How to solve the logarithms of an equation. Logarithmic equations. How to solve logarithmic equations? Solving the simplest logarithmic equations
In this lesson, we will review the basic theoretical facts about logarithms and consider solving the simplest logarithmic equations.
Let us recall the central definition - the definition of the logarithm. It is associated with the solution of the exponential equation. This equation has a single root, it is called the logarithm of b to base a:
Definition:
The logarithm of the number b to the base a is the exponent to which the base a must be raised to get the number b.
Recall basic logarithmic identity.
Expression (expression 1) is the root of the equation (expression 2). Substitute the value of x from expression 1 instead of x into expression 2 and get the basic logarithmic identity:
So we see that each value is assigned a value. We denote b by x (), c by y, and thus we obtain a logarithmic function:
For example: 
Let us recall the basic properties of the logarithmic function.
Let's pay attention once again, here, because under the logarithm there can be a strictly positive expression, as the base of the logarithm.
Rice. 1. Graph of the logarithmic function at various bases
The function graph for is shown in black. Rice. 1. If the argument increases from zero to infinity, the function increases from minus to plus infinity.
The function graph for is shown in red. Rice. 1.
Properties of this function:
Domain: ;
Range of values:;
The function is monotonic throughout its domain of definition. When monotonically (strictly) increases, a larger value of the argument corresponds to a larger value of the function. When monotonically (strictly) decreases, the larger value of the argument corresponds to the smaller value of the function.
The properties of the logarithmic function are the key to solving a variety of logarithmic equations.
Consider the simplest logarithmic equation, all other logarithmic equations, as a rule, are reduced to this form.
Since the bases of the logarithms and the logarithms themselves are equal, the functions under the logarithm are also equal, but we must not miss the domain of definition. Only a positive number can stand under the logarithm, we have:
We found out that the functions f and g are equal, so it is enough to choose any one inequality in order to comply with the DHS.
Thus, we got a mixed system in which there is an equation and inequality:
Inequality, as a rule, is not necessary to solve, it is enough to solve the equation and substitute the found roots into the inequality, thus performing a check.
Let us formulate a method for solving the simplest logarithmic equations:
Equalize the bases of logarithms;
Equate sub-logarithmic functions;
Check.
Let's consider specific examples.
Example 1 - Solve the equation:
The bases of the logarithms are initially equal, we have the right to equate sub-logarithmic expressions, do not forget about the ODZ, we will choose the first logarithm to compose the inequality:
Example 2 - Solve the equation:
This equation differs from the previous one in that the bases of the logarithms are less than one, but this does not affect the solution in any way:
Find the root and substitute it into the inequality:
We got the wrong inequality, which means that the found root does not satisfy the ODV.
Example 3 - Solve the equation:
The bases of the logarithms are initially equal, we have the right to equate sub-logarithmic expressions, do not forget about the ODZ, we will choose the second logarithm to compose the inequality:
Find the root and substitute it into the inequality:
Obviously, only the first root satisfies the ODV.
Examples:
\ (\ log_ (2) (x) = 32 \)
\ (\ log_3x = \ log_39 \)
\ (\ log_3 ((x ^ 2-3)) = \ log_3 ((2x)) \)
\ (\ log_ (x + 1) ((x ^ 2 + 3x-7)) = 2 \)
\ (\ lg ^ 2 ((x + 1)) + 10 = 11 \ lg ((x + 1)) \)
How to solve logarithmic equations:
When solving a logarithmic equation, you need to strive to transform it to the form \ (\ log_a (f (x)) = \ log_a (g (x)) \), then make the transition to \ (f (x) = g (x) \).
\ (\ log_a (f (x)) = \ log_a (g (x)) \) \ (⇒ \) \ (f (x) = g (x) \).
Example:\ (\ log_2 (x-2) = 3 \)
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Solution: |
ODZ: |
Very important! This transition can be done only if:
You wrote for the original equation, and at the end check to see if the ones found are included in the DHS. If this is not done, extra roots may appear, which means the wrong decision.
The number (or expression) on the left and right is the same;
The logarithms on the left and right are "pure", that is, there should be no multiplications, divisions, etc. - only lone logarithms on either side of the equal sign.
For example:
Note that equations 3 and 4 can be easily solved by applying the desired properties of logarithms.
