How to solve logarithmic equations. Logarithmic Equations
Logarithmic equations. From simple to complex.
Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")
What is a logarithmic equation?
This is an equation with logarithms. I was surprised, right?) Then I’ll clarify. This is an equation in which the unknowns (x) and expressions with them are inside logarithms. And only there! It is important.
Here are some examples logarithmic equations :
log 3 x = log 3 9
log 3 (x 2 -3) = log 3 (2x)
log x + 1 (x 2 + 3x-7) = 2
lg 2 (x+1)+10 = 11lg(x+1)
Well, you get the idea... )
Note! The most diverse expressions with x's are located only inside logarithms. If, suddenly, an x is found in the equation somewhere outside, For example:
log 2 x = 3+x,
this will be a mixed type equation. Such equations do not have clear rules for solving. We will not consider them for now. By the way, there are equations where inside the logarithms only numbers. For example:
What can I say? You're lucky if you come across this! The 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 specifically for solving logarithmic equations, not required here.
So, what is a logarithmic equation- figured it out.
How to solve logarithmic equations?
Solution logarithmic equations- a thing, in general, is not very simple. So the section we have is for four ... A decent supply of knowledge on all sorts of related topics is required. In addition, there is a special feature in these equations. And this feature is so important that it can be safely called the main problem in solving logarithmic equations. We will deal with this problem in detail in the next lesson.
Now, don't worry. We'll go the right way from simple to complex. On concrete examples. The main thing is to delve into simple things and do not be lazy to follow the links, I put them for a reason... And you will succeed. Necessarily.
Let's start with the most elementary, simplest equations. To solve them, it is desirable to have an idea about the logarithm, but nothing more. Just no idea logarithm take a decision logarithmic equations - somehow even embarrassing ... Very bold, I would say).
The simplest logarithmic equations.
These are equations of the form:
1. log 3 x = log 3 9
2. log 7 (2x-3) = log 7 x
3. log 7 (50x-1) = 2
Solution process any logarithmic equation consists in the transition from an equation with logarithms to an equation without them. In the simplest equations, this transition is carried out in one step. That's why it's simple.)
And such logarithmic equations are solved surprisingly simply. See for yourself.
Let's solve the first example:
log 3 x = log 3 9
To solve this example, you don’t need to know almost anything, yes ... Pure intuition!) What do we especially don't like this example? Something... I don't like logarithms! Right. Here we get rid of them. We closely look at the example, and a natural desire arises in us ... Downright irresistible! Take and throw out logarithms in general. And what pleases is Can do! Mathematics allows. The logarithms disappear the answer is:
It's great, right? This can (and should) always be done. Eliminating logarithms in this way is one of the main ways to solve logarithmic equations and inequalities. In mathematics, this operation is called potentiation. There are, of course, their own rules for such liquidation, but they are few. Remember:
You can eliminate logarithms without any fear if they have:
a) the same numerical bases
c) the left-right logarithms are clean (without any coefficients) and are in splendid isolation.
Let me explain the last point. In the equation, let's say
log 3 x = 2log 3 (3x-1)
logarithms cannot be removed. The deuce on the right does not allow. Coefficient, you know ... In the example
log 3 x + log 3 (x + 1) = log 3 (3 + x)
the equation cannot be potentiated either. There is no lone logarithm on the left side. There are two of them.
In short, you can remove logarithms if the equation looks like this and only this:
log a (.....) = log a (.....)
In parentheses, where the ellipsis can be any kind of expression. Simple, super complex, whatever. Whatever. The important thing is that after eliminating the logarithms, we are left with a simpler equation. It is assumed, of course, that you already know how to solve linear, quadratic, fractional, exponential and other equations without logarithms.)
Now you can easily solve the second example:
log 7 (2x-3) = log 7 x
Actually, it's in the mind. We potentiate, we get:
Well, is it very difficult?) As you can see, logarithmic part of the solution to the equation is only in the elimination of logarithms... And then comes the solution of the remaining equation already without them. Waste business.
We solve the third example:
log 7 (50x-1) = 2
We see that the logarithm is on the left:
We recall that this logarithm is some number to which the base (i.e. seven) must be raised in order to obtain a sublogarithmic expression, i.e. (50x-1).
But that number is two! According to the equation. That is:
That, in essence, is all. Logarithm disappeared the harmless equation remains:
We have solved this logarithmic equation based only on the meaning of the logarithm. Is it easier to eliminate logarithms?) I agree. By the way, if you make a logarithm out of two, you can solve this example through liquidation. You can take a logarithm from any number. And just the way we need it. Very useful technique in solving logarithmic equations and (especially!) inequalities.
Do you know how to make a logarithm from a number !? It's OK. Section 555 describes this technique in detail. You can master and apply it to its fullest! It greatly reduces the number of errors.
The fourth equation is solved in exactly the same way (by definition):
That's all there is to it.
Let's summarize this lesson. We considered the solution of the simplest logarithmic equations using examples. It is very important. And not only because such equations are on control-exams. The fact is that even the most evil and confused equations are necessarily reduced to the simplest ones!
