Examples on the topic of indicative equations. Power or demonstration equations
Do not be afraid of my words, you have already come across this method in grade 7, when I studied polynomials.
For example, if you needed:
Let's grip: the first and third terms, as well as the second and fourth.
It is clear that the first and third is the difference of squares:
and the second and fourth have a common factor Troika:
Then the initial expression is equivalent to this:
Where to make a general multiplier no longer poses Labor:
Hence,
That's about this way we will do when solving the indicative equations: search for "community" among the components and take it out for brackets, well, and then - if it will be, I believe that we will take \u003d))
Example number 14.
The right is not the degree of seven (I checked!) And on the left - a little better ...
You can, of course, "delay" from the first term factor and from the second, and then to understand with the received, but let's do it prudent.
I do not want to deal with fractions that are inevitably formed when "allocation", so is it not better to bear it?
Then the frains will not be: as they say, and wolves are full and sheep are safe:
Calculate the expression in brackets.
Magic, magically, it turns out that (surprisingly, although what else should we wait?).
Then reduce both parts of the equation to this multiplier. We get:, from where.
Here is an example more complicated (quite a bit, truth):
Here is the trouble! We do not have one common foundation here!
It is not entirely clear what to do now.
And let's do that we can: first we transfer the "four" one way, and the "fives" to another:
Now let's bring out "general" to the left and right:
So what now?
What is the benefit of such a stupid group? At first glance, it is not visible at all, but let's look deeper:
Well, now we will make it so that we have only an expression with, and on the right - everything else.
How do we do it?
But how: split both parts of the equation first on (so we will get rid of the degree of right), and then we split both parts on (so we get rid of the numerical factor on the left).
Finally get:
It's incredible!
Sleva we have an expression, and the right is simple.
Then immediately we conclude that
Example number 15.
I will give it summary (I do not particularly bother with explanations), try to figure it out in all the "subtleties" of solutions.
Now the final fastening of the material passed.
Self solving the following 7 tasks (with answers)
- I will summarize the brackets: where
- The first expression will be submitted in the form: we divide both parts on and get that
- , then the initial equation is converted to mind: Well, now the hint is looking for, where we have already solved this equation!
- Imagine how, as, and, and then, both parts are called on, so you will get the simplest indicative equation.
- I bring for brackets.
- I bring for brackets.
Indicative equations. AVERAGE LEVEL
I assume that after reading the first article in which it was told what is the indicative equations and how to solve them, you collapsed necessary minimum Knowledge required to solve the simplest examples.
Now I will figure out another method of solving the indicative equations, it ...
Method of introducing a new variable (or replacement)
They solve most of the "difficult" tasks, on the topic of indicative equations (and not only equations).
This method is one of most frequently used in practice. First I recommend getting acquainted with the topic.
As you already understood from the name, the essence of this method is to introduce such a replacement of the variable that your indicative equation will be miraculously transformed into such that you can easily solve.
All that you will remain after the solution of this very "simplified equation" is to make a "reverse replacement": that is, to return from the replaced to the replacement.
Let's illustrate just said on a very simple example:
Example 16. Method of simple replacement
This equation is solved with "Simple replacement"How it is negligiously called mathematics.
In fact, the replacement here is the most obvious. It is only worth seeing that
Then the initial equation will turn into such:
If you additionally imagine how it is absolutely clear that you need to replace ...
Of course, .
What will the initial equation turn into? But what:
You can easily find him roots without any problems :.
What should we do now?
It's time to return to the source variable.
What did I forget to specify?
It is: when replacing some extent to a new variable (i.e., when replacing the view), I will be interested only positive roots!
You yourself will easily answer why.
Thus, we are not interested with you, but the second root is quite suitable for us:
Then, from where.
Answer:
As you can see, in the previous example, the replacement was so asked to our hands. Unfortunately, it happens not always.
However, let's not go directly to the sad, but practice another example with a fairly simple replacement
Example 17. Method of simple replacement
It is clear that most likely to replace will have to be replaced (this is the smallest of degrees entering our equation).
However, before introducing a replacement, our equation needs to "prepare" to it, namely: ,.
Then you can replace, as a result, I will get the following expression:
Oh horror: cubic equation with perfectly terrible formulas for its solution (well, if we talk in general).
But let's not immediately despair, and think about what we do.
I will suggest something: we know that to get a "beautiful" answer, we need to get a threesome in the form of some degree (why would it, eh?).
And let's try to guess at least one root of our equation (I will start guessing the troika degrees).
First assumption. Not a root. Alas and ah ...
.
The left side is equal.
Right part: !
There is! Guess the first root. Now it will go easier!
Do you know about the division scheme "Corner"? Of course you know, you apply it when you share one number to another.
But few know that the same can be done with polynomials.
There is one wonderful theorem:
Applicable to my situation, it tells me that it is divided without a rest.
How is the division? That's how:
I look at which one I have to multiply to get
It is clear that on, then:
I submit the resulting expression from will receive:
Now, what do I need to multiply to get?
It is clear that on, then I will get:
and again the deduction expression from the remaining:
Well, the last step, Domain on, and the deduction from the remaining expression:
Hurray, the division is over! What did we accumulated in private?
By itself: .
Then they got this decomposition of the original polynomial:
Resolving the second equation:
It has roots:
Then the initial equation:
it has three roots:
The last root we, of course, throw it, because it is less than zero.
And the first two after the replacement will give us two roots:
Answer: ..
With this example, I didn't want to scare you!
Rather, on the contrary, I set it aim to show that at least we had a pretty simple replacement, nevertheless it led to a rather complex equation, the solution of which demanded some of us some special skills.
Well, no one is immune from this. But the replacement in this case was rather obvious.
Example №18 (with a less obvious replacement)
It is not at all clear what to do: the problem is that in our equation two different bases and one base is not obtained from another erection in any (reasonable, naturally) degree.
However, what do we see?
Both bases - differ only in the sign, and their work - there is a difference in squares equal to one:
Definition:
Thus, the numbers that are grounds in our example are conjugate.
In this case, a reasonable step will be draw both parts of the equation on the conjugate number.
For example, on, then left part The equations will be equal, and the right one.
If you make a replacement, our initial equation will become like this:
his roots, then, and remember that, we get that.
Answer: ,.
As a rule, the replacement method is enough to solve most of the "school" indicative equations.
Next tasks increased level The difficulties are taken from the EGE options.
