How to solve indicative equations with different bases. Solution of indicative equations. Examples

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 must!) Do such identical transformations that reduce 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 all types of indicative equations and methods of their solution are carried out, a general scheme is drawn up. (Used training computer program L.Ya. Borevsky "Mathematics Course - 2000", the author of the presentation in PowerPoint - so-called. Motversova.)

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.

By deciding this equation and making a replacement, you as a result come to the aggregate of the simplest demonstration equations that are solved in general with the help of 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)

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 now 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 at different degrees of the 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) \\]

By making this sign, I just did not perverted: and considered positive degrees, and negative, and even fractional ... so where is at least one 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 in 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 the equality sign.

Thus, we formulate the key conclusion: the simplest indicative equation of the type $ ((a) ^ (x)) \u003d b $ has a root then and only if $ b\u003e $ 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\u003e 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 worth the right, and $ b \u003d 2 $ is the most base of the indicative function to which we want to bring the right part so that we obtain 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 reduce the simple formulas that 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 various 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.

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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 I remember one of the most important rules for working 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:

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In the lower equations, be sure to get rid of decimal fractions, translate them into ordinary. This will allow you to see the same foundations of degrees and will significantly simplify the decision.

Let us now turn to more complex 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 degrees

Let 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 propose to download a set of equations for an independent solution on my site. To all equations there are answers, so you can always check yourself.

Lecture: "Methods for solving indicative equations."

1 . Indicative equations.

Equations containing unknown persons are called indicative equations. The simplest of them is the equation Ax \u003d B, where a\u003e 0, and ≠ 1.

1) at b< 0 и b = 0 это уравнение, согласно свойству 1 показательной функции, не имеет решения.

2) at b\u003e 0 using the monotony of the function and the theorem on the root, the equation has the only root. In order to find it, it is necessary to represent in the form B \u003d ax, Ax \u003d BC ó x \u003d C or x \u003d Logab.

Indicative equations by algebraic transformations lead to standard equations that are solved using the following methods:

1) the method of bringing to one base;

2) the assessment method;

3) graphic method;

4) the method of introducing new variables;

5) the method of decomposition of multipliers;

6) significantly - power equations;

7) indicative with the parameter.

2 . The method of bringing to one base.

The method is based on the following property of degrees: if two degrees are equal and their bases are equal, then their indicators are equal, i.e. the equation must be tried to reduce the form

Examples. Solve equation:

1 . 3x \u003d 81;

Imagine the right side of the equation in the form 81 \u003d 34 and install the equation, equivalent to the original 3 x \u003d 34; x \u003d 4. Reply: 4.

2. https://pandia.ru/text/80/142/images/image004_8.png "width \u003d" 52 "height \u003d" 49 "\u003e and move on to the equation for the indicators of 3x + 1 \u003d 3 - 5x; 8x \u003d 4; x \u003d 0.5. Answer: 0.5.

3. https://pandia.ru/text/80/142/images/image006_8.png "width \u003d" 105 "height \u003d" 47 "\u003e

Note that the numbers 0.2, 0.04, √5 and 25 are the degree of number 5. We use this and transform the initial equation as follows:

, from where 5-x-1 \u003d 5-2x-2 ó - x - 1 \u003d - 2x - 2, from which we find the solution x \u003d -1. Answer: -1.

5. 3X \u003d 5. By definition of logarithm X \u003d log35. Answer: log35.

6. 62x + 4 \u003d 33x. 2x + 8.

I rewrite the equation in the form of 32x + 4.22x + 4 \u003d 32x.2x + 8, t. E..png "width \u003d" 181 "height \u003d" 49 src \u003d "\u003e from here x - 4 \u003d 0, x \u003d 4. Answer: four.

7 . 2 ∙ 3X + 1 - 6 ∙ 3X-2 - 3X \u003d 9. Using the properties of degrees, write the equation in the form 6 ∙ 3x - 2 ∙ 3x - 3x \u003d 9 Next 3 ∙ 3x \u003d 9, 3x + 1 \u003d 32, t. e. X + 1 \u003d 2, x \u003d 1. Answer: 1.

