Equations with negative powers are examples of solutions. Power expressions (expressions with powers) and their transformation. Raising a number to a power

Negative exponentiation is one of the basic elements of mathematics that is often encountered when solving algebraic problems. Below is a detailed instruction.

How to raise to a negative power - theory

When we are a number to the usual power, we multiply its value several times. For example, 3 3 = 3 × 3 × 3 = 27. With a negative fraction, the opposite is true. The general view according to the formula will be as follows: a -n = 1 / a n. Thus, to raise a number to a negative power, you need to divide the unit by the given number, but already in a positive power.

How to raise to a negative power - examples on ordinary numbers

With the above rule in mind, let's solve a few examples.

4 -2 = 1/4 2 = 1/16
Answer: 4 -2 = 1/16

4 -2 = 1/-4 2 = 1/16.
The answer is -4 -2 = 1/16.

But why is the answer in the first and second examples the same? The fact is that when a negative number is raised to an even power (2, 4, 6, etc.), the sign becomes positive. If the degree were even, then the minus remained:

4 -3 = 1/(-4) 3 = 1/(-64)

How to raise to a negative power - numbers from 0 to 1

Recall that when you raise a number in the range from 0 to 1 to a positive power, the value decreases with increasing power. For example, 0.5 2 = 0.25. 0.25< 0,5. В случае с отрицательной степенью все обстоит наоборот. При возведении десятичного (дробного) числа в отрицательную степень, значение увеличивается.

Example 3: Calculate 0.5 -2
Solution: 0.5 -2 = 1/1/2 -2 = 1/1/4 = 1 × 4/1 = 4.
Answer: 0.5 -2 = 4

Analysis (sequence of actions):

• Convert the decimal 0.5 to 1/2. It's easier this way.
  Raise 1/2 to a negative power. 1 / (2) -2. Divide 1 by 1 / (2) 2, we get 1 / (1/2) 2 => 1/1/4 = 4

Example 4: Calculate 0.5 -3
Solution: 0.5 -3 = (1/2) -3 = 1 / (1/2) 3 = 1 / (1/8) = 8

Example 5: Calculate -0.5 -3
Solution: -0.5 -3 = (-1/2) -3 = 1 / (- 1/2) 3 = 1 / (- 1/8) = -8
Answer: -0.5 -3 = -8

Based on the 4th and 5th examples, we will draw several conclusions:

• For a positive number in the range from 0 to 1 (example 4), raised to a negative power, the evenness or oddness of the power is not important, the value of the expression will be positive. Moreover, the greater the degree, the greater the value.
• For a negative number in the range from 0 to 1 (example 5), raised to a negative power, the evenness or oddness of the power does not matter, the value of the expression will be negative. Moreover, the higher the degree, the lower the value.

How to raise to a negative power - a power as a fractional number

Expressions of this type have the following form: a -m / n, where a is an ordinary number, m is the numerator of the degree, n is the denominator of the degree.

Let's consider an example:
Calculate: 8 -1/3

Solution (sequence of actions):

• Remember the rule for raising a number to a negative power. We get: 8 -1/3 = 1 / (8) 1/3.
• Notice that the denominator is 8 as a fractional power. The general view of calculating a fractional power is as follows: a m / n = n √8 m.
• Thus, 1 / (8) 1/3 = 1 / (3 √8 1). We get the cube root of eight, which is 2. Based on this, 1 / (8) 1/3 = 1 / (1/2) = 2.
• Answer: 8 -1/3 = 2

Expressions, expression conversion

Power expressions (expressions with powers) and their conversion

In this article, we will talk about converting power expressions. First, we will focus on transformations that are performed with expressions of any kind, including exponential expressions, such as expanding parentheses, casting similar terms. And then we will analyze the transformations inherent precisely in expressions with powers: working with the base and exponent, using the properties of degrees, etc.

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What are exponential expressions?

The term "exponential expressions" is practically not found in school textbooks of mathematics, but it appears quite often in collections of problems, especially those designed to prepare for the exam and OGE, for example,. After analyzing the tasks in which you need to perform any actions with exponential expressions, it becomes clear that expres- sions are understood as expressions containing degrees in their records. Therefore, for yourself, you can accept the following definition:

Definition.

Power expressions Are expressions containing degrees.

Let us give examples of power expressions... Moreover, we will represent them according to how the development of views on occurs from a degree with a natural indicator to a degree with a real indicator.

