Rules for solving logarithmic equations. Logarithmic equation: basic formulas and techniques

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

The simplest equation is log a x = b, where a and b are some numbers, x is an unknown.
Solving the logarithmic equation is x = a b provided: a > 0, a 1.

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

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

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

Solving simple logarithmic equations

These include equations like log 2 x \u003d log 2 16. It can be seen with the naked eye that by omitting the sign of the logarithm we get x \u003d 16.

In order to solve a more complex logarithmic equation, it is usually led to the solution of an ordinary algebraic equation or to the solution of the simplest logarithmic equation log a x = b. In the simplest equations, this occurs in one movement, which is why they are called the simplest.

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

logarithms have the same numerical bases
logarithms in both parts of the equation are free, i.e. without any coefficients and other various kinds of expressions.

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

In general, you can remove logarithms only if the equation has the form:

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

Absolutely any expressions can be in brackets, this absolutely does not affect the potentiation operation. And after the elimination of logarithms, a simpler equation will remain - linear, quadratic, exponential, etc., which you already, I hope, know how to solve.

Let's take another example:

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

Applying potentiation, we get:

log 3 (2x-1) = 2

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

Again, we got a nice answer. Here we did without the elimination of logarithms, but potentiation is applicable here too, because the logarithm can be made from any number, and exactly the one that we need. This method is very helpful in solving logarithmic equations and especially inequalities.

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

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

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

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

In all that we have done above, we have overlooked one very important point, which will play a decisive role in the future. The fact is that the solution of any logarithmic equation, even the most elementary one, consists of two equivalent parts. The first is the solution of the equation itself, the second is work with the area of admissible values (ODV). That's just the first part we have mastered. In the above examples, the ODD does not affect the answer in any way, so we did not consider it.

Let's take another example:

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

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

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

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

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

We find the roots of the equation:

There are two roots.

Answer: 3 and -1

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

Let's start with x 1 = 3:

log 3 6 = log 3 6

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

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

Yes, stop! Externally, everything is perfect. One moment - there are no logarithms from negative numbers! And this means that the root x \u003d -1 is not suitable for solving our equation. And therefore the correct answer will be 3, not 2, as we wrote.

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

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

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

How could we get caught while solving a seemingly elementary example? And here it is at the moment of potentiation. The logarithms are gone, and with them all the limitations.

What to do in such a case? Refuse to eliminate logarithms? And completely abandon the solution of this equation?

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

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

Now let's decide how to write the ODZ. To do this, we carefully examine the original equation and look for suspicious places in it, such as division by x, the root of an even degree, etc. Until we have solved the equation, we do not know what x is equal to, but we know for sure that such x, which, when substituting, will give a division by 0 or the extraction of the square root of a negative number, are obviously not suitable for the answer. Therefore, such x's are unacceptable, while the rest will constitute the ODZ.

Let's use the same equation again:

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

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

Those. we have not solved anything yet, but we have already written down a mandatory condition for the entire sublogarithmic expression. The curly brace means that these conditions must be met at the same time.

The ODZ is written down, but it is also necessary to solve the resulting system of inequalities, which we will do. We get the answer x > v3. Now we know for sure which x will not suit us. And then we start solving the logarithmic equation itself, which we did above.

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

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

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

In the video, other examples of solving the log. equations and working out the method of intervals in practice.

To this on the subject, how to solve logarithmic equations until everything. If something according to the decision of the log. equations remained unclear or incomprehensible, write your questions in the comments.

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

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 past years shows that the logarithmic equations 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.

Pass the certification test successfully with the help of the educational portal "Shkolkovo"!

When preparing for the unified state exam, high school graduates need a reliable source that provides the most complete and accurate information for the successful solution of test problems. However, the textbook is not always at hand, and searching for the necessary rules and formulas on the Internet often takes time.

The educational portal "Shkolkovo" allows you to prepare for the exam anywhere at any time. Our site offers the most convenient approach to repeating and mastering a large amount of information on logarithms, as well as on one and several unknowns. Start with easy equations. If you coped with them without difficulty, move on to more difficult ones. If you're having trouble solving a particular inequality, you can add it to your Favorites so you can come back to it later.