Three tasks of increased complexity from the EGE options
You are already quite competent to solve these examples. I will only bring the required replacement.
- Solve the equation:
- Find the roots of the equation:
- Decide equation :. Find all the roots of this equation that belongs to the segment:
And now brief explanations and answers:
Example №19.
It is enough for us to notice that.
Then the initial equation will be equivalent to this:
This equation is solved by replacing
Further calculations do it yourself.
At the end of your task, it will be reduced to solving the simplest trigonometric (sinus or cosine-dependent). We will analyze such examples in other sections.
Example number 20.
Here you can even do without replacement ...
It is enough to transfer the ended to the right and present both bases through degree degree: and then immediately go to the square equation.
Example №21
Also solved quite standard: imagine how.
Then replacing the square equation: then
You already know what logarithm is? Not? Then urgently read the topic!
The first root, obviously, does not belong to the segment and the second - it is not clear!
But we will find out very soon!
Since, then (this is the property of the logarithm!)
Subscribe from both parts, then we get:
The left part can be represented as:
both parts for:
can be drawn on, then
Then compare:
since, then:
Then the second root belongs to the desired gap
Answer:
As you see, the selection of the roots of the indicative equations requires enough deep knowledge Properties of logarithmSo I advise you to be as close as possible when you decide the indicative equations.
As you understand, everything is interconnected in mathematics!
As my teacher in mathematics said: "Math, as a story, you will not read overnight."
As a rule, all the complexity in solving problems of an increased level of complexity is precisely the selection of the roots of the equation.
Another example for training ...
Example 22.
It is clear that the equation itself is pretty simple.
By making a replacement, we will reduce our original equation to the following:
First, let's consider the first root.
Compare and: since, then. (Property of the logarithmic function, when).
Then it is clear that the first root does not belong to our gap.
Now the second root :. It is clear that (since the function is increasing).
It remains to compare and.
since, then at the same time.
Thus, I can "knock the peg" between and.
This peg is the number.
The first expression is less, and the second is more.
Then the second expression more first And the root belongs to the gap.
Answer:.
At the end, let's consider another example of the equation where the replacement is quite non-standard.
Example №23 (equation with non-standard replacement!)
Let's immediately begin with what can be done, and what - in principle, it is possible, but it is better not to do.
You can - imagine everything through degrees of troika, twos and six.
Where it leads?
Yes, nothing will lead to anything: the mixture of degrees, and some will be quite difficult to get rid of.
And what then you need?
Let's notify that
And what will it give us?
And the fact that we can reduce the solution of this example to solve a fairly simple indicative equation!
First, let's rewrite our equation in the form:
Now we divide both parts of the resulting equation on:
Eureka! Now you can replace, we get:
Well, now your turn is to solve the challenges on the demonstration, and I will give them only a brief commentation so that you do not get away from the right track! Good luck!
Example number 24.
The most difficult!
Replace here to see Oh as Nemelko! Nevertheless, this example is quite solving with the help allocation of full square.
To solve it, it is enough to notice that:
Then here are you and replacement:
(Please note that here, with our replacement, we cannot discard negative root !!! And why, what do you think?)
Now for the solution of the example you remained to solve two equations:
Both are solved by the "standard replacement" (but the second one in one example!)
Example №25
2. Note that and make a replacement.
Example number 26.
3. Split a number on mutually simple factors and simplify the resulting expression.
Example number 27.
4. Put the numerator and denominator of the fraction on (or, if you like it more) and make a replacement or.
Example №28.
5. Note that numbers and - conjugate.
Solution of the indicative equations by the method of logarithming. ADVANCED LEVEL
In addition, let's consider another way - solution of the indicative equations by logarithming.
I can not say that the solution of the indicative equations by this method is very popular, but in some cases it is only able to bring us to right decision Our equation.
It is especially used to solve the so-called " mixed equations": That is, where functions of different types are found.
Example №29
in general, it is possible to solve only the logarithming of both parts (for example, for the base), in which the initial equation will turn into the following:
Let's consider the following example:
It is clear that according to OST logarithmic function, we are interested only.
However, this follows not only from OTZ Logarithm, but for another reason.
I think you will not be difficult to guess what exactly.
Let's prologate both parts of our equation based on:
As you can see, the logarithming of our original equation quickly led us to the correct (and beautiful!) Answer.
Let's take another example on one example.
Example number 30.
Here, too, there is nothing terrible: it is progriforming both sides of the equation on the basis, then we get:
We will replace:
However, we missed something! Did you notice where I did miss? After all, then:
what does not satisfy the requirement (think from where it came!)
Answer:
Try to independently write down the solution of the indicative equations below:
And now to take your decision with this:
Example number 31
Logarithming both parts on the ground, considering that:
(the second root does not suit us in view of the replacement)
Example number 32.
Logarithmium based on:
We transform the resulting expression to the following form:
Indicative equations. Brief description and basic formulas
Indicative equation
Equation of the form:
called the simplest indicative equation.
Properties of degrees
Approaches to the decision
- Breathing to the same base
- Bringing to the same indicator
- Replacing the variable
- Simplification of the expression and the use of one of the above.
At the stage of preparation for final testing of high school students, you need to tighten the knowledge on the topic "Indective Equations". The experience of past years indicates that such tasks cause certain difficulties from schoolchildren. Therefore, high school students, regardless of their level of preparation, it is necessary to carefully assimilate the theory, remember the formulas and understand the principle of solving such equations. Having managed to cope with this type of tasks, graduates will be able to count on high scores When passing the exam in mathematics.
Get ready for exam testing with "Shkolkovo"!
When repeating the materials passed, many students face the problem of finding the formulas necessary to solve the equations. The school textbook is not always at hand, and the selection of the necessary information on the topic on the Internet takes a long time.
The educational portal "Skolkovo" offers students to take advantage of our knowledge base. We implement a completely new method of preparation for final testing. When doing on our website, you can identify gaps in knowledge and pay attention to those assignments that cause the greatest difficulties.
Teachers "Shkolkovo" collected, systematized and outlined all necessary for successful surchase EGE Material as easy and accessible form.
The main definitions and formulas are presented in the "Theoretical Help" section.
For better assimilation of the material, we recommend practicing the tasks. Carefully view examples of exponential equations on this page to understand the calculation algorithm. After that, proceed to perform tasks in the "Catalogs" section. You can start with the easiest tasks or immediately move to solving complex indicative equations with several unknown or. The exercise base on our site is constantly complemented and updated.
