Mathematics is more than science is the language of science.
Danish physicist and public figure Niels Bohr
Logarithmic Equations
Among the typical tasks, offered at the entrance (competitive) tests, are tasks, related to the solution of logarithmic equations. To successfully solve such problems, it is necessary to have a good knowledge of the properties of logarithms and to have skills in their application.
In this article, we first present the basic concepts and properties of logarithms, and then examples of solving logarithmic equations are considered.
Basic concepts and properties
Initially, we present the main properties of logarithms, the use of which allows one to successfully solve relatively complex logarithmic equations.
The basic logarithmic identity is written as
, (1)
The most famous properties of logarithms include the following equalities:
1. If , , and , then , ,
2. If , , , and , then .
3. If , , and , then .
4. If , , and natural number, then
5. If , , and natural number, then
6. If , , and , then .
7. If , , and , then .
More complex properties logarithms are formulated through the following statements:
8. If , , , and , then
9. If , , and , then
10. If , , , and , then
The proof of the last two properties of logarithms is given in the author's textbook "Mathematics for High School Students: Additional Sections of School Mathematics" (Moscow: Lenand / URSS, 2014).
It should also be noted that function is increasing, if , and decreasing if .
Consider examples of problems for solving logarithmic equations, arranged in order of increasing complexity.
Examples of problem solving
Example 1. solve the equation
. (2)
Solution. From equation (2) we have . Let's transform the equation as follows: , or .
Because , then the root of equation (2) is.
Answer: .
Example 2. solve the equation
Solution. Equation (3) is equivalent to the equations
Or .
From here we get .
Answer: .
Example 3. solve the equation
Solution. Equation (4) implies, what . Using the basic logarithmic identity (1), can be written
or .
If we put , then from here we get quadratic equation , which has two roots and . However, therefore and a suitable root of the equation is only . Since , then or .
Answer: .
Example 4. solve the equation
Solution.Valid range of a variablein equation (5) are.
Let and . Since the functionon the domain of definition is decreasing, and the function increases throughout numerical axis , then the equation cannot have more than one root.
By selection we find the only root.
Answer: .
Example 5. solve the equation.
Solution. If both sides of the equation are taken as logarithms to base 10, then
Or .
Solving the quadratic equation for , we obtain and . Therefore, here we have and .
Answer: , .
Example 6. solve the equation
. (6)
Solution.We use identity (1) and transform equation (6) as follows:
Or .
Answer: , .
Example 7. solve the equation
. (7)
Solution. Taking property 9 into account, we have . In this regard, equation (7) takes the form
From here we get or .
Answer: .
Example 8. solve the equation
. (8)
Solution.Let us use property 9 and rewrite equation (8) in the equivalent form.
If we then designate, then we get the quadratic equation, where . Since the equationhas only one positive root, then or . This implies .
Answer: .
Example 9. solve the equation
. (9)
Solution. Since it follows from equation (9), then here . According to property 10, can be written down.
In this regard, equation (9) will be equivalent to the equations
Or .
From here we obtain the root of equation (9).
Example 10. solve the equation
. (10)
Solution. The range of acceptable values for the variable in equation (10) is . According to property 4, here we have
. (11)
Since , then equation (11) takes the form of a quadratic equation , where . The roots of the quadratic equation are and .
Since , then and . From here we get and .
Answer: , .
Example 11. solve the equation
. (12)
Solution. Let's denote then and equation (12) takes the form
Or
. (13)
It is easy to see that the root of equation (13) is . Let us show that this equation has no other roots. To do this, we divide both its parts by and get equivalent equation
. (14)
Since the function is decreasing, and the function is increasing on the entire real axis, equation (14) cannot have more than one root. Since equations (13) and (14) are equivalent, equation (13) has a single root .
Since , then and .
Answer: .
Example 12. solve the equation
. (15)
Solution. Let's denote and . Since the function is decreasing on the domain of definition, and the function is increasing for any values of , then the equation cannot have a Bode one root. By direct selection, we establish that the desired root of equation (15) is .
Answer: .
Example 13. solve the equation
. (16)
Solution. Using the properties of logarithms, we obtain
Since then and we have the inequality
The resulting inequality coincides with equation (16) only if or .
Value substitutioninto equation (16) we make sure that, what is its root.
Answer: .
Example 14. solve the equation
. (17)
Solution. Since here , then equation (17) takes the form .
