Solving equations with logarithms find roots examples. Logarithmic equation: basic formulas and techniques
Many students get stuck on equations of this kind. At the same time, the tasks themselves are by no means complex - it is enough just to perform a competent variable substitution, for which you should learn how to isolate stable expressions.
In addition to this lesson, you will find a rather voluminous independent work, consisting of two options with 6 tasks each.
Grouping method
Today we will analyze two logarithmic equations, one of which cannot be solved "throughout" and requires special transformations, and the second ... however, I will not tell everything at once. Watch the video, download independent work - and learn how to solve complex problems.
So, grouping and taking the common factors out of the bracket. In addition, I will tell you what pitfalls the domain of definition of logarithms carries, and how small remarks on the domain of definitions can significantly change both the roots and the whole solution.
Let's start with the grouping. We need to solve the following logarithmic equation:
log 2 x log 2 (x − 3) + 1 = log 2 (x 2 − 3x )
First of all, we note that x 2 − 3x can be factorized:
log 2 x (x − 3)
Then we remember the wonderful formula:
log a fg = log a f + log a g
Immediately a small note: this formula works fine when a, f and g are ordinary numbers. But when there are functions instead of them, these expressions cease to be equal in rights. Imagine this hypothetical situation:
f< 0; g < 0
In this case, the product fg will be positive, therefore, log a ( fg ) will exist, but log a f and log a g will not exist separately, and we cannot perform such a transformation.
Ignoring this fact will lead to a narrowing of the domain of definition and, as a result, to the loss of roots. Therefore, before performing such a transformation, it is necessary to make sure in advance that the functions f and g are positive.
In our case, everything is simple. Since there is a function log 2 x in the original equation, then x > 0 (after all, the variable x is in the argument). There is also log 2 (x − 3), so x − 3 > 0.
Therefore, in the function log 2 x (x − 3) each factor will be greater than zero. Therefore, we can safely decompose the product into the sum:
log 2 x log 2 (x − 3) + 1 = log 2 x + log 2 (x − 3)
log 2 x log 2 (x − 3) + 1 − log 2 x − log 2 (x − 3) = 0
At first glance, it may seem that it has not become easier. On the contrary: the number of terms only increased! To understand how to proceed further, we introduce new variables:
log 2 x = a
log 2 (x − 3) = b
a b + 1 − a − b = 0
And now we group the third term with the first:
(a b - a) + (1 - b) = 0
a (1 b - 1) + (1 - b ) = 0
Note that both the first and second brackets contain b − 1 (in the second case, you will have to take the “minus” out of the bracket). Let's factorize our construction:
a (1 b − 1) − (b − 1) = 0
(b − 1)(a 1 − 1) = 0
And now we recall our wonderful rule: the product is equal to zero when at least one of the factors is equal to zero:
b − 1 = 0 ⇒ b = 1;
a − 1 = 0 ⇒ a = 1.
Let's remember what b and a are. We get two simple logarithmic equations in which all that remains is to get rid of the signs of log and equate the arguments:
log 2 x = 1 ⇒ log 2 x = log 2 2 ⇒ x 1 =2;
log 2 (x − 3) = 1 ⇒ log 2 (x − 3) = log 2 2 ⇒ x 2 = 5
We got two roots, but this is not a solution to the original logarithmic equation, but only candidates for the answer. Now let's check the domain. For the first argument:
x > 0
Both roots satisfy the first requirement. Let's move on to the second argument:
x − 3 > 0 ⇒ x > 3
But here already x = 2 does not satisfy us, but x = 5 suits us quite well. Therefore, the only answer is x = 5.