As you know, first there is an acquaintance with the power of a number with a natural exponent, at this stage the first simplest power expressions of the type 3 2, 7 5 +1, (2 + 1) 5, (−0,1) 4, 3 a 2 −a + a 2, x 3−1, (a 2) 3, etc.

A little later, the power of a number with an integer exponent is studied, which leads to the appearance of power expressions with negative integer powers, like the following: 3 −2, , a −2 + 2 b −3 + c 2.

In high school, they return to degrees again. There, a degree with a rational exponent is introduced, which entails the appearance of the corresponding power expressions: , , etc. Finally, degrees with irrational indicators and expressions containing them are considered:,.

The matter is not limited to the listed power expressions: the variable penetrates further into the exponent, and, for example, such expressions 2 x 2 +1 or ... And after meeting with, expressions with powers and logarithms begin to occur, for example, x 2 · lgx −5 · x lgx.

So, we figured out the question of what are exponential expressions. Next, we will learn how to transform them.

Basic types of transformations of power expressions

With exponential expressions, you can perform any of the basic identical transformations of expressions. For example, you can expand parentheses, replace numeric expressions with their values, provide similar terms, etc. Naturally, in this case it is necessary to follow the accepted procedure for performing actions. Here are some examples.

Example.

Evaluate the value of the exponential expression 2 3 · (4 2 −12).

Solution.

According to the order of performing the actions, we first perform the actions in brackets. There, firstly, we replace the degree 4 2 with its value 16 (see if necessary), and secondly, we calculate the difference 16−12 = 4. We have 2 3 (4 2 −12) = 2 3 (16−12) = 2 3 4.

In the resulting expression, replace the power 2 3 with its value 8, and then calculate the product 8 4 = 32. This is the desired value.

So, 2 3 (4 2 −12) = 2 3 (16−12) = 2 3 4 = 8 4 = 32.

Answer:

2 3 (4 2 −12) = 32.

Example.

Simplify Power Expressions 3 a 4 b −7 −1 + 2 a 4 b −7.

Solution.

Obviously, this expression contains similar terms 3 · a 4 · b −7 and 2 · a 4 · b −7, and we can bring them:.

Answer:

3 a 4 b −7 −1 + 2 a 4 b −7 = 5 a 4 b −7 −1.

Example.

Imagine an expression with powers as a product.

Solution.

To cope with the task, the representation of the number 9 in the form of a power of 3 2 and the subsequent use of the formula for abbreviated multiplication is the difference of squares:

Answer:

There are also a number of identical transformations inherent in power expressions. Then we will analyze them.

Working with base and exponent

There are degrees, the base and / or exponent of which are not just numbers or variables, but some expressions. As an example, we present the entries (2 + 0.37) 5-3.7 and (a (a + 1) -a 2) 2 (x + 1).

When working with such expressions, you can replace both the expression based on the degree and the expression in the exponent with an identically equal expression on the ODZ of its variables. In other words, we can, according to the rules known to us, separately transform the base of the degree, and separately - the exponent. It is clear that as a result of this transformation, an expression will be obtained that is identically equal to the original one.

Such transformations allow us to simplify expres- sions with powers or achieve other goals we need. For example, in the above exponential expression (2 + 0.3 · 7) 5-3.7, you can perform actions with the numbers in the base and exponent, which will allow you to go to the power 4.1 1.3. And after expanding the parentheses and reducing similar terms in the base of the degree (a (a + 1) −a 2) 2 (x + 1), we get a power expression of a simpler form a 2 (x + 1).

Using power properties

One of the main tools for converting expres- sions with powers is equalities, reflecting. Let us recall the main ones. For any positive numbers a and b and arbitrary real numbers r and s, the following power properties are true:

• a r a s = a r + s;
• a r: a s = a r − s;
• (a b) r = a r b r;
• (a: b) r = a r: b r;
• (a r) s = a r s.

Note that for natural, integer, and also positive exponents, the restrictions on the numbers a and b may not be so strict. For example, for natural numbers m and n, the equality a m a n = a m + n is true not only for positive a, but also for negative ones, and for a = 0.