If we put , then from here we obtain the equation
, (18)
where . Equation (18) implies: or . Since , then the equation has one suitable root. However, therefore .
Example 15. solve the equation
. (19)
Solution. Denote , then equation (19) takes the form . If we take the logarithm of this equation in base 3, we get
Or
From this it follows that and . Since , then and . In this regard, and
Answer: , .
Example 16. solve the equation
. (20)
Solution. Let's introduce the parameterand rewrite equation (20) as a quadratic equation with respect to the parameter, i.e.
. (21)
The roots of equation (21) are
or , . Since , we have equations and . From here we get and .
Answer: , .
Example 17. solve the equation
. (22)
Solution. To establish the domain of definition of the variable in equation (22), it is necessary to consider a set of three inequalities: , and .
Applying property 2, from equation (22) we obtain
Or
. (23)
If in equation (23) we put, then we get the equation
. (24)
Equation (24) will be solved as follows:
Or
It follows from here that and , i.e., equation (24) has two roots: and .
Since , then , or , .
Answer: , .
Example 18. solve the equation
. (25)
Solution. Using the properties of logarithms, we transform equation (25) as follows:
, , .
From here we get .
Example 19. solve the equation
. (26)
Solution. Since , then .
Next, we have . Consequently , equality (26) is satisfied only if, when both sides of the equation are equal to 2 at the same time.
In this way , equation (26) is equivalent to the system of equations
From the second equation of the system we obtain
Or .
It's easy to see what's the meaning also satisfies the first equation of the system.
Answer: .
For a deeper study of methods for solving logarithmic equations, you can refer to teaching aids from the list of recommended literature.
1. Kushnir A.I. Masterpieces of school mathematics (problems and solutions in two books). - Kyiv: Astarta, book 1, 1995. - 576 p.
2. Collection of problems in mathematics for applicants to technical universities / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.
3. Suprun V.P. Mathematics for high school students: additional sections school curriculum. – M.: Lenand / URSS, 2014. - 216 p.
4. Suprun V.P. Mathematics for high school students: tasks of increased complexity. - M .: KD "Librocom" / URSS, 2017. - 200 p.
5. Suprun V.P. Mathematics for high school students: non-standard methods for solving problems. - M .: KD "Librocom" / URSS, 2017. - 296 p.
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With this video, I begin a long series of lessons about logarithmic equations. Now you have three examples at once, on the basis of which we will learn to solve the most simple tasks, which are called protozoa.
log 0.5 (3x - 1) = -3
lg (x + 3) = 3 + 2 lg 5
Let me remind you that the simplest logarithmic equation is the following:
log a f(x) = b
It is important that the variable x is present only inside the argument, i.e. only in the function f(x). And the numbers a and b are just numbers, and in no case are functions containing the variable x.
Basic solution methods
There are many ways to solve such structures. For example, most teachers at school suggest this way: Immediately express the function f ( x ) using the formula f( x ) = a b . That is, when you meet the simplest construction, you can immediately proceed to the solution without additional actions and constructions.
Yes, of course, the decision will turn out to be correct. However, the problem with this formula is that most students do not understand, where does it come from and why exactly we raise the letter a to the letter b.
As a result, I often observe very offensive errors, when, for example, these letters are interchanged. This formula must either be understood or memorized, and the second method leads to errors at the most inopportune and most crucial moments: in exams, tests, etc.
That is why I suggest to all my students to abandon the standard school formula and use the second approach to solve logarithmic equations, which, as you probably guessed from the name, is called canonical form.
The idea of the canonical form is simple. Let's look at our task again: on the left we have log a , while the letter a means exactly the number, and in no case the function containing the variable x. Therefore, this letter is subject to all restrictions that are imposed on the base of the logarithm. namely:
1 ≠ a > 0
On the other hand, from the same equation, we see that the logarithm must be is equal to the number b , and no restrictions are imposed on this letter, because it can take any value - both positive and negative. It all depends on what values the function f(x) takes.
And here we remember our wonderful rule that any number b can be represented as a logarithm in base a from a to the power of b:
b = log a a b
How to remember this formula? Yes, very simple. Let's write the following construction:
b = b 1 = b log a a
Of course, in this case, all the restrictions that we wrote down at the beginning arise. And now let's use the basic property of the logarithm, and enter the factor b as the power of a. We get:
b = b 1 = b log a a = log a a b
As a result, the original equation will be rewritten in the following form:
log a f (x) = log a a b → f (x) = a b
That's all. The new function no longer contains a logarithm and is solved by standard algebraic techniques.