Actually, the simplest equations are the final part of the solution any equations. And this finishing part must be understood ironically! And further. Be sure to read this page to the end. There is a surprise...
Let's decide on our own. We fill the hand, so to speak ...)
Find the root (or the sum of the roots, if there are several) of the equations:
ln(7x+2) = ln(5x+20)
log 2 (x 2 +32) = log 2 (12x)
log 16 (0.5x-1.5) = 0.25
log 0.2 (3x-1) = -3
ln (e 2 + 2x-3) \u003d 2
log 2 (14x) = log 2 7 + 2
Answers (in disarray, of course): 42; 12; 9; 25; 7; 1.5; 2; 16.
What doesn't work out? Happens. Do not grieve! In section 555, the solution to all these examples is described clearly and in detail. You will definitely find out there. And also useful practical techniques master.
Everything worked out!? All examples of "one left"?) Congratulations!
It's time to reveal the bitter truth to you. Successful solution of these examples does not at all guarantee success in solving all other logarithmic equations. Even simple ones like these. Alas.
The point is that the solution of any logarithmic equation (even the most elementary one!) consists of two equal parts. Solution of the equation, and work with ODZ. One part - the solution of the equation itself - we have mastered. It's not that hard right?
For this lesson, I specially selected such examples in which the ODZ does not affect the answer in any way. But not everyone is as kind as me, right?...)
Therefore, it is necessary to master the other part as well. ODZ. This is the main problem in solving logarithmic equations. And not because it is difficult - this part is even easier than the first. But because they simply forget about ODZ. Or they don't know. Or both). And they fall flat...
In the next lesson, we will deal with this problem. Then it will be possible to confidently decide any simple logarithmic equations and get close to quite solid tasks.
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
Introduction
Logarithms were invented to speed up and simplify calculations. The idea of the logarithm, that is, the idea of expressing numbers as a power of the same base, belongs to Mikhail Stiefel. But at the time of Stiefel, mathematics was not so developed and the idea of the logarithm did not find its development. Logarithms were invented later simultaneously and independently by the Scottish scientist John Napier (1550-1617) and the Swiss Jobst Burgi (1552-1632). Napier was the first to publish the work in 1614. titled "Description of the amazing table of logarithms", Napier's theory of logarithms was given in a fairly complete volume, the method for calculating logarithms was given in the simplest way, therefore Napier's merits in the invention of logarithms are greater than those of Burgi. Bürgi worked on the tables at the same time as Napier, but for a long time kept them secret and published only in 1620. Napier mastered the idea of the logarithm around 1594. although the tables were published 20 years later. At first, he called his logarithms "artificial numbers" and only then proposed to call these "artificial numbers" in one word "logarithm", which in Greek is "correlated numbers", taken one from an arithmetic progression, and the other from a geometric progression specially selected for it. progress. The first tables in Russian were published in 1703. with the participation of a remarkable teacher of the 18th century. L. F. Magnitsky. In the development of the theory of logarithms great importance had the work of the St. Petersburg academician Leonhard Euler. He was the first to consider logarithm as the inverse of exponentiation, he introduced the terms "base of the logarithm" and "mantissa" Briggs compiled tables of logarithms with base 10. Decimal tables are more convenient for practical use, their theory is simpler than that of Napier's logarithms . Therefore, decimal logarithms are sometimes called brigs. The term "characteristic" was introduced by Briggs.
In those distant times, when the wise men first began to think about equalities containing unknown quantities, there probably were no coins or wallets yet. But on the other hand, there were heaps, as well as pots, baskets, which were perfect for the role of caches-stores containing an unknown number of items. In the ancient mathematical problems of Mesopotamia, India, China, Greece, unknown quantities expressed the number of peacocks in the garden, the number of bulls in the herd, the totality of things taken into account when dividing property. Scribes, officials, and priests initiated into secret knowledge, well trained in the science of counting, coped with such tasks quite successfully.
Sources that have come down to us indicate that ancient scientists possessed some general methods for solving problems with unknown quantities. However, not a single papyrus, not a single clay tablet gives a description of these techniques. The authors only occasionally supplied their numerical calculations with mean comments like: "Look!", "Do it!", "You found it right." In this sense, the exception is the "Arithmetic" of the Greek mathematician Diophantus of Alexandria (III century) - a collection of problems for compiling equations with a systematic presentation of their solutions.
However, the work of the Baghdad scholar of the 9th century became the first manual for solving problems that became widely known. Muhammad bin Musa al-Khwarizmi. The word "al-jabr" from the Arabic title of this treatise - "Kitab al-jaber wal-muqabala" ("The Book of Restoration and Contrasting") - over time turned into the word "algebra" well known to everyone, and the work of al-Khwarizmi itself served as starting point in the development of the science of solving equations.
Logarithmic equations and inequalities
1. Logarithmic equations
An equation containing an unknown under the sign of the logarithm or at its base is called a logarithmic equation.