Example ... Solve the equation \ (2 \ log_8x = \ log_82,5 + \ log_810 \)
Solution :
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Let's write ODZ: \ (x> 0 \). |
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\ (2 \ log_8x = \ log_82,5 + \ log_810 \) ODZ: \ (x> 0 \) |
On the left in front of the logarithm is the coefficient, on the right is the sum of the logarithms. This disturbs us. We transfer two to the exponent \ (x \) by the property: \ (n \ log_b (a) = \ log_b (a ^ n) \). We represent the sum of the logarithms as one logarithm by the property: \ (\ log_ab + \ log_ac = \ log_a (bc) \) |
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\ (\ log_8 (x ^ 2) = \ log_825 \) |
We brought the equation to the form \ (\ log_a (f (x)) = \ log_a (g (x)) \) and wrote down the ODZ, so you can go to the form \ (f (x) = g (x) \ ). |
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Happened . We solve it and get the roots. |
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\ (x_1 = 5 \) \ (x_2 = -5 \) |
We check if the roots are suitable for ODZ. To do this, in \ (x> 0 \) instead of \ (x \) we substitute \ (5 \) and \ (- 5 \). This operation can be performed orally. |
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\(5>0\), \(-5>0\) |
The first inequality is true, the second is not. So \ (5 \) is the root of the equation, but \ (- 5 \) is not. We write down the answer. |
Answer : \(5\)
Example : Solve the equation \ (\ log ^ 2_2 (x) -3 \ log_2 (x) + 2 = 0 \)
Solution :
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Let's write ODZ: \ (x> 0 \). |
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\ (\ log ^ 2_2 (x) -3 \ log_2 (x) + 2 = 0 \) ODZ: \ (x> 0 \) |
A typical equation solved with. Replace \ (\ log_2x \) with \ (t \). |
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\ (t = \ log_2x \) |
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We got the usual. We are looking for its roots. |
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\ (t_1 = 2 \) \ (t_2 = 1 \) |
We do the reverse replacement |
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\ (\ log_2 (x) = 2 \) \ (\ log_2 (x) = 1 \) |
Transform the right-hand sides, representing them as logarithms: \ (2 = 2 \ cdot 1 = 2 \ log_22 = \ log_24 \) and \ (1 = \ log_22 \) |
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\ (\ log_2 (x) = \ log_24 \) \ (\ log_2 (x) = \ log_22 \) |
Now our equations are of the form \ (\ log_a (f (x)) = \ log_a (g (x)) \) and we can jump to \ (f (x) = g (x) \). |
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\ (x_1 = 4 \) \ (x_2 = 2 \) |
We check the correspondence of the roots of the ODZ. To do this, we substitute \ (4 \) and \ (2 \) into the inequality \ (x> 0 \) instead of \ (x \). |
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\(4>0\) \(2>0\) |
Both inequalities are true. Hence, both \ (4 \) and \ (2 \) are roots of the equation. |
Answer : \(4\); \(2\).
Logarithmic equations. From simple to complex.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very ..."
And for those who are "very even ...")
What is a logarithmic equation?
This is an equation with logarithms. I was surprised, right?) Then I'll clarify. This is an equation in which unknowns (x) and expressions with them are found inside logarithms. And only there! It is important.
Here are some examples logarithmic equations:
log 3 x = log 3 9
log 3 (x 2 -3) = log 3 (2x)
log x + 1 (x 2 + 3x-7) = 2
lg 2 (x + 1) +10 = 11lg (x + 1)
Well, you get the idea ... )
Note! A wide variety of expressions with x's are located exclusively inside the logarithms. If, suddenly, an x is found in the equation somewhere outside, for example:
log 2 x = 3 + x,
this will already be a mixed-type equation. Such equations do not have clear rules for solving. We will not consider them yet. By the way, there are equations where inside the logarithms only numbers... For example:
What can I say? Lucky you if you come across this! Logarithm with numbers is some number. And that's all. It is enough to know the properties of logarithms to solve such an equation. Knowledge of special rules, techniques adapted specifically for solving logarithmic equations, not required here.
So, what is logarithmic equation- figured it out.
How to solve logarithmic equations?
Solution logarithmic equations- the thing, in fact, is not very simple. So the section we have - for four ... Requires a decent stock of knowledge on all sorts of related topics. In addition, there is a special feature in these equations. And this feature is so important that it can be safely called the main problem in solving logarithmic equations. We will deal with this problem in detail in the next lesson.
