Solution of the fractions equations into degree x. What is the indicative equation and how to solve it
Solution of indicative equations. Examples.
Attention!
This topic has additional
Materials in a special section 555.
For those who are strongly "not very ..."
And for those who are "very ...")
What indicative equation? This equation in which unknown (Xers) and expressions with them are in indicators Some degrees. And only there! It is important.
There you are examples indicatory equations :
3 x · 2 x \u003d 8 x + 3
Note! In the grounds of degrees (below) - only numbers. IN indicators Degnese (at the top) - a wide variety of expressions with Xa. If, suddenly, the EX will come out in the equation somewhere, except for the indicator, for example:
it will already be a mixed type equation. Such equations do not have clear rules for solutions. We will not consider them yet. Here we will deal with by solving exponential equations in pure form.
In fact, even clean indicative equations are clearly solved far away. But there are certain types of indicative equations that can be solved and needed. Here are these types we will look at.
The solution of the simplest indicative equations.
To begin with, I decide something completely elementary. For example:
Even without any theories, it is clear to the simple selection that x \u003d 2. More, right, right!? No other value of the ICA rolls. And now we look at the record of the solution of this cunning indicative equation:
What did we do? In fact, we simply threw the same bases (three). They completely thrown away. And what pleases, got to the point!
Indeed, if in the indicative equation on the left and right the same Numbers in any degrees, these numbers can be removed and equated degrees. Mathematics allows. It remains to be expensive a much simpler equation. Great, right?)
However, remember Iron: you can remove the bases only when the left and right of the ground is in proud loneliness! Without any neighbors and coefficients. Say, in equations:
2 x +2 x + 1 \u003d 2 3, or
double can not be removed!
Well, the most important thing we have mastered. How to move from evil indicative expressions to simpler equations.
"That's the times!" - You will say. "Who will give such a primitive on the control and exams!?"
Forced to agree. No one will give. But now you know where to strive when solving free examples. It is necessary to bring it to the form when on the left - the same number is the same number. Further everything will be easier. Actually, this is the classic of mathematics. Take the original example and convert it to the desired us View. According to the rules of mathematics, of course.
Consider examples that require some additional efforts to bring them to the simplest. Let's call them simple indicative equations.
Solution of simple indicative equations. Examples.
When solving indicative equations, the main rules - actions with degrees. Without knowledge of these actions, nothing will work.
To actions with degrees it is necessary to add personal observation and smelting. We need the same foundations? Here we are looking for them in an example in a clear or encrypted form.
Let's see how this is done in practice?
Let us give us an example:
2 2x - 8 x + 1 \u003d 0
First angry look - on basis. They ... they are different! Two and eight. But to fall into the despondency - early. It's time to remember that
Two and eight - relative to the degree.) It is possible to write:
8 x + 1 \u003d (2 3) x + 1
If you recall the formula from action with degrees:
(a n) m \u003d a nm,
that generally it turns out:
8 x + 1 \u003d (2 3) x + 1 \u003d 2 3 (x + 1)
The initial example began to look like this:
2 2x - 2 3 (x + 1) \u003d 0
Transfer 2 3 (x + 1) To the right (nobody canceled elementary actions of mathematics!), We get:
2 2x \u003d 2 3 (x + 1)
Here, almost, and that's it. We remove the foundations:
Solve this monster and get
This is the correct answer.
In this example, we reinstate the knowledge of detects of two. we identified In the eight of the encrypted two. This technique (encryption of general bases under different numbers) is a very popular technique in the lower equations! Yes, and in logarithms too. It is necessary to be able to learn in the numbers of other numbers. This is extremely important for solving indicative equations.
The fact is that to build any number to any degree is not a problem. Multiply, even on a piece of paper, and that's it. For example, to build 3 to the fifth degree will be able to each. 243 It turns out if you know the multiplication table.) But in the lower equations, it is much more likely to not be erected, but on the contrary ... to find out what number to what extent Hiding for a number 243, or, say, 343 ... Here you will not help any calculator.
The degree of some numbers should be known in the face, yes ... do it?
