Solving complex logarithmic equations ege. Protection of personal information. Rules and some restrictions

Logarithmic expressions, solution of examples. In this article, we will consider problems related to solving logarithms. The tasks raise the question of finding the value of the expression. It should be noted that the concept of the logarithm is used in many tasks and it is extremely important to understand its meaning. As for the USE, the logarithm is used in solving equations, in applied problems, and also in tasks related to the study of functions.

Here are examples to understand the very meaning of the logarithm:

Basic logarithmic identity:

Properties of logarithms that you must always remember:

*The logarithm of the product is equal to the sum of the logarithms of the factors.

* * *

* The logarithm of the quotient (fraction) is equal to the difference of the logarithms of the factors.

* * *

* The logarithm of the degree is equal to the product of the exponent and the logarithm of its base.

* * *

*Transition to new base

* * *

More properties:

* * *

Computing logarithms is closely related to using the properties of exponents.

We list some of them:

The essence of this property is that when transferring the numerator to the denominator and vice versa, the sign of the exponent changes to the opposite. For example:

Consequence of this property:

* * *

When raising a power to a power, the base remains the same, but the exponents are multiplied.

* * *

As you can see, the very concept of the logarithm is simple. The main thing is that good practice is needed, which gives a certain skill. Certainly knowledge of formulas is obligatory. If the skill in converting elementary logarithms is not formed, then when solving simple tasks, one can easily make a mistake.

Practice, solve the simplest examples from the math course first, then move on to more complex ones. In the future, I will definitely show how the “ugly” logarithms are solved, there will be no such ones at the exam, but they are of interest, do not miss it!

That's all! Good luck to you!

Sincerely, Alexander Krutitskikh

P.S: I would be grateful if you tell about the site in social networks.

In this video tutorial, we will look at solving a rather serious logarithmic equation, in which you not only need to find the roots, but also select those that lie on a given segment.

Task C1. Solve the equation. Find all the roots of this equation that belong to the interval.

A note about logarithmic equations

However, from year to year, students come to me who try to solve such, frankly, difficult equations, but at the same time they cannot understand: where do they start at all and how to approach logarithms? Such a problem can arise even in strong, well-prepared students.

As a result, many begin to fear this topic, or even consider themselves stupid. So, remember: if you can’t solve such an equation, it doesn’t mean at all that you are stupid. Because, for example, you can deal with this equation almost verbally:

log 2 x = 4

And if this is not so, you would not be reading this text now, because you were busy with simpler and more mundane tasks. Of course, someone will now object: “What does this simplest equation have to do with our healthy design?” I answer: any logarithmic equation, no matter how complex it may be, eventually comes down to such simple, verbally solved constructions.

Of course, it is necessary to move from complex logarithmic equations to simpler ones not with the help of selection or dancing with a tambourine, but according to clear, long-defined rules, which are called so - rules for converting logarithmic expressions. Knowing them, you can easily figure out even the most sophisticated equations in the exam in mathematics.

And it is about these rules that we will talk in today's lesson. Go!

Solving the logarithmic equation in problem C1

So let's solve the equation:

First of all, when it comes to logarithmic equations, we recall the main tactic - if I may say, the basic rule for solving logarithmic equations. It consists in the following:

Canonical form theorem. Any logarithmic equation, no matter what it includes, no matter what logarithms, no matter what the base, and no matter what c has in itself, it is necessary to bring it to an equation of the form:

log a f (x ) = log a g (x )

If we look at our equation, we immediately notice two problems:

1. On the left we have the sum of two numbers, one of which is not a logarithm at all.
2. On the right is quite a logarithm, but at its base is a root. And the logarithm on the left has just 2, i.e. the bases of the logarithms on the left and on the right are different.

So we've come up with a list of issues that separate our equation from that canonical equation, to which you need to reduce any logarithmic equation in the process of solving. Thus, solving our equation at this stage boils down to eliminating the two problems described above.

Any logarithmic equation can be solved quickly and easily if reduced to its canonical form.