At school, the main attention when transforming power expressions is focused precisely on the ability to choose a suitable property and apply it correctly. In this case, the bases of degrees are usually positive, which allows using the properties of degrees without restrictions. The same applies to the transformation of expressions containing variables in the bases of degrees - the range of admissible values ​​of variables is usually such that the bases take only positive values ​​on it, which allows you to freely use the properties of degrees. In general, you need to constantly ask yourself whether it is possible in this case to apply any property of degrees, because inaccurate use of properties can lead to a narrowing of the ODV and other troubles. These points are discussed in detail and with examples in the article on conversion of expressions using degree properties. Here we restrict ourselves to a few simple examples.

Example.

Imagine the expression a 2.5 · (a 2) −3: a −5.5 as a power with base a.

Solution.

First, we transform the second factor (a 2) −3 by the property of raising a power to a power: (a 2) −3 = a 2 (−3) = a −6... The original exponential expression will then take the form a 2.5 · a −6: a −5.5. Obviously, it remains to use the properties of multiplication and division of powers with the same base, we have
a 2.5 a -6: a -5.5 =
a 2.5-6: a -5.5 = a -3.5: a -5.5 =
a −3.5 - (- 5.5) = a 2.

Answer:

a 2.5 (a 2) −3: a −5.5 = a 2.

Power properties are used both from left to right and from right to left when transforming exponential expressions.

Example.

Find the value of the exponential expression.

Solution.

Equality (a b) r = a r b r, applied from right to left, allows you to go from the original expression to the product of the form and further. And when multiplying degrees with the same bases, the indicators add up: .

It was possible to perform the transformation of the original expression in another way:

Answer:

.

Example.

Given the exponential expression a 1.5 −a 0.5 −6, enter the new variable t = a 0.5.

Solution.

The degree a 1.5 can be represented as a 0.5 · 3 and further, based on the property of the degree to the degree (a r) s = a r · s, applied from right to left, transform it to the form (a 0.5) 3. Thus, a 1.5 −a 0.5 −6 = (a 0.5) 3 −a 0.5 −6... Now it is easy to introduce a new variable t = a 0.5, we get t 3 −t − 6.

Answer:

t 3 −t − 6.

Converting fractions containing powers

Power expressions can contain fractions with powers or be such fractions. Any of the basic transformations of fractions that are inherent in fractions of any kind are fully applicable to such fractions. That is, fractions that contain powers can be canceled, reduced to a new denominator, worked separately with their numerator and separately with the denominator, etc. To illustrate the words spoken, consider the solutions of several examples.

Example.

Simplify exponential expression .

Solution.

This exponential expression is a fraction. Let's work with its numerator and denominator. In the numerator, we open the brackets and simplify the expression obtained after that using the properties of the powers, and in the denominator we give similar terms:

And we also change the sign of the denominator by placing a minus in front of the fraction: .

Answer:

.

The reduction of fractions containing powers to a new denominator is carried out similarly to the reduction of rational fractions to a new denominator. In this case, an additional factor is also found and the numerator and denominator of the fraction are multiplied by it. When performing this action, it is worth remembering that reduction to a new denominator can lead to a narrowing of the ODV. To prevent this from happening, it is necessary that the additional factor does not vanish for any values ​​of the variables from the ODZ variables for the original expression.

Example.

Reduce fractions to a new denominator: a) to the denominator a, b) to the denominator.

Solution.

a) In this case, it is quite easy to figure out which additional factor helps to achieve the desired result. This is a factor of a 0.3, since a 0.7 · a 0.3 = a 0.7 + 0.3 = a. Note that on the range of permissible values ​​of the variable a (this is the set of all positive real numbers) the degree a 0.3 does not vanish, therefore, we have the right to multiply the numerator and denominator of the given fraction by this additional factor:

b) Looking more closely at the denominator, you can find that

and multiplying this expression by will give the sum of the cubes and, that is,. And this is the new denominator to which we need to reduce the original fraction.

This is how we found an additional factor. On the range of valid values ​​of the variables x and y, the expression does not vanish, therefore, we can multiply the numerator and denominator of the fraction by it:

Answer:

a) , b) .

The reduction of fractions containing powers is also nothing new: the numerator and denominator are represented as a number of factors, and the same factors of the numerator and denominator are canceled.

Example.

Reduce the fraction: a) , b).

Solution.

a) First, the numerator and denominator can be reduced by the numbers 30 and 45, which is 15. Also, obviously, one can perform a reduction by x 0.5 +1 and by ... Here's what we have:

b) In this case, the same factors in the numerator and denominator are not immediately visible. To get them, you will have to perform preliminary transformations. In this case, they consist in factoring the denominator into factors according to the formula for the difference of squares:

Answer:

a)

b) .