The simplest logarithmic equation is the equation of the form
log a x = b . (1)
Statement 1. If a > 0, a≠ 1, equation (1) for any real b It has only decision x = a b .
Example 1. Solve equations:
a) log 2 x= 3, b) log 3 x= -1, c)
Solution. Using Statement 1, we obtain a) x= 2 3 or x= 8; b) x= 3 -1 or x= 1/3; c)
or x = 1.We present the main properties of the logarithm.
P1. Basic logarithmic identity:
Where a > 0, a≠ 1 and b > 0.
R2. The logarithm of the product of positive factors is equal to the sum of the logarithms of these factors:
log a N 1 · N 2 = log a N 1 + log a N 2 (a > 0, a ≠ 1, N 1 > 0, N 2 > 0).
Comment. If N 1 · N 2 > 0, then property P2 takes the form
log a N 1 · N 2 = log a |N 1 | +log a |N 2 | (a > 0, a ≠ 1, N 1 · N 2 > 0).
P3. The logarithm of the quotient of two positive numbers is equal to the difference between the logarithms of the dividend and the divisor
(a
> 0, a
≠ 1, N
1
> 0, N
2
> 0).
Comment. If
, (which is equivalent to N 1 N 2 > 0) then property P3 takes the form
(a
> 0, a
≠ 1, N
1
N
2
> 0).
P4. Logarithm of the power of a positive number is equal to the product exponent per logarithm of this number:
log a N k = k log a N (a > 0, a ≠ 1, N > 0).
Comment. If k- even number ( k = 2s), That
log a N 2s = 2s log a |N | (a > 0, a ≠ 1, N ≠ 0).
P5. The formula for moving to another base is:
(a
> 0, a
≠ 1, b
> 0, b
≠ 1, N
> 0),
in particular if N = b, we get
(a > 0, a ≠ 1, b > 0, b ≠ 1). (2)Using properties P4 and P5, it is easy to obtain the following properties
(a
> 0, a
≠ 1, b
> 0, c
≠ 0), (3)
(a
> 0, a
≠ 1, b
> 0, c
≠ 0), (4)
(a
> 0, a
≠ 1, b
> 0, c
≠ 0), (5)
and if in (5) c- even number ( c = 2n), occurs
(b
> 0, a
≠ 0, |a
| ≠ 1). (6)
We list the main properties of the logarithmic function f (x) = log a x :
1. The domain of the logarithmic function is the set of positive numbers.
2. The range of values of the logarithmic function is the set of real numbers.
3. When a> 1 the logarithmic function is strictly increasing (0< x 1 < x 2 log a x 1 < loga x 2), and at 0< a < 1, - строго убывает (0 < x 1 < x 2 log a x 1 > log a x 2).
4 log a 1 = 0 and log a a = 1 (a > 0, a ≠ 1).
5. If a> 1, then the logarithmic function is negative for x(0;1) and is positive for x(1;+∞), and if 0< a < 1, то логарифмическая функция положительна при x (0;1) and is negative for x (1;+∞).
6. If a> 1, then the logarithmic function is convex upwards, and if a(0;1) - convex down.
The following statements (see, for example, ) are used in solving logarithmic equations.
Today we will learn how to solve the simplest logarithmic equations, which do not require preliminary transformations and selection of roots. But if you learn how to solve such equations, then it will be much easier.
The simplest logarithmic equation is an equation of the form log a f (x) \u003d b, where a, b are numbers (a\u003e 0, a ≠ 1), f (x) is some function.
A distinctive feature of all logarithmic equations is the presence of the variable x under the sign of the logarithm. If such an equation is initially given in the problem, it is called the simplest one. Any other logarithmic equations are reduced to the simplest by special transformations (see "Basic properties of logarithms"). However, numerous subtleties must be taken into account: extra roots may appear, so complex logarithmic equations will be considered separately.
How to solve such equations? It is enough to replace the number to the right of the equal sign with a logarithm in the same base as on the left. Then you can get rid of the sign of the logarithm. We get:
log a f (x) \u003d b ⇒ log a f (x) \u003d log a a b ⇒ f (x) \u003d a b
We got the usual equation. Its roots are the roots of the original equation.
Pronouncement of degrees
Often, logarithmic equations, which outwardly look complicated and threatening, are solved in just a couple of lines without involving complex formulas. Today we will consider just such problems, where all that is required of you is to carefully reduce the formula to the canonical form and not get confused when searching for the domain of definition of logarithms.
Today, as you probably guessed from the title, we will solve logarithmic equations using the formulas for the transition to the canonical form. The main "trick" of this video lesson will be working with degrees, or rather, taking the degree from the base and argument. Let's look at the rule:
Similarly, you can take out the degree from the base:
As you can see, if when taking the degree out of the logarithm argument, we simply have an additional factor in front, then when taking the degree out of the base, it’s not just a factor, but an inverted factor. This must be remembered.