The sum of logarithms and the logarithm of the product

Let's proceed in order. First, let's deal with the structure that stands on the left. What can we say about the sum of two logarithms? Let's remember the wonderful formula:

log a f (x) + log a g (x) = log a f (x) g (x)

But it is worth considering that in our case the first term is not a logarithm at all. So, you need to represent the unit as a logarithm to base 2 (namely 2, because the logarithm to base 2 is on the left). How to do it? Again, remember the wonderful formula:

a = log b b a

Here you need to understand: when we say “Any base b”, then we mean that b still cannot be an arbitrary number. If we insert a number into the logarithm, certain numbers are immediately superimposed on it. restrictions, namely: the base of the logarithm must be greater than 0 and must not be equal to 1. Otherwise, the logarithm simply does not make sense. Let's write it down:

0 < b ≠ 1

Let's see what happens in our case:

1 = log 2 2 1 = log 2 2

Now let's rewrite our entire equation with this fact in mind. And immediately we apply another rule: the sum of logarithms is equal to the logarithm of the product of the arguments. As a result, we get:

We have a new equation. As you can see, it is already much closer to the canonical alignment that we are striving for. But there is one problem, we wrote it in the form of the second point: our logarithms, which are on the left and on the right, different grounds. Let's move on to the next step.

Rules for taking powers from the logarithm

So the logarithm on the left has a base of just 2, and the logarithm on the right has a root at the base. But this is not a problem either, if we remember that from the bases from the arguments of the logarithm can be taken out to a power. Let's write one of these rules:

log a b n = n log a b

Translating into human language: you can take out the degree from the base of the logarithm and put it in front as a factor. The number n "migrated" out of the logarithm and became a coefficient in front.

We might as well take the power out of the base of the logarithm. It will look like this:

In other words, if you take the power out of the logarithm argument, this power is also written as a factor in front of the logarithm, but not as a number, but as the reciprocal of 1/k.

However, that's not all! We can combine these two formulas and come up with the following formula:

When the exponent is in both the base and the argument of a logarithm, we can save time and simplify calculations by removing the exponents from both the base and the argument at once. In this case, what was in the argument (in our case, this is the coefficient n) will be in the numerator. And what was the degree at the base, a k , will go to the denominator.

And it is these formulas that we will now use in order to reduce our logarithms to the same base.

First of all, we will choose a more or less beautiful base. Obviously, the deuce at the base is much more pleasant to work with than with the root. So let's try to base 2 the second logarithm. Let's write this logarithm separately:

What can we do here? Recall the power formula with a rational exponent. In other words, we can write roots as a power with a rational exponent. And then we take out the power of 1/2 from both the argument and the base of the logarithm. We reduce the twos in the coefficients in the numerator and denominator in front of the logarithm:

Finally, we rewrite the original equation taking into account the new coefficients:

log 2 2(9x 2 + 5) = log 2 (8x 4 + 14)

We have obtained the canonical logarithmic equation. Both on the left and on the right we have a logarithm in the same base 2. In addition to these logarithms, there are no coefficients, no terms either on the left or on the right.

Consequently, we can get rid of the sign of the logarithm. Of course, taking into account the domain of definition. But before we do that, let's go back and make a little clarification about fractions.

Dividing a Fraction by a Fraction: Additional Considerations

Not all students understand where the factors in front of the right logarithm come from and where they go. Let's write it down again:

Let's understand what a fraction is. Let's write:

And now we recall the rule for dividing fractions: to divide by 1/2, you need to multiply by the inverted fraction:

Of course, for the convenience of further calculations, we can write the deuce as 2/1 - and this is exactly what we observe as the second coefficient in the solution process.

I hope now everyone understands where the second coefficient comes from, so we go directly to solving our canonical logarithmic equation.

Getting rid of the sign of the logarithm

I remind you that now we can get rid of the logarithms and leave the following expression:

2(9x2 + 5) = 8x4 + 14

Let's expand the brackets on the left. We get:

18x2 + 10 = 8x4 + 14

Let's move everything from the left side to the right:

8x4 + 14 - 18x2 - 10 = 0

We give similar ones and get:

8x4 - 18x2 + 4 = 0

We can divide both sides of this equation by 2 to simplify the coefficients, and we get:

4x4 - 9x2 + 2 = 0

Before us is the usual biquadratic equation, and its roots are easily calculated in terms of the discriminant. So let's write the discriminant:

D \u003d 81 - 4 4 2 \u003d 81 - 32 \u003d 49

Fine, the Discriminant is "beautiful", the root of it is 7. That's it, we consider the X's themselves. But in this case, the roots will turn out not x, but x 2, because we have a biquadratic equation. So our options are:

Please note: we extracted the roots, so there will be two answers, because. square - even function. And if we write only the root of two, then we will simply lose the second root.