Of course, someone will now object: why was it necessary to come up with some kind of canonical formula at all, why perform two additional unnecessary steps, if it was possible to immediately go from the original construction to the final formula? Yes, if only because most students do not understand where this formula comes from and, as a result, regularly make mistakes when applying it.
But such a sequence of actions, consisting of three steps, allows you to solve the original logarithmic equation, even if you do not understand where that final formula comes from. By the way, this entry is called the canonical formula:
log a f(x) = log a a b
The convenience of the canonical form also lies in the fact that it can be used to solve a very wide class of logarithmic equations, and not just the simplest ones that we are considering today.
Solution examples
Now let's look at real examples. So let's decide:
log 0.5 (3x - 1) = -3
Let's rewrite it like this:
log 0.5 (3x − 1) = log 0.5 0.5 −3
Many students are in a hurry and try to immediately raise the number 0.5 to the power that came to us from the original problem. And indeed, when you are already well trained in solving such problems, you can immediately perform this step.
However, if now you are just starting to study this topic, it is better not to rush anywhere so as not to make offensive mistakes. So we have the canonical form. We have:
3x - 1 = 0.5 -3
This is no longer a logarithmic equation, but a linear one with respect to the variable x. To solve it, let's first deal with the number 0.5 to the power of −3. Note that 0.5 is 1/2.
(1/2) −3 = (2/1) 3 = 8
Convert all decimals to fractions when you solve a logarithmic equation.
We rewrite and get:
3x − 1 = 8
3x=9
x=3
All we got the answer. The first task is solved.
Second task
Let's move on to the second task:
As you can see, this equation is no longer the simplest one. If only because the difference is on the left, and not a single logarithm in one base.
Therefore, you need to somehow get rid of this difference. In this case, everything is very simple. Let's take a closer look at the bases: on the left is the number under the root:
General recommendation: in all logarithmic equations, try to get rid of radicals, i.e., from entries with roots and move on to power functions, simply because the exponents of these powers are easily taken out of the sign of the logarithm and, ultimately, such a notation greatly simplifies and speeds up calculations. Let's write it like this:
Now we recall the remarkable property of the logarithm: from the argument, as well as from the base, you can take out degrees. In the case of bases, the following happens:
log a k b = 1/k loga b
In other words, the number that stood in the degree of the base is brought forward and at the same time turned over, that is, it becomes the reciprocal of the number. In our case, there was a degree of base with an indicator of 1/2. Therefore, we can take it out as 2/1. We get:
5 2 log 5 x − log 5 x = 18
10 log 5 x − log 5 x = 18
Please note: in no case should you get rid of logarithms at this step. Think back to grade 4-5 math and the order of operations: multiplication is performed first, and only then addition and subtraction are performed. In this case, we subtract one of the same elements from 10 elements:
9 log 5 x = 18
log 5 x = 2
Now our equation looks like it should. This is the simplest construction, and we solve it using the canonical form:
log 5 x = log 5 5 2
x = 5 2
x=25
That's all. The second problem is solved.
Third example
Let's move on to the third task:
lg (x + 3) = 3 + 2 lg 5
Recall the following formula:
log b = log 10 b
If for some reason you are confused by writing lg b , then when doing all the calculations, you can simply write log 10 b . You can work with decimal logarithms in the same way as with others: take out powers, add, and represent any number as lg 10.
It is precisely these properties that we will now use to solve the problem, since it is not the simplest one that we wrote down at the very beginning of our lesson.
To begin with, note that the factor 2 before lg 5 can be inserted and becomes a power of base 5. In addition, the free term 3 can also be represented as a logarithm - this is very easy to observe from our notation.
Judge for yourself: any number can be represented as log to base 10:
3 = log 10 10 3 = log 10 3
Let's rewrite the original problem taking into account the received changes:
lg (x − 3) = lg 1000 + lg 25
lg (x − 3) = lg 1000 25
lg (x - 3) = lg 25 000
Before us is again the canonical form, and we obtained it bypassing the stage of transformations, i.e., the simplest logarithmic equation did not come up anywhere with us.
That's what I was talking about at the very beginning of the lesson. The canonical form allows solving a wider class of problems than the standard school formula, which is given by most school teachers.
That's all, we get rid of the sign of the decimal logarithm, and we get a simple linear construction:
x + 3 = 25,000
x = 24997
All! Problem solved.