For now, don't worry. We'll go the right way from simple to complex. Using specific examples. The main thing is to delve into simple things and do not be lazy to follow the links, I did not put them just like that ... And you will succeed. Necessarily.
Let's start with the most elementary, simplest equations. To solve them, it is desirable to have an idea of the logarithm, but nothing more. Just no idea logarithm, tackle a solution logarithmic equations - somehow embarrassing even ... Very boldly, I would say).
The simplest logarithmic equations.
These are equations of the form:
1.log 3 x = log 3 9
2.log 7 (2x-3) = log 7x
3.log 7 (50x-1) = 2
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. Therefore, the simplest.)
And solving such logarithmic equations is surprisingly simple. See for yourself.
Solving the first example:
log 3 x = log 3 9
To solve this example, you don't need to know almost anything, yes ... Purely intuition!) especially don't like this example? What-what ... Logarithms are not pleasant! Right. So let's get rid of them. We look closely at an example, and we have a natural desire ... Downright irresistible! Take and throw out logarithms altogether. And what pleases me is can do! Mathematics allows. Logarithms disappear the answer is:
Great, isn't it? You can (and should) always do this. Eliminating logarithms in this way is one of the main ways to solve logarithmic equations and inequalities. In mathematics, this operation is called potentiation. There are, of course, their own rules for such liquidation, but they are few. Remember:
You can eliminate logarithms without any fear if they have:
a) identical numerical bases
c) left-right logarithms are pure (without any coefficients) and are in splendid isolation.
Let me explain the last point. In an equation, say
log 3 x = 2log 3 (3x-1)
you cannot remove logarithms. The deuce on the right does not allow. Coefficient, you know ... In the example
log 3 x + log 3 (x + 1) = log 3 (3 + x)
it is also impossible to potentiate the equation. There is no lone logarithm on the left. There are two of them.
In short, you can remove the logarithms if the equation looks like this and only like this:
log a (.....) = log a (.....)
In parentheses, where ellipsis can be any expressions. Simple, super complex, all sorts. Anything. The important thing is that after the elimination of logarithms, we still have a simpler equation. It is assumed, of course, that you already know how to solve linear, quadratic, fractional, exponential and other equations without logarithms.)
Now the second example can be easily solved:
log 7 (2x-3) = log 7 x
Actually, it is decided in the mind. Potentiating, we get:
Well, is it very difficult?) As you can see, logarithmic part of the solution to the equation is only in the elimination of logarithms ... And then the solution of the remaining equation goes without them. Trivial business.
Let's solve the third example:
log 7 (50x-1) = 2
We see that the logarithm is on the left:
We recall that this logarithm is some number to which the base (i.e., seven) must be raised in order to obtain a sub-logarithm expression, i.e. (50x-1).
But that number is two! According to the equation. That is:
That, in essence, is all. Logarithm disappeared, there is a harmless equation left:
We solved this logarithmic equation based only on the meaning of the logarithm. Is it easier to eliminate the logarithms?) I agree. By the way, if you make a logarithm of two, you can solve this example through liquidation. From any number, you can make a logarithm. Moreover, the way we need it. A very useful trick in solving logarithmic equations and (especially!) Inequalities.
Do not know how to make a logarithm from a number !? It's OK. Section 555 describes this technique in detail. You can master and apply it to its fullest! It greatly reduces the number of errors.
The fourth equation is solved in exactly the same way (by definition):
That's all there is to it.
Let's summarize this lesson. We have considered by examples the solution of the simplest logarithmic equations. It is very important. And not only because such equations are on the control exams. The fact is that even the most evil and confused equations must be reduced to the simplest ones!
Actually, the simplest equations are the finishing part of the solution. any equations. And this finishing part must be understood as a matter of course! And further. Be sure to read this page to the end. There is a surprise there ...)
Now we decide on our own. We fill our hand, so to speak ...)
Find the root (or the sum of the roots, if there are several) of the equations:
ln (7x + 2) = ln (5x + 20)
log 2 (x 2 +32) = log 2 (12x)
log 16 (0.5x-1.5) = 0.25
log 0.2 (3x-1) = -3
ln (e 2 + 2x-3) = 2
log 2 (14x) = log 2 7 + 2
Answers (in disarray, of course): 42; 12; nine; 25; 7; 1.5; 2; 16.