Those examples with the indicators that have trouble make it possible to add to favorites. So you can quickly find them and discuss the decision with the teacher.
To successfully pass the exam, engage in the "Shkolkovo" portal every day!
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First, let's remember the basic formulas of the degrees and their properties.
The work of the number a. The itself occurs n times, this expression we can write down as a a A ... a \u003d a n
1. A 0 \u003d 1 (A ≠ 0)
3. a n a m \u003d a n + m
4. (a n) m \u003d a nm
5. A N B n \u003d (AB) N
7. A N / A M \u003d A N - M
Power or demonstration equations - These are equations in which variables are in degrees (or indicators), and the basis is the number.
Examples of indicative equations:
IN this example Number 6 is the basis it always stands downstairs, but variable x. degree or indicator.
Let us give more examples of the indicative equations.
2 x * 5 \u003d 10
16 x - 4 x - 6 \u003d 0
Now we will analyze how the demonstration equations are solved?
Take a simple equation:
2 x \u003d 2 3
This example can be solved even in the mind. It can be seen that x \u003d 3. After all, so that the left and right part should be equal to the number 3 instead of x.
Now let's see how it is necessary to issue this decision:
2 x \u003d 2 3
x \u003d 3.
In order to solve such an equation, we removed same grounds (i.e. two) and recorded what remains, it is degrees. Received the desired answer.
Now summarize our decision.
Algorithm for solving an indicative equation:
1. Need to check the same Lee foundations at the equation on the right and left. If the bases are not the same as looking for options for solving this example.
2. After the foundations become the same, equal degrees and solve the resulting new equation.
Now rewrite a few examples:
Let's start with a simple.
The bases in the left and right part are equal to Number 2, which means we can reject and equate their degrees.
x + 2 \u003d 4 It turned out the simplest equation.
x \u003d 4 - 2
x \u003d 2.
Answer: x \u003d 2
In the following example, it can be seen that the bases are different. It is 3 and 9.
3 3x - 9 x + 8 \u003d 0
To begin with, we transfer the nine to the right side, we get:
Now you need to make the same foundation. We know that 9 \u003d 3 2. We use the degree formula (a n) m \u003d a nm.
3 3x \u003d (3 2) x + 8
We obtain 9 x + 8 \u003d (3 2) x + 8 \u003d 3 2x + 16
3 3x \u003d 3 2x + 16 Now it is clear that in the left and right side of the base the same and equal to the troika, which means we can discard them and equate degrees.
3x \u003d 2x + 16 Received the simplest equation
3x - 2x \u003d 16
x \u003d 16.
Answer: X \u003d 16.
We look at the following example:
2 2x + 4 - 10 4 x \u003d 2 4
First, we look at the base, the foundations are different two and four. And we need to be the same. We convert the four by the formula (a n) m \u003d a nm.
4 x \u003d (2 2) x \u003d 2 2x
And also use one formula a n a m \u003d a n + m:
2 2x + 4 \u003d 2 2x 2 4
Add to equation:
2 2x 2 4 - 10 2 2x \u003d 24
We led an example to the same reasons. But we interfere with other numbers 10 and 24. What to do with them? If you can see that it is clear that we have 2 2 2, that's the answer - 2 2, we can take out the brackets:
2 2x (2 4 - 10) \u003d 24
We calculate the expression in brackets:
2 4 — 10 = 16 — 10 = 6
All equation Delim to 6:
Imagine 4 \u003d 2 2:
2 2x \u003d 2 2 bases are the same, throwing out them and equate degrees.
2x \u003d 2 It turned out the simplest equation. We divide it on 2
x \u003d 1.
Answer: x \u003d 1.
Resolving equation:
9 x - 12 * 3 x + 27 \u003d 0
We transform:
9 x \u003d (3 2) x \u003d 3 2x
We get the equation:
3 2x - 12 3 x +27 \u003d 0
The foundations we have the same are equal to three. In this example, it can be seen that the first three degree twice (2x) is greater than that of the second (simply x). In this case, you can solve replacement method. The number with the smallest degree replace:
Then 3 2x \u003d (3 x) 2 \u003d T 2
We replace in equation all degrees with cavities on T:
t 2 - 12T + 27 \u003d 0
We get a square equation. We decide through the discriminant, we get:
D \u003d 144-108 \u003d 36
T 1 \u003d 9
T 2 \u003d 3
Return to the variable x..
Take T 1:
T 1 \u003d 9 \u003d 3 x
That is,
3 x \u003d 9
3 x \u003d 3 2
x 1 \u003d 2
One root found. We are looking for the second, from T 2:
T 2 \u003d 3 \u003d 3 x
3 x \u003d 3 1
x 2 \u003d 1
Answer: x 1 \u003d 2; x 2 \u003d 1.
On the site you can in the Help Resolve Decision to ask you ask questions. We will answer.
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Type of lesson
: A lesson of generalization and comprehensive applications of knowledge, skills and skills on the topic "Indicative equations and ways to solve them."Objectives lesson.
Equipment:
Computer and multimedia projector.At the lesson are used information Technology : methodical support To the lesson - presentation in the Microsoft Power Point program.
During the classes
Every skill with difficulty is given
I. Setting the purpose of the lesson(Slide number 2. )
In this lesson, we will summarize and summarize the "indicative equations, their solutions". Get acquainted with typical the tasks of the USE Different years on this topic.
The tasks for solving indicative equations may occur in any part of the tasks of the USE. In part IN " usually proposed to solve the simplest indicative equations. In part FROM " It is possible to meet more complex indicative equations, the solution of which is usually one of the stages of the task.
For example ( Slide number 3. ).
- Ege - 2007.
In 4 - find the greatest value of the expression x U.where ( x; W.) - Solution System:
In 1 - solve equations:
but) H. 6 3h. – 36 6 3h. = 0;
b) 4. h. +1 + 8 4 H.= 3.
In 4 - find the value of the expression x + U.where ( x; W.) - Solution System:
- Ege - 2010.
II. Actualization of reference knowledge. Reiteration
(Slides number 4 - 6 Presentations to the lesson)It is demonstrated on the screen support theoretical material on this topic.
The following questions are discussed:
- What equations are called indicative?
- Name the main ways to solve them. Create examples of their species ( Slide number 4. )
- What theorem is used when solving the simplest demonstration equations of the form: and f (x) \u003d a g (x)?