A note about scope
Here I would like to make an important remark about the domain of definition. Surely now there are students and teachers who will say: “When we solve expressions with logarithms, it is imperative to remember that the argument f (x) must be greater than zero!” In this regard, a logical question arises: why in none of the considered problems did we require that this inequality be satisfied?
Do not worry. No extra roots will appear in these cases. And this is another great trick that allows you to speed up the solution. Just know that if in the problem the variable x occurs only in one place (or rather, in the one and only argument of the one and only logarithm), and nowhere else in our case does the variable x, then write the domain no need because it will run automatically.
Judge for yourself: in the first equation, we got that 3x - 1, i.e., the argument should be equal to 8. This automatically means that 3x - 1 will be greater than zero.
With the same success, we can write that in the second case, x must be equal to 5 2, i.e., it is certainly greater than zero. And in the third case, where x + 3 = 25,000, i.e., again, obviously greater than zero. In other words, the scope is automatic, but only if x occurs only in the argument of only one logarithm.
That's all you need to know to solve simple problems. This rule alone, together with the transformation rules, will allow you to solve a very wide class of problems.
But let's be honest: in order to finally understand this technique, in order to learn how to apply the canonical form of the logarithmic equation, it is not enough just to watch one video lesson. Therefore, right now, download the options for an independent solution that are attached to this video tutorial and start solving at least one of these two independent works.
It will take you just a few minutes. But the effect of such training will be much higher compared to if you just watched this video tutorial.
I hope this lesson will help you understand logarithmic equations. Apply the canonical form, simplify expressions using the rules for working with logarithms - and you will not be afraid of any tasks. And that's all I have for today.
Scope consideration
Now let's talk about the domain of the logarithmic function, as well as how this affects the solution of logarithmic equations. Consider a construction of the form
log a f(x) = b
Such an expression is called the simplest - it has only one function, and the numbers a and b are just numbers, and in no case are a function that depends on the variable x. It is solved very simply. You just need to use the formula:
b = log a a b
This formula is one of the key properties of the logarithm, and when substituting into our original expression, we get the following:
log a f(x) = log a a b
f(x) = a b
This is already a familiar formula from school textbooks. Many students will probably have a question: since the function f ( x ) in the original expression is under the log sign, the following restrictions are imposed on it:
f(x) > 0
This restriction is valid because the logarithm of negative numbers does not exist. So, maybe because of this limitation, you should introduce a check for answers? Perhaps they need to be substituted in the source?
No, in the simplest logarithmic equations, an additional check is unnecessary. And that's why. Take a look at our final formula:
f(x) = a b
The fact is that the number a in any case is greater than 0 - this requirement is also imposed by the logarithm. The number a is the base. In this case, no restrictions are imposed on the number b. But this does not matter, because no matter what degree we raise a positive number, we will still get a positive number at the output. Thus, the requirement f (x) > 0 is fulfilled automatically.
What is really worth checking is the scope of the function under the log sign. There can be quite complex designs, and in the process of solving them, you must definitely follow them. Let's see.
First task:
First step: convert the fraction on the right. We get:
We get rid of the sign of the logarithm and get the usual irrational equation:
Of the obtained roots, only the first one suits us, since the second root is less than zero. The only answer will be the number 9. That's it, the problem is solved. No additional checks that the expression under the logarithm sign is greater than 0 are required, because it is not just greater than 0, but by the condition of the equation it is equal to 2. Therefore, the requirement "greater than zero" is automatically fulfilled.
Let's move on to the second task:
Everything is the same here. We rewrite the construction, replacing the triple:
We get rid of the signs of the logarithm and get an irrational equation:
We square both parts, taking into account the restrictions, and we get:
4 - 6x - x 2 = (x - 4) 2
4 - 6x - x 2 = x 2 + 8x + 16
x2 + 8x + 16 −4 + 6x + x2 = 0
2x2 + 14x + 12 = 0 |:2
x2 + 7x + 6 = 0
We solve the resulting equation through the discriminant:
D \u003d 49 - 24 \u003d 25
x 1 = −1
x 2 \u003d -6
But x = −6 does not suit us, because if we substitute this number into our inequality, we get:
−6 + 4 = −2 < 0
In our case, it is required that it be greater than 0 or, in extreme cases, equal. But x = −1 suits us:
−1 + 4 = 3 > 0
The only answer in our case is x = −1. That's all the solution. Let's go back to the very beginning of our calculations.
