Formulas for the solution of logarithmic equations. Some methods of solving logarithmic equations

This video I start a long series of lessons about logarithmic equations. Now there are three examples in front of you, on the basis of which we will learn to solve the simplest tasks that are so called - simplest.

log 0.5 (3x - 1) \u003d -3

lG (x + 3) \u003d 3 + 2 LG 5

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

log a f (x) \u003d b

It is important that the variable x is present only inside the argument, i.e. only in the function f (x). A and B numbers are precisely numbers, and in no case are not functions containing a variable x.

Basic methods of solution

There are many ways to solve such structures. For example, most teachers at school offer such a way: to immediately express the function f (x) by the formula f ( x) \u003d. a b. That is, when you meet the simplest design, immediately without additional actions and buildings can go to the solution.

Yes, definitely, the solution will be right. However, the problem of this formula is that most students do not understandWhere it comes from and why the letter and we are erected in the letter B.

As a result, I often observe very offensive errors when, for example, these letters are changing in places. This formula must be either understood or tooling, and the second method leads to errors in the most inappropriate and most responsible moments: on exams, control, etc.

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

The idea of \u200b\u200bcanonical form is simple. Let's look at our task again: on the left we have Log A, while under the letter A is meant exactly the number, and in no case is it not a function containing the variable x. Consequently, all the limitations are applied to this letter, which are superimposed on the basis of the logarithm. namely:

1 ≠ A\u003e 0

On the other hand, from the same equation, we see that the logarithm should be equal to the number B, and here it does not impose any restrictions on this letter, because it can take any values \u200b\u200b- both positive and negative. It all depends on what values \u200b\u200breceive the function f (x).

And here we remember our remarkable rule that any number B can be represented in the form of a logarithm on the base A of A to the degree B:

b \u003d Log A A B

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

b \u003d b · 1 \u003d b · log a a

Of course, while there are all the limitations that we recorded at the beginning. And now let's use the basic property of the logarithm, and make a multiplier B as a degree a. We get:

b \u003d B · 1 \u003d B · Log A A \u003d Log A A B

As a result, the initial equation will rewrite in the following form:

log a f (x) \u003d log a a b → f (x) \u003d a b

That's all. The new function no longer contains logarithm and is solved by standard algebraic techniques.

Of course, someone will now objected: why was it at all invented some canonical formula, why do two additional unnecessary steps, if you could immediately move from the original design to the final formula? Yes, at least then, then the majority of students do not understand where this formula comes from and, as a result, regularly allow errors when applied.

But this sequence of actions consisting of three steps allows you to solve the initial logarithmic equation, even if you do not understand where the very final formula is taken. By the way, the canonical formula is called this record:

lOG A F (X) \u003d Log A A B

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

Examples of solutions

Now let's consider real examples. So, we decide:

log 0.5 (3x - 1) \u003d -3

Let's rewrite it as follows:

lOG 0.5 (3X - 1) \u003d log 0.5 0.5 -3

Many students are in a hurry and try to immediately build the number 0.5 to the degree that came to us from the initial task. And indeed, when you are already well trained in solving such tasks, you can immediately perform this step.

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

3X - 1 \u003d 0.5 -3

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

(1/2) −3 = (2/1) 3 = 8

All decimal fractions translate into ordinary when you solve the logarithmic equation.

Rewrite and get:

3X - 1 \u003d 8
3X \u003d 9.
x \u003d 3.

All we got the answer. The first task is solved.

Second task

Go to the second task:

As we see, this equation is no longer the simplest. Already at least because on the left there is a difference, and not a single logarithm one for one base.

Consequently, it is necessary to somehow get rid of this difference. In this case, everything is very simple. Let's look attentively on the ground: left is the number under the root:

General Recommendation: In all logarithmic equations, try to get rid of the radicals, i.e., from records with roots and move to power functions, simply because the indicators of these degrees are easily carried out for the logarithm sign and in the final account, such a record simplifies significantly and speeds up the calculations. Here let's write and write down:

Now remember the remarkable property of the logarithm: from the argument, as well as from the base, you can endure degrees. In the case of bases, the following happens:

log A K B \u003d 1 / K Loga B

In other words, the number that stood in the degree of foundation is carried forward and at the same time turns over, i.e. it becomes a different number. In our case, there was a degree of foundation with an indicator 1/2. Consequently, we can take it as 2/1. We get:

5 · 2 log 5 x - log 5 x \u003d 18
10 log 5 x - log 5 x \u003d 18

Please note: in no case can you get rid of logarithms at this step. Remember the Mathematics of 4-5 Class and Procedure: First, multiplication is performed, but only then addition and subtraction. In this case, we subtract one of the same elements:

9 log 5 x \u003d 18
log 5 x \u003d 2

Now our equation looks like. This is the simplest design, and we solve it with the help of a canonical form:

lOG 5 x \u003d log 5 5 2
x \u003d 5 2
x \u003d 25.