Bank of tasks number 1.

Solve equation:

Test number 1.

	1) 0 2) 4 3) -2 4) -4
A2 32x-8 \u003d √3.	1)17/4 2) 17 3) 13/2 4) -17/4
A3.	1) 3; 1 2) -3; -1 3) 0; 2 4) no roots
	1) 7; 1 2) no roots 3) -7; 1 4) -1; -7
A5.	1) 0;2; 2) 0;2;3 3) 0 4) -2;-3;0
A6.	1) -1 2) 0 3) 2 4) 1

Test number 2.

A1.	1) 3 2) -1;3 3) -1;-3 4) 3;-1
A2.	1) 14/3 2) -14/3 3) -17 4) 11
A3.	1) 2; -1 2) no roots 3) 0 4) -2; 1
A4.	1) -4 2) 2 3) -2 4) -4;2
A5.	1) 3 2) -3;1 3) -1 4) -1;3

3 Evaluation method.

Root theorem: If the function f (x) increases (decreases) at the interval I, the number A -Unote value received by F at this gap, then the equation f (x) \u003d A has the only root at the interval I.

When solving equations, this theorem and the properties of the function monotonicity are used.

Examples. Solve equations: 1. 4x \u003d 5 - x.

Decision. I rewrite the equation in the form 4x + x \u003d 5.

1. If x \u003d 1, then 41 + 1 \u003d 5, 5 \u003d 5 is true, it means 1 - the root of the equation.

The function f (x) \u003d 4X - increases on R, and G (x) \u003d x-invents on R \u003d\u003e h (x) \u003d f (x) + g (x) increases on R, as the sum of increasing functions, then x \u003d 1 - the only root of equation 4x \u003d 5 - x. Answer: 1.

Decision. Rewrite the equation in the form .

1. If x \u003d -1, then , 3 \u003d 3rd, it means x \u003d -1 - the root of the equation.

2. We prove that he is the only one.

3. The function f (x) \u003d - decreases on R, and G (x) \u003d - x - decreases on R \u003d\u003e h (x) \u003d f (x) + g (x) - decreases on R, as the sum of decreasing functions . Hence the root theorem, x \u003d -1 is the only root of the equation. Answer: -1.

Bank of tasks number 2. Solve equation

a) 4x + 1 \u003d 6 - x;

c) 2x - 2 \u003d 1 - x;

4. Method of introducing new variables.

The method is described in paragraph 2.1. The introduction of a new variable (substitution) is usually made after transformations (simplification) of the equation members. Consider examples.

Examples. Rstick equation: 1. .

I rewrite the equation otherwise: https://pandia.ru/text/80/142/images/image030_0.png "width \u003d" 128 "height \u003d" 48 src \u003d "\u003e t. E..png" width \u003d "210" height \u003d "45"\u003e

Decision. I rewrite the equation otherwise:

Denote https://pandia.ru/text/80/142/images/image035_0.png "width \u003d" 245 "height \u003d" 57 "\u003e is not suitable.

t \u003d 4 \u003d\u003e https://pandia.ru/text/80/142/images/image037_0.png "width \u003d" 268 "height \u003d" 51 "\u003e - irrational equation. We note that

By solving the equation is x \u003d 2.5 ≤ 4, it means 2.5 - the root of the equation. Answer: 2.5.

Decision. We rewrite the equation in the form and split it both parts by 56x + 6 ≠ 0. We obtain the equation

2x2-6x-7 \u003d 2x2-6x-8 +1 \u003d 2 (x2-3x-4) +1, t..png "width \u003d" 118 "height \u003d" 56 "\u003e

Roots of the square equation - T1 \u003d 1 and T2<0, т. е..png" width="200" height="24">.

Decision . Rewrite the equation in the form

and we note that it is a homogeneous equation of the second degree.

We divide the 42x equation, we get

Replace https://pandia.ru/text/80/142/images/image049_0.png "width \u003d" 16 "height \u003d" 41 src \u003d "\u003e.

Answer: 0; 0.5.