We pass to the second logarithmic equation. At first glance, it is much simpler. However, in the process of solving it, we will consider subtle points related to the domain of definition, ignorance of which significantly complicates the life of novice students.
log 0.7 (x 2 - 6x + 2) = log 0.7 (7 - 2x)
Before us is the canonical form of the logarithmic equation. You don't need to convert anything - even the bases are the same. Therefore, we simply equate the arguments:
x 2 - 6x + 2 = 7 - 2x
x 2 - 6x + 2 - 7 + 2x = 0
x 2 - 4x - 5 = 0
Before us is the given quadratic equation, it is easily solved using the Vieta formulas:
(x − 5) (x + 1) = 0;
x − 5 = 0 ⇒ x = 5;
x + 1 = 0 ⇒ x = −1.
But these roots are not definitive answers yet. It is necessary to find the domain of definition, since there are two logarithms in the original equation, i.e. it is strictly necessary to take into account the domain of definition.
So, let's write out the domain of definition. On the one hand, the argument of the first logarithm must be greater than zero:
x 2 − 6x + 2 > 0
On the other hand, the second argument must also be greater than zero:
7 − 2x > 0
These requirements must be met at the same time. And here the most interesting begins. Of course, we can solve each of these inequalities, then intersect them and find the domain of the entire equation. But why make life so difficult for yourself?
Let's notice one subtlety. Getting rid of log signs, we equate arguments. This implies that the requirements x 2 − 6x + 2 > 0 and 7 − 2x > 0 are equivalent. As a consequence, either of the two inequalities can be crossed out. Let's cross out the most difficult, and leave the usual linear inequality for ourselves:
-2x > -7
x< 3,5
Since we were dividing both sides by a negative number, the sign of the inequality has changed.
So, we have found the ODZ without any square inequalities, discriminants and intersections. Now it remains just to choose the roots that lie on this interval. Obviously, only x = −1 will suit us, because x = 5 > 3.5.
You can write down the answer: x = 1 is the only solution to the original logarithmic equation.
The conclusions from this logarithmic equation are as follows:
- Don't be afraid to factor logarithms, and then factor the sum of logarithms. However, remember that by breaking the product into the sum of two logarithms, you thereby narrow the domain of definition. Therefore, before performing such a conversion, be sure to check what the scope requirements are. Most often, no problems arise, but it does not hurt to play it safe once again.
- When getting rid of the canonical form, try to optimize the calculations. In particular, if we are required that f > 0 and g > 0, but in the equation itself f = g , then we boldly cross out one of the inequalities, leaving only the simplest one for ourselves. In this case, the domain of definition and answers will not suffer in any way, but the amount of calculations will be significantly reduced.
That, in fact, is all that I wanted to tell about the grouping. :)
Typical mistakes in solving
Today we will analyze two typical logarithmic equations that many students stumble over. On the example of these equations, we will see what mistakes are most often made in the process of solving and transforming the original expressions.
Fractional-rational equations with logarithms
It should be noted right away that this is a rather insidious type of equation, in which a fraction with a logarithm somewhere in the denominator is not always immediately present. However, in the process of transformations, such a fraction will necessarily arise.
At the same time, be careful: in the process of transformations, the initial domain of definition of logarithms can change significantly!
We turn to even more rigid logarithmic equations containing fractions and variable bases. In order to do more in one short lesson, I will not tell an elementary theory. Let's go straight to the tasks:
4 log 25 (x − 1) − log 3 27 + 2 log x − 1 5 = 1
Looking at this equation, someone will ask: “What does the fractional rational equation have to do with it? Where is the fraction in this equation? Let's not rush and take a closer look at each term.
First term: 4 log 25 (x − 1). The base of the logarithm is a number, but the argument is a function of x . We can't do anything about this yet. Move on.