Reducing fractions to a new denominator and reducing fractions is mainly used to perform actions with fractions. Actions are performed according to known rules. When adding (subtracting) fractions, they are brought to a common denominator, after which the numerators are added (subtracted), and the denominator remains the same. The result is a fraction, the numerator of which is the product of the numerators, and the denominator is the product of the denominators. Division by a fraction is multiplication by the inverse of the fraction.

Example.

Follow the steps .

Solution.

First, we subtract the fractions in parentheses. To do this, we bring them to a common denominator, which is , after which we subtract the numerators:

Now we multiply the fractions:

Obviously, it is possible to cancel by a power of x 1/2, after which we have .

You can also simplify the exponential expression in the denominator by using the difference of squares formula: .

Answer:

Example.

Simplify exponential expression .

Solution.

Obviously, this fraction can be canceled by (x 2.7 +1) 2, this gives the fraction ... It is clear that something else needs to be done with the degrees of x. To do this, we transform the resulting fraction into a product. This gives us the opportunity to use the property of dividing degrees with the same bases: ... And at the end of the process, we pass from the last product to a fraction.

Answer:

.

And we also add that it is possible and in many cases desirable to transfer multipliers with negative exponents from the numerator to the denominator or from the denominator to the numerator, changing the sign of the exponent. Such transformations often simplify further actions. For example, an exponential expression can be replaced with.

Converting expressions with roots and powers

Often in expressions in which some transformations are required, along with powers with fractional exponents, roots are also present. To transform such an expression to the desired form, in most cases it is enough to go only to the roots or only to the powers. But since it is more convenient to work with degrees, they usually go from roots to degrees. However, it is advisable to carry out such a transition when the ODV of variables for the original expression allows you to replace the roots with powers without the need to refer to the module or split the ODV into several intervals (we analyzed this in detail in the article the transition from roots to powers and back. a degree with an irrational indicator is introduced, which makes it possible to talk about a degree with an arbitrary real indicator. exponential function, which is analytically set by the degree, at the base of which is the number, and in the indicator - the variable. So we are faced with exponential expressions containing numbers in the base of the degree, and in the exponent - expressions with variables, and naturally there is a need to perform transformations of such expressions.

It should be said that the transformation of expressions of this type usually has to be performed when solving exponential equations and exponential inequalities and these conversions are pretty simple. In the overwhelming majority of cases, they are based on the properties of the degree and are mainly aimed at introducing a new variable in the future. We can demonstrate them by the equation 5 2 x + 1 −3 5 x 7 x −14 7 2 x − 1 = 0.

First, the degrees in which the sum of a variable (or expressions with variables) and a number is found are replaced by products. This applies to the first and last terms of the expression on the left:
5 2 x 5 1 −3 5 x 7 x −14 7 2 x 7 −1 = 0,
5 5 2 x −3 5 x 7 x −2 7 2 x = 0.

Further, both sides of the equality are divided by the expression 7 2 x, which takes only positive values ​​on the ODZ of the variable x for the original equation (this is a standard technique for solving equations of this kind, we are not talking about it now, so focus on the subsequent transformations of expressions with powers ):

Fractions with powers are now canceled, which gives .

Finally, the ratio of degrees with the same exponents is replaced by the degrees of relations, which leads to the equation which is equivalent ... The performed transformations allow us to introduce a new variable, which reduces the solution of the original exponential equation to the solution of the quadratic equation

Lesson and presentation on the topic: "Degree with a negative indicator. Definition and examples of problem solving"

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Determination of the degree with a negative exponent

We well know that any number in the zero degree is equal to one. $ a ^ 0 = 1 $, $ a ≠ 0 $.
The question arises, what will happen if the number is raised to a negative power? For example, what is the number $ 2 ^ (- 2) $ equal to?
The first mathematicians who asked this question decided that reinventing the wheel was not worth it, and it was good that all the properties of the degrees remained the same. That is, when multiplying degrees with the same base, the exponents are added.
Let's consider this case: $ 2 ^ 3 * 2 ^ (- 3) = 2 ^ (3-3) = 2 ^ 0 = 1 $.
We got that the product of such numbers should give one. The unit in the product is obtained by multiplying the reciprocal numbers, that is, $ 2 ^ (- 3) = \ frac (1) (2 ^ 3) $.