The main conclusion from this lesson is that it is not required to check the limits for a function in the simplest logarithmic equations. Because in the process of solving all the constraints are executed automatically.
However, this by no means means that you can forget about verification altogether. In the process of working on a logarithmic equation, it may well turn into an irrational one, which will have its own limitations and requirements for the right side, which we have seen today in two different examples.
Feel free to solve such problems and be especially careful if there is a root in the argument.
Logarithmic equations with different bases
We continue to study logarithmic equations and analyze two more rather interesting tricks with which it is fashionable to solve more complex structures. But first, let's remember how the simplest tasks are solved:
log a f(x) = b
In this notation, a and b are just numbers, and in the function f (x) the variable x must be present, and only there, that is, x must be only in the argument. We will transform such logarithmic equations using the canonical form. For this, we note that
b = log a a b
And a b is just an argument. Let's rewrite this expression as follows:
log a f(x) = log a a b
This is exactly what we are trying to achieve, so that both on the left and on the right there is a logarithm to the base a. In this case, we can, figuratively speaking, cross out the signs of log, and from the point of view of mathematics, we can say that we simply equate the arguments:
f(x) = a b
As a result, we get a new expression that will be solved much easier. Let's apply this rule to our tasks today.
So the first design:
First of all, I note that there is a fraction on the right, the denominator of which is log. When you see an expression like this, it's worth remembering the wonderful property of logarithms:
Translated into Russian, this means that any logarithm can be represented as a quotient of two logarithms with any base c. Of course, 0< с ≠ 1.
So: this formula has one wonderful special case when the variable c is equal to the variable b. In this case, we get a construction of the form:
It is this construction that we observe from the sign on the right in our equation. Let's replace this construction with log a b , we get:
In other words, in comparison with the original task, we have swapped the argument and the base of the logarithm. Instead, we had to flip the fraction.
We recall that any degree can be taken out of the base according to the following rule:
In other words, the coefficient k, which is the degree of the base, is taken out as an inverted fraction. Let's take it out as an inverted fraction:
The fractional factor cannot be left in front, because in this case we will not be able to represent this entry as a canonical form (after all, in the canonical form, there is no additional factor in front of the second logarithm). Therefore, let's put the fraction 1/4 in the argument as a power:
Now we equate the arguments whose bases are the same (and we really have the same bases), and write:
x + 5 = 1
x = −4
That's all. We got the answer to the first logarithmic equation. Pay attention: in the original problem, the variable x occurs only in one log, and it is in its argument. Therefore, there is no need to check the domain, and our number x = −4 is indeed the answer.
Now let's move on to the second expression:
log 56 = log 2 log 2 7 − 3 log (x + 4)
Here, in addition to the usual logarithms, we will have to work with lg f (x). How to solve such an equation? It may seem to an unprepared student that this is some kind of tin, but in fact everything is solved elementarily.
Look closely at the term lg 2 log 2 7. What can we say about it? The bases and arguments of log and lg are the same, and this should give some clues. Let's remember once again how the degrees are taken out from under the sign of the logarithm:
log a b n = nlog a b
In other words, what was the power of the number b in the argument becomes a factor in front of log itself. Let's apply this formula to the expression lg 2 log 2 7. Don't be afraid of lg 2 - this is the most common expression. You can rewrite it like this:
For him, all the rules that apply to any other logarithm are valid. In particular, the factor in front can be introduced into the power of the argument. Let's write:
Very often, students point blank do not see this action, because it is not good to enter one log under the sign of another. In fact, there is nothing criminal in this. Moreover, we get a formula that is easy to calculate if you remember an important rule:
This formula can be considered both as a definition and as one of its properties. In any case, if you convert a logarithmic equation, you should know this formula in the same way as the representation of any number in the form of log.
We return to our task. We rewrite it taking into account the fact that the first term to the right of the equal sign will simply be equal to lg 7. We have:
lg 56 = lg 7 − 3lg (x + 4)
Let's move lg 7 to the left, we get:
lg 56 - lg 7 = -3lg (x + 4)
We subtract the expressions on the left because they have the same base:
lg (56/7) = -3lg (x + 4)
Now let's take a closer look at the equation we've got. It is practically the canonical form, but there is a factor −3 on the right. Let's put it in the right lg argument:
lg 8 = lg (x + 4) −3
Before us is the canonical form of the logarithmic equation, so we cross out the signs of lg and equate the arguments:
(x + 4) -3 = 8
x + 4 = 0.5
That's all! We have solved the second logarithmic equation. In this case, no additional checks are required, because in the original problem x was present in only one argument.
