Solving logarithmic equations. Part 1.
Logarithmic equation is an equation in which the unknown is contained under the sign of the logarithm (in particular, in the base of the logarithm).
The simplest logarithmic equation has the form:
Solving any logarithmic equation involves a transition from logarithms to expressions under the sign of logarithms. However, this action expands the range of permissible values of the equation and can lead to the appearance of extraneous roots. To avoid the appearance of foreign roots, you can do one of three ways:
1. Make an equivalent transition from the original equation to a system including
depending on which inequality or simpler.
If the equation contains an unknown in the base of the logarithm:
then we go to the system:
2. Separately find the range of acceptable values of the equation, then solve the equation and check whether the solutions found satisfy the equation.
3. Solve the equation, and then check: substitute the found solutions into the original equation and check whether we get the correct equality.
A logarithmic equation of any level of complexity always ultimately reduces to the simplest logarithmic equation.
All logarithmic equations can be divided into four types:
1 . Equations that contain logarithms only to the first power. With the help of transformations and use, they are brought to the form
Example. Let's solve the equation:
Let's equate the expressions under the logarithm sign:
Let's check whether our root of the equation satisfies:
Yes, it satisfies.
Answer: x=5
2 . Equations that contain logarithms to powers other than 1 (particularly in the denominator of a fraction). Such equations can be 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 can be solved using a change of variable.
Important! Before introducing a replacement, you need to “pull apart” the logarithms that are part of the equation into “bricks”, using the properties of logarithms.
When “pulling apart” logarithms, it is important to use the properties of logarithms very carefully:
In addition, there is one more subtle point here, and in order to avoid a common mistake, we will use an intermediate equality: we will write the degree of the logarithm in this form:
Likewise,
Let's substitute the resulting expressions into the original equation. We get:
Now we see that the unknown is contained in the equation as part of . Let's introduce the replacement: . Since it can take any real value, we do not impose any restrictions on the variable.
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 a logarithm. It involves solving an exponential equation. This equation has a single root, it is called the logarithm of b to base a:
Definition:
The logarithm of b to base a is the exponent to which base a must be raised to get b.
Let us remind you basic logarithmic identity.
The expression (expression 1) is the root of the equation (expression 2). Substitute the value x from expression 1 instead of x into expression 2 and get the main logarithmic identity:
So we see that each value is associated with a value. We denote b by x(), c by y, and thus obtain a logarithmic function:
For example: 
Let us recall the basic properties of the logarithmic function.
Let us pay attention once again, here, since under the logarithm there can be a strictly positive expression, as the base of the logarithm.
Rice. 1. Graph of a logarithmic function with different 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. 1.
Properties of this function:
Domain: ;
Range of values: ;
The function is monotonic throughout its entire 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, a larger value of the argument corresponds to a smaller value of the function.
The properties of the logarithmic function are the key to solving a variety of logarithmic equations.
Let's consider the simplest logarithmic equation; all other logarithmic equations, as a rule, are reduced to this form.
Since the bases of 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 appear 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 to comply with the ODZ.
Thus, we have a mixed system in which there is an equation and an inequality:
As a rule, it is not necessary to solve an inequality; 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;
Perform check.
Let's look at specific examples.
Example 1 - solve the equation:
The bases of logarithms are initially equal, we have the right to equate sublogarithmic expressions, do not forget about the ODZ, we 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:
Let's find the root and substitute it into the inequality:
We received an incorrect inequality, which means that the found root does not satisfy the ODZ.
Example 3 - solve the equation:
The bases of logarithms are initially equal, we have the right to equate sublogarithmic expressions, do not forget about the ODZ, we choose the second logarithm to compose the inequality:
Let's find the root and substitute it into the inequality:
Obviously, only the first root satisfies the ODZ.
Algebra 11th grade
Topic: “Methods for solving logarithmic equations”
Lesson objectives:
educational: developing knowledge about different ways to solve logarithmic equations, the ability to apply them in each specific situation and choose any method for solving;
developing: development of skills to observe, compare, apply knowledge in a new situation, identify patterns, generalize; developing skills of mutual control and self-control;
educational: fostering a responsible attitude to educational work, attentive perception of the material in the lesson, and careful note-taking.