- What other methods of solving indicative equations exist? ( Slide number 5. )
(Independently solve the proposed equations for each method and perform self-test using the slide)
- Method of decomposition of multipliers (based on the properties of degrees with the same bases, reception: the degree with the smallest indicator is taken out of the bracket).
- Reception (multiplication) on an indicative expression, different from zero, when solving homogeneous indicative equations .
- Tip:
-
Solving equations with the last two methods with subsequent comments
(Slide number 6. ).
. 4 h.+ 1 – 2 4 h.– 2 = 124, 4 h.– 2 (4 3 - 2) = 124, 4 h.– 2 62 = 124,4 h.– 2 = 2, 4 h.– 2 = 4 0,5 , h.– 2 = 0,5, x \u003d 2,5 .
2 2 2x - 3 2 h. 5 x - 5 5 2h. \u003d 0|: 5 2 h.0,2 (2/5) 2x - 3 (2/5) x - 5 = 0,
t \u003d (2/5) x, t. > 0, 2t. 2 - 3 T - 5 = 0, T.= -1(?...), t \u003d. 5/2; 5/2 \u003d (2/5) x, h.= ?...
III. Solution of the tasks of the EGE 2010
Students independently decide the tasks offered at the beginning of the lesson at the beginning of the lesson, using instructions for the solution, check their decision and answers to them using the presentation ( Slide number 7. ). During the work, options and ways of solutions are discussed, attention is paid to possible errors in solving.
: a) 7 h.- 2 \u003d 49, b) (1/6) 12 - 7 x = 36. Answer: but) h.\u003d 4, b) h. = 2. : 4 h.2 + 3h. – 2 - 0,5 2x2 + 2h. - 1 \u003d 0. (can be replaced with 0.5 \u003d 4 - 0.5)Decision. ,
h. 2 + 3h. – 2 = -h. 2 - 4h. + 0,5 …
Answer: h.= -5/2, h. = 1/2.
: 5 5 TG y. + 4 \u003d 5 -TG y. , with COS y.< 0.Note to the decision
. 5 5 TG. y. + 4 \u003d 5 -TG y. | 5 TG. y. 0,5 5 2G. y. + 4 5 TG y - 1 \u003d 0. Let h.\u003d 5 TG. y. , …
5 TG. y. = -1 (?...), 5 TG. y \u003d.1/5.
Since TG. y.\u003d -1 and COS y.< 0, T. w. II coordinate quarter
Answer: w.= 3/4 + 2k., k. N..
IV. Collaboration at the board
The task of a high level of training is considered - Slide number 8. . With this slide there is a dialogue of teacher and students who contribute to the development of the decision.
- with what parameter but equation 2 2. h. – 3 2 h. + but 2 – 4but \u003d 0 has two roots?Let be t.= 2 h. where t. > 0 . Receive t. 2 – 3t. + (but 2 – 4but) = 0 .
one). Since the equation has two roots, then D\u003e 0;
2). As t. 1,2\u003e 0, then t. 1 t. 2\u003e 0, that is but 2 – 4but> 0 (?...).
Answer: but(- 0.5; 0) or (4; 4.5).
V. Verification work
(Slide number 9. )Students are performed checking On leaves, carrying out self-control and self-esteem of completed work using the presentation, approved in the subject. Independently determines for themselves the program of regulation and correction of knowledge on assumed errors in working notebooks. Sheets with independent work are handed over to the teacher for verification.
Underlined numbers - basic level, with the stars - increased complexity.
Solution and answers.
3. 2 h.– 1 (5 2 4 - 4) = 19, 2 h.– 1 76 = 19, 2 h.– 1 = 1/4, 2 h.– 1 = 2 – 2 , h.– 1 = -2,
x \u003d -1.
4 * .3 9 x \u003d 2 3 h. 5 H.+ 5 25 h. | : 25 h. ,
3 (9/25) x \u003d 2 (3/5) H.+ 5,
3 (9/27) h. = 2 (3/5) h. + 5 = 0,
3 (3/5) 2h. – 2 (3/5) h. - 5 = 0,…, (3/5) h. = -1 (not suitable),
(3/5) h. = 5, x \u003d -1.
Vi. Task at home
(Slide number 10. )- Repeat § 11, 12.
- Of materials EGE 2008 - 2010 Select tasks on the topic and solve them.
- Home test work :
This lesson is designed for those who are just starting to study the indicative equations. As always, let's start with the definition and simplest examples.
If you read this lesson, I suspect that you already have at least a minimum idea of \u200b\u200bthe simplest equations - linear and square: $ 56x-11 \u003d 0 $; $ ((x) ^ (2)) + 5x + 4 \u003d 0 $; $ ((x) ^ (2)) - 12x + 32 \u003d 0 $, etc. To be able to solve such structures are absolutely necessary in order not to "hang" in the topic that we are talking about.
So, the indicative equations. Immediately I will give a couple of examples:
\\ [((2) ^ (x)) \u003d 4; \\ quad (((5) ^ (2x-3)) \u003d \\ FRAC (1) (25); \\ quad ((9) ^ (x)) \u003d - 3 \\]
Some of them may seem more complex, some - on the contrary, too simple. But all of them combines one important feature: in their records there is an indicative function $ F \\ left (x \\ right) \u003d ((a) ^ (x)) $. Thus, we introduce the definition:
The indicative equation is any equation containing an indicative function, i.e. Expression of the type $ ((a) ^ (x)) $. In addition to this function, such equations may contain any other algebraic designs - polynomials, roots, trigonometry, logarithms, etc.
Oh well. Defined figured out. Now the question is: how to solve all this crap? The answer is simultaneously simple, and complicated.
Let's start with good news: in your own experience, classes with many students I can say that most of them are indicative equations are much easier than the same logarithms and the more so the trigonometry.
But there is also bad news: sometimes there are "inspiration" tasks for all kinds of textbooks and exams, and their inflamed brain begins to issue such brutal equations that it becomes problematic not only to students - even many teachers stick to such tasks.
However, we will not be about sad. And back to those three equations that were presented at the very beginning of the narrative. Let's try to solve each of them.