Finally, the most interesting. These formulas can be combined, then we get:
Of course, when performing these transitions, there are certain pitfalls associated with the possible expansion of the domain of definition or, conversely, the narrowing of the domain of definition. Judge for yourself:
log 3 x 2 = 2 ∙ log 3 x
If in the first case, x could be any number other than 0, i.e., the requirement x ≠ 0, then in the second case, we will only be satisfied with x, which are not only not equal, but strictly greater than 0, because the domain of the logarithm is that the argument be strictly greater than 0. Therefore, I will remind you of a wonderful formula from the course of algebra in grades 8-9:
That is, we must write our formula as follows:
log 3 x 2 = 2 ∙ log 3 |x |
Then no narrowing of the domain of definition will occur.
However, in today's video tutorial there will be no squares. If you look at our tasks, you will see only the roots. Therefore, apply this rule we won’t, but it still needs to be kept in mind in order to right moment when you see quadratic function in the argument or base of the logarithm, you will remember this rule and perform all the transformations correctly.
So the first equation is:
To solve this problem, I propose to carefully look at each of the terms present in the formula.
Let's rewrite the first term as a power with a rational exponent:
We look at the second term: log 3 (1 − x ). You don't need to do anything here, everything is already being transformed.
Finally, 0, 5. As I said in previous lessons, when solving logarithmic equations and formulas, I highly recommend moving from decimal fractions to ordinary ones. Let's do this:
0,5 = 5/10 = 1/2
Let's rewrite our original formula taking into account the obtained terms:
log 3 (1 − x ) = 1
Now let's move on to the canonical form:
log 3 (1 − x ) = log 3 3
Get rid of the sign of the logarithm by equating the arguments:
1 − x = 3
-x = 2
x = −2
That's it, we've solved the equation. However, let's still play it safe and find the domain of definition. To do this, let's go back to the original formula and see:
1 − x > 0
-x > -1
x< 1
Our root x = −2 satisfies this requirement, so x = −2 is a solution to the original equation. Now we have a strict clear justification. Everything, the task is solved.
Let's move on to the second task:
Let's deal with each term separately.
We write out the first:
We have modified the first term. We work with the second term:
Finally, the last term, which is to the right of the equal sign:
We substitute the resulting expressions for the terms in the resulting formula:
log 3 x = 1
We pass to the canonical form:
log 3 x = log 3 3
We get rid of the sign of the logarithm by equating the arguments, and we get:
x=3
Again, just in case, let's play it safe, go back to the original equation and see. In the original formula, the variable x is present only in the argument, therefore,
x > 0
In the second logarithm, x is under the root, but again in the argument, therefore, the root must be greater than 0, that is, the root expression must be greater than 0. We look at our root x = 3. Obviously, it satisfies this requirement. Therefore, x = 3 is the solution to the original logarithmic equation. Everything, the task is solved.
There are two key points in today's video tutorial:
1) do not be afraid to convert logarithms and, in particular, do not be afraid to take degrees out of the sign of the logarithm, while remembering our basic formula: when taking the degree out of the argument, it is taken out simply without changes as a factor, and when taking the degree out of the base, this degree is reversed.
2) the second point is related to the self-canonical form. We performed the transition to the canonical form at the very end of the transformation of the formula of the logarithmic equation. Recall the following formula:
a = log b b a
Of course, by the expression "any number b", I mean those numbers that satisfy the requirements imposed on the base of the logarithm, i.e.
1 ≠ b > 0
For such b , and since we already know the base, this requirement will be fulfilled automatically. But for such b - any that satisfy this requirement - this transition can be performed, and we get a canonical form in which we can get rid of the sign of the logarithm.
Extension of the domain of definition and extra roots
In the process of transforming logarithmic equations, an implicit extension of the domain of definition may occur. Often, students do not even notice this, which leads to errors and incorrect answers.
Let's start with the simplest designs. The simplest logarithmic equation is the following:
log a f(x) = b
Note that x is present in only one argument of one logarithm. How do we solve such equations? We use the canonical form. To do this, we represent the number b \u003d log a a b, and our equation will be rewritten in the following form:
log a f(x) = log a a b
This notation is called the canonical form. It is to it that any logarithmic equation that you will meet not only in today's lesson, but also in any independent and control work should be reduced.
How to come to the canonical form, what techniques to use - this is already a matter of practice. The main thing to understand: as soon as you receive such a record, we can assume that the problem is solved. Because the next step is to write:
f(x) = a b
In other words, we get rid of the sign of the logarithm and simply equate the arguments.
Why all this talk? The fact is that the canonical form is applicable not only to the simplest problems, but also to any other. In particular, to those that we will address today. Let's get a look.
First task:
What is the problem with this equation? The fact that the function is in two logarithms at once. The problem can be reduced to the simplest by simply subtracting one logarithm from another. But there are problems with the domain of definition: extra roots may appear. So let's just move one of the logarithms to the right:
Here such a record is already much more similar to the canonical form. But there is one more nuance: in the canonical form, the arguments must be the same. And we have the logarithm to the base 3 on the left, and the logarithm to the base 1/3 on the right. You know, you need to bring these bases to the same number. For example, let's remember what negative exponents are:
And then we will use the exponent "-1" outside the log as a multiplier:
Please note: the degree that stood at the base is turned over and turns into a fraction. We got an almost canonical record by getting rid of different grounds, but instead received the multiplier "−1" on the right. Let's put this factor into the argument by turning it into a power:
Of course, having received the canonical form, we boldly cross out the sign of the logarithm and equate the arguments. At the same time, let me remind you that when raised to the power of “−1”, the fraction simply turns over - a proportion is obtained.