This reasoning led to the following definition.
Definition. If $ n $ is a natural number and $ a ≠ 0 $, then the equality holds: $ a ^ (- n) = \ frac (1) (a ^ n) $.

An important identity that is often used: $ (\ frac (a) (b)) ^ (- n) = (\ frac (b) (a)) ^ n $.
In particular, $ (\ frac (1) (a)) ^ (- n) = a ^ n $.

Solution examples

Solution.
Let's consider each term separately.
1. $ 2 ^ (- 3) = \ frac (1) (2 ^ 3) = \ frac (1) (2 * 2 * 2) = \ frac (1) (8) $.
2. $ (\ frac (2) (5)) ^ (- 2) = (\ frac (5) (2)) ^ 2 = \ frac (5 ^ 2) (2 ^ 2) = \ frac (25) (4) $.
3. $ 8 ^ (- 1) = \ frac (1) (8) $.
It remains to perform addition and subtraction operations: $ \ frac (1) (8) + \ frac (25) (4) - \ frac (1) (8) = \ frac (25) (4) = 6 \ frac (1) (4) $.
Answer: $ 6 \ frac (1) (4) $.

Example 2.
Represent a given number as a prime power $ \ frac (1) (729) $.

Solution.
Obviously, $ \ frac (1) (729) = 729 ^ (- 1) $.
But 729 is not a prime number ending in 9. It can be assumed that this number is a power of three. Let us sequentially divide 729 by 3.
1) $ \ frac (729) (3) = 243 $;
2) $ \ frac (243) (3) = 81 $;
3) $ \ frac (81) (3) = 27 $;
4) $ \ frac (27) (3) = 9 $;
5) $ \ frac (9) (3) = 3 $;
6) $ \ frac (3) (3) = 1 $.
Six operations have been performed, which means: $ 729 = 3 ^ 6 $.
For our task:
$729^{-1}=(3^6)^{-1}=3^{-6}$.
Answer: $ 3 ^ (- 6) $.

Example 3. Present the expression as a power: $ \ frac (a ^ 6 * (a ^ (- 5)) ^ 2) ((a ^ (- 3) * a ^ 8) ^ (- 1)) $.
Solution. The first action is always performed inside the parentheses, then the multiplication $ \ frac (a ^ 6 * (a ^ (- 5)) ^ 2) ((a ^ (- 3) * a ^ 8) ^ (- 1)) = \ frac (a ^ 6 * a ^ (- 10)) ((a ^ 5) ^ (- 1)) = \ frac (a ^ ((- 4))) (a ^ ((- 5))) = a ^ (-4 - (- 5)) = a ^ (- 4 + 5) = a $.
Answer: $ a $.

Example 4. Prove the identity:
$ (\ frac (y ^ 2 (xy ^ (- 1) -1) ^ 2) (x (1 + x ^ (- 1) y) ^ 2) * \ frac (y ^ 2 (x ^ (- 2 ) + y ^ (- 2))) (x (xy ^ (- 1) + x ^ (- 1) y))): \ frac (1-x ^ (- 1) y) (xy ^ (- 1 ) +1) = \ frac (xy) (x + y) $.