To determine what degrees and what numbers are numbers:
2; 8; 16; 27; 32; 64; 81; 100; 125; 128; 216; 243; 256; 343; 512; 625; 729, 1024.
Answers (in disarray, natural!):
5 4 ; 2 10 ; 7 3 ; 3 5 ; 2 7 ; 10 2 ; 2 6 ; 3 3 ; 2 3 ; 2 1 ; 3 6 ; 2 9 ; 2 8 ; 6 3 ; 5 3 ; 3 4 ; 2 5 ; 4 4 ; 4 2 ; 2 3 ; 9 3 ; 4 5 ; 8 2 ; 4 3 ; 8 3 .
If you look closely, you can see a strange fact. Replies are significantly more than tasks! Well, it happens ... for example, 2 6, 4 3, 8 2 is all 64.
Suppose you took note of the information about the acquaintance with numbers.) Let's remind you that to solve the indicative equations apply all The stock of mathematical knowledge. Including from the junior middle classes. You do not immediately go to senior classes, right?)
For example, when solving the indicative equations, the total multiplier of the brackets helps very often (hello grade 7!). Watch the following man:
3 2x + 4 -11 · 9 x \u003d 210
And again, first glance - on the ground! The foundations in degrees are different ... Troika and nine. And we want to be the same. Well, in this case, the desire is fulfilled!) Because:
9 x \u003d (3 2) x \u003d 3 2x
According to the same rules of action with degrees:
3 2x + 4 \u003d 3 2x · 3 4
So great, you can write:
3 2x · 3 4 - 11 · 3 2x \u003d 210
We led an example to the same reasons. So, what is next!? Troika can not throw out ... deadlock?
Not at all. Remember the most universal and powerful solution rule all Mathematical tasks:
You do not know what you need - do what you can!
You look, everything is formed).
What is in this indicative equation can do it? Yes, on the left side, it is directly asking for a bracket! The total multiplier of 3 2x clearly hints at it. Let's try, and then it will be visible:
3 2x (3 4 - 11) \u003d 210
3 4 - 11 = 81 - 11 = 70
The example is becoming better and better!
We remember that in order to eliminate grounds, we need a clean degree, without any coefficients. US number 70 interferes. So we divide both parts of the equation by 70, we get:
Op-Pa! Everything and settled!
This is the final answer.
It happens, however, that breaking on the same bases is obtained, but their liquidation is in any way. This happens in the indicative equations of another type. We will master this type.
Replacing the variable in solving indicative equations. Examples.
Resolving equation:
4 x - 3 · 2 x +2 \u003d 0
First - as usual. Go to one base. To twice.
4 x \u003d (2 2) x \u003d 2 2x
We get the equation:
2 2x - 3 · 2 x +2 \u003d 0
And here it will be dependent. Previous techniques will not work, no matter how sprinkling. We'll have to get another mighty and universal way from arsenal. Called O. replacing the variable.
The essence of the method is easy to surprise. Instead of one complex icon (in our case - 2 x) we write another, simpler (for example - T). This, it would seem, a meaningless replacement leads to awesome results!) Just everything becomes clear and understandable!
So, let
Then 2 2x \u003d 2 x2 \u003d (2 x) 2 \u003d T 2
We replace in our equation all degrees with cavities on T:
Well, insens?) Square equations have not forgotten yet? We decide through the discriminant, we get:
Here, most importantly, do not stop, as it happens ... This is not a response, we are needed, and not t. We return to the ICCAM, i.e. We make a replacement. First for T 1:
That is,
One root found. We are looking for the second, from T 2:
Um ... left 2 x, right 1 ... no problem? Yes no! Enough to remember (from action with degrees, yes ...) that one is anyone Number to zero degree. Any. What you need, and put it. We need a two. So:
Now everything is. Received 2 roots:
This is the answer.
For solving indicative equations In the end sometimes it turns out some inconvenient expression. Type:
From the seven deuce through a simple degree does not work. Do not relatives they ... how to be here? Someone, maybe confused ... And here is a person who read the topic on this site "What is a logarithm?" , only Skupo smile and will make a solid right answer to the hard hand:
There may be no such an answer in the tasks "in". There is a specific number required. But in the tasks "C" - easily.