What, not everything is working out? It happens. Do not grieve! Section 555 describes the solution to all these examples in a clear and detailed manner. You will certainly figure it out there. Moreover, master useful practical techniques.
Everything worked out!? All examples are "one left"?) Congratulations!
The time has come to reveal to you the bitter truth. Successful solution of these examples does not at all guarantee success in solving all other logarithmic equations. Even the simplest ones like these. Alas.
The fact is that the solution to any logarithmic equation (even the most elementary one!) Consists of two equal parts. Solving the equation, and working with the ODZ. One part - solving the equation itself - we have mastered. It's not that hard right?
For this lesson, I have specially selected such examples in which the LDO does not affect the answer in any way. But not everyone is as kind as me, right? ...)
Therefore, it is imperative to master the other part. ODZ. This is the main problem in solving logarithmic equations. And not because it's difficult - this part is even easier than the first. But because they simply forget about ODZ. Or they don’t know. Or both). And fall out of the blue ...
In the next lesson, we will deal with this problem. Then you can confidently decide any simple logarithmic equations and get to quite solid tasks.
If you like this site ...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Instant validation testing. Learning - with interest!)
you can get acquainted with functions and derivatives.
This article contains a systematic presentation of methods for solving logarithmic equations in one variable. This will help the teacher, primarily in the didactic sense: the selection of exercises allows you to compose individual assignments for students, taking into account their capabilities. These exercises can be used for a generalization lesson and to prepare for the exam.
Brief theoretical information and problem solving allow students to independently develop their skills in solving logarithmic equations.
Solving logarithmic equations.
Logarithmic equations - equations containing the unknown under the sign logarithm. When solving logarithmic equations, theoretical information is often used:
Usually, solving logarithmic equations begins with determining the ODV. In logarithmic equations, it is recommended to transform all logarithms so that their bases are equal. Then the equations are either expressed in terms of one logarithm, which is denoted by a new variable, or the equation is transformed to a form convenient for potentiation.
Transformations of logarithmic expressions should not lead to a narrowing of the ODV, but if the applied solution method narrows the ODV, leaving out individual numbers from consideration, then these numbers at the end of the problem must be checked by substituting into the original equation, since with a narrowing of the ODZ, loss of roots is possible.
1.
Equations of the form- an expression containing an unknown number, but a number.
1) use the definition of the logarithm:;
2) make a check or find the range of admissible values for an unknown number and select the corresponding roots (solutions).
If ) .
2. Equations of the first degree with respect to the logarithm, in the solution of which the properties of the logarithms are used.
To solve such equations, you need:
1) using the properties of logarithms, transform the equation;
2) solve the resulting equation;
3) make a check or find the range of admissible values for an unknown number and select the corresponding roots (solutions).
).
3. Equation of the second and higher degree with respect to the logarithm.
To solve such equations, you need:
- make a change to a variable;
- solve the resulting equation;
- make a reverse replacement;
- solve the resulting equation;
- check or find the range of admissible values for an unknown number and select the corresponding roots (solutions).
4. Equations containing the unknown in the base and in the exponent.
To solve such equations, you need:
- logarithm the equation;
- solve the resulting equation;
- check or find the range of acceptable values for an unknown number and select the corresponding
roots (solutions).
5. Equations that have no solution.
- To solve such equations, it is necessary to find the ODZ equations.
- Analyze the left and right sides of the equation.
- Make the appropriate conclusions.
The original equation is equivalent to the system:
Prove that the equation has no solution.
The ODZ of the equation is determined by the inequality x ≥ 0. On the ODZ we have
The sum of a positive number and a non-negative number is not zero, so the original equation has no solutions.
Answer: There are no solutions.
Only one root x = 0 gets into ODZ. Answer: 0.
We will make a reverse replacement.
Found roots belong to ODZ.
ODZ equations - the set of all positive numbers.
Insofar as
These equations are solved in a similar way:
Tasks for an independent solution:
Used Books.
- Beschetnov V.M. Maths. Moscow Demiurge 1994
- Borodulya I.T. Exponential and logarithmic functions. (tasks and exercises). Moscow "Education" 1984
- Vavilov V.V., Melnikov I.I., Olekhnik S.N., Pasichenko P.I. Mathematics tasks. Equations and inequalities. Moscow "Science" 1987
- Merzlyak A.G., Polonskiy V.B., Yakir M.S. Algebraic simulator. Moscow "Ileksa" 2007
- Saakyan S.M., Goldman A.M., Denisov D.V. Problems in algebra and the principles of analysis. Moscow "Education" 2003
Logarithmic equation is called an equation 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 unknown.