That's all. The second task is solved.

Third example

Go to the third task:

lG (x + 3) \u003d 3 + 2 LG 5

Let me remind you to the following formula:

lG B \u003d Log 10 B

If for some reason you are confused by the LG B record, then when performing all the calculations, you can only write Log 10 B. With decimal logarithms, you can work in the same way as with others: to make degrees, fold and represent any numbers in the form of LG 10.

That's it with these properties, we will now use to solve the problem, since it is not the simplest, which we recorded at the very beginning of our lesson.

To begin with, we note that the multiplier 2 facing LG 5 can be made and becomes a degree of foundation 5. In addition, the free term 3 also represents in the form of a logarithm - it is very easy to observe from our record.

Judge for yourself: any number can be represented in the form of LOG based on 10:

3 \u003d log 10 10 3 \u003d lg 10 3

We rewrite the source task, taking into account the changes received:

lG (x - 3) \u003d LG 1000 + LG 25
lG (x - 3) \u003d LG 1000 · 25
lG (x - 3) \u003d LG 25 000

We are again a canonical form, and we got it, bypassing the stage of transformations, i.e. the simplest logarithmic equation did not come across anywhere.

That was what I said at the very beginning of the lesson. The canonical form allows you to solve a wider class of tasks than the standard school formula that most school teachers give.

Well, all, get rid of the sign of the decimal logarithm, and we get a simple linear design:

x + 3 \u003d 25 000
x \u003d 24 997

Everything! The task is solved.

Note on the area of \u200b\u200bdefinition

Here I would like to bring an important remark about the area of \u200b\u200bdefinition. Surely now there will be students and teachers who will say: "When we solve expressions with logarithms, it is necessary to remember that the argument f (x) should be greater than zero!" In this regard, there is a logical question: why didn't we require in one of the considered tasks that this inequality is carried out?

Do not worry. No extra roots in these cases will arise. And this is another wonderful trick that allows you to speed up the decision. Just know that if the task variable x is found only in one place (or rather - in a single argument of a single logarithm), and nowhere else in our case there is no variable, then write the definition area not necessarybecause it will be performed automatically.

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

With the same success, we can write down that in the second case x should be 5 2, that is, he is obviously more zero. And in the third case, where x + 3 \u003d 25,000, i.e., again, more than more zero. In other words, the definition area is performed automatically, but only under the condition that X is found only in the argument of only one logarithm.

That's all you need to know to solve the simplest tasks. Already one of this rule, together with the conversion rules, will allow you to solve a very wide class of tasks.

But let's be honest: in order to finally deal with this technique to learn how to apply the canonical form of the logarithmic equation, it is not enough to simply see one video tutorial. Therefore, right now download options for an independent solution that are attached to this video language and start solving at least one of these two independent work.

You will leave you literally a few minutes. But the effect of such training will be much higher compared to the time you simply viewed this video tutorial.

I hope this lesson will help you deal with logarithmic equations. Use the canonical form, simplify expressions using the rules of working with logarithms - and no tasks will not be terrible. And I have everything today.

Accounting of the definition area

Now let's talk about the field of determining the logarithmic function, as well as how it affects the solution of logarithmic equations. Consider the design of the view

log a f (x) \u003d b

This expression is called the simplest - in it only one function, and the number A and B is precisely numbers, and in no case is not a function depending on the variable x. It is solved very simple. Just use the formula only:

b \u003d Log A A B

This formula is one of the key properties of the logarithm, and when substituting in our initial expression, we will receive the following:

lOG A F (X) \u003d Log A A B

f (x) \u003d a b

This is already familiar formula from school textbooks. Many students will certainly have a question: since in the initial expression, the function f (x) is under the Log sign, the following restrictions are superimposed on it:

f (x)\u003e 0

This restriction acts because the logarithm from negative numbers does not exist. So, maybe, as a result, the restrictions should be imposed on the answers? Maybe they need to be substituted into the source?

No, in the simplest logarithmic equations, an additional check of excess. And that's why. Take a look at our final formula:

f (x) \u003d a b

The fact is that the number A in any case is more than 0 - this requirement is also superimposed by logarithm. The number A is the basis. At the same time, the number b no restrictions is superimposed. But it does not matter, because in which a degree we would have erected a positive number, we still receive a positive number at the exit. Thus, the requirement F (x)\u003e 0 is performed automatically.

What is really worth checking is the area of \u200b\u200bdefining a function under the Log sign. There may be pretty difficult designs, and in the process of solving it is necessary to follow them. Let's see.