In this lesson, examples of solving the most common indicative equations are given. We highlight the main one.
1. The first thing we look at basis degrees. We think whether it is impossible to make them same. Try to do it, actively using actions with degrees. Do not forget that the numbers without ICs can also be turned into a degree!
2. We try to bring the indicative equation to the form when the left and right are the same Numbers in any degrees. Using actions with degrees and factorization.What can I consider in numbers - believe.
3. If the second board did not work, we try to apply the replacement of the variable. As a result, an equation can turn out that is easily solved. Most often - square. Or fractional, which also comes down to square.
4. To successfully solve the indicative equations, it is necessary to know the degree of some numbers "in the face".
As usual, at the end of the lesson you are offered to clean a little.) Alone. From simple - to complex.
Decide indicative equations:
More complied with:
2 x + 3 - 2 x + 2 - 2 x \u003d 48
9 x - 8 · 3 x \u003d 9
2 x - 2 0,5x + 1 - 8 \u003d 0
Find the product of the roots:
2 3 + 2 x \u003d 9
Happened?
Well, then the most complex example (solved, however, in the mind ...):
7 0.13x + 13 0.7x + 1 + 2 0,5x + 1 \u003d -3
What is more interesting? Then you have an evil example. It is quite pulling on increased difficulty. Nickname that in this example savings savings and the most universal rule of solving all mathematical tasks.)
2 5x-1 · 3 3x-1 · 5 2x-1 \u003d 720 x
Example simpler, for rest):
9 · 2 x - 4 · 3 x \u003d 0
And for dessert. Find the number of roots equation:
x · 3 x - 9x + 7 · 3 x - 63 \u003d 0
Yes Yes! This is a mixed type equation! Which we did not consider in this lesson. And what to consider them, it is necessary to solve!) This lesson is quite enough to solve the equation. Well, the cutter is needed ... and let it help you with the seventh class (this is a hint!).
Answers (in disorder, through a comma point):
one; 2; 3; four; no solutions; 2; -2; -five; four; 0.
All successful? Excellent.
There is a problem? No problem! In a special section 555, all these indicative equations are solved with detailed explanations. What, why, and why. And, of course, there is additional valuable information on working with all sorts of indicative equations. Not only with these.)
The last fun question for a consideration. In this lesson, we worked with accurate equations. Why didn't I say a word here about OTZ? In equations, this is a very important thing, by the way ...
If you like this site ...
By the way, I have another couple of interesting sites for you.)
It can be accessed in solving examples and find out your level. Testing with instant check. Learn - with interest!)
You can get acquainted with features and derivatives.
Equipment:
- a computer,
- multimedia projector,
- screen,
- Attachment 1(Slide presentation in PowerPoint) "Methods for solving indicative equations"
- Appendix 2. (Solving the equation of type "three different bases of degrees" in Word)
- Appendix 3. (distribution material in Word for practical work).
- Appendix 4. (distribution material in Word for homework).
During the classes
1. Organizational stage
- message Topics lesson (recorded on the board),
- the need for a generalizing lesson in grades 10-11:
Stage of training students for active learning knowledge
Reiteration
Definition.
An indicative equation is called an equation containing a variable in an indicator of the degree (the student is answered).
Teacher's remark. The indicative equations belong to the class of transcendental equations. This difficult-acting name suggests that such equations generally speaking are not solved as a formula.
They can only be solved by approximately numerical methods on computers. But what about the examination tasks? All the trick is that the examiner is this task that it just admits an analytical solution. In other words, you can (and should!) Do such identical transformationswhich reduces this indicative equation to the simplest indicative equation. This is the easiest equation so called: the simplest indicative equation. It is solved logarithming.
The situation with the solution of the indicative equation resembles a journey through a labyrinth, which is specially invented by the compiler of the task. Of these very common reasoning, there are quite concrete recommendations.