The first equation: $ ((2) ^ (x)) \u003d $ 4. Well, what extent you need to build a number 2 to get the number 4? Probably in the second? After all, $ ((2) ^ (2)) \u003d 2 \\ Cdot 2 \u003d 4 $ - and we obtained the right numerical equality, i.e. really $ x \u003d $ 2. Well, thanks, Cap, but this equation was so simple that I would even solve my cat. :)
Let's look at the following equation:
\\ [((5) ^ (2x-3)) \u003d \\ FRAC (1) (25) \\]
And here is already a little more difficult. Many students know that $ ((5) ^ (2)) \u003d $ 25 is a multiplication table. Some also suspect that $ ((5) ^ (- 1)) \u003d \\ FRAC (1) (5) $ is essentially the definition of negative degrees (by analogy with the $ formula ((a) ^ (- n)) \u003d \\ FRAC (1) (((a) ^ (n))) $).
Finally, only the favorites guess that these facts can be combined and at the output to get the following result:
\\ [\\ FRAC (1) (25) \u003d \\ FRAC (1) (((5) ^ (2))) \u003d ((5) ^ (- 2)) \\]
Thus, our initial equation will rewrite as follows:
\\ [(((5) ^ (2x-3)) \u003d \\ FRAC (1) (25) \\ RIGHTARROW ((5) ^ (2x-3)) \u003d ((5) ^ (- 2)) \\]
But this is already quite solved! On the left in the equation there is an indicative function, the right in the equation is the indicative function, nothing but they are no longer anywhere. Consequently, it is possible to "discard" the foundations and stupidly equate the indicators:
Received the simplest linear equation, which any student will decide literally in a couple of lines. Well, in four lines:
\\ [\\ begin (align) & 2x-3 \u003d -2 \\\\ & 2x \u003d 3-2 \\\\ & 2x \u003d 1 \\\\ & x \u003d \\ FRAC (1) (2) \\\\ End (Align) \\]
If you do not understand what happened in the last four lines - be sure to return to the topic " linear equations"And repeat it. Because without a clear assimilation of this topic, it is too early for the indicative equations.
\\ [((9) ^ (x)) \u003d - 3 \\]
Well, how to solve this? The first thought: $ 9 \u003d 3 \\ Cdot 3 \u003d (((3) ^ (2)) $, so the initial equation can be rewritten so:
\\ [((\\ left (((3) ^ (2)) \\ right)) ^ (x)) \u003d - 3 \\]
Then you remember that when the degree is raised into the degree, the indicators are variable:
\\ [((\\ left (((3) ^ (2)) \\ RIGHT)) ^ (x)) \u003d ((3) ^ (2x)) \\ Rightarrow ((3) ^ (2x)) \u003d - (( 3) ^ (1)) \\]
\\ [\\ begin (align) & 2x \u003d -1 \\\\ & x \u003d - \\ FRAC (1) (2) \\\\\\ End (Align) \\]
And here for such a decision we will get honestly deserved two. For we with the calm of the Pokemon sent a "minus" sign, facing the top three, to the degree of this troika. And so do it is impossible. And that's why. Take a look different degrees Troika:
\\ [\\ begin (Matrix) ((3) ^ (1)) \u003d 3 ° ((3) ^ (- 1)) \u003d \\ FRAC (1) (3) & ((3) ^ (\\ FRAC (1) ( 2))) \u003d \\ sqrt (3) \\\\ ((3) ^ (2)) \u003d 9 Δ (3) ^ (- 2)) \u003d \\ FRAC (1) (9) & ((3) ^ (\\ 3) ^ (- \\ FRAC (1) (2))) \u003d \\ FRAC (1) (\\ SQRT (3)) \\\\\\ End (Matrix) \\]
Composing this sign, I just did not pervert: and considered positive degrees, and negative, and even fractional ... Well, where is at least one a negative number? His not! And can not be because the indicative function is $ y \u003d (((a) ^ (x)) $, first, always takes only positive values \u200b\u200b(how many units do not multiply or not delivered to a twice - there will still be a positive number), And secondly, the basis of such a function is the number $ a $ - by definition is a positive number!
Well, how then to solve the equation $ ((9) ^ (x)) \u003d - $ 3? But in no way: there are no roots. And in this sense, the indicative equations are very similar to square - there may also not be roots. But if B. square equations The number of roots is determined by the discriminant (discriminant positive - 2 roots, negative - no roots), then everything depends on what is worth the right of equality.
Thus, we formulate a key conclusion: the simplest indicative equation of the type $ ((a) ^ (x)) \u003d b $ has a root then and only if $ B \\ gt 0 $. Knowing this simple fact, you can easily determine: there is a root equation proposed for you or not. Those. Is it worth it to solve it or immediately write down that there are no roots.
This knowledge will still repeatedly help us when you have to solve more complex tasks. In the meantime, the lyrics are enough - it's time to study the main algorithm for solving indicative equations.
How to solve exponential equations
So, we formulate the task. It is necessary to solve the indicative equation:
\\ [((a) ^ (x)) \u003d b, \\ quad a, b \\ gt 0 \\]
According to the "naive" algorithm, through which we have earlier, it is necessary to present the number $ b $ as the degree of $ A $:
In addition, if there will be any expression instead of the $ x $ variable, we get a new equation that can be solved already. For example:
\\ [\\ begin (align) & ((2) ^ (x)) \u003d 8 \\ rightarrow ((2) ^ (x)) \u003d ((2) ^ (3)) \\ rightarrow x \u003d 3; \\\\ & ((3) ^ (- x)) \u003d 81 \\ rightarrow ((3) ^ (- x)) \u003d ((3) ^ (4)) \\ rightarrow -x \u003d 4 \\ rightarrow x \u003d -4; \\\\ & ((5) ^ (2x)) \u003d 125 \\ rightarrow ((5) ^ (2x)) \u003d ((5) ^ (3)) \\ Rightarrow 2x \u003d 3 \\ Rightarrow x \u003d \\ FRAC (3) ( 2). \\\\\\ End (Align) \\]
And oddly enough, this scheme works in about 90% of cases. And then with the rest of 10%? The remaining 10% is a bit "schizophrenic" indicative equations of the form:
\\ [((2) ^ (x)) \u003d 3; \\ quad ((5) ^ (x)) \u003d 15; \\ quad ((4) ^ (2x)) \u003d 11 \\]
Well, what extent you need to build 2 to get 3? First? And here is not: $ ((2) ^ (1)) \u003d 2 $ - not enough. In the second? There is also no: $ ((2) ^ (2)) \u003d $ 4 - a bit too much. And in which then?
