:

Log 3 9 = 2 since 3 2 = 9

Logarithm properties:

Special cases of logarithms:

We will solve the problems. In the first example, we will do a check. In subsequent checks, do it yourself.

Find the root of the equation: log 3 (4 – x) = 4

Since log b a = x b x = a, then

3 4 = 4 - x

x = 4 - 81

x = - 77

Examination:

log 3 (4 - (- 77)) = 4

log 3 81 = 4

3 4 = 81 Correct.

Answer: - 77

Decide for yourself:

Find the root of the equation: log 2 (4 - x) = 7

Find the root of the equation log 5(4 + x) = 2

We use the basic logarithmic identity.

Since log a b = x b x = a, then

5 2 = 4 + x

x = 5 2 - 4

x = 21

Examination:

log 5 (4 + 21) = 2

log 5 25 = 2

5 2 = 25 Correct.

Answer: 21

Find the root of the equation log 3 (14 - x) = log 3 5.

Occurs next property, its meaning is as follows: if on the left and right sides of the equation we have logarithms with on the same basis, then we can equate the expressions under the signs of the logarithms.

14 - x = 5

x = 9

Check it out.

Answer: 9

Decide for yourself:

Find the root of the equation log 5 (5 - x) = log 5 3.

Find the root of the equation: log 4 (x + 3) = log 4 (4x - 15).

If log c a = log c b then a = b

x + 3 = 4x - 15

3x = 18

x = 6

Check it out.

Answer: 6

Find the root of the equation log 1/8 (13 - x) = - 2.

(1/8) –2 = 13 - x

8 2 = 13 - x

x = 13 - 64

x = - 51

Check it out.

A small addition - the property is used here

degree ().

Answer: - 51

Decide for yourself:

Find the root of the equation: log 1/7 (7 - x) = - 2

Find the root of the equation log 2 (4 - x) = 2 log 2 5.

Let's transform the right side. let's use the property:

log a b m = m ∙ log a b

log 2 (4 - x) = log 2 5 2

If log c a = log c b then a = b

4 - x = 5 2

4 - x = 25

x = - 21

Check it out.

Answer: - 21

Decide for yourself:

Find the root of the equation: log 5 (5 - x) = 2 log 5 3

Solve the equation log 5 (x 2 + 4x) = log 5 (x 2 + 11)

If log c a = log c b then a = b

x 2 + 4x = x 2 + 11

4x = 11

x = 2.75

Check it out.

Answer: 2.75

Decide for yourself:

Find the root of the equation log 5 (x 2 + x) = log 5 (x 2 + 10).

Solve the equation log 2 (2 - x) = log 2 (2 - 3x) +1.

It is necessary to obtain an expression of the form on the right side of the equation:

log 2 (......)

We represent 1 as a logarithm to base 2:

1 = log 2 2

log with (ab) = log with a + log with b

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

We get:

log 2 (2 - x) = log 2 2 (2 - 3x)

If log c a = log c b, then a = b, then

2 - x = 4 - 6x

5x = 2

x = 0.4

Check it out.

Answer: 0.4

Decide for yourself: Next, you need to solve the quadratic equation. By the way,

roots are 6 and - 4.

Root "-4 "is not a solution, since the base of the logarithm must be greater than zero, and when"– 4 "it is equal to" – 5". The solution is root 6.Check it out.

Answer: 6.

R Eat yourself:

Solve the equation log x –5 49 = 2. If the equation has more than one root, fill in the answer with the smaller root.

As you can see, no complicated transformations with logarithmic equationsno. It is enough to know the properties of the logarithm and be able to apply them. In the tasks of the exam related to the transformation logarithmic expressions, more serious transformations are being performed and deeper solution skills are required. We will consider such examples, do not miss!I wish you success!!!

Best regards, Alexander Krutitskikh.

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