To successfully solve the indicative equations, it is necessary:
1. Not only actively know all demonstration identities, but also to find many variable values \u200b\u200bon which these identities are determined that when using these identities, do not acquire extra roots, and even more so, do not lose solutions of the equation.
2. Actively know all demonstration identities.
3. Clearly, in detail and without errors to do mathematical transformations of equations (to transfer the components from one part of the equation to another, without forgetting about the shift of the sign, lead to the general denominator of the fraction and the like). This is called mathematical culture. At the same time, the calculations themselves should be made automatically with their hands, and the head should think about the overall tracking thread of the solution. Making conversions should be as close as possible and more. Only this will give a guarantee of the right unmistakable solution. And remember: a small arithmetic error may simply create a transcendental equation, which in principle is not solved analytically. It turns out, you got off the way and rested into the wall of the labyrinth.
4. Know Methods for solving problems (that is, know all the ways of passing the labyrinth solution). For proper orientation at each stage you will have (consciously or intuitive!):
- determine type of equation;
- remember the corresponding type decision method tasks.
Stage of generalization and systematization of the studied material.
A teacher, together with students with the involvement of a computer, a review of the repetition of all types of indicative equations and methods for their solution is drawn up, is drawn up general scheme. (Education is used computer program L.Ya. Borevsky "Mathematics Course - 2000", Presentation by PowerPoint - so-called Matsov.)
Fig. one.The figure shows the general diagram of all types of indicative equations.
As can be seen from this scheme, the strategy for solving the indicative equations is to bring this indicative equation to the equation, first of all, with the same bases of degrees and then - and with the same indicators of degrees.
After receiving the equation with the same bases and indicators of degrees, you replace this degree to a new variable and get a simple algebraic equation (usually, fractional rational or square) relative to this new variable.
Deciding this equation and making a replacement, you as a result come to the aggregate of the simplest indicative equations that are solved in general With logarithmation.
The equations are located, in which only works (private) degrees are found. Taking advantage of the indicative identities, these equations manage to bring immediately to one base, in particular, to the simplest indicative equation.
Consider how the indicative equation is solved with three different bases of degrees.
(If the teacher has a training computer program L.Ya. Borevsky "Mathematics - 2000 course, we naturally work with a disk if not - you can make a printout of this type of equation from it, presented below.)
Fig. 2. The solution plan for the equation.
Fig. 3. The beginning of the solution of the equation
Fig. four. Elimination of the solution of the equation.
Performing practical work
Determine the type of equation and solve it.
1. 2. 3. 0,125 4. 5. 6.
Summing up the lesson
Installing estimates for the lesson.
End of lesson
For teacher
Diagram of responses of practical work.
The task: From the list of equations, select the equations of the specified type (№ respond to the table):
- Three different bases of degrees
- Two different bases - different indicators of degree
- The foundations of degrees - the degree of one number
- Same bases - different indicators of degrees
- The same bases of degrees - the same indicators of degrees
- The work of degrees
- Two different bases of degrees - the same indicators
- The simplest indicative equations
1. 
(Work of degrees)
2. 
(The same foundations are different indicators of degrees)
Examples:
\\ (4 ^ x \u003d 32 \\)
\\ (5 ^ (2x-1) -5 ^ (2x-3) \u003d 4.8 \\)
\\ ((\\ sqrt (7)) ^ (2x + 2) -50 \\ CDOT (\\ SQRT (7)) ^ (x) + 7 \u003d 0 \\)
How to solve exponential equations
When solving, any indicative equation, we strive to lead to the form \\ (a ^ (f (x)) \u003d a ^ (G (x)) \\), and then make the transition to equality of indicators, that is:
\\ (a ^ (f (x)) \u003d a ^ (G (x)) \\) \\ (⇔ \\) \\ (f (x) \u003d g (x) \\)
For example: \\ (2 ^ (x + 1) \u003d 2 ^ 2 \\) \\ (⇔ \\) \\ (x + 1 \u003d 2 \\)
Important! From the same logic follows two requirements for such a transition:
- number B. on the left and right should be the same;
- degrees on the left and right should be "clean"that is, there should be no, multiplications, divisions, etc.