Solution.
On the left, we will consider each factor in parentheses separately.
1. $ \ frac (y ^ 2 (xy ^ (- 1) -1) ^ 2) (x (1 + x ^ (- 1) y) ^ 2) = \ frac (y ^ 2 (\ frac (x ) (y) -1) ^ 2) (x (1+ \ frac (y) (x)) ^ 2) = \ frac (y ^ 2 (\ frac (x ^ 2) (y ^ 2) -2 \ frac (x) (y) +1)) (x (1 + 2 \ frac (y) (x) + \ frac (y ^ 2) (x ^ 2))) = \ frac (x ^ 2-2xy + y ^ 2) (x + 2y + \ frac (y ^ 2) (x)) = \ frac (x ^ 2-2xy + y ^ 2) (\ frac (x ^ 2 + 2xy + y ^ 2) (x) ) = \ frac (x (x ^ 2-2xy + y ^ 2)) ((x ^ 2 + 2xy + y ^ 2)) $.
2. $ \ frac (y ^ 2 (x ^ (- 2) + y ^ (- 2))) (x (xy ^ (- 1) + x ^ (- 1) y)) = \ frac (y ^ 2 (\ frac (1) (x ^ 2) + \ frac (1) (y ^ 2))) (x (\ frac (x) (y) + \ frac (y) (x))) = \ frac (\ frac (y ^ 2) (x ^ 2) +1) (\ frac (x ^ 2) (y) + y) = \ frac (\ frac (y ^ 2 + x ^ 2) (x ^ 2) ) ((\ frac (x ^ 2 + y ^ 2) (y))) = \ frac (y ^ 2 + x ^ 2) (x ^ 2) * \ frac (y) (x ^ 2 + y ^ 2 ) = \ frac (y) (x ^ 2) $.
3. $ \ frac (x (x ^ 2-2xy + y ^ 2)) ((x ^ 2 + 2xy + y ^ 2)) * \ frac (y) (x ^ 2) = \ frac (y (x ^ 2-2xy + y ^ 2)) (x (x ^ 2 + 2xy + y ^ 2)) = \ frac (y (xy) ^ 2) (x (x + y) ^ 2) $.
4. Let's move on to the fraction by which we divide.
$ \ frac (1-x ^ (- 1) y) (xy ^ (- 1) +1) = \ frac (1- \ frac (y) (x)) (\ frac (x) (y) +1 ) = \ frac (\ frac (xy) (x)) (\ frac (x + y) (y)) = \ frac (xy) (x) * \ frac (y) (x + y) = \ frac ( y (xy)) (x (x + y)) $.
5. Let's do the division.
$ \ frac (y (xy) ^ 2) (x (x + y) ^ 2): \ frac (y (xy)) (x (x + y)) = \ frac (y (xy) ^ 2) ( x (x + y) ^ 2) * \ frac (x (x + y)) (y (xy)) = \ frac (xy) (x + y) $.
We got the correct identity, which was required to prove.

At the end of the lesson, we will once again write down the rules for operating with powers, here the exponent is an integer.
$ a ^ s * a ^ t = a ^ (s + t) $.
$ \ frac (a ^ s) (a ^ t) = a ^ (s-t) $.
$ (a ^ s) ^ t = a ^ (st) $.
$ (ab) ^ s = a ^ s * b ^ s $.
$ (\ frac (a) (b)) ^ s = \ frac (a ^ s) (b ^ s) $.

Tasks for independent solution

In this article we will figure out what is degree of... Here we will give definitions of the degree of a number, while considering in detail all possible exponents, starting with a natural exponent, ending with an irrational one. In the material you will find a lot of examples of degrees, covering all the subtleties that arise.

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Degree with natural exponent, square of number, cube of number

Let's start with. Looking ahead, we say that the definition of the degree of a number a with natural exponent n is given for a, which we will call basis degree, and n, which we will call exponent... Also note that the degree with a natural exponent is determined through the product, so to understand the material below, you need to have an idea of ​​the multiplication of numbers.

Definition.

Power of number a with natural exponent n is an expression of the form a n, the value of which is equal to the product of n factors, each of which is equal to a, that is,.
In particular, the power of a number a with exponent 1 is the number a itself, that is, a 1 = a.

Immediately it is worth mentioning the rules for reading degrees. The universal way to read a record a n is as follows: "a to the power of n". In some cases, the following options are also acceptable: "a to the n-th power" and "n-th power of the number a". For example, take the power of 8 12, which is "eight to the power of twelve" or "eight to the twelfth degree" or "the twelfth power of eight".

The second degree of a number, as well as the third degree of a number, have their own names. The second power of a number is called square number for example, 7 2 reads “seven squared” or “the square of the number seven”. The third power of a number is called cube numbers for example, 5 3 can be read as "cube five" or say "cube of number 5".

It's time to lead examples of degrees with natural indicators... Let's start with the power of 5 7, here 5 is the base of the power, and 7 is the exponent. Let's give another example: 4.32 is the base, and the natural number 9 is the exponent (4.32) 9.

Please note that in the last example, the base of the 4.32 degree is written in parentheses: to avoid confusion, we will put in parentheses all bases of the degree that are different from natural numbers. As an example, we give the following degrees with natural indicators , their bases are not natural numbers, so they are written in parentheses. Well, for complete clarity in this moment, we will show the difference between the entries of the form (−2) 3 and −2 3. The expression (−2) 3 is the power of −2 with a natural exponent of 3, and the expression −2 3 (it can be written as - (2 3)) corresponds to the number, the value of the power 2 3.