Bank Tasks number 3. Solve equation

Test number 3. With the choice of response. Minimum level.

A1.	1) -0.2; 2 2) log52 3) -Log52 4) 2
A2 0.52X - 3 0.5X +2 \u003d 0.	1) 2; 1 2) -1; 0 3) no roots 4) 0
	1) 0 2) 1; -1/3 3) 1 4) 5
A4 52x-5x - 600 \u003d 0.	1) -24;25 2) -24,5; 25,5 3) 25 4) 2
	1) no roots 2) 2; 4 3) 3 4) -1; 2

Test number 4. With the choice of response. Common level.

A1.	1) 2; 1 2) ½; 0 3) 2; 0 4) 0
A2 2X - (0.5) 2x - (0.5) x + 1 \u003d 0	1) -1;1 2) 0 3) -1;0;1 4) 1
	1) 64 2) -14 3) 3 4) 8
	1)-1 2) 1 3) -1;1 4) 0
A5.	1) 0 2) 1 3) 0; 1 4) no roots

5. Method of decomposition on multipliers.

1. Decide equation: 5x + 1 - 5x-1 \u003d 24.

Solution..png "width \u003d" 169 "height \u003d" 69 "\u003e, from where

2. 6x + 6x + 1 \u003d 2x + 2x + 1 + 2x + 2.

Decision. I submit for brackets in the left side of the 6x equation, and in the right part - 2x. We obtain equation 6x (1 + 6) \u003d 2x (1 + 2 + 4) ó 6x \u003d 2x.

Since 2x\u003e 0 for all x, both parts of this equation can be divided by 2x, without fear of loss of solutions. We obtain 3x \u003d 1ó x \u003d 0.

Decision. I solve the equation by decomposition by multipliers.

We highlight the square of the bouncer

4. https://pandia.ru/text/80/142/images/image067_0.png "width \u003d" 500 "height \u003d" 181 "\u003e

x \u003d -2 - the root of the equation.

Equation x + 1 \u003d 0 "Style \u003d" Border-Collapse: Collapse; Border: None "\u003e

A1 5x-1 + 5x -5x + 1 \u003d -19.

1) 1 2) 95/4 3) 0 4) -1

A2 3X + 1 + 3X-1 \u003d 270.

1) 2 2) -4 3) 0 4) 4

A3 32X + 32X + 1 -108 \u003d 0. x \u003d 1.5

1) 0,2 2) 1,5 3) -1,5 4) 3

1) 1 2) -3 3) -1 4) 0

A5 2X -2x-4 \u003d 15. x \u003d 4

1) -4 2) 4 3) -4;4 4) 2

Test No. 6. Common level.

A1 (22x-1) (24x + 22x + 1) \u003d 7.	1) ½ 2) 2 3) -1; 3 4) 0,2
A2.	1) 2.5 2) 3; 4 3) log43 / 2 4) 0
A3 2X-1-3X \u003d 3X-1-2X + 2.	1) 2 2) -1 3) 3 4) -3
A4.	1) 1,5 2) 3 3) 1 4) -4
A5.	1) 2 2) -2 3) 5 4) 0

6. Ensure - power equations.

The indicative equations are adjacent to the so-called significant - power equations, i.e. the equations of the form (F (x)) G (x) \u003d (f (x)) h (x).

If it is known that f (x)\u003e 0 and f (x) ≠ 1, the equation, as indicative, is solved by equating the indicators G (x) \u003d f (x).

If the condition does not exceed the possibility of f (x) \u003d 0 and f (x) \u003d 1, then it is necessary to consider these cases when solving an indicative - power equation.

1..png "width \u003d" 182 "height \u003d" 116 src \u003d "\u003e

Decision. x2 + 2x-8 - it makes sense at any X, because the polynomial means equation is equivalent to the totality

https://pandia.ru/text/80/142/images/image078_0.png "width \u003d" 137 "height \u003d" 35 "\u003e

7. Indicative equations with parameters.

1. At what values \u200b\u200bof the parameter p, equation 4 (5 - 3) 2 + 4p2-3p \u003d 0 (1) has a single solution?

Decision. We introduce the replacement 2x \u003d T, T\u003e 0, then the equation (1) will take the form T2 - (5p - 3) T + 4P2 - 3P \u003d 0. (2)

Discriminant equation (2) d \u003d (5p - 3) 2 - 4 (4p2 - 3p) \u003d 9 (p - 1) 2.