For example:
To enjoy the equation to the form \\ (a ^ (f (x)) \u003d a ^ (G (x)) \\) apply and.
Example
. Decide the indicative equation \\ (\\ sqrt (27) · 3 ^ (x - 1) \u003d ((\\ FRAC (1) (3))) ^ (2x) \\)
Decision:
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\\ (\\ sqrt (27) · 3 ^ (x - 1) \u003d ((\\ FRAC (1) (3))) ^ (2x) \\) |
We know that \\ (27 \u003d 3 ^ 3 \\). With this in mind, we transform the equation. |
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\\ (\\ sqrt (3 ^ 3) · 3 ^ (x - 1) \u003d ((\\ FRAC (1) (3))) ^ (2x) \\) |
By the property of the root \\ (\\ sqrt [n] (a) \u003d a ^ (\\ FRAC (1) (N)) \\) we obtain that \\ (\\ sqrt (3 ^ 3) \u003d ((3 ^ 3)) ^ ( \\ FRAC (1) (2)) \\). Next, using the degree of degree \\ ((a ^ b) ^ C \u003d a ^ (Bc) \\), we obtain \\ (((3 ^ 3)) ^ (\\ FRAC (1) (2)) \u003d 3 ^ (3 \\ \\ (3 ^ (\\ FRAC (3) (2)) \\ Cdot 3 ^ (x - 1) \u003d (\\ FRAC (1) (3)) ^ (2x) \\) |
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We also know that \\ (a ^ b · a ^ C \u003d a ^ (b + c) \\). Applying this to the left side, we get: \\ (3 ^ (\\ FRAC (3) (2)) · 3 ^ (x - 1) \u003d 3 ^ (\\ FRAC (3) (2) + x - 1) \u003d 3 ^ (1.5 + x - 1) \u003d 3 ^ (x + 0.5) \\). |
\\ (3 ^ (x + 0.5) \u003d (\\ FRAC (1) (3)) ^ (2x) \\) |
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Now remember that: \\ (a ^ (- n) \u003d \\ FRAC (1) (a ^ n) \\). This formula can also be used in the opposite direction: \\ (\\ FRAC (1) (a ^ n) \u003d a ^ (- n) \\). Then \\ (\\ FRAC (1) (3) \u003d \\ FRAC (1) (3 ^ 1) \u003d 3 ^ (- 1) \\). |
\\ (3 ^ (x + 0.5) \u003d (3 ^ (- 1)) ^ (2x) \\) |
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Applying the property \\ ((a ^ b) ^ C \u003d a ^ (Bc) \\) to the right part, we obtain: \\ ((3 ^ (- 1)) ^ (2x) \u003d 3 ^ ((- 1) · 2x) \u003d 3 ^ (- 2x) \\). |
\\ (3 ^ (x + 0.5) \u003d 3 ^ (- 2x) \\) |
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And now we have the foundations equal and there are no interfering coefficients, etc. So we can make the transition. |
. Solve the indicative equation \\ (4 ^ (x + 0.5) -5 · 2 ^ x + 2 \u003d 0 \\) |
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Example
\\ (4 ^ (x + 0.5) -5 · 2 ^ x + 2 \u003d 0 \\)
The question remains - how to understand when which method is applied? It comes with experience. In the meantime you did not work out, use : \(-1; 1\). General recommendation to solve complex tasks, "you do not know what to do - do what you can". That is, look for how you can convert the equation in principle, and try to do it - suddenly what will come out? The main thing about to make only mathematically reasonable transformations. Indicative equations that do not have solutions We will analyze two more situations that are often put in the Student's deadlock: Let's try to solve the bust. If X is a positive number, then the increasing degree \\ (2 ^ x \\) will only grow: \\ (x \u003d 1 \\); \\ (2 ^ 1 \u003d 2 \\) \\ (x \u003d 0 \\); \\ (2 ^ 0 \u003d 1 \\) Also by. There are negative canes. Remembering the property \\ (a ^ (- n) \u003d \\ FRAC (1) (A ^ n) \\), check: \\ (x \u003d -1 \\); \\ (2 ^ (- 1) \u003d \\ FRAC (1) (2 ^ 1) \u003d \\ FRAC (1) (2) \\) Despite the fact that the number with each step becomes smaller, it will never reach zero. So and the negative degree did not save us. We come to logical conclusion: A positive number to any extent will remain a positive number.Thus, both equations above have no solutions. Indicative equations with different basesIn practice, sometimes there are indicative equations with different bases that are not reduced to each other, and at the same time with the same indicators. They look like this: \\ (a ^ (f (x)) \u003d b ^ (f (x)) \\), where \\ (a \\) and \\ (b \\) are positive numbers. For example: \\ (7 ^ (x) \u003d 11 ^ (x) \\) Such equations can be easily solved by dividing on any of the parts of the equation (usually divided to the right side, that is, on \\ (b ^ (f (x)) \\). So you can divide, because a positive number is in any extent positive (that is, We are not divided by zero). We get: \\ (\\ FRAC (a ^ (f (x))) (b ^ (f (x))) \\) \\ (\u003d 1 \\) Example