First task:

The first step: we transform the fraction on the right. We get:

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

From the resulting roots we are satisfied with only the first, since the second root is less than zero. The only answer will be the number 9. All, the task is solved. No additional checks that the expression under the logarithm sign 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. Consequently, the requirement "greater than zero" is performed automatically.

Go to the second task:

Here is all the same. Rewrite the design by replacing the top three:

Get rid of logarithm signs and get an irrational equation:

We build both parts into a square with regard to restrictions and get:

4 - 6x - x 2 \u003d (x - 4) 2

4 - 6x - x 2 \u003d x 2 + 8x + 16

x 2 + 8x + 16 -4 + 6x + x 2 \u003d 0

2x 2 + 14x + 12 \u003d 0 |: 2

x 2 + 7x + 6 \u003d 0

We solve the obtained equation through the discriminant:

D \u003d 49 - 24 \u003d 25

x 1 \u003d -1

x 2 \u003d -6

But x \u003d -6 does not suit us, because if we substitute this number in our inequality, we will get:

−6 + 4 = −2 < 0

In our case, it is required that there is more than 0 or in a pinch is equal. But x \u003d -1 suits us:

−1 + 4 = 3 > 0

The only answer in our case will be x \u003d -1. That's all the decision. Let's return to the very beginning of our calculations.

The main conclusion from this lesson: Checking the limitations for a function in the simplest logarithmic equations is not required. Because in the process of solving, all restrictions are performed automatically.

However, this does not mean that you can forget about checking. In the process of working on the logarithmic equation, it may well go to the irrational, in which there will be its limitations and requirements for the right part, in which we have been sent to two different examples.

Boldly solve such tasks and be particularly attentive if the root stands in the argument.

Logarithmic equations with different bases

We continue to study the logarithmic equations and we will analyze two more fairly interesting techniques, with the help of which it is fashionable to solve more complex structures. But first, let's remember how simple tasks are solved:

log a f (x) \u003d b

In this record, A and B are precisely numbers, and in the function f (x) the variable x should be present, and only there, i.e. x should be only in the argument. We will convert such logarithmic equations using a canonical form. To do this, we note that

b \u003d Log A A B

And a B is the argument. Let's rewrite this expression as follows:

lOG A F (X) \u003d Log A A B

We just do this and seek on the left, and on the right there was a logarithm on the basis of a. In this case, we can, figuratively speaking, cross the signs of Log, and from the point of view of mathematics we can say that we simply equate the arguments:

f (x) \u003d a b

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

So, the first design:

First of all, I note that the right is a fraction, in the denominator which is Log. When you see such an expression, it will not be superfluous to recall the remarkable property of logarithms:

Transferring to Russian, this means that any logarithm can be represented as a private two logarithms with any base with. Of course, 0< с ≠ 1.

So: this formula has one wonderful case when the variable C is equal to the variable b. In this case, we will get the design of the form:

It is such a design that we observe from the sign to the right in our equation. Let's replace this design on Log A B, we get:

In other words, in comparison with the initial task, we changed the argument and the base of the logarithm. In exchange, we had to turn the fraction.

We remember that any degree can be made from the ground according to the following rule:

In other words, the K coefficient, which is the degree of foundation, is carried out as an inverted fraction. Let's bring it out as an inverted fraction:

The fractional multiplier cannot be left in front, because in this case we will not be able to submit this entry as a canonical form (because in canonical form before the second logarithm is no additional multiplier). Consequently, let's make a fraction of 1/4 into an argument in the form of a degree:

Now we equate the arguments, the foundations of which are the same (and the foundations we are really the same), and write:

x + 5 \u003d 1

x \u003d -4.

That's all. We received an answer to the first logarithmic equation. Please note: in the source task, the variable x is found only in one LOG, and it is in its argument. Consequently, checking the definition area is not required, and our number x \u003d -4 is indeed the answer.

Now go to the second expression:

lG 56 \u003d LG 2 LOG 2 7 - 3LG (X + 4)

Here, in addition to ordinary logarithms, we will have to work with LG F (X). How to solve such an equation? An unprepared student may seem that it is some kind of tin, but in fact everything is solved elementary.

Carefully look at the term LG 2 log 2 7. What can we say about it? The bases and arguments of the LOG and LG coincide, and it should make some thoughts. Let's remember once again how degrees from the logarithm sign:

log A B n \u003d nlog a b

In other words, what was the degree from number B in the argument becomes a multiplier to the Log itself. Let's apply this formula to expressions LG 2 log 2 7. Let you not scare LG 2 - this is the most common expression. You can rewrite it as follows:

All rules that act for any other logarithm are fair for him. In particular, the multiplier standing in front can be made to the degree of argument. Let's write:

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

This formula can be considered as a definition, and as one of its properties. In any case, if you convert the logarithmic equation, this formula you should know exactly the same as the representation of any number in the form of log.

