3 to varying degrees. Raising to an irrational degree. Maximum signed number

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First, let's recall the basic formulas of degrees and their properties.

Product of a number a happens on itself n times, we can write this expression as a a … a=a n

1. a 0 = 1 (a ≠ 0)

3. a n a m = a n + m

4. (a n) m = a nm

5. a n b n = (ab) n

7. a n / a m \u003d a n - m

Power or exponential equations- these are equations in which the variables are in powers (or exponents), and the base is a number.

Examples of exponential equations:

In this example, the number 6 is the base, it is always at the bottom, and the variable x degree or measure.

Let us give more examples of exponential equations.
2 x *5=10
16x-4x-6=0

Now let's look at how exponential equations are solved?

Let's take a simple equation:

2 x = 2 3

Such an example can be solved even in the mind. It can be seen that x=3. After all, in order for the left and right sides to be equal, you need to put the number 3 instead of x.
Now let's see how this decision should be made:

2 x = 2 3
x = 3

To solve this equation, we removed same grounds(that is, deuces) and wrote down what was left, these are degrees. We got the answer we were looking for.

Now let's summarize our solution.

Algorithm for solving the exponential equation:
1. Need to check the same whether the bases of the equation on the right and on the left. If the grounds are not the same, we are looking for options to solve this example.
2. After the bases are the same, equate degree and solve the resulting new equation.

Now let's solve some examples:

Let's start simple.

The bases on the left and right sides are equal to the number 2, which means we can discard the base and equate their degrees.

x+2=4 The simplest equation has turned out.
x=4 - 2
x=2
Answer: x=2

In the following example, you can see that the bases are different, these are 3 and 9.

3 3x - 9 x + 8 = 0

To begin with, we transfer the nine to the right side, we get:

Now you need to make the same bases. We know that 9=3 2 . Let's use the power formula (a n) m = a nm .

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

We get 9 x + 8 \u003d (3 2) x + 8 \u003d 3 2 x + 16

3 3x \u003d 3 2x + 16 now it is clear that the bases on the left and right sides are the same and equal to three, which means we can discard them and equate the degrees.

3x=2x+16 got the simplest equation
3x-2x=16
x=16
Answer: x=16.

Let's look at the following example:

2 2x + 4 - 10 4 x \u003d 2 4

First of all, we look at the bases, the bases are different two and four. And we need to be the same. We transform the quadruple according to the formula (a n) m = a nm .

4 x = (2 2) x = 2 2x

And we also use one formula a n a m = a n + m:

2 2x+4 = 2 2x 2 4

Add to the equation:

2 2x 2 4 - 10 2 2x = 24

We gave an example for the same reasons. But other numbers 10 and 24 interfere with us. What to do with them? If you look closely, you can see that on the left side we repeat 2 2x, here is the answer - we can put 2 2x out of brackets:

2 2x (2 4 - 10) = 24

Let's calculate the expression in brackets:

2 4 — 10 = 16 — 10 = 6

We divide the whole equation by 6:

Imagine 4=2 2:

2 2x \u003d 2 2 bases are the same, discard them and equate the degrees.
2x \u003d 2 turned out to be the simplest equation. We divide it by 2, we get
x = 1
Answer: x = 1.

Let's solve the equation:

9 x - 12*3 x +27= 0

Let's transform:
9 x = (3 2) x = 3 2x

We get the equation:
3 2x - 12 3 x +27 = 0

Our bases are the same, equal to three. In this example, it is clear that the first triple has a degree twice (2x) than the second (just x). In this case, you can decide substitution method. The number with the smallest degree is replaced by:

Then 3 2x \u003d (3 x) 2 \u003d t 2

We replace all degrees with x's in the equation with t:

t 2 - 12t + 27 \u003d 0
We get a quadratic equation. We solve through the discriminant, we get:
D=144-108=36
t1 = 9
t2 = 3

Back to Variable x.

We take t 1:
t 1 \u003d 9 \u003d 3 x

That is,

3 x = 9
3 x = 3 2
x 1 = 2

One root was found. We are looking for the second one, from t 2:
t 2 \u003d 3 \u003d 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 \u003d 2; x 2 = 1.

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primary goal

To acquaint students with the properties of degrees with natural indicators and teach them to perform actions with degrees.