Note that there is a notation for the degree of a number a with exponent n of the form a ^ n. Moreover, if n is a multivalued natural number, then the exponent is taken in parentheses. For example, 4 ^ 9 is another notation for the power of 4 9. And here are some more examples of writing degrees using the "^" symbol: 14 ^ (21), (−2,1) ^ (155). In what follows, we will mainly use the notation of the degree of the form a n.

One of the tasks, the inverse of raising to a power with a natural exponent, is the problem of finding the base of a degree from a known value of the degree and a known exponent. This task leads to.

It is known that the set of rational numbers consists of integers and fractional numbers, and each fractional number can be represented as a positive or negative ordinary fraction. We defined the degree with an integer exponent in the previous paragraph, therefore, in order to complete the definition of the degree with a rational exponent, you need to give the meaning of the degree of a number a with a fractional exponent m / n, where m is an integer and n is a natural number. Let's do it.

Consider a degree with a fractional exponent of the form. For the property of degree to degree to be valid, the equality must be fulfilled ... If we take into account the obtained equality and the way we determined it, then it is logical to accept, provided that for the given m, n and a, the expression makes sense.

It is easy to check that for all properties of a degree with an integer exponent (this is done in the section on properties of a degree with a rational exponent).

The above reasoning allows us to do the following. output: if for the given m, n and a the expression makes sense, then the power of the number a with the fractional exponent m / n is the nth root of a to the power of m.

This statement brings us very close to determining the degree with a fractional exponent. It remains only to describe for which m, n and a the expression makes sense. There are two main approaches depending on the constraints on m, n and a.

The easiest way is to restrict a by assuming a≥0 for positive m and a> 0 for negative m (since for m≤0 the degree 0 m is not defined). Then we get the following definition of a fractional exponent.

Definition.

The power of a positive number a with a fractional exponent m / n, where m is an integer and n is a natural number, is called the nth root of a to the power of m, that is,.

A fractional power of zero is also determined with the only proviso that the indicator must be positive.

Definition.

Power of zero with positive fractional exponent m / n, where m is a positive integer and n is a natural number, is defined as .
When the degree is not determined, that is, the degree of a number zero with a fractional negative exponent does not make sense.

It should be noted that with such a definition of a degree with a fractional exponent, there is one nuance: for some negative a and some m and n, the expression makes sense, and we discarded these cases by introducing the condition a≥0. For example, it makes sense to write or, and the definition given above forces us to say that degrees with a fractional exponent of the form do not make sense, since the base should not be negative.

Another approach to determining the exponent with a fractional exponent m / n is to consider separately the odd and even exponents of the root. This approach requires an additional condition: the degree of the number a, the indicator of which is, is considered the power of the number a, the indicator of which is the corresponding irreducible fraction (the importance of this condition will be explained below). That is, if m / n is an irreducible fraction, then for any natural number k, the degree is preliminarily replaced by.

For even n and positive m, the expression makes sense for any non-negative a (an even root of a negative number does not make sense), for negative m, the number a must still be nonzero (otherwise there will be division by zero). And for odd n and positive m, the number a can be any (an odd root is defined for any real number), and for negative m, the number a must be nonzero (so that there is no division by zero).

The above reasoning leads us to such a definition of the degree with a fractional exponent.

Definition.

Let m / n be an irreducible fraction, m an integer, and n a natural number. For any cancellable fraction, the exponent is replaced by. The power of a number with an irreducible fractional exponent m / n is for



Let us explain why a degree with a reducible fractional exponent is previously replaced by a degree with an irreducible exponent. If we simply defined the degree as, and did not make a reservation about the irreducibility of the fraction m / n, then we would be faced with situations similar to the following: since 6/10 = 3/5, then the equality should hold , but , a .

In one of the previous articles, we already mentioned the degree of number. Today we will try to orient ourselves in the process of finding its meaning. Scientifically speaking, we will be figuring out how to properly raise to a power. We will figure out how this process is carried out, at the same time we will touch upon all possible indicators of degree: natural, irrational, rational, whole.

So let's take a closer look at the solutions of the examples and find out what it means:

1. Definition of the concept.
2. Elevation to negative art.
3. Whole indicator.
4. Raising a number to an irrational power.

Definition of the concept

Here is a definition that accurately reflects the meaning: "Exponentiation is the definition of the meaning of the power of a number."