Return to our task. Rewrite it, taking into account the fact that the first term to the right of the equality sign will be just LG 7. We have:

lG 56 \u003d LG 7 - 3LG (X + 4)

Let's transfer LG 7 to the left, we get:

lG 56 - LG 7 \u003d -3LG (X + 4)

We subtract the expression on the left, because they have the same base:

lG (56/7) \u003d -3lg (x + 4)

Now let's look at the equation that we got. It is practically a canonical form, but the multiplier is present to the right. Let's make it in the right LG argument:

lG 8 \u003d LG (x + 4) -3

We have the canonical form of the logarithmic equation, so we strike the LG signs and equate the arguments:

(x + 4) -3 \u003d 8

x + 4 \u003d 0.5

That's all! We solved the second logarithmic equation. At the same time, no additional checks are required, because in the initial task it was visited only in one argument.

I will list the key points of this lesson again.

The main formula that is studied in all lessons on this page dedicated to the solution of logarithmic equations is a canonical form. And let you not frighten that in most school textbooks you are taught to solve such tasks in a different way. This tool works very effectively and allows you to solve a much wider class of tasks, rather than the simplest, which 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 transition formula to one base and a special case when we turn the Log (it was very useful to us in the first task);
2. The formula for making and making degrees from under the sign of the logarithm. Here, many students hang and emphasize that the submission and contribution itself may contain Log F (x). Nothing wrong with that. We can make one log on another sign and at the same time to 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 definition area in each of these cases, because everywhere the variable x is present only in one LOG sign, and at the same time it is in its argument. As a result, all the requirements of the definition area are performed automatically.

Variable base tasks

Today we will look at the logarithmic equations that for many students seem non-standard, and even unresolved at all. We are talking about expressions, at the base of which are not numbers, but variables and even functions. We will solve such structures with the help of our standard reception, namely through the canonical form.

To begin with, let's remember how the simplest tasks are solved, at the base of which there are ordinary numbers. So, the simplest is called the design of the type

log a f (x) \u003d b

To solve such tasks, we can use the following formula:

b \u003d Log A A B

We rewrite our initial expression and get:

lOG A F (X) \u003d Log A A B

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

f (x) \u003d a b

Thus, we get rid of the sign of the Log and we decide the usual task. At the same time, the roots obtained in solving the root and will be the roots of the original logarithmic equation. In addition, the entry, when to the left, and the right is one and the same logarithm with the same base, just 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) \u003d 1

We replace 1 on log x - 2 (x - 2) 1. The extent that we observe the argument is, in fact, the number B that stood to the right of the sign of equality. Thus, we rewrite our expression. We get:

lOG X - 2 (2X 2 - 13X + 18) \u003d LOG X - 2 (X - 2)

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

2x 2 - 13x + 18 \u003d x - 2

But this decision does not end, because this equation is not equivalent to the initial one. After all, the resulting design consists of functions that are defined on the entire numerical line, and our initial logarithms are not defined everywhere and not always.

Therefore, we must separately write the definition area. Let's not wise and write all the requirements for the beginning:

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

2x 2 - 13x + 18\u003e 0

x - 2\u003e 0

Secondly, the base should be not only more than 0, but also different from 1:

x - 2 ≠ 1

As a result, we get the system:

But you do not be afraid: when processing logarithmic equations, such a system can be significantly simplified.

Judge for yourself: on the one hand, we require the quadratic function to be greater than zero, and on the other hand, this quadratic function is equal to a linear expression, which also requires it to be greater than zero.

In this case, if we require x - 2\u003e 0, then the requirement 2x 2 - 13x + 18\u003e 0 will automatically be performed. Therefore, we can safely cross the inequality containing a quadratic function. Thus, the number of expressions that is contained in our system will decrease to three.

Of course, with the same success, we could cross the linear inequality, i.e., delete x - 2\u003e 0 and demand that 2x 2 - 13x + 18\u003e 0. But agree that it is much faster to solve the simplest linear inequality much faster and easier, The quadratic, even if, if, as a result of the solution of the whole system, we will get the same roots.

In general, if possible, try to optimize calculations. And in the case of logarithmic equations, strike out the most complex inequalities.

Let's rewrite our system:

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

2x 2 - 14x + 20 \u003d 0

x 2 - 7x + 10 \u003d 0

We have a given square three-half-one and, therefore, we can take advantage of the formulas of the Vieta. We get:

(x - 5) (x - 2) \u003d 0

x 1 \u003d 5

x 2 \u003d 2

And now we return to our system and we find that x \u003d 2 does not suit us, because we need from it to be strictly greater than 2.

But x \u003d 5 we are quite satisfied with: the number 5 is greater than 2, and at the same time 5 is not 3. Therefore, the only solution of this system will be x \u003d 5.

