The equation containing the variable in the indicator is called. Solution of indicative equations. Examples

Solution of indicative equations. Examples.

Attention!
This topic has additional
Materials in a special section 555.
For those who are strongly "not very ..."
And for those who are "very ...")

What indicative equation? This equation in which unknown (Xers) and expressions with them are in indicators Some degrees. And only there! It is important.

There you are examples of indicative equations:

3 x · 2 x \u003d 8 x + 3

Note! In the grounds of degrees (below) - only numbers. IN indicators Degnese (at the top) - a wide variety of expressions with Xa. If, suddenly, the EX will come out in the equation somewhere, except for the indicator, for example:

it will already be a mixed type equation. Such equations do not have clear rules for solutions. We will not consider them yet. Here we will deal with by solving exponential equations in pure form.

In fact, even clean indicative equations are clearly solved far away. But there are certain types of indicative equations that can be solved and needed. Here are these types we will look at.

The solution of the simplest indicative equations.

To begin with, I decide something completely elementary. For example:

Even without any theories, it is clear to the simple selection that x \u003d 2. More, right, right!? No other value of the ICA rolls. And now we look at the record of the solution of this cunning indicative equation:

What did we do? In fact, we simply threw the same bases (three). They completely thrown away. And what pleases, got to the point!

Indeed, if in the indicative equation on the left and right the same Numbers in any degrees, these numbers can be removed and equated degrees. Mathematics allows. It remains to be expensive a much simpler equation. Great, right?)

However, remember Iron: you can remove the bases only when the left and right of the ground is in proud loneliness! Without any neighbors and coefficients. Say, in equations:

2 x +2 x + 1 \u003d 2 3, or

double can not be removed!

Well, the most important thing we have mastered. How to move from evil indicative expressions to simpler equations.

"That's the times!" - You will say. "Who will give such a primitive on the control and exams!?"

Forced to agree. No one will give. But now you know where to strive when solving free examples. It is necessary to bring it to the form when on the left - the same number is the same number. Further everything will be easier. Actually, this is the classic of mathematics. Take the original example and convert it to the desired us View. According to the rules of mathematics, of course.

Consider examples that require some additional efforts to bring them to the simplest. Let's call them simple indicative equations.

Solution of simple indicative equations. Examples.

When solving indicative equations, the main rules - actions with degrees. Without knowledge of these actions, nothing will work.

To actions with degrees it is necessary to add personal observation and smelting. We need the same foundations? Here we are looking for them in an example in a clear or encrypted form.

Let's see how this is done in practice?

Let us give us an example:

2 2x - 8 x + 1 \u003d 0

First angry look - on basis. They ... they are different! Two and eight. But to fall into the despondency - early. It's time to remember that

Two and eight - relative to the degree.) It is possible to write:

8 x + 1 \u003d (2 3) x + 1

If you recall the formula from action with degrees:

(a n) m \u003d a nm,

that generally it turns out:

8 x + 1 \u003d (2 3) x + 1 \u003d 2 3 (x + 1)

The initial example began to look like this:

2 2x - 2 3 (x + 1) \u003d 0

Transfer 2 3 (x + 1) To the right (nobody canceled elementary actions of mathematics!), We get:

2 2x \u003d 2 3 (x + 1)

Here, almost, and that's it. We remove the foundations:

Solve this monster and get

This is the correct answer.

In this example, we reinstate the knowledge of detects of two. we identified In the eight of the encrypted two. This technique (encryption of general bases under different numbers) is a very popular technique in the lower equations! Yes, and in logarithms too. It is necessary to be able to learn in the numbers of other numbers. This is extremely important for solving indicative equations.

The fact is that to build any number to any degree is not a problem. Multiply, even on a piece of paper, and that's it. For example, to build 3 to the fifth degree will be able to each. 243 It turns out if you know the multiplication table.) But in the lower equations, it is much more likely to not be erected, but on the contrary ... to find out what number to what extent Hiding for a number 243, or, say, 343 ... Here you will not help any calculator.

The degree of some numbers should be known in the face, yes ... do it?

To determine what degrees and what numbers are numbers:

2; 8; 16; 27; 32; 64; 81; 100; 125; 128; 216; 243; 256; 343; 512; 625; 729, 1024.

Answers (in disarray, natural!):

5 4 ; 2 10 ; 7 3 ; 3 5 ; 2 7 ; 10 2 ; 2 6 ; 3 3 ; 2 3 ; 2 1 ; 3 6 ; 2 9 ; 2 8 ; 6 3 ; 5 3 ; 3 4 ; 2 5 ; 4 4 ; 4 2 ; 2 3 ; 9 3 ; 4 5 ; 8 2 ; 4 3 ; 8 3 .

If you look closely, you can see a strange fact. Replies are significantly more than tasks! Well, it happens ... for example, 2 6, 4 3, 8 2 is all 64.

Suppose you took note of the information about the acquaintance with numbers.) Let's remind you that to solve the indicative equations apply all The stock of mathematical knowledge. Including from the junior middle classes. You do not immediately go to senior classes, right?)

For example, when solving the indicative equations, the total multiplier of the brackets helps very often (hello grade 7!). Watch the following man:

3 2x + 4 -11 · 9 x \u003d 210

And again, first glance - on the ground! The foundations in degrees are different ... Troika and nine. And we want to be the same. Well, in this case, the desire is fulfilled!) Because:

9 x \u003d (3 2) x \u003d 3 2x

According to the same rules of action with degrees:

3 2x + 4 \u003d 3 2x · 3 4

So great, you can write:

3 2x · 3 4 - 11 · 3 2x \u003d 210

We led an example to the same reasons. So, what is next!? Troika can not throw out ... deadlock?