You can find the necessary formulas to complete the task, repeat special cases and methods for calculating the root of a standard logarithmic equation by looking at the "Theoretical Reference" section. Teachers of "Shkolkovo" collected, systematized and presented all the materials necessary for successful delivery in the most simple and understandable form.

In order to easily cope with 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 presented a large number of examples, including those with equations of the profile level of the Unified State Examination in mathematics.

Students from schools all over Russia can use our portal. To get started, just register in the system and start solving equations. To consolidate the results, we advise you to return to the Shkolkovo website daily.

Instruction

Write down the given logarithmic expression. If the expression uses the logarithm of 10, then its notation is shortened and looks like this: lg b is the decimal logarithm. If the logarithm has the number e as the base, then the expression is written: ln b is the natural logarithm. It is understood that the result of any is the power to which the base number must be raised to get the number b.

When finding two functions from the sum, you just need to differentiate them one by one, and add the results: (u+v)" = u"+v";

When finding the derivative of the product of two functions, it is necessary to multiply the derivative of the first function by the second and add the derivative of the second function, multiplied by the first function: (u*v)" = u"*v+v"*u;

In order to find the derivative of the quotient of two functions, it is necessary, from the product of the derivative of the dividend multiplied by the divisor function, to subtract the product of the derivative of the divisor multiplied by the divisor function, and divide all this by the divisor function squared. (u/v)" = (u"*v-v"*u)/v^2;

If a complex function is given, then it is necessary to multiply the derivative of the inner function and the derivative of the outer one. Let y=u(v(x)), then y"(x)=y"(u)*v"(x).

Using the obtained above, you can differentiate almost any function. So let's look at a few examples:

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

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

2) Calculate the value of the function at the given point y"(1)=8*e^0=8

General principles of solution

Repeat from a textbook on mathematical analysis or higher mathematics, which is a definite integral. As you know, the solution of a definite integral is a function whose derivative will give an integrand. This function is called antiderivative. According to this principle, the basic integrals are constructed.
Determine by the form of the integrand which of the table integrals is suitable in this case. It is not always possible to determine this immediately. Often, the tabular form becomes noticeable only after several transformations to simplify the integrand.

Variable substitution method

If the integrand is a trigonometric function whose argument is some polynomial, then try using the change of variables method. To do this, replace the polynomial in the argument of the integrand with some new variable. Based on the ratio between the new and old variable, determine the new limits of integration. By differentiating this expression, find a new differential in . Thus, you will get a new form of the old integral, close or even corresponding to any tabular one.

Solution of integrals of the second kind

If the integral is an integral of the second kind, the vector form of the integrand, then you will need to use the rules for moving from these integrals to scalar ones. One such rule is the Ostrogradsky-Gauss ratio. This law makes it possible to pass from the rotor flow of some vector function to a triple integral over the divergence of a given vector field.

Substitution of limits of integration

After finding the antiderivative, it is necessary to substitute the limits of integration. First, substitute the value of the upper limit into the expression for the antiderivative. You will receive some number. Next, subtract from the resulting number another number, the resulting lower limit to the antiderivative. If one of the integration limits is infinity, then when substituting it into the antiderivative function, it is necessary to go to the limit and find what the expression tends to.
If the integral is two-dimensional or three-dimensional, then you will have to represent the geometric limits of integration in order to understand how to calculate the integral. Indeed, in the case of, say, a three-dimensional integral, the limits of integration can be entire planes that limit the volume to be integrated.

Algebra Grade 11

Topic: "Methods for solving logarithmic equations"

Lesson Objectives:

educational: the formation of knowledge about different ways of solving logarithmic equations, the ability to apply them in each specific situation and choose any method for solving;

developing: development of skills to observe, compare, apply knowledge in a new situation, identify patterns, generalize; formation of skills of mutual control and self-control;

educational: education of a responsible attitude to educational work, careful perception of the material in the lesson, accuracy of record keeping.