Let's use the main property of the proportion and multiply it crosswise:
(x - 4) (2x - 1) = (x - 5) (3x - 4)
2x 2 - x - 8x + 4 = 3x 2 - 4x - 15x + 20
2x2 - 9x + 4 = 3x2 - 19x + 20
x2 − 10x + 16 = 0
Before us is the quadratic equation, so we solve it using the Vieta formulas:
(x − 8)(x − 2) = 0
x 1 = 8; x2 = 2
That's all. Do you think the equation is solved? No! For such a solution, we will get 0 points, because in the original equation there are two logarithms with the variable x at once. Therefore, it is necessary to take into account the domain of definition.
And this is where the fun begins. Most students are confused: what is the domain of the logarithm? Of course, all arguments (we have two of them) must be Above zero:
(x − 4)/(3x − 4) > 0
(x − 5)/(2x − 1) > 0
Each of these inequalities must be solved, marked on a straight line, crossed - and only then see what roots lie at the intersection.
I'll be honest: this technique has the right to exist, it is reliable, and you will get the right answer, but there are too many extra steps in it. So let's go through our solution again and see: where exactly do you want to apply scope? In other words, you need to clearly understand exactly when extra roots appear.
- Initially, we had two logarithms. Then we moved one of them to the right, but this did not affect the definition area.
- Then we remove the power from the base, but there are still two logarithms, and each of them contains the variable x .
- Finally, we cross out the signs of log and get the classic fractional rational equation.
It is at the last step that the domain of definition is expanded! As soon as we switched to a fractional rational equation, getting rid of the signs of log, the requirements for the x variable changed dramatically!
Therefore, the domain of definition can be considered not at the very beginning of the solution, but only at the mentioned step - before we directly equate the arguments.
This is where the opportunity for optimization lies. On the one hand, we are required that both arguments be greater than zero. On the other hand, we further equate these arguments. Therefore, if at least one of them is positive, then the second one will also be positive!
So it turns out that requiring the fulfillment of two inequalities at once is an overkill. It is enough to consider only one of these fractions. Which one? The one that is easier. For example, let's look at the right fraction:
(x − 5)/(2x − 1) > 0
This is a typical fractional rational inequality, we solve it using the interval method:
How to place signs? Take a number that is obviously greater than all our roots. For example, 1 billion. And we substitute its fraction. We get a positive number, i.e. to the right of the root x = 5 there will be a plus sign.
Then the signs alternate, because there are no roots of even multiplicity anywhere. We are interested in intervals where the function is positive. Hence x ∈ (−∞; −1/2)∪(5; +∞).
Now let's remember the answers: x = 8 and x = 2. Strictly speaking, these are not answers yet, but only candidates for an answer. Which one belongs to the specified set? Of course, x = 8. But x = 2 does not suit us in terms of the domain of definition.
The total answer to the first logarithmic equation will be x = 8. Now we have received a competent, an informed decision taking into account the domain of definition.
Let's move on to the second equation:
log 5 (x - 9) = log 0.5 4 - log 5 (x - 5) + 3
I remind you that if there is a decimal fraction in the equation, then you should get rid of it. In other words, let's rewrite 0.5 as a regular fraction. We immediately notice that the logarithm containing this base is easily considered:
This is a very important moment! When we have degrees in both the base and the argument, we can take out the indicators of these degrees using the formula:
We return to our original logarithmic equation and rewrite it:
log 5 (x - 9) = 1 - log 5 (x - 5)
We got a construction that is quite close to the canonical form. However, we are confused by the terms and the minus sign to the right of the equals sign. Let's represent unity as a logarithm to base 5:
log 5 (x - 9) = log 5 5 1 - log 5 (x - 5)
Subtract the logarithms on the right (while their arguments are divided):
log 5 (x − 9) = log 5 5/(x − 5)
Wonderful. So we got the canonical form! We cross out the log signs and equate the arguments:
(x − 9)/1 = 5/(x − 5)
This is a proportion that is easily solved by cross-multiplication:
(x − 9)(x − 5) = 5 1
x 2 - 9x - 5x + 45 = 5
x2 − 14x + 40 = 0
Obviously, we have a given quadratic equation. It is easily solved using the Vieta formulas:
(x − 10)(x − 4) = 0
x 1 = 10
x 2 = 4
We have two roots. But these are not final answers, but only candidates, because the logarithmic equation also requires checking the domain.
I remind you: do not look when every of the arguments will be greater than zero. It suffices to require that one argument, either x − 9 or 5/(x − 5) be greater than zero. Consider the first argument:
x − 9 > 0
x > 9
Obviously, only x = 10 satisfies this requirement. This is the final answer. All problem solved.