Now we paint the second root of our biquadratic equation:

Again, we take the arithmetic square root of both sides of our equation and get two roots. However, remember:

It is not enough to simply equate the arguments of the logarithms in canonical form. Remember the scope!

In total, we got four roots. All of them are indeed solutions to our original equation. Take a look: in our original logarithmic equation, inside the logarithms is either 9x 2 + 5 (this function is always positive), or 8x 4 + 14 - it is also always positive. Therefore, the domain of definition of logarithms is satisfied in any case, no matter what root we get, which means that all four roots are solutions to our equation.

Great, now let's move on to the second part of the problem.

Selection of roots of a logarithmic equation on a segment

We select from our four roots those that lie on the interval [−1; 8/9]. We return to our roots, and now we will carry out their selection. To begin with, I propose to draw a coordinate axis and mark the ends of the segment on it:

Both points will be shaded. Those. by the condition of the problem, we are interested in the shaded segment. Now let's deal with the roots.

Irrational roots

Let's start with irrational roots. Note that 8/9< 9/9 = 1. С другой стороны, корень из двух явно больше единицы. Следовательно, наши корни будут находиться на отрезке в таком положении:

It follows from this that the root of two does not fall into the segment of interest to us. Similarly, we get with a negative root: it is less than −1, i.e., lies to the left of the segment of interest to us.

rational roots

There are two roots left: x = 1/2 and x = −1/2. Let's notice that the left end of the segment (−1) is negative, and the right end (8/9) is positive. Therefore, somewhere between these ends lies the number 0. The root x = −1/2 will be between −1 and 0, i.e. will be included in the final answer. We do the same with the root x = 1/2. This root also lies on the segment under consideration.

It is very easy to make sure that the number 8/9 is greater than 1/2. Let's subtract these numbers from each other:

We got the fraction 7/18 > 0, which by definition means that 8/9 > 1/2.

Let's mark suitable roots on the coordinate axis:

The final answer will be two roots: 1/2 and −1/2.

Comparison of irrational numbers: a universal algorithm

In conclusion, I would like to return to irrational numbers once again. Using their example, we will now see how to compare rational and irrational quantities in mathematics. To begin with, there is such a tick V between them - the sign "more" or "less", but we do not yet know in which direction it is directed. Let's write:

Why do we need any comparison algorithms at all? The fact is that in this problem we were very lucky: in the process of solving, a separating number 1 arose, about which we can definitely say:

However, you will not always see such a number on the move. Therefore, let's try to compare our numbers head-on, directly.

How it's done? We do the same as with the usual inequalities:

1. First, if we had negative coefficients somewhere, then we would multiply both sides of the inequality by −1. Of course changing the sign. Such a tick V would change to such a - Λ.
2. But in our case, both sides are already positive, so there is no need to change anything. What is really needed is square both sides to get rid of the radical.

If, when comparing irrational numbers, it is not possible to pick up a separating element on the go, I recommend performing such a comparison "on the forehead" - describing it as an ordinary inequality.

When solving it, it looks like this:

Now it's all easy to compare. The fact is that 64/81< 81/81 = 1 < 2. На основании той цепочки преобразований мы заключаем, что 64/81 < 2 и, следовательно, корень больше 8/9.

That's it, we have received a rigorous proof that all numbers are marked on the number line x correctly and in exactly the sequence in which they should actually be. No one will complain about such a decision, so remember: if you don’t immediately see the separating number (in our case, it’s 1), then feel free to write out the above construction, multiply, square - and in the end you will get a beautiful inequality. From this inequality it will be clear exactly which number is larger and which is smaller.

Returning to our problem, I would like to once again draw your attention to what we did at the very beginning when solving our equation. Namely, we looked closely at our original logarithmic equation and tried to reduce it to canonical logarithmic equation. Where there are only logarithms on the left and right - without any additional terms, coefficients in front, etc. We need not two logarithms to the base a or b, namely a logarithm equal to another logarithm.