Everything, the task is solved, including, taking into account the OTZ. Go to the second equation. Here we are waiting for more interesting and meaningful calculations:

The first step: as last time, we give all this case to canonical form. For this, the number 9 we can write as follows:

The base with the root can not be touched, but the argument is better to convert. Let's go from the root into a degree with a rational indicator. We write:

Let me not rewrite all our large logarithmic equation, but simply equalize the arguments immediately:

x 3 + 10x 2 + 31x + 30 \u003d x 3 + 9x 2 + 27x + 27

x 2 + 4x + 3 \u003d 0

We have a newly reduced three-half-one in front of us, we use the formulas of the Vieta and write:

(x + 3) (x + 1) \u003d 0

x 1 \u003d -3

x 2 \u003d -1

So, we got roots, but no one guarantees us that they would suit the initial logarithmic equation. After all, Log signs impose additional restrictions (here we would have to write down the system, but because of the bulkness of the whole design, I decided to calculate the definition area separately).

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

These are the requirements imposed by the definition area.

Immediately note that since we equate the first two expressions of the system to each other, then any of them we can delete. Let us cross the first because it looks more threatening than the second.

In addition, we note that the decision of the second and third inequality will be the same (cube of some number more than zero, if the very number is greater than zero; similarly and with the root of the third degree - these inequalities are completely similar, so we can cross it out).

But with the third inequality it will not pass. Get rid of the sign of the radical standing on the left, for which they erect both parts in the cube. We get:

So, we obtain the following requirements:

- 2 ≠ x\u003e -3

Which of our roots: x 1 \u003d -3 or x 2 \u003d -1 meet these requirements? Obviously, only x \u003d -1, because x \u003d -3 does not satisfy the first inequality (for inequality we have strict). Total returning to our task, we get one root: x \u003d -1. That's all, the task is solved.

Once again the key points of this task:

1. Feel free to apply and solve logarithmic equations using a canonical form. Students who make such a record, and do not pass directly from the initial problem to the design of the Log A f (x) \u003d b design, allow much less errors than those that are in a rush somewhere, passing intermediate steps of calculations;
2. As soon as an alternating base appears in logarithm, the task stops being the simplest. Consequently, when it is decided, it is necessary to take into account the field of definition: the arguments must be greater than zero, and the base is not only greater than 0, but they should not be equal to 1.

You can enter the latest requirements for final answers in different ways. For example, you can solve a whole system containing all the requirements for the definition area. On the other hand, you can first solve the task itself, and then remember about the definition area, separately work it in the form of the system and impose on the roots received.

What is the way to choose when solving a specific logarithmic equation, solve only you. In any case, the answer will turn out the same.

Solving logarithmic equations. Part 1.

Logarithmic equation An equation is called in which the unknown is contained under the sign of the logarithm (in particular, at the base of the logarithm).

Simplest logarithmic equation It has the form:

Solution of any logarithmic equation It assumes the transition from logarithms to expressions under the sign of logarithms. However, this action expands the area of \u200b\u200bpermissible values \u200b\u200bof the equation and can lead to the appearance of foreign roots. To avoid the appearance of foreign roots, You can do one of three ways:

1. Make an equivalent transmission from the initial equation to the system including

depending on what kind of inequality is or easier.

If the equation contains an unknown at the base of the logarithm:

then we go to the system:

2. Separately find the area of \u200b\u200bpermissible values \u200b\u200bof the equation, then solve the equation and check whether the solutions found are satisfying.

3. Solve the equation, and then check:substitute the solutions found to the original equation, and check whether we will get faithful equality.

The logarithmic equation of any level of complexity ultimately always comes down to the simplest logarithmic equation.

All logarithmic equations can be divided into four types:

1 . Equations that contain logarithms only in the first degree. They are given by transformations and use

Example. Resolving equation:

We equate expressions under the sign of the logarithm:

Check whether our root satisfies the equation:

Yes, satisfies.

Answer: x \u003d 5

2 . Equations that contain logarithms to a degree other than 1 (in particular, in the denomoter denominator). Such equations are solved with introduction of the change of the variable.

Example. Resolving equation:

Find the OTZ equations:

The equation contains logarithms in the square, so it is solved by replacing the variable.

Important! Before entering the replacement, you need to "remove" logarithms that are part of the equation on "bricks" using the properties of logarithms.

When "collapse" logarithms, it is important to very accurately apply the properties of logarithms:

In addition, there is one more subtle place here, and to avoid a common mistake, we use the intermediate equality: We write the logarithm degree in this form:

Similarly,

We substitute the expressions obtained into the original equation. We get:

Now we see that the unknown is contained in the equation in the composition. We introduce a replacement:. Since it can take any actual value, we do not impose any restrictions on the variable.