Equation (1) has a single solution if equation (2) has one positive root. This is possible in the following cases.

1. If D \u003d 0, that is, P \u003d 1, then the equation (2) will take the form T2 - 2T + 1 \u003d 0, hence T \u003d 1, therefore, equation (1) has a single solution x \u003d 0.

2. If P1, then 9 (p - 1) 2\u003e 0, then equation (2) has two different roots T1 \u003d P, T2 \u003d 4P - 3. The problem of the problem satisfies the set of systems

Substituting T1 and T2 in the system, we have

https://pandia.ru/text/80/142/images/image084_0.png "alt \u003d" (! Lang: NO35_11" width="375" height="54"> в зависимости от параметра a?!}

Decision. Let be then equation (3) will take the type T2 - 6T - a \u003d 0. (4)

Find the parameter A values \u200b\u200bof A, at which at least one root of equation (4) satisfies the condition T\u003e 0.

We introduce the function f (t) \u003d T2 - 6T - a. The following cases are possible.

https://pandia.ru/text/80/142/images/image087.png "Alt \u003d" (! Lang: http: //1september.ru/ru/mat/2002/35/no35_14.gif" align="left" width="215" height="73 src=">где t0 - абсцисса вершины параболы и D - дискриминант квадратного трехчлена f(t);!}

https://pandia.ru/text/80/142/images/image089.png "Alt \u003d" (! Lang: http: //1september.ru/ru/mat/2002/35/no35_16.gif" align="left" width="60" height="51 src=">!}

Case 2. Equation (4) has a single positive decision if

D \u003d 0, if A \u003d - 9, then equation (4) will take the form (T - 3) 2 \u003d 0, t \u003d 3, x \u003d - 1.

Case 3. Equation (4) has two roots, but one of them does not satisfy the inequality T\u003e 0. This is possible if

https://pandia.ru/text/80/142/images/image092.png "alt \u003d" (! Lang: NO35_17" width="267" height="63">!}

Thus, with A 0, equation (4) has a single positive root . Then equation (3) has a single solution

With A.< – 9 уравнение (3) корней не имеет.

if A.< – 9, то корней нет; если – 9 < a < 0, то
if a \u003d - 9, then x \u003d - 1;

if a  0, then

Compare methods for solving equations (1) and (3). Note that when solving, equation (1) was reduced to a square equation, the discriminant of which is a full square; Thus, the roots of equation (2) were immediately calculated by the formula of the roots of the square equation, and then conclusions were made relative to these roots. Equation (3) was reduced to the square equation (4), the discriminant of which is not a complete square, therefore, when solving equation (3), it is advisable to use the theorems on the location of the roots of the square three declests and the graphical model. Note that equation (4) can be solved using the Vieta theorem.

We solve more complex equations.

Task 3. Decide the equation

Decision. OTZ: X1, X2.

We introduce a replacement. Let 2x \u003d T, T\u003e 0, then as a result of the transformations, the equation will take the type T2 + 2t - 13 - a \u003d 0. (*) Find the values \u200b\u200bof A, at which at least one root of the equation (*) satisfies the condition T\u003e 0.

https://pandia.ru/text/80/142/images/image098.png "Alt \u003d" (! Lang: http: //1september.ru/ru/mat/2002/35/no35_23.gif" align="left" width="71" height="68 src=">где t0 - абсцисса вершины f(t) = t2 + 2t – 13 – a, D - дискриминант квадратного трехчлена f(t).!}

https://pandia.ru/text/80/142/images/image100.png "Alt \u003d" (! Lang: http: //1september.ru/ru/mat/2002/35/no35_25.gif" align="left" width="360" height="32 src=">!}

https://pandia.ru/text/80/142/images/image102.png "alt \u003d" (! Lang: http: //1september.ru/ru/mat/2002/35/no35_27.gif" align="left" width="218" height="42 src=">!}

Answer: If a\u003e - 13, a  11, a  5, then if A - 13,

a \u003d 11, a \u003d 5, then no roots.