. Solve the indicative equation \\ (5 ^ (x + 7) \u003d 3 ^ (x + 7) \\)
The question remains - how to understand when which method is applied? It comes with experience. In the meantime you did not work out, use : \(-7\). Sometimes "the same" indicators of the degree is not obvious, but the skillful use of the degree of degree solves this issue. Example
. Solve the indicative equation \\ (7 ^ (2x-4) \u003d (\\ FRAC (1) (3)) ^ (- x + 2) \\)
The question remains - how to understand when which method is applied? It comes with experience. In the meantime you did not work out, use : \(2\). |
In this lesson, we will consider the solution of more complex demonstration equations, remember the main theoretical provisions regarding indicative function.
1. Definition and properties of the indicative function, method of solving the simplest indicative equations
Recall the definition and basic properties of the indicative function. It is on the properties that the solution of all the indicative equations and inequalities is based.
Exponential function - this is a function of the form where the basis of the degree and here x is an independent variable, the argument; y - dependent variable, function.
Fig. 1. schedule indicative function
The graph shows increasing and decreasing exhibitors illustrating an indicative function at the basis of a larger unit and a smaller unit, but a large zero, respectively.
Both curves pass through the point (0; 1)
Properties of the indicative function:
Domain: ;
Value area:;
The function of the monotonne, with increasing, with decreases.
The monotone feature takes each of its own value with the only value of the argument.
When the argument increases from minus to plus infinity, the function increases from zero not inclusive to plus infinity. With the opposite, when the argument increases from minus to plus infinity, the function decreases from infinity to zero not inclusive.
2. Solution of typical indicative equations
Recall how to solve the simplest demonstration equations. Their decision is based on the monotony of the indicative function. To such equations, almost all complex indicative equations are reduced.
Equality of indicators of the degree with equal bases is due to the property of the indicative function, namely its monotony.
Technique Solution:
Equalize the bases of degrees;
Equate degrees.
Let us turn to the consideration of more complex exponential equations, our goal is to reduce each of them to the simplest.
We free from the root in the left side and give a degree to the same basis:
In order to reduce the complex indicative equation to the simplest, the variable replacement is often used.
We use the degree property:
We introduce the replacement. Let, then 
Multiply the resulting equation for two and move all the components to the left side:
The first root does not satisfy the gap of the values \u200b\u200bof y, throwing it off. We get:
Let the extent to the same indicator:
We introduce a replacement:
Let, then 
. With this replacement, it is obvious that it takes strictly positive values. We get:
We know how to solve such square equations, we drank the answer:
To make sure that the roots are correcting, you can perform a check on the WESET Theorem, i.e. to find the amount of the roots and their work and verify with the corresponding coefficients of the equation.