Instruction

Write down the given logarithmic expression. If logarithm 10 is used in the expression, then its recording is shortened and looks like this: LG B is a decimal logarithm. If the logarithm has the number E in the form of base, then the expression is recorded: LN B is a natural logarithm. It is understood that the result of any degree in which the number of foundations should be erected in order to obtain the number B.

When there are two functions from the sum of two functions, it is simply necessary to predlect them, and the results are folded: (U + V) "\u003d U" + V ";

When the derivative of the product of two functions is derived, a derivative from the first function is to multiply onto the second and add a derivative of the second function multiplied by the first function: (U * V) "\u003d U" * V + V "* U;

In order to find a derivative of private two functions, it is necessary, from the product of the derivative divide, multiplied to the function of the divider, to deduct the product of the derivative of the divider multiplied by the function of the dividery, and all this is divided into the function of the divider erected into the square. (U / V) "\u003d (U" * V-V "* U) / V ^ 2;

If a complex function is given, it is necessary to multiply the derivative from the internal function and the derivative of the external one. Let y \u003d u (V (x)), then y "(x) \u003d y" (u) * v "(x).

Using the above, you can directly indifferentizing any function. So, consider a few examples:

y \u003d x ^ 4, y "\u003d 4 * x ^ (4-1) \u003d 4 * x ^ 3;

y \u003d 2 * x ^ 3 * (E ^ xx ^ 2 + 6), y "\u003d 2 * (3 * x ^ 2 * (E ^ xx ^ 2 + 6) + x ^ 3 * (E ^ x-2 * x));
There are also tasks for calculating the derivative at the point. Let the function y \u003d e ^ (x ^ 2 + 6x + 5) be given, you need to find the value of the function at the point x \u003d 1.
1) Find the function derivative: Y "\u003d E ^ (x ^ 2-6x + 5) * (2 * x +6).

2) Calculate the value of the function at a specified point Y "(1) \u003d 8 * E ^ 0 \u003d 8

Video on the topic

Helpful advice

Learn the table of elementary derivatives. It will noticeably save time.

Sources:

• derived Constanta

So, how is the irrational equation from rational? If an unknown variable is under a sign of a square root, the equation is considered irrational.

Instruction

The main method of solving such equations is the method of erection of both parts equations in a square. However. This is natural, the first thing you need to get rid of the sign. Technically, this method is not complicated, but sometimes it can lead to trouble. For example, equation V (2x-5) \u003d V (4x-7). Establishing both sides in the square, you will receive 2x-5 \u003d 4x-7. Such an equation will not be difficult to solve; x \u003d 1. But the number 1 will not be this equations. Why? Submold the unit into the equation instead of the value of X. and in the right and in the left part will contain expressions that do not make sense, that is. This value is not permissible for square root. Therefore, 1 is an extraneous root, and therefore this equation does not have roots.

So, the irrational equation is solved using the construction method of both parts into the square. And solving the equation, it is necessary to cut off foreign roots. To do this, substitute the roots found in the original equation.

Consider another one.
2x + Vx-3 \u003d 0
Of course, this equation can be solved by the same as the previous one. Transfer composure equationsnot having a square root, on the right side and further use the method of exercise into the square. solve the resulting rational equation and roots. But another, more elegant. Enter a new variable; Vx \u003d y. Accordingly, you will receive the equation of the 2Y2 + Y-3 \u003d 0 equation. That is the usual square equation. Find it the roots; Y1 \u003d 1 and Y2 \u003d -3 / 2. Next, decide two equations Vx \u003d 1; Vx \u003d -3 / 2. The second equation of roots does not have, from the first we find that x \u003d 1. Do not forget about the need to check the roots.

It is easy to solve identities. To do this, you need to make identical conversions until the target is reached. Thus, with the help of simple arithmetic actions, the task will be solved.

You will need

• - paper;
• - a pen.

Instruction

The simplest of such transformations is algebraic abbreviated multiplication (such as the sum of the sum (difference), the difference of squares, the amount (difference), the cube amount (difference)). In addition, there are many and trigonometric formulas that are inherently those identities.

Indeed, the square of the sum of the two components is equal to the square of the first plus a twisted product of the first to the second and plus the square of the second, that is, (a + b) ^ 2 \u003d (a + b) (a + b) \u003d a ^ 2 + ab + ba + b ^ 2 \u003d a ^ 2 + 2ab + b ^ 2.

Simplify both

General principles of solution

Repeat on the textbook on mathematical analysis or higher mathematics, which is a specific integral. As you know, the solution of a specific integral is a function, the derivative of which will give a source expression. This feature is called primitive. According to this principle, the main integrals are built.
Determine the view of the integrated function, which of the table integrals is suitable in this case. It is not always possible to determine this immediately. Often, a tabular view becomes noticeable after several transformations to simplify the integrand function.