The next term is log 3 27. Recall that 27 = 3 3 . Therefore, we can rewrite the entire logarithm as follows:
log 3 27 = 3 3 = 3
So the second term is just a three. The third term: 2 log x − 1 5. Not everything is simple here either: the base is a function, the argument is an ordinary number. I propose to flip the whole logarithm according to the following formula:
log a b = 1/log b a
Such a transformation can only be performed if b ≠ 1. Otherwise, the logarithm that will be obtained in the denominator of the second fraction simply will not exist. In our case, b = 5, so everything is fine:
2 log x − 1 5 = 2/log 5 (x − 1)
Let's rewrite the original equation taking into account the obtained transformations:
4 log 25 (x − 1) − 3 + 2/ log 5 (x − 1) = 1
We have log 5 (x − 1) in the denominator of the fraction, and log 25 (x − 1) in the first term. But 25 \u003d 5 2, so we take out the square from the base of the logarithm according to the rule:
In other words, the exponent at the base of the logarithm becomes the fraction at the front. And the expression will be rewritten like this:
4 1/2 log 5 (x − 1) − 3 + 2/ log 5 (x − 1) − 1 = 0
We ended up with a long equation with a bunch of identical logarithms. Let's introduce a new variable:
log 5 (x − 1) = t;
2t − 4 + 2/t = 0;
But this is already a fractional-rational equation, which is solved by means of algebra of grades 8-9. First, let's break it down into two:
t − 2 + 1/t = 0;
(t 2 − 2t + 1)/t = 0
The exact square is in brackets. Let's roll it up:
(t − 1) 2 /t = 0
A fraction is zero when its numerator is zero and its denominator is non-zero. Never forget this fact:
(t − 1) 2 = 0
t=1
t ≠ 0
Let's remember what t is:
log 5 (x − 1) = 1
log 5 (x − 1) = log 5 5
We get rid of the log signs, equate their arguments, and we get:
x − 1 = 5 ⇒ x = 6
Everybody. Problem solved. But let's go back to the original equation and remember that there were two logarithms with the x variable at once. Therefore, you need to write out the domain of definition. Since x − 1 is in the logarithm argument, this expression must be greater than zero:
x − 1 > 0
On the other hand, the same x − 1 is also present in the base, so it must differ from one:
x − 1 ≠ 1
Hence we conclude:
x > 1; x ≠ 2
These requirements must be met at the same time. The value x = 6 satisfies both requirements, so x = 6 is the final solution to the logarithmic equation.
Let's move on to the second task:
Again, let's not rush and look at each term:
log 4 (x + 1) - there is a four at the base. The usual number, and you can not touch it. But last time we stumbled upon an exact square at the base, which had to be taken out from under the sign of the logarithm. Let's do the same now:
log 4 (x + 1) = 1/2 log 2 (x + 1)
The trick is that we already have a logarithm with variable x , albeit in the base - it is the inverse of the logarithm that we just found:
8 log x + 1 2 = 8 (1/log 2 (x + 1)) = 8/log 2 (x + 1)
The next term is log 2 8. This is a constant, since both the argument and the base are ordinary numbers. Let's find the value:
log 2 8 = log 2 2 3 = 3
We can do the same with the last logarithm:
Now let's rewrite the original equation:
1/2 log 2 (x + 1) + 8/log 2 (x + 1) − 3 − 1 = 0;
log 2 (x + 1)/2 + 8/log 2 (x + 1) − 4 = 0
Let's bring everything to a common denominator:
Before us is again a fractional-rational equation. Let's introduce a new variable:
t = log 2 (x + 1)
Let's rewrite the equation taking into account the new variable:
Be careful: at this step, I swapped the terms. The numerator of the fraction is the square of the difference:
Like last time, a fraction is zero when its numerator is zero and its denominator is non-zero:
(t − 4) 2 = 0 ⇒ t = 4;
t ≠ 0
We got one root that satisfies all the requirements, so we return to the x variable:
log 2 (x + 1) = 4;
log 2 (x + 1) = log 2 2 4;
x + 1 = 16;
x=15
That's it, we've solved the equation. But since there were several logarithms in the original equation, it is necessary to write out the domain of definition.