Lesson type: a lesson of familiarization with new material.

"The invention of logarithms, by shortening the work of the astronomer, has lengthened his life."
French mathematician and astronomer P.S. Laplace

During the classes

I. Setting the goal of the lesson

The studied definition of the logarithm, the properties of logarithms and the logarithmic function will allow us to solve logarithmic equations. All logarithmic equations, no matter how complex they are, are solved using the same algorithms. We will consider these algorithms today in the lesson. There are few of them. If you master them, then any equation with logarithms will be feasible for each of you.

Write in your notebook the topic of the lesson: "Methods for solving logarithmic equations." I invite everyone to cooperation.

II. Updating of basic knowledge

Let's get ready to study the topic of the lesson. You solve each task and write down the answer, you can not write the condition. Work in pairs.

1) For what values of x does the function make sense:

(Answers are checked for each slide and errors are sorted out)

2) Do the function graphs match?

3) Rewrite the equalities as logarithmic equalities:

4) Write the numbers as logarithms with base 2:

5) Calculate:

6) Try to restore or complete the missing elements in these equalities.

III. Introduction to new material

The statement is shown on the screen:

"The equation is the golden key that unlocks all mathematical sesame."
Modern Polish mathematician S. Koval

Try to formulate the definition of a logarithmic equation. (An equation containing the unknown under the sign of the logarithm).

Consider the simplest logarithmic equation:logax = b(where a>0, a ≠ 1). Since the logarithmic function increases (or decreases) on the set of positive numbers and takes all real values, it follows from the root theorem that for any b, this equation has, and moreover, only one solution, and a positive one.

Remember the definition of a logarithm. (The logarithm of the number x to the base a is the exponent to which the base a must be raised to get the number x). It immediately follows from the definition of the logarithm that ain is such a solution.

Write down the title: Methods for solving logarithmic equations

1. By definition of the logarithm.

This is how simple equations of the form are solved.

Consider No. 514(a): Solve the equation

How do you propose to solve it? (By definition of logarithm)

Decision. , Hence 2x - 4 = 4; x = 4.

In this task, 2x - 4 > 0, since > 0, therefore, extraneous roots cannot appear, and there is no need to check. The condition 2x - 4 > 0 is not necessary to write out in this task.

2. Potentiation(transition from the logarithm of the given expression to this expression itself).

Consider No. 519(g): log5(x2+8)-log5(x+1)=3log5 2

What feature did you notice? (The bases are the same and the logarithms of the two expressions are equal). What can be done? (potentiate).

In this case, it should be taken into account that any solution is contained among all x for which the logarithm expressions are positive.

Solution: ODZ:

X2+8>0 extra inequality

log5(x2+8) = log5 23+ log5(x+1)

log5(x2+8)=log5(8x+8)

Potentiate the original equation

we get the equation x2+8= 8x+8

We solve it: x2-8x=0

Answer: 0; eight

In general transition to an equivalent system:

The equation

(The system contains a redundant condition - one of the inequalities can be ignored).

Question to the class: Which of these three solutions did you like the most? (Discussion of methods).

You have the right to decide in any way.

3. Introduction of a new variable.

Consider No. 520(g). .

What did you notice? (This is a quadratic equation for log3x) Any suggestions? (Introduce new variable)

Decision. ODZ: x > 0.

Let , then the equation will take the form:. Discriminant D > 0. Roots by Vieta's theorem:.

Let's return to the replacement: or .

Solving the simplest logarithmic equations, we get:

Answer: 27;

4. Logarithm of both sides of the equation.

Solve the equation:.

Solution: ODZ: x>0, take the logarithm of both sides of the equation in base 10:

Apply the property of the logarithm of the degree:

(lgx + 3) lgx = 4

Let lgx = y, then (y + 3)y = 4

, (D > 0) the roots according to the Vieta theorem: y1 = -4 and y2 = 1.