Let me recap the key points of this lesson.
The main formula that is studied in all the lessons on this page devoted to solving logarithmic equations is the canonical form. And don't be put off by the fact that most school textbooks teach you how to solve these kinds of problems differently. This tool works very efficiently and allows you to solve a much wider class of problems than the simplest ones that we studied at the very beginning of our lesson.
In addition, to solve logarithmic equations, it will be useful to know the basic properties. Namely:
- The formula for moving to one base and a special case when we flip log (this was very useful to us in the first task);
- The formula for bringing in and taking out powers from under the sign of the logarithm. Here, many students get stuck and do not see point-blank that the power taken out and brought in can itself contain log f (x). Nothing wrong with that. We can introduce one log according to the sign of another and at the same time significantly simplify the solution of the problem, which is what we observe in the second case.
In conclusion, I would like to add that it is not required to check the scope in each of these cases, because everywhere the variable x is present in only one sign of log, and at the same time is in its argument. As a consequence, all domain requirements are met automatically.
Problems with variable base
Today we will consider logarithmic equations, which for many students seem non-standard, if not completely unsolvable. We are talking about expressions that are based not on numbers, but on variables and even functions. We will solve such constructions using our standard technique, namely, through the canonical form.
To begin with, let's recall how the simplest problems are solved, which are based on ordinary numbers. So, the simplest construction is called
log a f(x) = b
To solve such problems, we can use the following formula:
b = log a a b
We rewrite our original expression and get:
log a f(x) = log a a b
Then we equate the arguments, i.e. we write:
f(x) = a b
Thus, we get rid of the log sign and solve the usual problem. In this case, the roots obtained in the solution will be the roots of the original logarithmic equation. In addition, the record, when both the left and the right are on the same logarithm with the same base, is called the canonical form. It is to this record that we will try to reduce today's constructions. So let's go.
First task:
log x − 2 (2x 2 − 13x + 18) = 1
Replace 1 with log x − 2 (x − 2) 1 . The degree that we observe in the argument is, in fact, the number b , which was to the right of the equal sign. So let's rewrite our expression. We get:
log x - 2 (2x 2 - 13x + 18) = log x - 2 (x - 2)
What do we see? Before us is the canonical form of the logarithmic equation, so we can safely equate the arguments. We get:
2x2 - 13x + 18 = x - 2
But the solution does not end there, because this equation is not equivalent to the original one. After all, the resulting construction consists of functions that are defined on the entire number line, and our original logarithms are not defined everywhere and not always.
Therefore, we must write down the domain of definition separately. Let's not be wiser and first write down all the requirements:
First, the argument of each of the logarithms must be greater than 0:
2x 2 − 13x + 18 > 0
x − 2 > 0
Secondly, the base must not only be greater than 0, but also different from 1:
x − 2 ≠ 1
As a result, we get the system:
But don't be alarmed: when processing logarithmic equations, such a system can be greatly simplified.
Judge for yourself: on the one hand, we are required that the quadratic function be greater than zero, and on the other hand, this quadratic function is equated to a certain linear expression, which is also required that it be greater than zero.
In this case, if we require that x − 2 > 0, then the requirement 2x 2 − 13x + 18 > 0 will also be automatically satisfied. Therefore, we can safely cross out the inequality containing a quadratic function. Thus, the number of expressions contained in our system will be reduced to three.
Of course, we might as well cross out linear inequality, i.e. cross out x − 2 > 0 and require that 2x 2 − 13x + 18 > 0. But you must agree that it is much faster and easier to solve the simplest linear inequality than this system we get the same roots.
In general, try to optimize calculations whenever possible. And in the case of logarithmic equations, cross out the most difficult inequalities.
Let's rewrite our system:
Here is such a system of three expressions, two of which we, in fact, have already figured out. Let's separately write out the quadratic equation and solve it:
2x2 - 14x + 20 = 0
x2 − 7x + 10 = 0
Before us is a reduced square trinomial and, therefore, we can use the Vieta formulas. We get:
(x − 5)(x − 2) = 0
x 1 = 5
x2 = 2
Now, back to our system, we find that x = 2 doesn't suit us, because we're required to have x strictly greater than 2.