So, the expression x + 1 is in the argument of the logarithm. Therefore, x + 1 > 0. On the other hand, x + 1 is also present in the base, i.e. x + 1 ≠ 1. Total:
0 ≠ x > −1
Does the found root satisfy these requirements? Undoubtedly. Therefore, x = 15 is the solution to the original logarithmic equation.
Finally, I would like to say the following: if you look at the equation and understand that you have to solve something complex and non-standard, try to highlight stable structures, which will later be denoted by another variable. If some terms do not contain the variable x at all, they can often be simply calculated.
That's all I wanted to talk about today. I hope this lesson will help you in solving complex logarithmic equations. Watch other video tutorials, download and solve independent work, and see you in the next video!
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 of 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" (M.: 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)
Decision. 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
Decision. Equation (3) is equivalent to the equations
Or .
From here we get .
Answer: .
Example 3. solve the equation
Decision. Equation (4) implies, what . Using the basic logarithmic identity (1), can be written
or .
If we put , then from here we get the 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
Decision.Valid range of a variablein equation (5) are.
Let and . Since the functionon the domain of definition is decreasing, and the function increases on the entire number axis, then the equation cannot have more than one root.
By selection we find the only root.
Answer: .
Example 5. solve the equation.
Decision. 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)
Decision.We use identity (1) and transform equation (6) as follows:
Or .
Answer: , .
Example 7. solve the equation
. (7)
Decision. 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)
Decision.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)
Decision. 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)
Decision. 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)
Decision. 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 obtain an 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)
Decision. 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 single root. By direct selection, we establish that the desired root of equation (15) is .
Answer: .
Example 13. solve the equation
. (16)
Decision. 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)
Decision. 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)
Decision. 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)
Decision. 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)
Decision. 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)
Decision. Using the properties of logarithms, we transform equation (25) as follows:
, , .
From here we get .
Example 19. solve the equation
. (26)
Decision. 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.
Thus , 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 the tutorials from the list of recommended literature.
1. Kushnir A.I. Masterpieces of school mathematics (problems and solutions in two books). – Kyiv: Astarte, 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 of the 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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Preparation for the final test in mathematics includes an important section - "Logarithms". Tasks from this topic are necessarily contained in the exam. The experience of past years shows that the logarithmic equations caused difficulties for many schoolchildren. Therefore, students with different levels of training should understand how to find the correct answer and quickly cope with them.
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Solution of logarithmic equations. Part 1.
Logarithmic equation called an equation in which the unknown is contained under the sign of the logarithm (in particular, in the base of the logarithm).
Protozoa logarithmic equation looks like:
Solving any logarithmic equation involves the transition from logarithms to expressions under the sign of logarithms. However, this action expands the range of valid values of the equation and can lead to the appearance of extraneous roots. To avoid the appearance of extraneous roots you can do it in one of three ways:
1. Make an equivalent transition from the original equation to a system including
depending on which inequality or easier.
If the equation contains an unknown at the base of the logarithm:
then we go to the system:
2. Separately find the range of admissible values of the equation, then solve the equation and check if the solutions found satisfy the equation.
3. Solve the equation, and then do a check: substitute the found solutions into the original equation, and check if we get the correct equality.
A logarithmic equation of any level of complexity always eventually reduces to the simplest logarithmic equation.
All logarithmic equations can be divided into four types:
1 . Equations that contain logarithms to the first power only. With the help of transformations and use, they are reduced to the form
Example. Let's solve the equation:
Equate the expressions under the sign of the logarithm:
Let's check if our root of the equation satisfies:
Yes, it satisfies.
Answer: x=5
2 . Equations that contain logarithms to a power other than 1 (in particular, in the denominator of a fraction). These equations are solved using introducing a change of variable.
Example. Let's solve the equation:
Let's find the ODZ equation:
The equation contains logarithms squared, so it is solved using a change of variable.
Important! Before introducing a replacement, you need to "pull" the logarithms that are part of the equation into "bricks" using the properties of logarithms.