Accordingly, raising the number a to Art. r and the process of finding the value of the exponent a with exponent r are identical concepts. For example, if the task is to calculate the value of the power (0.6) 6 ″, then it can be simplified to the expression “Raise the number 0.6 to the power of 6”.

After that, you can proceed directly to the construction rules.

Negative exponentiation

For clarity, you should pay attention to the following chain of expressions:

110 = 0.1 = 1 * 10 in minus 1 st.,

1100 = 0.01 = 1 * 10 in minus 2 steps.,

11000 = 0.0001 = 1 * 10 minus 3 st.,

110000 = 0.00001 = 1 * 10 in minus 4 degrees.

Thanks to these examples, you can clearly see the ability to instantly calculate 10 to any minus power. For this purpose, it is quite corny to shift the decimal component:

• 10 to -1 degrees - before the unit 1 zero;
• at -3 - three zeros before one;
• in -9 is 9 zeros and so on.

It is just as easy to understand according to this scheme, how much will be 10 to minus 5 tbsp. -

1100000=0,000001=(1*10)-5.

How to raise a natural number

Recalling the definition, we take into account that the natural number a in Art. n is equal to the product of n factors, each of which is equal to a. To illustrate: (a * a * ... a) n, where n is the number of numbers that are multiplied. Accordingly, in order to raise a to n, it is necessary to calculate the product of the following form: a * a *… and divide by n times.

From this it becomes obvious that erection in natural art. relies on the ability to multiply(This material is covered in the section on multiplying real numbers). Let's take a look at the problem:

Erect -2 in the 4th st.

We are dealing with a natural indicator. Accordingly, the course of the decision will be as follows: (-2) in art. 4 = (-2) * (- 2) * (- 2) * (- 2). Now it remains only to carry out the multiplication of whole numbers: (- 2) * (- 2) * (- 2) * (- 2). We get 16.

Answer to the problem:

(-2) in art. 4 = 16.

Example:

Calculate the value: three point two sevens squared.

This example is equal to the following product: three point two sevenths multiplied by three point two sevenths. Remembering how the multiplication of mixed numbers is carried out, we complete the construction:

• 3 point 2 sevenths multiply by themselves;
• equals 23 sevenths multiplied by 23 sevenths;
• equal to 529 forty-ninth;
• we cut and get 10 thirty-nine forty-nine.

Answer: 10 39/49

With regard to the issue of raising to an irrational indicator, it should be noted that calculations begin to be carried out after the completion of the preliminary rounding of the basis of the degree to any category that would allow obtaining a value with a given accuracy. For example, we need to square the number P (pi).

We start by rounding P to hundredths and get:

P squared = (3.14) 2 = 9.8596. However, if we reduce P to ten thousandths, we get P = 3.14159. Then squaring gets a completely different number: 9.8695877281.

It should be noted here that in many problems there is no need to raise irrational numbers to a power. As a rule, the answer is written either in the form of a degree, for example, the root of 6 to the power of 3, or, if the expression allows, its transformation is carried out: root of 5 in power 7 = 125 root of 5.

How to raise a number to a whole power

This algebraic manipulation is appropriate take into account for the following cases:

• for whole numbers;
• for a zero indicator;
• for a whole positive indicator.

Since practically all positive integers coincide with the mass of natural numbers, then setting to a positive integer power is the same process as setting in Art. natural. We described this process in the previous paragraph.

Now let's talk about calculating Art. null. We have already found out above that the zero degree of the number a can be determined for any nonzero a (real), while a in Art. 0 will equal 1.

Accordingly, raising any real number to zero st. will give one.

For example, 10 in st. 0 = 1, (-3.65) 0 = 1, and 0 in st. 0 cannot be determined.

In order to complete the raising to an integer power, it remains to decide on the options for integer negative values. We remember that Art. from a with integer exponent -z will be defined as a fraction. The denominator of the fraction is Art. with a positive integer value, the value of which we have already learned to find. Now it remains only to consider an example of construction.

Example:

Calculate the value of the number 2 in a cube with a negative integer exponent.

Solution process:

According to the definition of a degree with a negative indicator, we denote: two at minus 3 tbsp. equals one to two in the third degree.

The denominator is calculated simply: two cubed;

3 = 2*2*2=8.

Answer: two in minus 3rd tbsp. = one eighth.

Video

This video will show you what to do if the degree is negative.