Lesson type : lesson on introducing new material.
“The invention of logarithms, while reducing the work of the astronomer, extended his life.”
French mathematician and astronomer P.S. Laplace
During the classes
I. Setting the lesson goal
The studied definition of logarithm, properties of logarithms and logarithmic function will allow us to solve logarithmic equations. All logarithmic equations, no matter how complex they are, are solved using uniform algorithms. We'll look at these algorithms in today's lesson. There are not many of them. If you master them, then any equation with logarithms will be feasible for each of you.
Write down the topic of the lesson in your notebook: “Methods for solving logarithmic equations.” I invite everyone to cooperate.
II. Updating of reference knowledge
Let's prepare to study the topic of the lesson. You solve each task and write down the answer; you don’t have to write the condition. Work in pairs.
1) For what values of x does the function make sense:
A)
b)
V)
d)
(Answers are checked for each slide and errors are sorted out)
2) Do the graphs of the functions coincide?
a) y = x and
b)And
3) Rewrite the equalities as logarithmic equalities:
4) Write the numbers as logarithms with base 2:
4 =
2 =
0,5 =
1 =
5) Calculate :
6) Try to restore or supplement the missing elements in these equalities.
III. Introduction to new material
The following statement is displayed on the screen:
“The equation is the golden key that opens all mathematical sesames.”
Modern Polish mathematician S. Kowal
Try to formulate the definition of a logarithmic equation. (Equation containing an unknown under the logarithm sign ).
Let's considerthe simplest logarithmic equation: log A x = b (where a>0, a ≠ 1). Since the logarithmic function increases (or decreases) on the set of positive numbers and takes all real values, then by the root theorem it follows that for any b this equation has, and only one, solution, and a positive one.
Remember the definition of logarithm. (The logarithm of a number x to the base a is an indicator of the power to which the base a must be raised to obtain the number x ). From the definition of logarithm it immediately follows thatA V is such a solution.
Write down the title:Methods for solving logarithmic equations
1. By definition of logarithm .
This is how the simplest equations of the form are solved.
Let's considerNo. 514(a)
): Solve the equation
How do you propose to solve it? (By definition of logarithm )
Solution
.
, Hence 2x – 4 = 4; x = 4.
Answer: 4.
In this task 2x – 4 > 0, since> 0, so no extraneous roots can appear, andno need to check . There is no need to write out the condition 2x – 4 > 0 in this task.
2. Potentization (transition from the logarithm of a given expression to this expression itself).
Let's considerNo. 519(g): log 5 ( x 2 +8)- log 5 ( x+1)=3 log 5 2
What feature did you notice?(The bases are the same and the logarithms of the two expressions are equal) . What can be done?(Potentize).
It should be taken into account that any solution is contained among all x for which the logarithmic expressions are positive.
Solution: ODZ:
X 2 +8>0 unnecessary inequality
log 5 ( x 2 +8) = log 5 2 3 + log 5 ( x+1)
log 5 ( x 2 +8)= log 5 (8 x+8)
Let's potentiate the original equation
x 2 +8= 8 x+8
we get the equationx 2 +8= 8 x+8
Let's solve it:x 2 -8 x=0
x=0, x=8
Answer: 0; 8
In generaltransition to an equivalent system :
The equation
(The system contains a redundant condition - one of the inequalities need not be considered).
Question for the class : Which of these three solutions did you like best? (Discussion of methods).
You have the right to decide in any way.
3. Introduction of a new variable .
Let's considerNo. 520(g) . .
What did you notice? (This is a quadratic equation with respect to log3x) Your suggestions? (Introduce a new variable)
Solution . ODZ: x > 0.
Let, then the equation will take the form:
. Discriminant D > 0. Roots according to Vieta’s theorem:
.
Let's go back to the replacement:or.
Having solved the simplest logarithmic equations, we get:
; .