Not at all. Remember the most universal and powerful solution rule all Mathematical tasks:

You do not know what you need - do what you can!

You look, everything is formed).

What is in this indicative equation can do it? Yes, on the left side, it is directly asking for a bracket! The total multiplier of 3 2x clearly hints at it. Let's try, and then it will be visible:

3 2x (3 4 - 11) \u003d 210

3 4 - 11 = 81 - 11 = 70

The example is becoming better and better!

We remember that in order to eliminate grounds, we need a clean degree, without any coefficients. US number 70 interferes. So we divide both parts of the equation by 70, we get:

Op-Pa! Everything and settled!

This is the final answer.

It happens, however, that breaking on the same bases is obtained, but their liquidation is in any way. This happens in the indicative equations of another type. We will master this type.

Replacing the variable in solving indicative equations. Examples.

Resolving equation:

4 x - 3 · 2 x +2 \u003d 0

First - as usual. Go to one base. To twice.

4 x \u003d (2 2) x \u003d 2 2x

We get the equation:

2 2x - 3 · 2 x +2 \u003d 0

And here it will be dependent. Previous techniques will not work, no matter how sprinkling. We'll have to get another mighty and universal way from arsenal. Called O. replacing the variable.

The essence of the method is easy to surprise. Instead of one complex icon (in our case - 2 x) we write another, simpler (for example - t). This, it would seem, a meaningless replacement leads to awesome results!) Just everything becomes clear and understandable!

So, let

Then 2 2x \u003d 2 x2 \u003d (2 x) 2 \u003d T 2

We replace in our equation all degrees with cavities on T:

Well, insens?) Square equations have not forgotten yet? We decide through the discriminant, we get:

Here, most importantly, do not stop, as it happens ... This is not a response, we are needed, and not t. We return to the ICCAM, i.e. We make a replacement. First for T 1:

That is,

One root found. We are looking for the second, from T 2:

Um ... left 2 x, right 1 ... no problem? Yes no! Enough to remember (from action with degrees, yes ...) that one is anyone Number to zero degree. Any. What you need, and put it. We need a two. So:

Now everything is. Received 2 roots:

This is the answer.

For solving indicative equations In the end sometimes it turns out some inconvenient expression. Type:

From the seven deuce through a simple degree does not work. Do not relatives they ... how to be here? Someone, maybe confused ... And here is a person who read the topic on this site "What is a logarithm?" , only Skupo smile and will make a solid right answer to the hard hand:

There may be no such an answer in the tasks "in". There is a specific number required. But in the tasks "C" - easily.

In this lesson, examples of solving the most common indicative equations are given. We highlight the main one.

Practical tips:

1. The first thing we look at basis degrees. We think whether it is impossible to make them same. Try to do it, actively using actions with degrees. Do not forget that the numbers without ICs can also be turned into a degree!

2. We try to bring the indicative equation to the form when the left and right are the same Numbers in any degrees. Using actions with degrees and factorization.What can I consider in numbers - believe.

3. If the second board did not work, we try to apply the replacement of the variable. As a result, an equation can turn out that is easily solved. Most often - square. Or fractional, which also comes down to square.

4. To successfully solve the indicative equations, it is necessary to know the degree of some numbers "in the face".

As usual, at the end of the lesson you are offered to clean a little.) Alone. From simple - to complex.

Decide indicative equations:

More complied with:

2 x + 3 - 2 x + 2 - 2 x \u003d 48

9 x - 8 · 3 x \u003d 9

2 x - 2 0,5x + 1 - 8 \u003d 0

Find the product of the roots:

2 3 + 2 x \u003d 9

Happened?

Well, then the most complex example (solved, however, in the mind ...):

7 0.13x + 13 0.7x + 1 + 2 0,5x + 1 \u003d -3

What is more interesting? Then you have an evil example. It is quite pulling on increased difficulty. Nickname that in this example savings savings and the most universal rule of solving all mathematical tasks.)

2 5x-1 · 3 3x-1 · 5 2x-1 \u003d 720 x

Example simpler, for rest):

9 · 2 x - 4 · 3 x \u003d 0

And for dessert. Find the number of roots equation:

x · 3 x - 9x + 7 · 3 x - 63 \u003d 0

Yes Yes! This is a mixed type equation! Which we did not consider in this lesson. And what to consider them, it is necessary to solve!) This lesson is quite enough to solve the equation. Well, the cutter is needed ... and let it help you with the seventh class (this is a hint!).

Answers (in disorder, through a comma point):

one; 2; 3; four; no solutions; 2; -2; -five; four; 0.

All successful? Excellent.

There is a problem? No problem! In a special section 555, all these indicative equations are solved with detailed explanations. What, why, and why. And, of course, there is additional valuable information on working with all sorts of indicative equations. Not only with these.)

The last fun question for a consideration. In this lesson, we worked with accurate equations. Why didn't I say a word here about OTZ? In equations, this is a very important thing, by the way ...

If you like this site ...

By the way, I have another couple of interesting sites for you.)

It can be accessed in solving examples and find out your level. Testing with instant check. Learn - with interest!)

You can get acquainted with features and derivatives.

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.

{!LANG-32d9e9f16c699aabf977b037dfd1d2fd!}

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 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:

{!LANG-93ec33fb5380d8325169a9196e5d47eb!}

{!LANG-6f3783030a206c981b03424bbb936b7b!}

\\ [\\ 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 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.

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.

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

Example \\ (4 ^ (x + 0.5) -5 · 2 ^ x + 2 \u003d 0 \\)
Decision:

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 \\).		{!LANG-dfa53a3c948beada20a38f677927ec84!}
\\ ((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 : $2$.

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 points when passing the exam in mathematics.

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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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