Let's return to the replacement, we get: lgx = -4,; logx = 1, .

Answer: 0.0001; ten.

5. Reduction to one base.

No. 523(c). Solve the equation:

Solution: ODZ: x>0. Let's move on to base 3.

6. Functional-graphical method.

№ 509(d). Solve graphically the equation: = 3 - x.

How do you propose to solve? (Construct graphs of two functions y \u003d log2x and y \u003d 3 - x by points and look for the abscissa of the intersection points of the graphs).

See your solution on the slide.

Is there a way to avoid plotting . It is as follows : if one of the functions y = f(x) increases and the other y = g(x) decreases on the interval X, then the equation f(x)=g(x) has at most one root on the interval X.

If there is a root, then it can be guessed.

In our case, the function increases for x>0, and the function y \u003d 3 - x decreases for all values of x, including x>0, which means that the equation has no more than one root. Note that for x = 2, the equation turns into a true equality, since .

“The correct application of methods can be learned,
only by applying them to various examples.
Danish historian of mathematics G. G. Zeiten

IV. Homework

P. 39 consider example 3, solve No. 514 (b), No. 529 (b), No. 520 (b), No. 523 (b)

V. Summing up the lesson

What methods for solving logarithmic equations did we consider in the lesson?

In the next lessons, we will look at more complex equations. To solve them, the studied methods are useful.

Showing the last slide:

“What is more than anything in the world?
Space.
What is the wisest?
Time.
What is the most enjoyable?
Achieve what you want."
Thales

I want everyone to achieve what they want. Thank you for your cooperation and understanding.

With this video, I begin a long series of lessons about logarithmic equations. Now you have three examples at once, on the basis of which we will learn to solve the simplest tasks, 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

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

Basic solution methods

There are many ways to solve such structures. For example, most teachers at school suggest this way: Immediately express the function f ( x ) using the formula f( x ) = a b . That is, when you meet the simplest construction, you can immediately proceed 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 does it come from and why exactly we raise the letter a to the letter b.

As a result, I often observe very offensive errors, when, for example, these letters are interchanged. This formula must either be understood or memorized, and the second method leads to errors at the most inopportune and most crucial moments: in exams, tests, etc.

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

The idea of the canonical form is simple. Let's look at our task again: on the left we have log a , while the letter a means exactly the number, and in no case the function containing the 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 must be equal to the number b, and no restrictions are imposed on this letter, because it can take any value - both positive and negative. It all depends on what values the function f(x) takes.

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

b = log a a b

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

b = b 1 = b log a a

Of course, in this case, all the restrictions that we wrote down at the beginning arise. And now let's use the basic property of the logarithm, and enter 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 in the following form:

log a f (x) = log a a b → f (x) = a b

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

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

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

log a f(x) = log a a b

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

Solution examples

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

log 0.5 (3x - 1) = -3

Let's rewrite it like this:

log 0.5 (3x − 1) = log 0.5 0.5 −3

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

However, if now you are just starting to study this topic, it is better not to rush anywhere so as not to make offensive mistakes. So we have 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 power of −3. Note that 0.5 is 1/2.

(1/2) −3 = (2/1) 3 = 8

Convert all decimals to fractions when you solve a logarithmic equation.

We rewrite and get:

3x − 1 = 8
3x=9
x=3

All we got the answer. The first task is 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 a 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 closer look at the bases: on the left is the number under the root:

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

Now we recall the remarkable property of the logarithm: from the argument, as well as from the base, you can take out degrees. In the case of bases, the following happens:

log a k b = 1/k loga b

In other words, the number that stood in the degree of the base is brought forward and at the same time turned over, that is, it becomes the reciprocal of the number. In our case, there was a degree of base with an indicator of 1/2. Therefore, we can take it out as 2/1. We get:

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

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

9 log 5 x = 18
log 5 x = 2

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

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

That's all. The second problem is solved.

Third example

Let's move on to the third task:

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

Recall the following formula:

log b = log 10 b

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

It is precisely 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 inserted and becomes a power of base 5. In addition, the free term 3 can also be represented as a logarithm - this is very easy to observe from our notation.