In addition, the bases of the logarithms must also be equal. At the same time, if the equation is composed correctly, then with the help of elementary logarithmic transformations (the sum of logarithms, converting a number to a logarithm, etc.), we will reduce this equation to the canonical one.

Therefore, henceforth, when you see a logarithmic equation that is not immediately solved “on the forehead”, you should not get lost or try to find an answer. It is enough to follow these steps:

1. Bring all free elements to the logarithm;
2. Then add these logarithms;
3. In the resulting construction, all logarithms lead to the same base.

As a result, you will get a simple equation that can be solved by elementary means of algebra from materials of grades 8-9. In general, go to my site, practice solving logarithms, solve logarithmic equations like me, solve them better than me. And that's all for me. Pavel Berdov was with you. See you soon!

As you know, when multiplying expressions with powers, their exponents always add up (a b * a c = a b + c). This mathematical law was derived by Archimedes, and later, in the 8th century, the mathematician Virasen created a table of integer indicators. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where it is required to simplify cumbersome multiplication to simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. Simple and accessible language.

Definition in mathematics

The logarithm is an expression of the following form: log ab=c, that is, the logarithm of any non-negative number (that is, any positive) "b" by its base "a" is considered the power of "c", to which the base "a" must be raised, so that in the end get the value "b". Let's analyze the logarithm using examples, let's say there is an expression log 2 8. How to find the answer? It's very simple, you need to find such a degree that from 2 to the required degree you get 8. Having done some calculations in your mind, we get the number 3! And rightly so, because 2 to the power of 3 gives the number 8 in the answer.

Varieties of logarithms

For many pupils and students, this topic seems complicated and incomprehensible, but in fact, logarithms are not so scary, the main thing is to understand their general meaning and remember their properties and some rules. There are three distinct kinds of logarithmic expressions:

1. Natural logarithm ln a, where the base is the Euler number (e = 2.7).
2. Decimal a, where the base is 10.
3. The logarithm of any number b to the base a>1.

Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to one logarithm using logarithmic theorems. To obtain the correct values ​​​​of logarithms, one should remember their properties and the order of actions in their decisions.

Rules and some restrictions

In mathematics, there are several rules-limitations that are accepted as an axiom, that is, they are not subject to discussion and are true. For example, it is impossible to divide numbers by zero, and it is also impossible to extract the root of an even degree from negative numbers. Logarithms also have their own rules, following which you can easily learn how to work even with long and capacious logarithmic expressions:

• the base "a" must always be greater than zero, and at the same time not be equal to 1, otherwise the expression will lose its meaning, because "1" and "0" to any degree are always equal to their values;
• if a > 0, then a b > 0, it turns out that "c" must be greater than zero.

How to solve logarithms?

For example, the task was given to find the answer to the equation 10 x \u003d 100. It is very easy, you need to choose such a power, raising the number ten to which we get 100. This, of course, is 10 2 \u003d 100.

Now let's represent this expression as a logarithmic one. We get log 10 100 = 2. When solving logarithms, all actions practically converge to finding the degree to which the base of the logarithm must be entered in order to obtain a given number.

To accurately determine the value of an unknown degree, you must learn how to work with a table of degrees. It looks like this:

As you can see, some exponents can be guessed intuitively if you have a technical mindset and knowledge of the multiplication table. However, larger values ​​will require a power table. It can be used even by those who do not understand anything at all in complex mathematical topics. The left column contains numbers (base a), the top row of numbers is the value of the power c, to which the number a is raised. At the intersection in the cells, the values ​​of the numbers are determined, which are the answer (a c =b). Let's take, for example, the very first cell with the number 10 and square it, we get the value 100, which is indicated at the intersection of our two cells. Everything is so simple and easy that even the most real humanist will understand!

Equations and inequalities

It turns out that under certain conditions, the exponent is the logarithm. Therefore, any mathematical numerical expressions can be written as a logarithmic equation. For example, 3 4 =81 can be written as the logarithm of 81 to base 3, which is four (log 3 81 = 4). For negative powers, the rules are the same: 2 -5 = 1/32 we write as a logarithm, we get log 2 (1/32) = -5. One of the most fascinating sections of mathematics is the topic of "logarithms". We will consider examples and solutions of equations a little lower, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.