Answer : 27;
4. Logarithm both sides of the equation.
Solve the equation:
.
Solution : ODZ: x>0, let's take the logarithm of both sides of the equation in base 10:
. Let's apply the property of the logarithm of a power:
(lgx + 3) lgx =
(logx + 3) logx = 4
Let logx = y, then (y + 3)y = 4
, (D > 0) roots according to Vieta’s theorem: y1 = -4 and y2 = 1.
Let's go back to the replacement, we get: lgx = -4,
; logx = 1,
.
.
It is as follows:
if one of the functions
y = f(x)
increases, and the other
y = g(x)
decreases on the interval X, then the equation
f(x)= g(x)
has at most one root on the interval X
.
If there is a root, then it can be guessed. .
Answer : 2
“The correct application of methods can be learned by
only by applying them to various examples.”
Danish historian of mathematics G. G. Zeiten
I V. Homework
P. 39 consider example 3, solve No. 514(b), No. 529(b), No. 520(b), No. 523(b)
V. Summing up the lesson
What methods of solving logarithmic equations did we look at in class?
In the next lessons we will look at more complex equations. To solve them, the studied methods will be useful.
Last slide shown:
“What is more than anything in the world?
Space.
What is the wisest thing?
Time.
What's the best part?
Achieve what you want."
Thales
I wish everyone to achieve what they want. Thank you for your cooperation and understanding.
Introduction
Logarithms were invented to speed up and simplify calculations. The idea of a logarithm, that is, the idea of expressing numbers as powers of the same base, belongs to Mikhail Stiefel. But in Stiefel’s time, mathematics was not so developed and the idea of the logarithm was not developed. Logarithms were later invented simultaneously and independently of each other by the Scottish scientist John Napier (1550-1617) and the Swiss Jobst Burgi (1552-1632). Napier was the first to publish the work in 1614. under the title “Description of an amazing table of logarithms”, Napier’s theory of logarithms was given in a fairly complete volume, the method of calculating logarithms was given the simplest, therefore Napier’s merits in the invention of logarithms were greater than those of Bürgi. Burgi worked on the tables at the same time as Napier, but kept them secret for a long time and published them only in 1620. Napier mastered the idea of the logarithm around 1594. although the tables were published 20 years later. At first he called his logarithms “artificial numbers” and only then proposed to call these “artificial numbers” in one word “logarithm”, which translated from Greek means “correlated numbers”, taken one from an arithmetic progression, and the other from a geometric progression specially selected for it. progress. The first tables in Russian were published in 1703. with the participation of a wonderful teacher of the 18th century. L. F. Magnitsky. The works of St. Petersburg academician Leonhard Euler were of great importance in the development of the theory of logarithms. He was the first to consider logarithms as the inverse of raising to a power; he introduced the terms “logarithm base” and “mantissa.” Briggs compiled tables of logarithms with base 10. Decimal tables are more convenient for practical use, their theory is simpler than that of Napier’s logarithms . Therefore, decimal logarithms are sometimes called Briggs logarithms. The term "characterization" was introduced by Briggs.
In those distant times, when the sages first began to think about equalities containing unknown quantities, there were probably no coins or wallets. But there were heaps, as well as pots and baskets, which were perfect for the role of storage caches that could hold an unknown number of items. In the ancient mathematical problems of Mesopotamia, India, China, Greece, unknown quantities expressed the number of peacocks in the garden, the number of bulls in the herd, and the totality of things taken into account when dividing property. Scribes, officials and priests initiated into secret knowledge, well trained in the science of accounts, coped with such tasks quite successfully.
Sources that have reached us indicate that ancient scientists had some general techniques for solving problems with unknown quantities. However, not a single papyrus or clay tablet contains a description of these techniques. The authors only occasionally supplied their numerical calculations with skimpy comments such as: “Look!”, “Do this!”, “You found the right one.” In this sense, the exception is the “Arithmetic” of the Greek mathematician Diophantus of Alexandria (III century) - a collection of problems for composing equations with a systematic presentation of their solutions.