Topic “Degree and its properties” includes three questions:

• Determination of the degree with a natural indicator.
• Multiplication and division of powers.
• Exponentiation of product and degree.

test questions

1. Formulate the definition of a degree with a natural exponent greater than 1. Give an example.
2. Formulate a definition of the degree with an indicator of 1. Give an example.
3. What is the order of operations when evaluating the value of an expression containing powers?
4. Formulate the main property of the degree. Give an example.
5. Formulate a rule for multiplying powers with the same base. Give an example.
6. Formulate a rule for dividing powers with the same bases. Give an example.
7. Formulate the rule for exponentiation of a product. Give an example. Prove the identity (ab) n = a n b n .
8. Formulate a rule for raising a degree to a power. Give an example. Prove the identity (a m) n = a m n .

Definition of degree.

degree of number a with a natural indicator n, greater than 1, is called the product of n factors, each of which is equal to a. degree of number a with exponent 1 the number itself is called a.

Degree with base a and indicator n is written like this: a n. It reads " a to the extent n”; “ n-th power of a number a ”.

By definition of degree:

a 4 = a a a a

. . . . . . . . . . . .

Finding the value of the degree is called exponentiation .

1. Examples of exponentiation:

3 3 = 3 3 3 = 27

0 4 = 0 0 0 0 = 0

(-5) 3 = (-5) (-5) (-5) = -125

25 ; 0,09 ;

25 = 5 2 ; 0,09 = (0,3) 2 ; .

27 ; 0,001 ; 8 .

27 = 3 3 ; 0,001 = (0,1) 3 ; 8 = 2 3 .

4. Find expression values:

a) 3 10 3 = 3 10 10 10 = 3 1000 = 3000

b) -2 4 + (-3) 2 = 7
2 4 = 16
(-3) 2 = 9
-16 + 9 = 7

Option 1

a) 0.3 0.3 0.3

c) b b b b b b b

d) (-x) (-x) (-x) (-x)

e) (ab) (ab) (ab)

2. Square the numbers:

3. Cube the numbers:

4. Find expression values:

c) -1 4 + (-2) 3

d) -4 3 + (-3) 2

e) 100 - 5 2 4

Multiplication of powers.

For any number a and arbitrary numbers m and n, the following is true:

a m a n = a m + n .

Proof:

rule : When multiplying powers with the same base, the bases remain the same, and the exponents are added.

a m a n a k = a m + n a k = a (m + n) + k = a m + n + k

a) x 5 x 4 = x 5 + 4 = x 9

b) y y 6 = y 1 y 6 = y 1 + 6 = y 7

c) b 2 b 5 b 4 \u003d b 2 + 5 + 4 \u003d b 11

d) 3 4 9 = 3 4 3 2 = 3 6

e) 0.01 0.1 3 = 0.1 2 0.1 3 = 0.1 5

a) 2 3 2 = 2 4 = 16

b) 3 2 3 5 = 3 7 = 2187

Option 1

1. Present as a degree:

a) x 3 x 4 e) x 2 x 3 x 4

b) a 6 a 2 g) 3 3 9

c) y 4 y h) 7 4 49

d) a a 8 i) 16 2 7

e) 2 3 2 4 j) 0.3 3 0.09

2. Present as a degree and find the value in the table:

a) 2 2 2 3 c) 8 2 5

b) 3 4 3 2 d) 27 243

Division of degrees.

For any number a0 and arbitrary natural numbers m and n such that m>n, the following holds:

a m: a n = a m - n

Proof:

a m - n a n = a (m - n) + n = a m - n + n = a m

by definition of private:

a m: a n \u003d a m - n.

rule: When dividing powers with the same base, the base is left the same, and the exponent of the divisor is subtracted from the exponent of the dividend.

Definition: The degree of a non-zero number with a zero exponent is equal to one:

because a n: a n = 1 for a0 .

a) x 4: x 2 \u003d x 4 - 2 \u003d x 2

b) y 8: y 3 = y 8 - 3 = y 5

c) a 7: a \u003d a 7: a 1 \u003d a 7 - 1 \u003d a 6

d) s 5:s 0 = s 5:1 = s 5

a) 5 7:5 5 = 5 2 = 25

b) 10 20:10 17 = 10 3 = 1000

in)

G)

e)

Option 1

1. Express the quotient as a power:

2. Find the values ​​of expressions:

Raising to the power of a product.

For any a and b and an arbitrary natural number n:

(ab) n = a n b n

Proof:

By definition of degree

(ab) n =

Grouping the factors a and factors b separately, we get:

=

The proved property of the degree of the product extends to the degree of the product of three or more factors.

For example:

(a b c) n = a n b n c n ;

(a b c d) n = a n b n c n d n .

rule: When raising a product to a power, each factor is raised to that power and the result is multiplied.