An expression of the following form is given: log 2 (x-1) > 3 - it is a logarithmic inequality, since the unknown value "x" is under the sign of the logarithm. And also in the expression two quantities are compared: the logarithm of the desired number in base two is greater than the number three.

The most important difference between logarithmic equations and inequalities is that equations with logarithms (for example, the logarithm of 2 x = √9) imply one or more specific numerical values ​​in the answer, while when solving the inequality, both the range of acceptable values ​​and the points breaking this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer of the equation, but a continuous series or set of numbers.

Basic theorems about logarithms

When solving primitive tasks on finding the values ​​of the logarithm, its properties may not be known. However, when it comes to logarithmic equations or inequalities, first of all, it is necessary to clearly understand and apply in practice all the basic properties of logarithms. We will get acquainted with examples of equations later, let's first analyze each property in more detail.

1. The basic identity looks like this: a logaB =B. It only applies if a is greater than 0, not equal to one, and B is greater than zero.
2. The logarithm of the product can be represented in the following formula: log d (s 1 * s 2) = log d s 1 + log d s 2. In this case, the prerequisite is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this formula of logarithms, with examples and a solution. Let log as 1 = f 1 and log as 2 = f 2 , then a f1 = s 1 , a f2 = s 2. We get that s 1 *s 2 = a f1 *a f2 = a f1+f2 (degree properties ), and further by definition: log a (s 1 *s 2)= f 1 + f 2 = log a s1 + log as 2, which was to be proved.
3. The logarithm of the quotient looks like this: log a (s 1 / s 2) = log a s 1 - log a s 2.
4. The theorem in the form of a formula takes the following form: log a q b n = n/q log a b.

This formula is called "property of the degree of the logarithm". It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics rests on regular postulates. Let's look at the proof.

Let log a b \u003d t, it turns out a t \u003d b. If you raise both parts to the power m: a tn = b n ;

but since a tn = (a q) nt/q = b n , hence log a q b n = (n*t)/t, then log a q b n = n/q log a b. The theorem has been proven.

Examples of problems and inequalities

The most common types of logarithm problems are examples of equations and inequalities. They are found in almost all problem books, and are also included in the mandatory part of exams in mathematics. To enter a university or pass entrance tests in mathematics, you need to know how to solve such tasks correctly.

Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, however, certain rules can be applied to each mathematical inequality or logarithmic equation. First of all, you should find out whether the expression can be simplified or reduced to a general form. You can simplify long logarithmic expressions if you use their properties correctly. Let's get to know them soon.

When solving logarithmic equations, it is necessary to determine what type of logarithm we have before us: an example of an expression may contain a natural logarithm or a decimal one.

Here are examples ln100, ln1026. Their solution boils down to the fact that you need to determine the degree to which the base 10 will be equal to 100 and 1026, respectively. For solutions of natural logarithms, one must apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various types.

How to Use Logarithm Formulas: With Examples and Solutions

So, let's look at examples of using the main theorems on logarithms.

1. The property of the logarithm of the product can be used in tasks where it is necessary to decompose a large value of the number b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4*128) = log 2 512. The answer is 9.
2. log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, by applying the fourth property of the degree of the logarithm, we managed to solve at first glance a complex and unsolvable expression. It is only necessary to factorize the base and then take the exponent values ​​out of the sign of the logarithm.

Tasks from the exam

Logarithms are often found in entrance exams, especially a lot of logarithmic problems in the Unified State Exam (state exam for all school graduates). Usually these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most difficult and voluminous tasks). The exam implies an accurate and perfect knowledge of the topic "Natural logarithms".

Examples and problem solving are taken from the official versions of the exam. Let's see how such tasks are solved.

Given log 2 (2x-1) = 4. Solution:
let's rewrite the expression, simplifying it a little log 2 (2x-1) = 2 2, by the definition of the logarithm, we get that 2x-1 = 2 4, therefore 2x = 17; x = 8.5.

• All logarithms are best reduced to the same base so that the solution is not cumbersome and confusing.
• All expressions under the sign of the logarithm are indicated as positive, therefore, when taking out the exponent of the exponent of the expression, which is under the sign of the logarithm and as its base, the expression remaining under the logarithm must be positive.

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