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

3 = log 10 10 3 = log 10 3

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

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

Before us is again the canonical form, and we obtained it bypassing the stage of transformations, i.e., the simplest logarithmic equation did not come up anywhere with us.

That's what I was talking about at the very beginning of the lesson. The canonical form allows solving a wider class of problems than the standard school formula, which is given by most school teachers.

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 = 24997

All! Problem solved.

A note about scope

Here I would like to make an important remark about the domain 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 that this inequality be satisfied?

Do not worry. No extra roots will appear in these cases. And this is another great trick that allows you to speed up the solution. Just know that if in the problem the variable x occurs only in one place (or rather, in the one and only argument of the one and only logarithm), and nowhere else in our case does the 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, i.e., the argument should be equal to 8. This automatically means that 3x - 1 will be greater than zero.

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

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

But let's be honest: in order to finally understand this technique, in order to learn how to apply the canonical form of the logarithmic equation, it is not enough just to watch one video lesson. Therefore, right now, download the options for an independent solution 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 lesson will help you understand logarithmic equations. Apply the canonical form, simplify expressions using the rules for working with logarithms - and you will not be afraid of any tasks. And that's all I have for today.

Scope consideration

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 - it has only one function, and the numbers a and b are just numbers, and in no case are 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 substituting into our original expression, we get the following:

log a f(x) = log a a b

f(x) = a b

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

f(x) > 0

This restriction is valid 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 in any case is 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, we will still get a positive number at the output. 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 can be quite complex designs, and in the process of solving them, you must definitely follow them. Let's get a look.

First task:

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

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

Of the obtained roots, only the first one suits us, since the second root is less than zero. The only answer will be the number 9. That's it, the problem is solved. No additional checks that the expression under the logarithm sign is greater than 0 are 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 automatically satisfied.

Let's move on to the second task:

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

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

We square both parts, taking into account the restrictions, and we get:

4 - 6x - x 2 = (x - 4) 2

4 - 6x - x 2 = x 2 + 8x + 16

x2 + 8x + 16 −4 + 6x + x2 = 0

2x2 + 14x + 12 = 0 |:2

x2 + 7x + 6 = 0

We solve the resulting equation through the discriminant:

D \u003d 49 - 24 \u003d 25

x 1 = −1

x 2 \u003d -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 all the solution. Let's go back to the very beginning of our calculations.

The main conclusion from this lesson is that it is not required to check the limits for a function in the simplest logarithmic equations. Because in the process of solving all the constraints are executed automatically.

However, this by no means means that you can forget about verification 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 side, which we have seen today in two different examples.

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

Logarithmic equations with different bases

We continue to study logarithmic equations and 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 just 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. For this, we note that

b = log a a b

And a b is just an argument. Let's rewrite this expression as follows:

log a f(x) = log a a b

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

f(x) = a b

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

So the first design:

First of all, I note that there is a fraction on the right, the denominator of which is log. When you see an expression like this, it's worth remembering 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 c. Of course, 0< с ≠ 1.

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

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

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

We recall that any degree can be taken out of 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 entry as a canonical form (after all, in the canonical form, there is no additional factor in front of the second logarithm). Therefore, let's put the fraction 1/4 in the argument as a power:

Now we equate the 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. Pay attention: 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:

log 56 = log 2 log 2 7 − 3 log (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 unprepared student that this is some kind of tin, but in fact everything is solved elementarily.

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

log a b n = n log 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 apply this formula to the expression lg 2 log 2 7. Don't be afraid of lg 2 - this is the most common expression. You can rewrite it like this:

For him, all the rules that apply to any other logarithm are valid. In particular, the factor in front can be introduced into the power of the argument. Let's write:

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

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

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

lg (56/7) = -3lg (x + 4)

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

lg 8 = lg (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.

Let me recap the key points of this lesson.