1. Raise to a power:

a) (a b) 4 = a 4 b 4

b) (2 x y) 3 \u003d 2 3 x 3 y 3 \u003d 8 x 3 y 3

c) (3 a) 4 = 3 4 a 4 = 81 a 4

d) (-5 y) 3 \u003d (-5) 3 y 3 \u003d -125 y 3

e) (-0.2 x y) 2 \u003d (-0.2) 2 x 2 y 2 \u003d 0.04 x 2 y 2

f) (-3 a b c) 4 = (-3) 4 a 4 b 4 c 4 = 81 a 4 b 4 c 4

2. Find the value of the expression:

a) (2 10) 4 = 2 4 10 4 = 16 1000 = 16000

b) (3 5 20) 2 = 3 2 100 2 = 9 10000= 90000

c) 2 4 5 4 = (2 5) 4 = 10 4 = 10000

d) 0.25 11 4 11 = (0.25 4) 11 = 1 11 = 1

e)

Option 1

1. Raise to a power:

b) (2 a c) 4

e) (-0.1 x y) 3

2. Find the value of the expression:

b) (5 7 20) 2

Exponentiation.

For any number a and arbitrary natural numbers m and n:

(a m) n = a m n

Proof:

By definition of degree

(a m) n =

Rule: When raising a power to a power, the base is left the same, and the exponents are multiplied.

1. Raise to a power:

(a 3) 2 = a 6 (x 5) 4 = x 20

(y 5) 2 = y 10 (b 3) 3 = b 9

2. Simplify expressions:

a) a 3 (a 2) 5 = a 3 a 10 = a 13

b) (b 3) 2 b 7 \u003d b 6 b 7 \u003d b 13

c) (x 3) 2 (x 2) 4 \u003d x 6 x 8 \u003d x 14

d) (y y 7) 3 = (y 8) 3 = y 24

a)

b)

Option 1

1. Raise to a power:

a) (a 4) 2 b) (x 4) 5

c) (y 3) 2 d) (b 4) 4

2. Simplify expressions:

a) a 4 (a 3) 2

b) (b 4) 3 b 5+

c) (x 2) 4 (x 4) 3

d) (y y 9) 2

3. Find the meaning of expressions:

Application

Definition of degree.

Option 2

1st Write the product in the form of a degree:

a) 0.4 0.4 0.4

c) a a a a a a a a a

d) (-y) (-y) (-y) (-y)

e) (bc) (bc) (bc)

2. Square the numbers:

3. Cube the numbers:

4. Find expression values:

c) -1 3 + (-2) 4

d) -6 2 + (-3) 2

e) 4 5 2 – 100

Option 3

1. Write the product as a degree:

a) 0.5 0.5 0.5

c) c c c c c c c c c

d) (-x) (-x) (-x) (-x)

e) (ab) (ab) (ab)

2. Present in the form of a square of the number: 100; 0.49; .

3. Cube the numbers:

4. Find expression values:

c) -1 5 + (-3) 2

d) -5 3 + (-4) 2

e) 5 4 2 - 100

Option 4

1. Write the product as a degree:

a) 0.7 0.7 0.7

c) x x x x x x

d) (-а) (-а) (-а)

e) (bc) (bc) (bc) (bc)

2. Square the numbers:

3. Cube the numbers:

4. Find expression values:

c) -1 4 + (-3) 3

d) -3 4 + (-5) 2

e) 100 - 3 2 5

Multiplication of powers.

Option 2

1. Present as a degree:

a) x 4 x 5 e) x 3 x 4 x 5

b) a 7 a 3 g) 2 3 4

c) y 5 y h) 4 3 16

d) a a 7 i) 4 2 5

e) 2 2 2 5 j) 0.2 3 0.04

2. Present as a degree and find the value in the table:

a) 3 2 3 3 c) 16 2 3

b) 2 4 2 5 d) 9 81

Option 3

1. Present as a degree:

a) a 3 a 5 e) y 2 y 4 y 6

b) x 4 x 7 g) 3 5 9

c) b 6 b h) 5 3 25

d) y 8 i) 49 7 4

e) 2 3 2 6 j) 0.3 4 0.27

2. Present as a degree and find the value in the table:

a) 3 3 3 4 c) 27 3 4

b) 2 4 2 6 d) 16 64

Option 4

1. Present as a degree:

a) a 6 a 2 e) x 4 x x 6

b) x 7 x 8 g) 3 4 27

c) y 6 y h) 4 3 16

d) x x 10 i) 36 6 3

e) 2 4 2 5 j) 0.2 2 0.008

2. Present as a degree and find the value in the table:

a) 2 6 2 3 c) 64 2 4

b) 3 5 3 2 d) 81 27

Division of degrees.

