How to solve equations with fractional degrees. Methods for solving indicative equations

Indicatively called equations in which the unknown is contained in an indicator. The simplest indicative equation is: a x \u003d a b, where a\u003e 0, and 1, x - unknown.

The main properties of degrees, with which the indicative equations are converted: a\u003e 0, b\u003e 0.

When solving indicatory equations Enjoy also the following properties indicative function: y \u003d a x, a\u003e 0, a1:

To represent the number of degree, the main logarithmic identity is used: B \u003d, A\u003e 0, A1, B\u003e \u200b\u200b0.

Tasks and tests on the topic "Indicative equations"

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• §2.1. Solution of indicative equations

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• §7 Indicative and logarithmic equations and inequalities - section 5. Indicative and logarithmic functions 10 class

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To successfully solve the indicative equations, you should know the basic properties of degrees, the properties of the indicative function, the main logarithmic identity.

When solving indicative equations, two main methods are used:

1. transition from equation a f (x) \u003d a g (x) to equation f (x) \u003d g (x);
2. introduction new straight lines.

Examples.

1. Equations reduced to the simplest. Resolved by bringing both parts of the equation to the degree with the same base.

3 x \u003d 9 x - 2.

Decision:

3 x \u003d (3 2) x - 2;
3 x \u003d 3 2x - 4;
x \u003d 2x -4;
x \u003d 4.

Answer: 4.

2. Equations solved by making a general multiplier for parentheses.

Decision:

3 x - 3 x - 2 \u003d 24
3 x - 2 (3 2 - 1) \u003d 24
3 x - 2 × 8 \u003d 24
3 x - 2 \u003d 3
x - 2 \u003d 1
x \u003d 3.

Answer: 3.

3. Equations solved by replacing the variable.

Decision:

2 2x + 2 x - 12 \u003d 0
We indicate 2 x \u003d y.
Y 2 + Y - 12 \u003d 0
y 1 \u003d - 4; y 2 \u003d 3.
a) 2 x \u003d - 4.The does not have solutions, because 2 x\u003e 0.
b) 2 x \u003d 3; 2 x \u003d 2 log 2 3; x \u003d log 2 3.

Answer: Log 2 3.

4. Equations containing degrees with two different (not reduced to each other) grounds.

3 × 2 x + 1 - 2 × 5 x - 2 \u003d 5 x + 2 x - 2.

3 × 2 x + 1 - 2 x - 2 \u003d 5 x - 2 × 5 x - 2
2 x - 2 × 23 \u003d 5 x - 2
× 23.
2 x - 2 \u003d 5 x - 2
(5/2) x- 2 \u003d 1
x - 2 \u003d 0
x \u003d 2.

Answer:2.

5. Equations are homogeneous relative to the x and b x.

General form: .

9 x + 4 x \u003d 2.5 × 6 x.

Decision:

3 2x - 2.5 × 2 x × 3 x +2 2x \u003d 0 |: 2 2x\u003e 0
(3/2) 2x - 2.5 × (3/2) x + 1 \u003d 0.
Denote (3/2) x \u003d y.
y 2 - 2.5y + 1 \u003d 0,
y 1 \u003d 2; Y 2 \u003d ½.

Answer: log 3/2 2; - Log 3/2 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 Number 6 is the basis it always stands downstairs, but variable x. degree or indicator.

Let us give more examples of the indicative equations.
2 x * 5 \u003d 10
16 x - 4 x - 6 \u003d 0

Now we will analyze how the demonstration equations are solved?

Take a simple equation:

2 x \u003d 2 3

This example can be solved even in the mind. It can be seen that x \u003d 3. After all, so that the left and right part should be equal to the number 3 instead of x.
Now let's see how it is necessary to issue this decision:

2 x \u003d 2 3
x \u003d 3.

In order to solve such an equation, we removed same grounds (i.e. two) and recorded what remains, it is degrees. Received the desired answer.

Now summarize our decision.

Algorithm for solving an indicative equation:
1. Need to check the same Lee foundations at the equation on the right and left. If the bases are not the same as looking for options for solving this example.
2. After the foundations become the same, equal degrees and solve the resulting new equation.

Now rewrite a few examples:

Let's start with a simple.

The bases in the left and right part are equal to Number 2, which means we can reject and equate their degrees.

x + 2 \u003d 4 It turned out the simplest equation.
x \u003d 4 - 2
x \u003d 2.
Answer: x \u003d 2

IN next example It can be seen that the foundations are different this 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. Number C. the smallest degree We 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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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 common grounds under different numbers) - Very popular technique in the indicative 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. Will have to get out of arsenal another mighty and universal way. 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, autumn?) Quadratic equations Didn't forget 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. Make 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 two through 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 advice:

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 difficult example (solved, truth, in 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. Hint that in this example saves the sedzalka and the most universal rule Solutions of 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.

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.

Get ready for exam testing with "Shkolkovo"!

When repeating the materials passed, many students face the problem of finding the formulas necessary to solve the equations. The school textbook is not always at hand, and the selection of the necessary information on the topic on the Internet takes a long time.

The educational portal "Skolkovo" offers students to take advantage of our knowledge base. We are implementing perfect new method Preparation for final testing. When doing on our website, you can identify gaps in knowledge and pay attention to those assignments that cause the greatest difficulties.

Teachers "Shkolkovo" collected, systematized and outlined all necessary for successful surchase EGE Material as easy and accessible form.

The main definitions and formulas are presented in the "Theoretical Help" section.

For better assimilation of the material, we recommend practicing the tasks. Carefully view examples of exponential equations on this page to understand the calculation algorithm. After that, proceed to perform tasks in the "Catalogs" section. You can start with the easiest tasks or immediately move to solving complex indicative equations with several unknown or. The exercise base on our site is constantly complemented and updated.

Those examples with the indicators that have trouble make it possible to add to favorites. So you can quickly find them and discuss the decision with the teacher.

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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 be used in		reverse side : \\ (\\ 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 \\). \\ ((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 \\) {!LANG-1a3f89cb61279b2939a2c4203ccc17c5!} 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 general recommendation To solve complex tasks, "you do not know what to do - do what you can". That is, look for how you can convert the equation in principle, and try to do it - suddenly what will come out? The main thing about to make only mathematically reasonable transformations. Indicative equations that do not have solutions We will analyze two more situations that are often put in the Student's deadlock: - 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 basesnot be 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 : \(2\).