We get:
3. Methods of solving homogeneous indicative equations of the second degree
Let's study the following important type of indicative equations:
Equations of this type are called homogeneous second degree relative to the functions f and g. In the left part there is a square three-step relative to F with a G parameter or a square three decreased relative to G with the F parameter.
Technique Solution:
This equation can be solved as a square, but it is easier to do differently. Two cases should be considered:
In the first case we get
In the second case, we have the right to divide on the older degree and get:
It should be replaced by variables, we obtain a square equation relative to:
We note that the functions f and g can be any, but we are interested in the case when it is indicative functions.
4. Examples of solving homogeneous equations
We transfer all the terms in the left part of the equation:
Since the indicative functions acquire strictly positive values, we have the right to immediately share the equation on, without considering the case when:
We get:
We introduce a replacement: 
(According to the properties of the indicative function)
Received a square equation:
We define the roots on the Vieta Theorem:
The first root does not satisfy the gap of the values \u200b\u200bof y, throwing it off, we get:
We use the degree properties and give it all degrees to simple grounds:
It is easy to notice the functions F and G:
Since the indicative functions acquire strictly positive values, we have the right to immediately share the equation on, without considering the case when.
The solution of most mathematical problems is one way or another associated with the transformation of numerical, algebraic or functional expressions. Specified in particular applies to the decision. In the variants of the EGE in mathematics to such a type of tasks, the task is, in particular, the C3 problem. Learn how to solve the tasks C3 is important not only for the purpose of successful surchase EGEBut for the reason that this skill is useful when studying the course of mathematics in the highest school.
Performing C3 tasks, you have to decide different kinds equations and inequalities. Among them are rational, irrational, indicative, logarithmic, trigonometrics containing modules (absolute values), as well as combined. This article discusses the main types of indicative equations and inequalities, as well as various methods of their solutions. Read about the decision of the remaining types of equations and inequalities in the heading "" in articles on the methods of solving problems C3 from esmer options mathematics.
Before proceeding with the analysis of specific indicative equations and inequalitiesAs a tutor in mathematics, I suggest you refreshing some theoretical material that we need.
Exponential function
What is an indicative function?
Function of type y. = a X.where a. \u003e 0 I. a. ≠ 1, called indicative function.
Maintenance properties of the indicative function y. = a X.:
Graph indicative function
The graph of the indicative function is exhibitor:
Graphs of indicative functions (exhibitors)
Solution of indicative equations
Indicative They are called equations in which an unknown variable is only in the indicators of any degrees.
For solutions indicatory equations You need to know and be able to use the following simple theorem:
Theorem 1. Indicative equation a. f.(x.) = a. g.(x.) (where a. > 0, a. ≠ 1) equivalent equation f.(x.) = g.(x.).
In addition, it is useful to remember the basic formulas and actions with degrees:
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Example 1. Solve the equation:
Decision: We use the above formulas and substitution:
The equation then takes the form:
Discriminant received square equation Positive:
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This means that this equation has two roots. We find them:
Turning to the return substitution, we get:
The second root equation does not have, since the indicative function is strictly positive throughout the entire definition area. We solve the second:
Taking into account the said in Theorem 1, go to the equivalent equation: x. \u003d 3. This will be the answer to the task.
Answer: x. = 3.
Example 2. Solve the equation:
Decision: There are no restrictions on the area of \u200b\u200bpermissible values \u200b\u200bat the equation, since the feeding expression makes sense at any meaning x. (exponential function y. = 9 4 -X. positive and not equal to zero).
We solve the equation by equivalent transformations using the rules of multiplication and division of degrees:
The last transition was carried out in accordance with Theorem 1.
Answer:x.= 6.
Example 3. Solve the equation:
Decision: Both parts of the source equation can be divided by 0.2 x. . This transition will be equivalent, since this expression is greater than zero in any value. x. (The indicative function is strictly positive on its definition area). Then the equation takes the form:
Answer: x. = 0.
Example 4. Solve the equation:
Decision:we simplify the equation to elementary by equivalent transformations using the rules of division and multiplication of degrees given at the beginning of the article:
Division of both parts of the equation for 4 x. , as in the previous example, is an equivalent transformation, since this expression is not zero at no matter what values x..