However, the first manual for solving problems that became widely known was the work of the Baghdad scientist of the 9th century. Muhammad bin Musa al-Khwarizmi. The word "al-jabr" from the Arabic name of this treatise - "Kitab al-jaber wal-mukabala" ("Book of restoration and opposition") - over time turned into the well-known word "algebra", and the work of al-Khwarizmi itself served the starting point in the development of the science of solving equations.
Logarithmic equations and inequalities
1. Logarithmic equations
An equation containing an unknown under the logarithm sign or at its base is called a logarithmic equation.
The simplest logarithmic equation is an equation of the form
log a x = b . (1)
Statement 1. If a > 0, a≠ 1, equation (1) for any real b has a unique solution x = a b .
Example 1. Solve the equations:
a)log 2 x= 3, b) log 3 x= -1, c)
Solution. Using Statement 1, we obtain a) x= 2 3 or x= 8; b) x= 3 -1 or x= 1 / 3 ; c)
or x = 1.Let us present the basic properties of the logarithm.
P1. Basic logarithmic identity:
Where a > 0, a≠ 1 and b > 0.
P2. The logarithm of the product of positive factors is equal to the sum of the logarithms of these factors:
log a N 1 · N 2 = log a N 1 + log a N 2 (a > 0, a ≠ 1, N 1 > 0, N 2 > 0).
Comment. If N 1 · N 2 > 0, then property P2 takes the form
log a N 1 · N 2 = log a |N 1 | + log a |N 2 | (a > 0, a ≠ 1, N 1 · N 2 > 0).
P3. The logarithm of the quotient of two positive numbers is equal to the difference between the logarithms of the dividend and the divisor
(a
> 0, a
≠ 1, N
1
> 0, N
2
> 0).
Comment. If
, (which is equivalent N 1 N 2 > 0) then property P3 takes the form
(a
> 0, a
≠ 1, N
1
N
2
> 0).
P4. The logarithm of the power of a positive number is equal to the product of the exponent and the logarithm of this number:
log a N k = k log a N (a > 0, a ≠ 1, N > 0).
Comment. If k- even number ( k = 2s), That
log a N 2s = 2s log a |N | (a > 0, a ≠ 1, N ≠ 0).
P5. Formula for moving to another base:
(a
> 0, a
≠ 1, b
> 0, b
≠ 1, N
> 0),
in particular if N = b, we get
(a > 0, a ≠ 1, b > 0, b ≠ 1). (2)Using properties P4 and P5, it is easy to obtain the following properties
(a
> 0, a
≠ 1, b
> 0, c
≠ 0), (3)
(a
> 0, a
≠ 1, b
> 0, c
≠ 0), (4)
(a
> 0, a
≠ 1, b
> 0, c
≠ 0), (5)
and, if in (5) c- even number ( c = 2n), occurs
(b
> 0, a
≠ 0, |a
| ≠ 1). (6)
Let us list the main properties of the logarithmic function f (x) = log a x :
1. The domain of definition of a logarithmic function is the set of positive numbers.
2. The range of values of the logarithmic function is the set of real numbers.
3. When a> 1 logarithmic function is strictly increasing (0< x 1 < x 2log a x 1 < loga x 2), and at 0< a < 1, - строго убывает (0 < x 1 < x 2log a x 1 > log a x 2).
4. log a 1 = 0 and log a a = 1 (a > 0, a ≠ 1).
5. If a> 1, then the logarithmic function is negative when x(0;1) and positive at x(1;+∞), and if 0< a < 1, то логарифмическая функция положительна при x (0;1) and negative at x (1;+∞).
6. If a> 1, then the logarithmic function is convex upward, and if a(0;1) - convex downwards.
The following statements (see, for example,) are used when solving logarithmic equations.
Preparation for the final test in mathematics includes an important section - “Logarithms”. Tasks from this topic are necessarily contained in the Unified State Examination. Experience from past years shows that logarithmic equations caused difficulties for many schoolchildren. Therefore, students with different levels of training must understand how to find the correct answer and quickly cope with them.
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