Knowing students already probably guessed: in such cases, when "beautifully" cannot be solved, "heavy artillery" - logarithms are connected. Let me remind you that with the help of logarithms, any positive number can be represented as a degree of any other positive number (except for one):
Remember this formula? When I tell my students about the logarithm, I always warn it: this formula (it is the main logarithmic identity or, if you like, the definition of logarithm) will chase it for a very long time and "pop up" in the most unexpected places. Well, she pops up. Let's look at our equation and for this formula:
\\ [\\ begin (Align) & ((2) ^ (x)) \u003d 3 \\\\ & \u003d (((b) ^ (((\\ log) _ (b)) a)) \\\\\\ End (Align) \\]
If we assume that $ a \u003d $ 3 is our original number, which is standing on the right, and $ b \u003d $ 2 is the very basis indicative functionTo which we want to bring the right side, we will get the following:
\\ [\\ begin (align) & a \u003d ((b) ^ (((\\ log) _ (b)) a)) \\ rightarrow 3 \u003d (((2) ^ (((\\ log) _ (2)) 3 )); \\\\ & ((2) ^ (x)) \u003d 3 \\ rightarrow ((2) ^ (x)) \u003d (((2) ^ (((\\ log) _ (2)) 3)) \\ Rightarrow x \u003d ( (\\ log) _ (2)) 3. \\\\\\ End (Align) \\]
Received a little strange answer: $ x \u003d ((\\ log) _ (2)) $ 3. In some other task, many would be laughed in such an answer and began to recheck their solution: suddenly there was a mistake somewhere? I hurry to refress you: no error is not here, and logarithm in the roots of the indicative equations is a completely typical situation. So get used to. :)
Now we decide by the analogy of the remaining two equations:
\\ [\\ begin (align) & ((5) ^ (x)) \u003d 15 \\ rightarrow ((5) ^ (x)) \u003d ((5) ^ (((\\ log) _ (5)) 15)) \\ Rightarrow x \u003d ((\\ log) _ (5)) 15; \\\\ Δ ((4) ^ (2x)) \u003d 11 \\ rightarrow ((4) ^ (2x)) \u003d ((4) ^ (((\\ log) _ (4)) 11)) \\ Rightarrow 2x \u003d ( (\\ log) _ (4)) 11 \\ RIGHTARROW X \u003d \\ FRAC (1) (2) ((\\ Log) _ (4)) 11. \\\\\\ End (Align) \\]
That's all! By the way, the last answer can be written otherwise:
This we made a multiplier to the argument of Logarithm. But no one prevents us from making this multiplier to the ground:
In this case, all three options are correct - these are simply different forms of recording of the same number. Which one to choose and write down in the present decision - to solve only you.
Thus, we learned how to solve any indicative equations of the type $ ((a) ^ (x)) \u003d b $, where the numbers $ a $ and $ b $ are strictly positive. However, the harsh reality of our world is such that such simple tasks will meet you very and very rarely. Much more often you will come across something like this:
\\ [\\ begin (align) & ((4) ^ (x)) + ((4) ^ (x - 1)) \u003d ((4) ^ (x + 1)) - 11; \\\\ & ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((21) ^ (3x)); \\\\ & ((100) ^ (x - 1)) \\ Cdot ((2.7) ^ (1-x)) \u003d 0.09. \\\\\\ End (Align) \\]
Well, how to solve this? Is it possible to solve? And if so, how?
Without panic. All these equations quickly and simply go down to those simple formulasthat we have already considered. Just need to know remember a couple of techniques from the course of algebra. And of course, here is nowhere without rules for working with degrees. About this I will tell you now. :)
Transformation of indicative equations
The first thing to remember is: any indicative equation, no matter how difficult it is, anyway, should be reduced to the simplest equations - thereby we have already considered and which we know how to solve. In other words, the scheme of solving any indicative equation is as follows:
- Record the source equation. For example: $ ((4) ^ (x)) + ((4) ^ (x-1)) \u003d ((4) ^ (x + 1)) - 11 $;
- Make some incomprehensible crap. Or even a few horses, which are called "convert equation";
- At the output to obtain the simplest expressions of the type $ ((4) ^ (x)) \u003d $ 4 or something else in this spirit. Moreover, one initial equation can give several such expressions at once.
With the first item, everything is clear - even my cat will be able to record the equation on the leaf. With the third point, too, it seems, more or less clearly - we have already groaned such equations.
But how to be with the second item? What kind of transformation? What to convert in? And How?
Well, let's understand. First of all, I will note the following. All indicative equations are divided into two types:
- The equation is composed of indicative functions with the same base. Example: $ ((4) ^ (x)) + ((4) ^ (x - 1)) \u003d ((4) ^ (x + 1)) - 11 $;
- The formula has demonstration functions with different bases. Examples: $ ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((21) ^ (3x)) $ and $ ((100) ^ (x - 1) ) \\ Cdot ((2.7) ^ (1-x)) \u003d 0.09 $.
Let's start with the equations of the first type - they are solved the easiest. And in their solution, we will help such a reception as the allocation of sustainable expressions.