Method of replacement of variables

If the integrand is a trigonometric function, in the argument which is some polynomial, then try using the method of replacement of variables. In order to do this, replace the polynomial standing in the argument of the integrand function, on some new variable. By the ratio between the new and old variable, determine the new integration limits. Differentiation of this expression Find a new differential in. Thus, you will get a new type of previous integral, close or even corresponding to any table.

The solution of the integrals of the second kind

If the integral is the integral of the second kind, the vector view of the integrated function, then you will need to use the transition rules from these integrals to scalar. One of these rules is the ratio of Ostrograd Gauss. This law allows you to move from the flux of the rotor of some vector function to the triple integral on the divergence of this vector field.

Entering integration limits

After finding a primary need to substitute the integration limits. First substitute the value of the upper limit in the expression for the primitive. You will get a number. Next, delete another number from the resulting number, the resulting lower limit in the primordial one. If one of the integration limits is infinity, then when substituting it in a primitive function, you must go to the limit and find what the expression seeks.
If the integral is a two-dimensional or three-dimensional, then you will have to portray geometrically integration limits to understand how to count the integral. Indeed, in the case, say, the three-dimensional integral of the integration limits may be whole planes that limit the integrable volume.

Preparation for final testing in mathematics includes an important section - "logarithms". Tasks from this topic are necessarily found in the USE. The experience of past years shows that the logarithmic equations caused difficulties from many schoolchildren. Therefore, to understand how to find the right answer, and students with different levels of preparation must be operatively to fully cope with them.

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When preparing for a single state examination, graduates of high schools require a reliable source that provides the most complete and accurate information for the successful solution of test tasks. However, the textbook does not always turn out to be at hand, and the search for the necessary rules and formulas on the Internet often takes time.

The educational portal "Shkolkovo" allows you to prepare for the exam in any place at any time. You can find the most convenient approach to the repetition and assimilation of a large number of information on logarithms, as well as with one and several unknown. Start with light equations. If you coped with them without difficulty, go to more complex. If you have any problems with solving a certain inequality, you can add it to "Favorites" to return to it later.

Find the necessary formulas to perform the task, repeat the special cases and methods for calculating the root of the standard logarithmic equation, you can, viewing the "Theoretical Help" section. Teachers "Shkolkovo" collected, systematized and outlined all the materials necessary for the successful delivery in the most simple and understandable form.

To cope without difficulty with the tasks of any complexity, on our portal you can familiarize yourself with the solution of some typical logarithmic equations. To do this, go to the "Catalogs" section. We have a large number of examples, including with the equations of the profile level of the exam in mathematics.

Pupils from schools throughout Russia can take advantage of our portal. To start classes, simply register in the system and proceed to solve equations. To secure the results, we advise you to return to the site "Skolkovo" daily.

Logarithmic equations. From simple - to complex.

Attention!
This topic has additional
Materials in a special section 555.
For those who are strongly "not very ..."
And for those who are "very ...")

What is a logarithmic equation?

This is the equation with logarithms. So surprised, yes?) Then I will clarify. This is an equation in which unknown (Xers) and expressions with them are inside logarithms. And only there! It is important.

Here are examples logarithmic equations:

log 3 x \u003d Log 3 9

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

log X + 1 (x 2 + 3x-7) \u003d 2

lG 2 (x + 1) +10 \u003d 11lg (x + 1)

Well, you understood ... )

Note! A variety of expressions with cavities are located exceptionally inside logarithms. If, suddenly, the equation will be detected by X somewhere outside, eg:

log 2 x \u003d 3 + x,

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

What to say here? Lucky you, if it got like that! Logarithm with numbers is some number.And that's it. It is enough to know the properties of logarithms to solve such an equation. Knowledge of special rules, techniques adapted to solve logarithmic equations Not required here.

So, what is a logarithmic equation - figured out.

How to solve logarithmic equations?

Decision logarithmic equations - The thing, actually, not very simple. So and the section with us - on the fourth ... It requires a decent knowledge supply for every adjacent topics. In addition, there is a special feature in these equations. And the chip is so important that it can be safely called the main problem in solving logarithmic equations. We will understand in detail in detail this problem in the next lesson.

And now - do not worry. We will go right way from simple to complex. On specific examples. The main thing is to get into simple things and do not be lazy to walk along the links, I didn't just put them so much ... and everything will turn out. Necessarily.

Let's start with the most elementary, simplest equations. To solve them, it is desirable to have an idea of \u200b\u200blogarithm, but no more. Just without concept logarithm To take a decision logarithmic Equations - somehow awkwardly even ... Very boldly, I would say).

The simplest logarithmic equations.