The main formula that is studied in all the lessons on this page devoted to solving logarithmic equations is the canonical form. And don't be put off by the fact that most school textbooks teach you how to solve these kinds of problems differently. This tool works very efficiently and allows you to solve a much wider class of problems than the simplest ones that we studied at the very beginning of our lesson.

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

The formula for moving to one base and a special case when we flip log (this was very useful to us in the first task);
The formula for bringing in and taking out powers from under the sign of the logarithm. Here, many students get stuck and do not see point-blank that the power taken out and brought in can itself contain log f (x). Nothing wrong with that. We can introduce one log according to the sign of another and at the same time significantly simplify the solution of the problem, which is what we observe in the second case.

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

Problems with variable base

Today we will consider logarithmic equations, which for many students seem non-standard, if not completely unsolvable. We are talking about expressions that are 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 recall how the simplest problems are solved, which are based on ordinary numbers. So, the simplest construction is called

log a f(x) = b

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

b = log a a b

We rewrite our original expression and get:

log a f(x) = log a a b

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

f(x) = a b

Thus, we get rid of the log sign and solve the usual problem. In this case, the roots obtained 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 this 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 , which was to the right of the equal sign. So let's rewrite our expression. We get:

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

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

2x2 - 13x + 18 = x - 2

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

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

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

2x 2 − 13x + 18 > 0

x − 2 > 0

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

x − 2 ≠ 1

As a result, we get the system:

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

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

In this case, if we require that x − 2 > 0, then the requirement 2x 2 − 13x + 18 > 0 will also be automatically satisfied. Therefore, we can safely cross out the inequality containing a 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 the linear inequality, i.e. cross out 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 quadratic, even if as a result of solving this entire system we get the same roots.

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

Let's rewrite our system:

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

2x2 - 14x + 20 = 0

x2 − 7x + 10 = 0

Before us is a reduced square trinomial and, therefore, we can use the Vieta formulas. We get:

(x − 5)(x − 2) = 0

x 1 = 5

x2 = 2

Now, back to our system, we find that x = 2 doesn't suit us, because we're required to have x strictly greater than 2.

But x \u003d 5 suits us quite well: 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 \u003d 5.

Everything, the task is solved, including taking into account the ODZ. Let's move on to the second equation. Here we are waiting for more interesting and meaningful calculations:

The first step: as well as last time, we bring all this business to a canonical form. To do this, we can write the number 9 as follows:

The base with the root can not be touched, but it is better to transform the argument. Let's move from the root to the power with a rational exponent. Let's write:

Let me not rewrite our whole big logarithmic equation, but just immediately equate the arguments:

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

x 2 + 4x + 3 = 0

Before us is the again reduced square trinomial, we will use the Vieta 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, log signs impose additional restrictions (here we would have to write down the system, but due to the cumbersomeness of the whole construction, I decided to calculate the domain of definition separately).

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

These are the requirements imposed by the domain of definition.

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

In addition, note that the solutions of 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 the root of the third degree - these inequalities are completely similar, so one of them we can cross it out).

But with the third inequality, this will not work. Let's get rid of the sign of the radical on the left, for which we raise both parts to a cube. We get:

So we get the following requirements:

−2 ≠ x > −3

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

Once again, the key points of this task:

Feel free to apply and solve logarithmic equations using 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 errors than those who are in a hurry somewhere, skipping intermediate steps of calculations;
As soon as a variable base appears in the logarithm, the problem ceases to be the simplest. Therefore, when solving it, it is necessary to take into account the domain of definition: the arguments must be greater than zero, and the bases must not only be greater than 0, but they must also not be equal to 1.

You can impose the last requirements on the final answers in different ways. For example, it is possible to solve a whole system containing all domain requirements. 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 apply it to the obtained roots.

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

Portal for the student. Self-training

Solving simple logarithmic equations

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General principles of solution

Variable substitution method

Solution of integrals of the second kind

Substitution of limits of integration

Basic solution methods

Solution examples

Second task

Third example

A note about scope

Scope consideration

Logarithmic equations with different bases

Problems with variable base

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