Allocation of a stable expression
Let's look at this equation again:
\\ [((4) ^ (x)) + ((4) ^ (x - 1)) \u003d ((4) ^ (x + 1)) - 11 \\]
What do we see? The fourthkee is erected in different degrees. But all these degrees are the simple amounts of the $ x $ variable with other numbers. Therefore, it is necessary to recall the rules for working with degrees:
\\ [\\ begin (align) & ((a) ^ (x + y)) \u003d ((a) ^ (x)) \\ Cdot ((a) ^ (y)); \\\\ & ((a) ^ (xy)) \u003d ((a) ^ (x)): ((a) ^ (y)) \u003d \\ FRAC (((a) ^ (x))) ((( ) ^ (y))). \\\\\\ End (Align) \\]
Simply put, the addition of indicators can be converted into the work of degrees, and the subtraction is easily converted into division. Let's try to apply these formulas to degrees from our equation:
\\ [\\ Begin (Align) & ((4) ^ (x - 1)) \u003d \\ FRAC (((4) ^ (x))) (((4) ^ (1))) \u003d ((4) ^ (x)) \\ Cdot \\ FRAC (1) (4); \\\\ · ((4) ^ (x + 1)) \u003d ((4) ^ (x)) \\ Cdot ((4) ^ (1)) \u003d ((4) ^ (x)) \\ CDOT 4. \\ I rewrite the original equation, taking into account this fact, and then collect all the components on the left:
\\ [\\ begin (align) & ((4) ^ (x)) + ((4) ^ (x)) \\ Cdot \\ FRAC (1) (4) \u003d ((4) ^ (x)) \\ CDOT 4 -eleven; \\\\ & ((4) ^ (x)) + ((4) ^ (x)) \\ Cdot \\ FRAC (1) (4) - ((4) ^ (x)) \\ Cdot 4 + 11 \u003d 0. \\\\\\ End (Align) \\]
In the first four components there is an element $ ((4) ^ (x)) $ - I will bring it for the bracket:
\\ [\\ begin (align) & ((4) ^ (x)) \\ cdot \\ left (1+ \\ FRAC (1) (4) -4 \\ RIGHT) + 11 \u003d 0; \\\\ & ((4) ^ (x)) \\ CDOT \\ FRAC (4 + 1-16) (4) + 11 \u003d 0; \\\\ & ((4) ^ (x)) \\ Cdot \\ left (- \\ FRAC (11) (4) \\ Right) \u003d - 11. \\\\\\ End (Align) \\]
It remains to divide both parts of the equation for the fraction of $ - \\ FRAC (11) (4) $, i.e. Essentially multiply to the overtook fraction - $ - \\ FRAC (4) (11) $. We get:
\\ [\\ Begin (Align) & ((4) ^ (x)) \\ Cdot \\ left (- \\ FRAC (11) (4) \\ RIGHT) \\ CDOT \\ LEFT (- \\ FRAC (4) (11) \\ RIGHT ) \u003d - 11 \\ Cdot \\ left (- \\ FRAC (4) (11) \\ RIGHT); \\\\ & ((4) ^ (x)) \u003d 4; \\\\ & ((4) ^ (x)) \u003d ((4) ^ (1)); \\\\ & x \u003d 1. \\\\\\ End (Align) \\]
That's all! We reduced the initial equation to the simplest and got the final answer.
At the same time, in the process of solutions, we found (and even carried out for the bracket) the total multiplier $ ((4) ^ (x)) $ is a stable expression. It can be denoted by a new variable, and you can simply gently express and get the answer. In any case, the key principle of solving the following:
Find a stable expression in the source equation containing a variable that is easily highlighted from all the indicative functions.
The good news is that almost every indicative equation allows the allocation of such a stable expression.
But there are bad news: such expressions may be very cunning, and it is quite difficult to allocate them. Therefore, we will analyze another task:
\\ [((5) ^ (x + 2)) + ((0.2) ^ (- x - 1)) + 4 \\ Cdot ((5) ^ (x + 1)) \u003d 2 \\]
Perhaps someone will now have a question: "Pasha, what did you whistle? Here, different bases - 5 and 0.2 ". But let's try to convert a degree with a base of 0.2. For example, get rid of decimal fractions, bringing it to normal:
\\ [((0.2) ^ (- x - 1)) \u003d (((0.2) ^ (- \\ left (x + 1 \\ right))) \u003d ((\\ left (\\ FRAC (2) (10 ) \\ Right)) ^ (- \\ left (x + 1 \\ right))) \u003d ((\\ left (\\ FRAC (1) (5) \\ RIGHT)) ^ (- \\ left (x + 1 \\ right)) ) \\]
As you can see, the number 5 after all appeared, let it both in the denominator. At the same time rewrote the indicator in the form of a negative. And now you remember one of the most important rules Work with degrees:
\\ [((a) ^ (- n)) \u003d \\ FRAC (1) (((a) ^ (n))) \\ Rightarrow ((\\ left (\\ FRAC (1) (5) \\ Right)) ^ ( - \\ left (x + 1 \\ right))) \u003d ((\\ left (\\ FRAC (5) (1) \\ RIGHT)) ^ (x + 1)) \u003d ((5) ^ (x + 1)) \\ Here I, of course, slightly rushed. Because for a complete understanding of the formula of deliverance from negative indicators, it was necessary to record like this:
\\ [((a) ^ (- n)) \u003d \\ FRAC (1) (((a) ^ (n))) \u003d ((\\ left (\\ FRAC (1) (A) \\ RIGHT)) ^ (n )) \\ Rightarrow ((\\ left (\\ FRAC (1) (5) \\ RIGHT)) ^ (- \\ left (x + 1 \\ right))) \u003d ((\\ left (\\ FRAC (5) (1) \\ On the other hand, nothing prevented us to work with one shot:
\\ [((\\ left (\\ FRAC (1) (5) \\ RIGHT)) ^ (- \\ left (x + 1 \\ right))) \u003d ((\\ left (((5) ^ (- 1)) \\ )) \u003d ((5) ^ (x + 1)) \\]
But in this case, you need to be able to erect a degree to another degree (remind you: the indicators are folded). But I did not have to "turn over" the fractions - perhaps for someone it will be easier. :)
In any case, the initial indicative equation will be rewritten as:
\\ [\\ begin (align) & ((5) ^ (x + 2)) + ((5) ^ (x + 1)) + 4 \\ cdot ((5) ^ (x + 1)) \u003d 2; \\\\ & ((5) ^ (x + 2)) + 5 \\ Cdot ((5) ^ (x + 1)) \u003d 2; \\\\ & ((5) ^ (x + 2)) + ((5) ^ (1)) \\ Cdot ((5) ^ (x + 1)) \u003d 2; \\\\ & ((5) ^ (x + 2)) + ((5) ^ (x + 2)) \u003d 2; \\\\ & 2 \\ Cdot ((5) ^ (x + 2)) \u003d 2; \\\\ & ((5) ^ (x + 2)) \u003d 1. \\\\\\ End (Align) \\]
So it turns out that the initial equation is even easier than the previously considered: there is no need to allocate a steady expression - everything itself has decreased. It remains only to recall that $ 1 \u003d ((5) ^ (0)) $, from where we get:
\\ [\\ begin (align) & ((5) ^ (x + 2)) \u003d ((5) ^ (0)); \\\\ & x + 2 \u003d 0; \\\\ & x \u003d -2. \\\\\\ End (Align) \\]
That's all the decision! We got the final answer: $ x \u003d -2 $. At the same time I would like to note one reception, which greatly simplified us all the calculations:
In the indicative equations, be sure to get rid of
decimal fractions
, Translate them to ordinary. This will allow you to see the same foundations of degrees and will significantly simplify the decision. moving now to morecomplex equations
In which there are different foundations that are not at all reduced to each other with the help of degrees. use the properties of degreesLet me remind you that we have two more particularly harsh equations:
\\ [\\ begin (align) & ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((21) ^ (3x)); \\\\ & ((100) ^ (x - 1)) \\ Cdot ((2.7) ^ (1-x)) \u003d 0.09. \\\\\\ End (Align) \\]
The main difficulty here is not clear what to bring to what basis. Where are the stable expressions? Where are the same foundations? There is no need for it.