Answer: x. = 0.
Example 5. Solve the equation:
Decision: function y. = 3 X., standing in the left side of the equation, is increasing. Function y. = —x.-2/3, standing in the right part of the equation, is descending. This means that if the graphs of these functions intersect, then not more than one point. In this case, it is not difficult to guess that the graphs intersect at the point x. \u003d -1. There will be no other roots.
Answer: x. = -1.
Example 6. Solve the equation:
Decision: We simplify the equation by equivalent transformations, bearing in mind everywhere that the indicative function is strictly greater than zero in any meaning x.and using the rules for calculating the work and private degrees given at the beginning of the article:
Answer: x. = 2.
Solution of indicative inequalities
Indicative It is called inequalities in which an unknown variable is contained only in the indicators of any degrees.
For solutions indicative inequalities Knowledge requires the following theorem:
Theorem 2. If a a. \u003e 1, then inequality a. f.(x.) > a. g.(x.) It is equivalent to the inequality of the same meaning: f.(x.) > g.(x.). If 0< a. < 1, то показательное неравенство a. f.(x.) > a. g.(x.) It is equivalent to the inequality of the opposite sense: f.(x.) < g.(x.).
Example 7.Solve inequality:
Decision: Imagine initial inequality in the form:
We divide both parts of this inequality for 3 2 x. at the same time (due to the positivity of the function y.= 3 2x.) The sign of inequality will not change:
We use the substitution:
Then the inequality will take the form:
So, the solution of inequality is the gap:
turning to the return substitution, we get:
The left inequality due to the positiveness of the indicative function is performed automatically. Taking advantage of the well-known property of the logarithm, proceed to the equivalent inequality:
Since there will be a transition to the following inequality to the following in the degree of degree, there will be a transition to the following inequality:
So, finally get answer:
Example 8. Solve inequality:
Decision: Using the properties of multiplication and dividing degrees, rewrite inequality in the form:
We introduce a new variable:
Taking into account this substitution, inequality takes the form:
Multiply the numerator and denominator of the fraction on 7, we get the following equivalent inequality:
So, inequality satisfy the following values \u200b\u200bof the variable t.:
Then, passing to the return substitution, we get:
Since the foundation of the degree here is more than a unit, equivalent (by Theorem 2) will transition to inequality:
Finally get answer:
Example 9. Solve inequality:
Decision:
We divide both parts of inequality to the expression:
It is always greater than zero (due to the positiveness of the indicative function), so the sign of inequality is not necessary. We get:
t, located in the interval:
Turning to the return substitution we obtain that the initial inequality disintegrates into two cases:
The first inequality of solutions does not have the relevance of the indicative function. We solve the second:
Example 10. Solve inequality:
Decision:
Parabola branches y. = 2x.+2-x. 2 are directed down, therefore it is limited from above the value that it reaches in its vertex:
Parabola branches y. = x. 2 -2x.+2, standing in the indicator, are directed up, it means it is limited to the bottom with the value that it reaches in its vertex:
Together with this limited bottom, the function is also y. = 3 x. 2 -2x.+2, standing in the right part of the equation. It reaches its smallest value at the same point as the parabola standing in the indicator, and this value is 3 1 \u003d 3. So, the initial inequality may be correct only if the function on the left and the function on the right is taken at one point the value equal to 3 (the intersection of areas of these functions is only this number). This condition is performed in a single point. x. = 1.
Answer: x.= 1.
In order to learn to decide indicative equations and inequalities, It is necessary to constantly train in their decision. In this difficult case you can help various methodical manuals, teachers in elementary mathematics, collections of competitive tasks, classes in mathematics at school, as well individual sessions With a professional tutor. I sincerely wish you success in the preparation and brilliant results on the exam.
Sergey Valerievich
P. S. Dear guests! Please do not write applications in the comments on solving your equations. Unfortunately, I have absolutely no time for this. Such messages will be deleted. Please read the article. Perhaps you will find answers to questions that did not allow you to solve your task yourself.