Bibliography.

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2. Guzeyev Technology: from reception to philosophy.

M. "School Director" №4, 1996

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Mathematics at school №2, 1987, P.9 - 11.

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M. "Enlightenment", 1990

8. Ivanov prepare lessons - workshops.

Mathematics at school №6, 1990 s. 37 - 40.

9. Smirnova Mathematics learning model.

Mathematics at school №1, 1997 with. 32 - 36.

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Mathematics at school №1, 1993 s. 27 - 28.

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Mathematics at school №2, 1994, S.63 - 64.

12. Khazankin creative abilities of schoolchildren.

Mathematics at school №2, 1989 with. 10.

13. Scanavi. Publisher, 1997

14. And others. Algebra and the beginning of the analysis. Didactic materials for

15. Mathematics tasks krivonogs.

M. "First of September", 2002

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entering universities. "And with T - Press School", 2002

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Minsk and the Russian Federation "Review", 1996

18. Written D. Preparing for the exam in mathematics. M. Rolf, 1999

19. And others learn to solve equations and inequalities.

M. "Intellect - Center", 2003

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M. "Intellect - Center", 2003 and 2004

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Mathematics, 1997 №3.

24 Okunev for the lesson, children! M. Enlightenment, 1988

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26. Liimets work in the lesson. M. Knowledge, 1975

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:

\\ (\\ 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.

\\ (\\ 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) \\)

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) \\)

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) \\)

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) \\)

And now we have the foundations equal and there are no interfering coefficients, etc. So we can make the transition.

Example . Solve the indicative equation \\ (4 ^ (x + 0.5) -5 · 2 ^ x + 2 \u003d 0 \\)
Decision:

\\ (4 ^ (x + 0.5) -5 · 2 ^ x + 2 \u003d 0 \\)		We again use the degree of degree \\ (a ^ b \\ cdot a ^ c \u003d a ^ (b + c) \\) in the opposite direction.
\\ (4 ^ x · 4 ^ (0.5) -5 · 2 ^ x + 2 \u003d 0 \\)		Now you remember that \\ (4 \u003d 2 ^ 2 \\).
\\ ((2 ^ 2) ^ x · (2 \u200b\u200b^ 2) ^ (0.5) -5 · 2 ^ x + 2 \u003d 0 \\)		Using the degree properties, we convert: \\ ((2 ^ 2) ^ x \u003d 2 ^ (2x) \u003d 2 ^ (x · 2) \u003d (2 ^ x) ^ 2 \\) \\ ((2 ^ 2) ^ (0.5) \u003d 2 ^ (2 · 0.5) \u003d 2 ^ 1 \u003d 2. \\)
\\ (2 · (2 \u200b\u200b^ x) ^ 2-5 · 2 ^ x + 2 \u003d 0 \\)		We look carefully on the equation, and we see that it suggests the replacement \\ (t \u003d 2 ^ x \\).

\\ (T_1 \u003d 2 \\) \\ (T_2 \u003d \\ FRAC (1) (2) \\)		However, we found the values \u200b\u200b\\ (t \\), and we need \\ (x \\). We return to the ICS, making the reverse replacement.
\\ (2 ^ x \u003d 2 \\) \\ (2 ^ x \u003d \\ FRAC (1) (2) \\)		We transform the second equation using the property of a negative degree ...
\\ (2 ^ x \u003d 2 ^ 1 \\) \\ (2 ^ x \u003d 2 ^ (- 1) \\)		... and exist before the answer.
\\ (x_1 \u003d 1 \\) \\ (x_2 \u003d -1 \\)

Answer : $-1; 1$.

The question remains - how to understand when which method is applied? It comes with experience. In the meantime, you did not work out, use the general recommendation for solving 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:
- a positive number to a degree is zero, for example, \\ (2 ^ x \u003d 0 \\);
- A positive number is to a degree equal to a negative number, for example, \\ (2 ^ x \u003d -4 \\).