Once again, the main ideas of today's lesson:
- As soon as the variable x appears in several logarithms, the equation ceases to be elementary, and for it it is necessary to calculate the domain of definition. Otherwise, you can easily write extra roots in response.
- Working with the domain of definition itself can be greatly simplified if the inequality is not written immediately, but exactly at the moment when we get rid of the signs of log. After all, when the arguments are equated to each other, it is enough to require that only one of them be greater than zero.
Of course, we ourselves choose from which argument to make an inequality, so it is logical to choose the simplest one. For example, in the second equation, we chose the argument (x − 9) − linear function, as opposed to the fractionally rational second argument. Agree, solving the inequality x − 9 > 0 is much easier than 5/(x − 5) > 0. Although the result is the same.
This remark greatly simplifies the search for ODZ, but be careful: you can use one inequality instead of two only when the arguments are precisely equate to each other!
Of course, someone will now ask: what happens differently? Yes, sometimes. For example, in the step itself, when we multiply two arguments containing a variable, there is a danger of extra roots.
Judge for yourself: at first it is required that each of the arguments be greater than zero, but after multiplication it is sufficient that their product be greater than zero. As a result, the case when each of these fractions is negative is missed.
Therefore, if you are just starting to deal with complex logarithmic equations, in no case do not multiply logarithms containing the variable x - too often this will lead to extra roots. Better take one extra step, transfer one term to the other side, make up the canonical form.
Well, what to do if you cannot do without multiplying such logarithms, we will discuss in the next video tutorial. :)
Once again about the powers in the equation
Today we will analyze a rather slippery topic regarding logarithmic equations, or rather, the removal of powers from the arguments and bases of logarithms.
I would even say that we will talk about taking out even powers, because it is with even powers that most of the difficulties arise when solving real logarithmic equations.
Let's start with the canonical form. Let's say we have an equation like log a f (x) = b. In this case, we rewrite the number b according to the formula b = log a a b . It turns out the following:
log a f(x) = log a a b
Then we equate the arguments:
f(x) = a b
The penultimate formula is called the canonical form. It is to her that they try to reduce any logarithmic equation, no matter how complicated and terrible it may seem at first glance.
Here, let's try. Let's start with the first task:
Preliminary remark: as I said, all decimals in a logarithmic equation, it is better to translate it into ordinary ones:
0,5 = 5/10 = 1/2
Let's rewrite our equation with this fact in mind. Note that both 1/1000 and 100 are powers of 10, and then we take the powers out of wherever they are: from the arguments and even from the base of the logarithms:
And here the question arises for many students: “Where did the module come from on the right?” Indeed, why not just write (x − 1)? Of course, now we will write (x − 1), but the right to such a record gives us the account of the domain of definition. After all, the other logarithm already contains (x − 1), and this expression must be greater than zero.
But when we take out the square from the base of the logarithm, we must leave the module at the base. I'll explain why.
The fact is that from the point of view of mathematics, taking a degree is tantamount to taking a root. In particular, when the expression (x − 1) 2 is squared, we are essentially extracting the root of the second degree. But the square root is nothing more than a modulus. Exactly module, because even if the expression x - 1 is negative, when squaring "minus" will still burn. Further extraction of the root will give us a positive number - already without any minuses.
In general, in order to avoid offensive mistakes, remember once and for all:
The root of an even degree from any function that is raised to the same power is equal not to the function itself, but to its modulus:
We return to our logarithmic equation. Speaking about the module, I argued that we can painlessly remove it. This is true. Now I will explain why. Strictly speaking, we had to consider two options:
- x − 1 > 0 ⇒ |x − 1| = x − 1
- x − 1< 0 ⇒ |х − 1| = −х + 1
Each of these options would need to be addressed. But there is one catch: the original formula already contains the function (x − 1) without any modulus. And following the domain of definition of logarithms, we can immediately write down that x − 1 > 0.
This requirement must be satisfied regardless of any modules and other transformations that we perform in the solution process. Therefore, it is pointless to consider the second option - it will never arise. Even if, when solving this branch of the inequality, we get some numbers, they still will not be included in the final answer.
Now we are literally one step away from the canonical form of the logarithmic equation. Let's represent the unit as follows:
1 = log x − 1 (x − 1) 1
In addition, we introduce the factor −4, which is on the right, into the argument:
log x − 1 10 −4 = log x − 1 (x − 1)
Before us is the canonical form of the logarithmic equation. Get rid of the sign of the logarithm:
10 −4 = x − 1
But since the base was a function (and not a prime number), we additionally require that this function be greater than zero and not equal to one. Get the system:
Since the requirement x − 1 > 0 is automatically satisfied (because x − 1 = 10 −4), one of the inequalities can be deleted from our system. The second condition can also be crossed out because x − 1 = 0.0001< 1. Итого получаем:
x = 1 + 0.0001 = 1.0001
This is the only root that automatically satisfies all the requirements for the domain of definition of the logarithm (however, all requirements were eliminated as knowingly fulfilled in the conditions of our problem).