But x \u003d 5 suits us quite well: the number 5 is greater than 2, and at the same time 5 is not equal to 3. Therefore, the only solution to this system will be x \u003d 5.
Everything, the task is solved, including taking into account the ODZ. Let's move on to the second equation. Here we are waiting for more interesting and meaningful calculations:
The first step: as well as last time, we bring all this business to a canonical form. To do this, we can write the number 9 as follows:
The base with the root can not be touched, but it is better to transform the argument. Let's move from the root to the power with a rational exponent. Let's write:
Let me not rewrite our whole big logarithmic equation, but just immediately equate the arguments:
x 3 + 10x 2 + 31x + 30 = x 3 + 9x 2 + 27x + 27
x 2 + 4x + 3 = 0
Before us is the again reduced square trinomial, we will use the Vieta formulas and write:
(x + 3)(x + 1) = 0
x 1 = -3
x 2 = -1
So, we got the roots, but no one guaranteed us that they would fit the original logarithmic equation. After all, log signs impose additional restrictions (here we would have to write down the system, but due to the cumbersomeness of the whole construction, I decided to calculate the domain of definition separately).
First of all, remember that the arguments must be greater than 0, namely:
These are the requirements imposed by the domain of definition.
We note right away that since we equate the first two expressions of the system to each other, we can cross out any of them. Let's cross out the first one because it looks more menacing than the second one.
In addition, note that the solutions of the second and third inequalities will be the same sets (the cube of some number is greater than zero, if this number itself is greater than zero; similarly with the root of the third degree - these inequalities are completely similar, so one of them we can cross it out).
But with the third inequality, this will not work. Let's get rid of the sign of the radical on the left, for which we raise both parts to a cube. We get:
So we get the following requirements:
−2 ≠ x > −3
Which of our roots: x 1 = -3 or x 2 = -1 meets these requirements? Obviously, only x = −1, because x = −3 does not satisfy the first inequality (because our inequality is strict). In total, returning to our problem, we get one root: x = −1. That's it, problem solved.
Once again, the key points of this task:
- Feel free to apply and solve logarithmic equations using canonical form. Students who make such a record, and do not go directly from the original problem to a construction like log a f ( x ) = b , make much fewer errors than those who are in a hurry somewhere, skipping intermediate steps of calculations;
- As soon as in the logarithm appears variable base, the task ceases to be simple. Therefore, when solving it, it is necessary to take into account the domain of definition: the arguments must be greater than zero, and the bases must not only be greater than 0, but they must also not be equal to 1.
You can impose the last requirements on the final answers in different ways. For example, it is possible to solve a whole system containing all domain requirements. On the other hand, you can first solve the problem itself, and then remember about the domain of definition, work it out separately in the form of a system and apply it to the obtained roots.
Which way to choose when solving a particular logarithmic equation is up to you. In any case, the answer will be the same.
Logarithmic equations. We continue to consider tasks from part B of the Unified State Examination in mathematics. We have already considered the solutions of some equations in the articles "", "". In this article, we will consider logarithmic equations. I must say right away that there will be no complex transformations when solving such equations at the USE. They are simple.
It is enough to know and understand the basic logarithmic identity, to know the properties of the logarithm. Pay attention to the fact that after the decision, it is MANDATORY to do a check - substitute the obtained value into the original equation and calculate, as a result, the correct equality should be obtained.
Definition:
The logarithm of the number a to the base b is the exponent,to which b must be raised to get a.
For example:
Log 3 9 = 2 since 3 2 = 9
Properties of logarithms:
Special cases of logarithms:
We solve problems. In the first example, we will do a check. Do the following check yourself.
Find the root of the equation: log 3 (4–x) = 4
Since log b a = x b x = a, then
3 4 \u003d 4 - x
x = 4 - 81
x = -77
Examination:
log 3 (4–(–77)) = 4
log 3 81 = 4
3 4 = 81 Correct.
Answer: - 77
Decide for yourself:
Find the root of the equation: log 2 (4 - x) = 7
Find the root of the log 5 equation(4 + x) = 2
We use the basic logarithmic identity.
Since log a b = x b x = a, then
5 2 = 4 + x
x =5 2 – 4
x=21
Examination:
log 5 (4 + 21) = 2
log 5 25 = 2
5 2 = 25 Correct.
Answer: 21
Find the root of the equation log 3 (14 - x) = log 3 5.