By 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 = x-2, then such an equation is already called mixed and a special approach is needed to solve it.
The ideal case is a situation when you come across an equation in which only numbers are under the sign of the logarithm, for example x + 2 = log 2 2. Here it is enough to know the properties of logarithms to solve it. But this kind of luck doesn't happen often, so get ready for the harder things.
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 the simplest logarithmic equations
These include equations of the type log 2 x = log 2 16. The naked eye can see that dropping the sign of the logarithm, we get x = 16.
In order to solve a more complex logarithmic equation, it is usually reduced to solving an ordinary algebraic equation or to solving the simplest logarithmic equation log a x = b. In the simplest equations, this happens in one motion, which is why they are called the simplest ones.
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:
- identical numerical bases for logarithms
- logarithms in both sides of the equation are found freely, i.e. without any coefficients and other various kinds of expressions.
Let's say in the equation log 2 x = 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) also fails one of the constraints - on the left there are two logarithms. 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 found in brackets, this has absolutely no effect on the operation of potentiation. And after the elimination of logarithms, a simpler equation will remain - linear, quadratic, exponential, etc., which, I hope, you already know how to solve.
Let's take another example:
log 3 (2x-5) = log 3 x
We apply 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:
We got a nice answer again. Here we have dispensed with the elimination of logarithms, but potentiation is applicable here, because a 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 examined how to solve the simplest logarithmic equations, which are actually very important, because solving logarithmic equations, even the most terrible and twisted, in the end always comes down to solving the simplest equations.
In everything that we did above, we lost sight of one very important point, which in the future will have a decisive role. 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 the work with the range of permissible values (ADV). That's just the first part we have mastered. In the above examples, the DHS 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, we will, of course, solve it, but most likely it will be wrong, because there is a small ambush in it, into which both C-students and excellent students are immediately caught. 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 of them:
log 3 (x 2 -3) = log 3 (2x)
We use potentiation, here it is permissible. As a result, we get the usual quadratic equation.
Find the roots of the equation:
It turned out two roots.
Answer: 3 and -1
At first glance, everything is correct. But let's check the result and plug 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)
So stop! Outwardly, everything is perfect. One point - there are no logarithms of negative numbers! This means that the root x = -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 ODZ played its fatal role, which we forgot about.
Let me remind you that under the range of valid values, such values of x are accepted that are allowed or make sense for the original example.
Without ODZ, any solution, even the absolutely correct one, of any equation turns into a lottery - 50/50.
How could we get caught while solving a seemingly elementary example? But exactly at the moment of potentiation. Logarithms disappeared, and with them all restrictions.
What, then, to do? Refuse to eliminate logarithms? And completely refuse to solve 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 away those roots that are not included in our LDZ, 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, an even root, etc. Until we solve the equation, we do not know what x equals, but we firmly know that such x, which, when substituted, will give division by 0 or taking the square root of a negative number, will certainly not fit in the answer. Therefore, such x are unacceptable, while the rest will constitute the ODZ.
Let's use the same equation again:
log 3 (x 2 -3) = log 3 (2x)
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 remember that the expression inside the logarithm must always be> 0. We write this condition in the form of the ODZ:
Those. we haven’t decided anything yet, but we have already written down a prerequisite for the entire sub-logarithmic expression. The curly brace means that these conditions must be met at the same time.
ODZ is written down, but it is also necessary to solve the resulting system of inequalities, which is what we will do. We get the answer x> v3. Now we know for sure which x won't suit us. And then we are already starting to solve the logarithmic equation itself, which we did above.
Having received the answers x 1 = 3 and x 2 = -1, it is easy to see that only x1 = 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 do the solution of any logarithmic equation in 2 stages. The first one - we solve the equation itself, the second one - we solve the condition of ODD. Both stages are performed independently of each other and are compared only when writing an answer, i.e. discard all unnecessary and write down the correct answer.
To consolidate the material, we strongly recommend watching the video:
On the video, there are other examples of solving the log. equations and working out the method of intervals in practice.
On this question, how to solve logarithmic equations, for now. If something is decided by the log. equations remained unclear or incomprehensible, write your questions in the comments.
Note: The Academy of Social Education (KSUI) is ready to accept new students.