Option 2

1. Express the quotient as a power:

2. Find the meaning of expressions.

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We figured out what the degree of a number is in general. Now we need to understand how to correctly calculate it, i.e. raise numbers to powers. In this material, we will analyze the basic rules for calculating the degree in the case of an integer, natural, fractional, rational and irrational exponent. All definitions will be illustrated with examples.

Yandex.RTB R-A-339285-1

The concept of exponentiation

Let's start with the formulation of basic definitions.

Definition 1

Exponentiation is the calculation of the value of the power of some number.

That is, the words "calculation of the value of the degree" and "exponentiation" mean the same thing. So, if the task is "Raise the number 0 , 5 to the fifth power", this should be understood as "calculate the value of the power (0 , 5) 5 .

Now we give the basic rules that must be followed in such calculations.

Recall what a power of a number with a natural exponent is. For a power with base a and exponent n, this will be the product of the nth number of factors, each of which is equal to a. This can be written like this:

To calculate the value of the degree, you need to perform the operation of multiplication, that is, multiply the bases of the degree the specified number of times. The very concept of a degree with a natural indicator is based on the ability to quickly multiply. Let's give examples.

Example 1

Condition: Raise - 2 to the power of 4 .

Solution

Using the definition above, we write: (− 2) 4 = (− 2) (− 2) (− 2) (− 2) . Next, we just need to follow these steps and get 16 .

Let's take a more complicated example.

Example 2

Calculate the value 3 2 7 2

Solution

This entry can be rewritten as 3 2 7 · 3 2 7 . Earlier we looked at how to correctly multiply the mixed numbers mentioned in the condition.

Perform these steps and get the answer: 3 2 7 3 2 7 = 23 7 23 7 = 529 49 = 10 39 49

If the task indicates the need to raise irrational numbers to a natural power, we will need to first round their bases to a digit that will allow us to get an answer of the desired accuracy. Let's take an example.

Example 3

Perform the squaring of the number π .

Solution

Let's round it up to hundredths first. Then π 2 ≈ (3, 14) 2 = 9, 8596. If π ≈ 3 . 14159, then we will get a more accurate result: π 2 ≈ (3, 14159) 2 = 9, 8695877281.

Note that the need to calculate the powers of irrational numbers in practice arises relatively rarely. We can then write the answer as the power itself (ln 6) 3 or convert if possible: 5 7 = 125 5 .

Separately, it should be indicated what the first power of a number is. Here you can just remember that any number raised to the first power will remain itself:

This is clear from the record. .

It does not depend on the basis of the degree.

Example 4

So, (− 9) 1 = − 9 , and 7 3 raised to the first power remains equal to 7 3 .

For convenience, we will analyze three cases separately: if the exponent is a positive integer, if it is zero, and if it is a negative integer.

In the first case, this is the same as raising to a natural power: after all, positive integers belong to the set of natural numbers. We have already described how to work with such degrees above.

Now let's see how to properly raise to the zero power. With a base that is non-zero, this calculation always produces an output of 1 . We have previously explained that the 0th power of a can be defined for any real number not equal to 0 , and a 0 = 1 .

Example 5

5 0 = 1 , (- 2 , 56) 0 = 1 2 3 0 = 1

0 0 - not defined.

We are left with only the case of a degree with a negative integer exponent. We have already discussed that such degrees can be written as a fraction 1 a z, where a is any number, and z is a negative integer. We see that the denominator of this fraction is nothing more than an ordinary degree with a positive integer, and we have already learned how to calculate it. Let's give examples of tasks.

Example 6

Raise 3 to the -2 power.

Solution

Using the definition above, we write: 2 - 3 = 1 2 3

We calculate the denominator of this fraction and get 8: 2 3 \u003d 2 2 2 \u003d 8.

Then the answer is: 2 - 3 = 1 2 3 = 1 8

Example 7

Raise 1, 43 to the -2 power.

Solution

Reformulate: 1 , 43 - 2 = 1 (1 , 43) 2

We calculate the square in the denominator: 1.43 1.43. Decimals can be multiplied in this way:

As a result, we got (1, 43) - 2 = 1 (1, 43) 2 = 1 2 , 0449 . It remains for us to write this result in the form of an ordinary fraction, for which it is necessary to multiply it by 10 thousand (see the material on the conversion of fractions).