The following property takes place, its meaning is as follows: if on the left and right sides of the equation we have logarithms with the same base, then we can equate the expressions under the signs of logarithms.
14 - x = 5
x=9
Make a check.
Answer: 9
Decide for yourself:
Find the root of the equation log 5 (5 - x) = log 5 3.
Find the root of the equation: log 4 (x + 3) = log 4 (4x - 15).
If log c a = log c b, then a = b
x + 3 = 4x - 15
3x = 18
x=6
Make a check.
Answer: 6
Find the root of the equation log 1/8 (13 - x) = - 2.
(1/8) -2 = 13 - x
8 2 \u003d 13 - x
x = 13 - 64
x = -51
Make a check.
A small addition - here the property is used
degree().
Answer: - 51
Decide for yourself:
Find the root of the equation: log 1/7 (7 - x) = - 2
Find the root of the equation log 2 (4 - x) = 2 log 2 5.
Let's transform the right side. use the property:
log a b m = m∙ log a b
log 2 (4 - x) = log 2 5 2
If log c a = log c b, then a = b
4 – x = 5 2
4 - x = 25
x = -21
Make a check.
Answer: - 21
Decide for yourself:
Find the root of the equation: log 5 (5 - x) = 2 log 5 3
Solve the equation log 5 (x 2 + 4x) = log 5 (x 2 + 11)
If log c a = log c b, then a = b
x2 + 4x = x2 + 11
4x = 11
x=2.75
Make a check.
Answer: 2.75
Decide for yourself:
Find the root of the equation log 5 (x 2 + x) = log 5 (x 2 + 10).
Solve the equation log 2 (2 - x) = log 2 (2 - 3x) +1.
On the right side of the equation, you need to get an expression of the form:
log 2 (......)
Representing 1 as a base 2 logarithm:
1 = log 2 2
log c (ab) = log c a + log c b
log 2 (2 - x) = log 2 (2 - 3x) + log 2 2
We get:
log 2 (2 - x) = log 2 2 (2 - 3x)
If log c a = log c b, then a = b, then
2 – x = 4 – 6x
5x = 2
x=0.4
Make a check.
Answer: 0.4
Decide for yourself: Next, you need to solve a quadratic equation. By the way,
the roots are 6 and -4.
Root "-4" is not a solution, since the base of the logarithm must be greater than zero, and with "– 4" is equal to " – 5". The solution is root 6.Make a check.
Answer: 6.
R eat on your own:
Solve the equation log x –5 49 = 2. If the equation has more than one root, answer the smaller one.
As you can see, no complex transformations with logarithmic equationsno. It is enough to know the properties of the logarithm and be able to apply them. In the USE tasks related to the transformation of logarithmic expressions, more serious transformations are performed and deeper skills in solving are required. We will consider such examples, do not miss it!I wish you success!!!
Sincerely, Alexander Krutitskikh.
P.S: I would be grateful if you tell about the site in social networks.
On the this lesson we will repeat the basic theoretical facts about logarithms and consider the solution of the simplest logarithmic equations.
Recall the central definition - the definition of the logarithm. It is connected with the solution of the exponential equation. This equation has a single root, it is called the logarithm of b to the 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.
The expression (expression 1) is the root of the equation (expression 2). We substitute the value of x from expression 1 instead of x in expression 2 and we get the basic logarithmic identity:
So we see that each value is assigned a value. We denote b for x (), c for y, and thus we get the logarithmic function:
For example: 
Recall the basic properties of the logarithmic function.
Let us 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 for various bases
The graph of the function at is shown in black. Rice. 1. If the argument increases from zero to infinity, the function increases from minus to plus infinity.
The graph of the function at is shown in red. Rice. one.
Properties of this function:
Domain: ;
Range of values: ;
The function is monotone throughout its domain of definition. When monotonically (strictly) increases, the larger value of the argument corresponds to the 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 various 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 lose the scope. 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 ODZ.
Thus, we got a mixed system in which there is an equation and an 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 sublogarithmic functions;
Run a check.
Let's consider specific examples.
Example 1 - solve the equation:
The bases of the logarithms are initially equal;
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:
Let's find the root and substitute it into the inequality:
We got an incorrect inequality, which means that the root found does not satisfy the ODZ.
Example 3 - solve the equation:
The bases of the logarithms are initially equal;
Let's find the root and substitute it into the inequality:
Obviously, only the first root satisfies the ODZ.
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