So the second equation is:
3 log 3 x x = 2 log 9 x x 2
How is this equation fundamentally different from the previous one? Already at least by the fact that the bases of logarithms - 3x and 9x - are not natural powers of each other. Therefore, the transition we used in the previous solution is not possible.
Let's at least get rid of the degrees. In our case, the only power is in the second argument:
3 log 3 x x = 2 ∙ 2 log 9 x |x |
However, the modulus sign can be removed, because the variable x is also in the base, i.e. x > 0 ⇒ |x| = x. Let's rewrite our logarithmic equation:
3 log 3 x x = 4 log 9 x x
We got logarithms in which the same arguments, but different bases. How to proceed? There are many options here, but we will consider only two of them, which are the most logical, and most importantly, these are quick and understandable tricks for most students.
We have already considered the first option: in any incomprehensible situation, translate logarithms with a variable base to some constant base. For example, to a deuce. The conversion formula is simple:
Of course, a normal number should act as a variable c: 1 ≠ c > 0. Let c = 2 in our case. Now we have an ordinary fractional rational equation. We collect all the elements on the left:
Obviously, the factor log 2 x is better to take out, since it is present in both the first and second fractions.
log 2 x = 0;
3 log 2 9x = 4 log 2 3x
We break each log into two terms:
log 2 9x = log 2 9 + log 2 x = 2 log 2 3 + log 2 x;
log 2 3x = log 2 3 + log 2 x
Let's rewrite both sides of the equality taking into account these facts:
3 (2 log 2 3 + log 2 x ) = 4 (log 2 3 + log 2 x )
6 log 2 3 + 3 log 2 x = 4 log 2 3 + 4 log 2 x
2 log 2 3 = log 2 x
Now it remains to add a deuce under the sign of the logarithm (it will turn into a power: 3 2 \u003d 9):
log 2 9 = log 2 x
Before us is the classical canonical form, we get rid of the sign of the logarithm and get:
As expected, this root turned out to be greater than zero. It remains to check the domain of definition. Let's look at the bases:
But the root x = 9 satisfies these requirements. Therefore, it is the final decision.
The conclusion from this solution is simple: do not be afraid of long calculations! It's just that at the very beginning we chose a new base at random - and this significantly complicated the process.
But then the question arises: what basis is optimal? I will talk about this in the second way.
Let's go back to our original equation:
3 log 3x x = 2 log 9x x 2
3 log 3x x = 2 ∙ 2 log 9x |x |
x > 0 ⇒ |x| = x
3 log 3 x x = 4 log 9 x x
Now let's think a little: what number or function will be optimal foundation? It's obvious that the best option will be c = x - what is already in the arguments. In this case, the formula log a b = log c b / log c a will take the form:
In other words, the expression is simply reversed. In this case, the argument and the basis are reversed.
This formula is very useful and very often used in solving complex logarithmic equations. However, when using this formula, there is one very serious pitfall. If instead of the base we substitute the variable x, then restrictions are imposed on it that were not previously observed:
There was no such restriction in the original equation. Therefore, we should separately check the case when x = 1. Substitute this value in our equation:
3 log 3 1 = 4 log 9 1
We get the correct numerical equality. Therefore, x = 1 is a root. We found exactly the same root in the previous method at the very beginning of the solution.
But now, when we separately considered this special case, we safely assume that x ≠ 1. Then our logarithmic equation will be rewritten in the following form:
3 log x 9x = 4 log x 3x
We expand both logarithms according to the same formula as before. Note that log x x = 1:
3 (log x 9 + log x x ) = 4 (log x 3 + log x x )
3 log x 9 + 3 = 4 log x 3 + 4
3 log x 3 2 − 4 log x 3 = 4 − 3
2 log x 3 = 1
Here we come to the canonical form:
log x 9 = log x x 1
x=9
We got the second root. It satisfies the requirement x ≠ 1. Therefore, x = 9 along with x = 1 is the final answer.
As you can see, the volume of calculations has slightly decreased. But when solving a real logarithmic equation, the number of steps will be much less also because you are not required to describe each step in such detail.
The key rule of today's lesson is as follows: if there is an even degree in the problem, from which the root of the same degree is extracted, then at the output we will get a module. However, this module can be removed if you pay attention to the domain of definition of logarithms.
But be careful: most students after this lesson think that they understand everything. But when solving real problems, they cannot reproduce the entire logical chain. As a result, the equation acquires extra roots, and the answer is wrong.
We are all familiar with equations. primary school. Even there we learned to solve the simplest examples, and it must be admitted that they find their application even in higher mathematics. Everything is simple with equations, including square ones. If you have problems with this theme, we strongly recommend that you retry it.
Logarithms you probably already passed too. Nevertheless, we consider it important to tell what it is for those who do not know yet. The logarithm equates to the power to which the base must be raised to get the number to the right of the sign of the logarithm. Let's give an example, based on which, everything will become clear to you.