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 2 \\); \\ (2 ^ 2 \u003d 4 \\)
\\ (x \u003d 3 \\); \\ (2 ^ 3 \u003d 8 \\).

\\ (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) \\)
\\ (x \u003d -2 \\); \\ (2 ^ (- 2) \u003d \\ FRAC (1) (2 ^ 2) \u003d \\ FRAC (1) (4) \\)
\\ (x \u003d -3 \\); \\ (2 ^ (- 3) \u003d \\ FRAC (1) (2 ^ 3) \u003d \\ FRAC (1) (8) \\)

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 bases

In 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) \\)
\\ (5 ^ (x + 2) \u003d 3 ^ (x + 2) \\)
\\ (15 ^ (2x-1) \u003d (\\ FRAC (1) (7)) ^ (2x-1) \\)

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) \\)
Decision:

\\ (5 ^ (x + 7) \u003d 3 ^ (x + 7) \\)		Here we can not turn out the top five in the top three, nor the opposite (at least without use). So we cannot come to the form \\ (a ^ (f (x)) \u003d a ^ (G (x)) \\). At the same time, the indicators are the same. Let's divide the equation on the right side, that is, on \\ (3 ^ (x + 7) \\) (we can do it, as we know that the top one will not be zero).
\\ (\\ FRAC (5 ^ (x + 7)) (3 ^ (x + 7)) \\) \\ (\u003d \\) \\ (\\ FRAC (3 ^ (x + 7)) (3 ^ (x + 7) ) \\)		Now you remember the property \\ ((\\ FRAC (A) (B)) ^ C \u003d \\ FRAC (A ^ C) (B ^ C) \\) and use it on the left in the opposite direction. To the right we simply cut the fraction.
\\ ((\\ FRAC (5) (3)) ^ (x + 7) \\) \\ (\u003d 1 \\)		It would seem better not. But remember another property of the degree: \\ (A ^ 0 \u003d 1 \\), in other words: "Any number to zero degree is equal to \\ (1 \\)". True and inverse: "The unit can be represented as any number to zero degree." We use this by making the base to the right as the left.
\\ ((\\ FRAC (5) (3)) ^ (x + 7) \\) \\ (\u003d \\) \\ (((\\ FRAC (5) (3)) ^ 0 \\)		Voila! Get rid of the grounds.
		We write an answer.

Answer : $-7$.

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) \\)
Decision:

\\ (7 ^ (2x-4) \u003d (\\ FRAC (1) (3)) ^ (- x + 2) \\)		The equation looks quite sad ... Not only cannot be reduced to the same number (the seven will not be equal to the same \\ (\\ FRAC (1) (3) \\)), so also different indicators ... however, let's in the indicator of the left degree Two.
\\ (7 ^ (2 (x-2)) \u003d (\\ FRAC (1) (3)) ^ (- x + 2) \\)		I remember the property \\ ((a ^ b) ^ c \u003d a ^ (b · c) \\), we convert left: \\ (7 ^ (2 (x-2)) \u003d 7 ^ (2 · (x-2)) \u003d (7 ^ 2) ^ (x - 2) \u003d 49 ^ (x-2) \\).
\\ (49 ^ (x-2) \u003d (\\ FRAC (1) (3)) ^ (- x + 2) \\)		Now, remembering the property of a negative degree \\ (a ^ (- n) \u003d \\ FRAC (1) (a) ^ n \\), we translate right: \\ ((\\ FRAC (1) (3)) ^ (- x + 2) \u003d (3 ^ (- 1)) ^ (- x + 2) \u003d 3 ^ (- 1 (-x + 2)) \u003d 3 ^ (x-2) \\)
\\ (49 ^ (x-2) \u003d 3 ^ (x-2) \\)		Hallelujah! The indicators became the same! Acting the scheme already familiar to us, we decide before the answer.

Answer : $2$.

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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, the number 6 is the basis it always stands downstairs, and the 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.

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