These are the equations of the form:

1. Log 3 x \u003d Log 3 9

2. Log 7 (2x-3) \u003d log 7 x

3. Log 7 (50x-1) \u003d 2

Process solution any logarithmic equation It is to transition from the equation with logarithms to the equation without them. In the simplest equations, this transition is carried out in one step. Therefore, the simplest.)

And such logarithmic equations are surprisingly solved. See yourself.

We solve the first example:

log 3 x \u003d Log 3 9

To solve this example, nothing to know and do not need, yes ... purely intuition!) What do we especially Do not like this example? What ... Logarithms do not like! Right. So get rid of them. We look closely for example, and we have a natural desire ... continuously insurmountable! Take and throw out logarithms at all. And what pleases it can To do! Mathematics allows. Logarithms disappear It turns out the answer:

Great, right? So it is possible (and necessary) to do always. The elimination of logarithms is similarly similar - one of the basic 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 there are few of them. Remember:

Liquidation of logarithms without any concerns, if they have:

a) identical numeric bases

c) the logarithms on the left-right clean (without any coefficients) and are in proud loneliness.

I will explain the last item. In the equation, let's say

log 3 x \u003d 2log 3 (3x-1)

it is impossible to remove logarithms. Two right does not allow. Coefficient, you understand ... In the example

log 3 x + Log 3 (x + 1) \u003d log 3 (3 + x)

also can not be potential equation. There is no lonely logarithm in the left side. There are two of them.

In short, it is possible to remove logarithms if the equation looks like this and just like this:

log A (.....) \u003d Log A (.....)

In brackets where the ellipsis may be any expression. Simple, superstrate, all sorts. Any please. It is important that after the elimination of logarithms we have remain more simple equation.Of course, it is expected to solve linear, square, fractional, indicative and other equations without logarithms you already know how.)

Now you can easily solve the second example:

log 7 (2x-3) \u003d log 7 x

Actually, in mind is solved. We will potentiate, we get:

Well, very difficult?) As you can see logarithmic part of the solution of the equation is only in the elimination of logarithms ... And then there is a decision of the remaining equation already without them. Trivial.

We solve the third example:

log 7 (50x-1) \u003d 2

We see that the left is the logarithm:

We remember that this logarithm is some number in which the base should be undertaken (ie, seven) to get a legitimate expression, i.e. (50x-1).

But this number is two! By equation. That is:

Here, in essence, and that's it. Logarithm disappeared The harmless equation remains:

We solved this logarithmic equation based only on the meaning of logarithm. What, eliminating logarithms is still easier?) I agree. By the way, if you make from the two logarithm, you can solve this example and through the elimination. From any number you can make logarithm. And, what we need. Very useful technique in solving logarithmic equations and (especially!) Inequalities.

Do not know how to do the logarithm!? Nothing wrong. In section 555, this admission is described in detail. You can master and apply it to the full coil! It greatly reduces the number of errors.

Completely similar to (by definition), the fourth equation is solved:

That's all things.

Let's summarize this lesson. We looked at the examples of the simplest logarithmic equations. It is very important. And not only because such equations are in testing exams. The fact is that even the most evil and freezed equations are necessarily reduced to the simplest!

Actually, the simplest equations are the finish part of the decision any equations. And this finish part must be understood iron! And further. Be sure to read this page to the end. There is a surprise ...)

We decide now yourself. Put your hand, so to speak ...)

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

ln (7x + 2) \u003d ln (5x + 20)

log 2 (x 2 +32) \u003d log 2 (12x)

log 16 (0,5x-1,5) \u003d 0.25

log 0.2 (3x-1) \u003d -3

ln (e 2 + 2x-3) \u003d 2

log 2 (14x) \u003d log 2 7 + 2

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

What, not everything turns out? It happens. Do not grieve! In section 555, the solution of all these examples is painted and in detail. There will definitely understand. Yes, and useful practical techniques are mastered.

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

It's time to open the bitter truth to you. Successful solution of these examples does not guarantee success in solving all other logarithmic equations. Even the simplest like this. Alas.

The fact is that the solution of any logarithmic equation (even the most elementary!) Consists of two equal parts. Solution of the equation, and work with OTZ. One part is the solution of the equation itself - we have mastered. Not so hard right?

For this lesson, I specifically picked such examples in which OTZ does not affect the response. But not all so kind of good, how are I, right? ...)

Therefore, it is necessary to master the other part. Odd This is the main problem in solving logarithmic equations. And not because difficult - this part is even easier. And because about OTZ just forget. Or do not know. Or both). And fall in the same place ...

In the next lesson, we will deal with this problem. Then it will be possible to confidently decide any Uncomplicated logarithmic equations and seamless to quite solid tasks.

If you like this site ...

By the way, I have another couple of interesting sites for you.)

It can be accessed in solving examples and find out your level. Testing with instant check. Learn - with interest!)

You can get acquainted with features and derivatives.