If you raise 3 to the fourth power, you get 81. Now substitute the numbers by analogy, and you will finally understand how logarithms are solved. Now it remains only to combine the two considered concepts. Initially, the situation seems extremely difficult, but upon closer examination, the weight falls into place. We are sure that after this short article you will have no problems in this part of the exam.
Today, there are many ways to solve such structures. We will talk about the simplest, most effective and most applicable in the case of USE tasks. Solving logarithmic equations must start from the very beginning. a simple example. The simplest logarithmic equations consist of a function and one variable in it.
It is important to note that x is inside the argument. A and b must be numbers. In this case, you can simply express the function in terms of a number in a power. It looks like this.
Of course, solving a logarithmic equation in this way will lead you to the correct answer. But the problem of the vast majority of students in this case is that they do not understand what and where it comes from. As a result, you have to put up with mistakes and not get the desired points. The most offensive mistake will be if you mix up the letters in places. To solve an equation in this way, you need to memorize this standard school formula because it's hard to understand.
To make it easier, you can resort to another method - the canonical form. The idea is extremely simple. Pay attention to the task again. Remember that the letter a is a number, not a function or a variable. A is not equal to one and is greater than zero. There are no restrictions on b. Now of all the formulas, we recall one. B can be expressed as follows.
From this it follows that all the original equations with logarithms can be represented as:
Now we can discard the logarithms. It turns out simple design, which we have seen before.
The convenience of this formula lies in the fact that it can be used in the most different occasions and not just for the simplest designs.
Don't worry about OOF!
Many experienced mathematicians will notice that we have not paid attention to the domain of definition. The rule boils down to the fact that F(x) is necessarily greater than 0. No, we have not missed this point. Now we are talking about another serious advantage of the canonical form.
There will be no extra roots here. If the variable will only occur in one place, then scope is not necessary. It runs automatically. To verify this judgment, consider solving a few simple examples.
How to solve logarithmic equations with different bases
These are already complex logarithmic equations, and the approach to their solution should be special. Here it is rarely possible to confine ourselves to the notorious canonical form. Let's start our detailed story. We have the following construction.
Notice the fraction. It contains the logarithm. If you see this in the task, it is worth remembering one interesting trick.
What does it mean? Each logarithm can be expressed as a quotient of two logarithms with a convenient base. And this formula has a special case that is applicable to this example (we mean if c=b).
This is exactly what we see in our example. Thus.
In fact, they turned the fraction over and got a more convenient expression. Remember this algorithm!
Now we need that the logarithmic equation did not contain different bases. Let's represent the base as a fraction.
In mathematics, there is a rule, based on which, you can take out the degree from the base. It turns out the following construction.
It would seem that now what prevents us from turning our expression into a canonical form and elementarily solving it? Not so simple. There should be no fractions before the logarithm. Let's fix this situation! A fraction is allowed to be taken out as a degree.
Respectively.
If the bases are the same, we can remove the logarithms and equate the expressions themselves. So the situation will become many times easier than it was. There will be an elementary equation that each of us knew how to solve back in 8th or even 7th grade. You can do the calculations yourself.
We got the only true root of this logarithmic equation. Examples of solving a logarithmic equation are quite simple, right? Now you will be able to independently deal with even the most difficult tasks for preparing and passing the exam.
What is the result?
In the case of any logarithmic equations, we start from one very important rule. It is necessary to act in such a way as to bring the expression to the maximum plain sight. In this case, you will have more chances not only to solve the problem correctly, but also to do it in the simplest and most logical way. That's how mathematicians always work.
We strongly do not recommend that you look for difficult paths, especially in this case. Remember a few simple rules, which will allow you to transform any expression. For example, bring two or three logarithms to the same base, or take a power from the base and win on it.
It is also worth remembering that in solving logarithmic equations you need to constantly train. Gradually you will move on to more and more complex structures, and this will lead you to a confident solution of all variants of problems on the exam. Prepare for your exams well in advance, and good luck!
Your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please read our privacy policy and let us know if you have any questions.
Collection and use of personal information
Personal information refers to data that can be used to identify or contact a specific person.
You may be asked to provide your personal information at any time when you contact us.
The following are some examples of the types of personal information we may collect and how we may use such information.
What personal information we collect:
- When you submit an application on the site, we may collect various information including your name, phone number, address Email etc.
How we use your personal information:
- Collected by us personal information allows us to contact you and inform you about unique offers, promotions and other events and upcoming events.
- From time to time, we may use your personal information to send you important notices and messages.
- We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
- If you enter a prize draw, contest or similar incentive, we may use the information you provide to administer such programs.
Disclosure to third parties
We do not disclose information received from you to third parties.
Exceptions:
- If necessary - in accordance with the law, judicial order, in legal proceedings, and/or based on public requests or requests from government agencies on the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public important occasions.
- In the event of a reorganization, merger or sale, we may transfer the personal information we collect to the relevant third party successor.
Protection of personal information
We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as from unauthorized access, disclosure, alteration and destruction.
Maintaining your privacy at the company level
To ensure that your personal information is secure, we communicate privacy and security practices to our employees and strictly enforce privacy practices.