But let's try to go to another way. If there are no ready-made values, you can try to find, laying out the reasons for multipliers.
Let's start with the first equation:
\\ [\\ begin (align) & ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((21) ^ (3x)); \\\\ & 21 \u003d 7 \\ CDOT 3 \\ RIGHTARROW ((21) ^ (3X)) \u003d ((\\ left (7 \\ CDOT 3 \\ RIGHT)) ^ (3x)) \u003d ((7) ^ (3x)) \\ \\\\\\ End (Align) \\]
But after all, you can proceed on the contrary - make up from numbers 7 and 3 number 21. Especially it is easy to do on the left, since the indicators and both degrees are the same:
\\ [\\ begin (align) & ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((\\ left (7 \\ CDOT 3 \\ RIGHT)) ^ (X + 6)) \u003d ((21) ^ (x + 6)); \\\\ & ((21) ^ (x + 6)) \u003d ((21) ^ (3x)); \\\\ & x + 6 \u003d 3x; \\\\ & 2x \u003d 6; \\\\ & x \u003d 3. \\\\\\ End (Align) \\]
That's all! You made an indicator of the degree outside the work and immediately got a beautiful equation, which is solved in a couple of lines.
Now we will deal with the second equation. Everything is much more difficult here:
\\ [((100) ^ (x - 1)) \\ CDOT ((2.7) ^ (1-x)) \u003d 0.09 \\]
\\ [((100) ^ (x - 1)) \\ CDOT ((\\ left (\\ FRAC (27) (10) \\ RIGHT)) ^ (1-x)) \u003d \\ FRAC (9) (100) \\]
In this case, the fractions were disracotized, but if something could be reduced - be sure to reduce. Often, at the same time, interesting grounds will appear with which you can already work.
Also, unfortunately, nothing really appeared. But we see that the indicators of degrees standing in the work on the left are opposite:
Let me remind you: to get rid of the "minus" sign in the indicator, it is enough to "turn over" the fraction. Well, rewrite the original equation:
\\ [\\ begin (align) & ((100) ^ (x - 1)) \\ CDOT ((\\ left (\\ FRAC (10) (27) \\ RIGHT)) ^ (x - 1)) \u003d \\ FRAC (9 )(100); \\\\ \\ (\\ left (100 \\ CDOT \\ FRAC (10) (27) \\ RIGHT)) ^ (x - 1)) \u003d \\ FRAC (9) (100); \\\\ & ((\\ left (\\ FRAC (1000) (27) \\ RIGHT)) ^ (X - 1)) \u003d \\ FRAC (9) (100). \\\\\\ End (Align) \\]
In the second line, we simply carried out a general figure from the work for a bracket according to the rule $ ((a) ^ (x)) \\ Cdot ((b) ^ (x)) \u003d ((\\ left (a \\ cdot b \\ right)) ^ (x)) $, and in the latter just multiplied the number 100 by fraction.
Now we note that the numbers standing on the left (at the base) and on the right, are alike. Than? Yes, obviously: they are degrees of the same number! We have:
\\ [\\ begin (align) \\ FRAC (1000) (27) \u003d \\ FRAC (((10) ^ (3))) (((3) ^ (3))) \u003d ((\\ left (\\ FRAC ( 10) (3) \\ RIGHT)) ^ (3)); \\\\ \\ FRAC (9) (100) \u003d \\ FRAC (((3) ^ (2))) (((10) ^ (3))) \u003d ((\\ left (\\ FRAC (3) (10) \\ Right)) ^ (2)). \\\\\\ End (Align) \\]
Thus, our equation will rewrite as follows:
\\ [((\\ left ((\\ left (\\ FRAC (10) (3) \\ RIGHT)) ^ (3)) \\ Right)) ^ (x - 1)) \u003d ((\\ left (\\ FRAC (3 ) (10) \\ RIGHT)) ^ (2)) \\]
\\ [((\\ left (((\\ left (\\ FRAC (10) (3) \\ RIGHT)) ^ (3)) \\ Right)) ^ (x - 1)) \u003d ((\\ left (\\ FRAC (10 ) (3) \\ Right)) ^ (3 \\ left (x - 1 \\ right))) \u003d ((\\ left (\\ FRAC (10) (3) \\ RIGHT)) ^ (3x-3)) \\]
At the same time, you can also get a degree with the same basis, for which it is enough to "turn over" the fraction:
\\ [((\\ left (\\ FRAC (3) (10) \\ RIGHT)) ^ (2)) \u003d ((\\ left (\\ FRAC (10) (3) \\ Right)) ^ (- 2)) \\]
Finally, our equation will take the form:
\\ [\\ begin (Align) & ((\\ Left (\\ FRAC (10) (3) \\ RIGHT)) ^ (3x-3)) \u003d ((\\ left (\\ FRAC (10) (3) \\ RIGHT)) ^ (- 2)); \\\\ & 3x-3 \u003d -2; \\\\ & 3x \u003d 1; \\\\ & X \u003d \\ FRAC (1) (3). \\\\\\ End (Align) \\]
That's the whole decision. His main idea is reduced to the fact that even under different reasons, we are trying by any truths and inconsistencies to reduce these grounds for the same. This is helped by elementary transformations of equations and rules for working with degrees.
But what are the rules and when to use? How to understand that in one equation you need to share both sides for something, and in the other - to lay the basis of the indicative function on multipliers?
The answer to this question will come with experience. Try your hand at first on ordinary equations, and then gradually complicate the tasks - and very soon your skills will be enough to solve any indicative equation from the same USE or any independent / test work.
And to help you in this hard matter, I suggest downloading a set of equations for my site. self-decide. To all equations there are answers, so you can always check yourself.
In general, I wish you a good workout. And you will see you in the next lesson - there we will disassemble the truly complex lower equations, where the methods described above are not enough. And the simple training will also be not enough. :)
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