When "pulling" logarithms, it is important to apply the properties of logarithms very carefully:
In addition, there is one more subtle place here, and in order to avoid a common mistake, we will use an intermediate equality: we write the degree of the logarithm in this form:
Likewise,
We substitute the obtained expressions into the original equation. We get:
Now we see that the unknown is contained in the equation as part of . We introduce the replacement: . Since it can take any real value, we do not impose any restrictions on the variable.
Let's consider some types of logarithmic equations that are not so often considered in mathematics lessons at school, but are widely used in the preparation of competitive tasks, including for the USE.
1. Equations solved by the logarithm method
When solving equations containing a variable both in the base and in the exponent, the logarithm method is used. If, in addition, the exponent contains a logarithm, then both sides of the equation must be logarithmized to the base of this logarithm.
Example 1
Solve the equation: x log 2 x + 2 = 8.
Decision.
We take the logarithm of the left and right sides of the equation in base 2. We get
log 2 (x log 2 x + 2) = log 2 8,
(log 2 x + 2) log 2 x = 3.
Let log 2 x = t.
Then (t + 2)t = 3.
t 2 + 2t - 3 = 0.
D \u003d 16. t 1 \u003d 1; t 2 \u003d -3.
So log 2 x \u003d 1 and x 1 \u003d 2 or log 2 x \u003d -3 and x 2 \u003d 1/8
Answer: 1/8; 2.
2. Homogeneous logarithmic equations.
Example 2
Solve the equation log 2 3 (x 2 - 3x + 4) - 3log 3 (x + 5) log 3 (x 2 - 3x + 4) - 2log 2 3 (x + 5) = 0
Decision.
Equation domain
(x 2 - 3x + 4 > 0,
(x + 5 > 0. → x > -5.
log 3 (x + 5) = 0 for x = -4. By checking, we determine that the given value of x is not 
is the root of the original equation. Therefore, we can divide both sides of the equation by log 2 3 (x + 5).
We get log 2 3 (x 2 - 3x + 4) / log 2 3 (x + 5) - 3 log 3 (x 2 - 3x + 4) / log 3 (x + 5) + 2 = 0.
Let log 3 (x 2 - 3x + 4) / log 3 (x + 5) = t. Then t 2 - 3 t + 2 = 0. The roots of this equation are 1; 2. Returning to the original variable, we obtain a set of two equations
But taking into account the existence of the logarithm, only the values \u200b\u200bof (0; 9] should be considered. This means that the expression on the left side takes the largest value 2 at x \u003d 1. Now consider the function y \u003d 2 x-1 + 2 1-x. If we take t \u003d 2 x -1, then it will take the form y = t + 1/t, where t > 0. Under these conditions, it has a single critical point t = 1. This is the minimum point. Y vin = 2. And it is reached at x = 1.
It is now obvious that the graphs of the considered functions can intersect only once at the point (1; 2). It turns out that x \u003d 1 is the only root of the equation being solved.
Answer: x = 1.
Example 5. Solve the equation log 2 2 x + (x - 1) log 2 x \u003d 6 - 2x
Decision.
Let's solve this equation for log 2 x. Let log 2 x = t. Then t 2 + (x - 1) t - 6 + 2x \u003d 0.
D \u003d (x - 1) 2 - 4 (2x - 6) \u003d (x - 5) 2. t 1 \u003d -2; t 2 \u003d 3 - x.
We get the equation log 2 x \u003d -2 or log 2 x \u003d 3 - x.
The root of the first equation is x 1 = 1/4. 
The root of the equation log 2 x \u003d 3 - x will be found by selection. This number is 2. This root is unique, since the function y \u003d log 2 x is increasing over the entire domain of definition, and the function y \u003d 3 - x is decreasing.
By checking it is easy to make sure that both numbers are the roots of the equation
Answer: 1/4; 2.
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