Answer: (1, 43) - 2 = 10000 20449

A separate case is raising a number to the minus first power. The value of such a degree is equal to the number opposite to the original value of the base: a - 1 \u003d 1 a 1 \u003d 1 a.

Example 8

Example: 3 − 1 = 1 / 3

9 13 - 1 = 13 9 6 4 - 1 = 1 6 4 .

How to raise a number to a fractional power

To perform such an operation, we need to recall the basic definition of a degree with a fractional exponent: a m n \u003d a m n for any positive a, integer m and natural n.

Definition 2

Thus, the calculation of a fractional degree must be performed in two steps: raising to an integer power and finding the root of the nth degree.

We have the equality a m n = a m n , which, given the properties of the roots, is usually used to solve problems in the form a m n = a n m . This means that if we raise the number a to a fractional power m / n, then first we extract the root of the nth degree from a, then we raise the result to a power with an integer exponent m.

Let's illustrate with an example.

Example 9

Calculate 8 - 2 3 .

Solution

Method 1. According to the basic definition, we can represent this as: 8 - 2 3 \u003d 8 - 2 3

Now let's calculate the degree under the root and extract the third root from the result: 8 - 2 3 = 1 64 3 = 1 3 3 64 3 = 1 3 3 4 3 3 = 1 4

Method 2. Let's transform the basic equality: 8 - 2 3 \u003d 8 - 2 3 \u003d 8 3 - 2

After that, we extract the root 8 3 - 2 = 2 3 3 - 2 = 2 - 2 and square the result: 2 - 2 = 1 2 2 = 1 4

We see that the solutions are identical. You can use any way you like.

There are cases when the degree has an indicator expressed as a mixed number or decimal fraction. For ease of calculation, it is better to replace it with an ordinary fraction and count as indicated above.

Example 10

Raise 44.89 to the power of 2.5.

Solution

Let's convert the value of the indicator into an ordinary fraction - 44, 89 2, 5 = 49, 89 5 2.

And now we perform all the actions indicated above in order: 44 , 89 5 2 = 44 , 89 5 = 44 , 89 5 = 4489 100 5 = 4489 100 5 = 67 2 10 2 5 = 67 10 5 = = 1350125107 100000 = 13 501, 25107

Answer: 13501, 25107.

If there are large numbers in the numerator and denominator of a fractional exponent, then calculating such exponents with rational exponents is a rather difficult job. It usually requires computer technology.

Separately, we dwell on the degree with a zero base and a fractional exponent. An expression of the form 0 m n can be given the following meaning: if m n > 0, then 0 m n = 0 m n = 0 ; if m n< 0 нуль остается не определен. Таким образом, возведение нуля в дробную положительную степень приводит к нулю: 0 7 12 = 0 , 0 3 2 5 = 0 , 0 0 , 024 = 0 , а в целую отрицательную - значения не имеет: 0 - 4 3 .

How to raise a number to an irrational power

The need to calculate the value of the degree, in the indicator of which there is an irrational number, does not arise so often. In practice, the task is usually limited to calculating an approximate value (up to a certain number of decimal places). This is usually calculated on a computer due to the complexity of such calculations, so we will not dwell on this in detail, we will only indicate the main provisions.

If we need to calculate the value of the degree a with an irrational exponent a , then we take the decimal approximation of the exponent and count from it. The result will be an approximate answer. The more accurate the decimal approximation taken, the more accurate the answer. Let's show with an example:

Example 11

Compute an approximate value of 21 , 174367 ....

Solution

We restrict ourselves to the decimal approximation a n = 1 , 17 . Let's do the calculations using this number: 2 1 , 17 ≈ 2 , 250116 . If we take, for example, the approximation a n = 1 , 1743 , then the answer will be a little more precise: 2 1 , 174367 . . . ≈ 2 1 . 1743 ≈ 2 . 256833 .

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When the number multiplies itself to myself, work called degree.

So 2.2 = 4, square or second power of 2
2.2.2 = 8, cube or third power.
2.2.2.2 = 16, fourth degree.

Also, 10.10 = 100, the second power is 10.
10.10.10 = 1000, third degree.
10.10.10.10 = 10000 fourth degree.

And a.a = aa, the second power of a
a.a.a = aaa, the third power of a
a.a.a.a = aaaa, fourth power of a

The original number is called root degrees of that number, because that is the number from which the degrees were created.

However, it is not very convenient, especially in the case of high powers, to write down all the factors that make up the powers. Therefore, an abbreviated notation method is used. The root of the degree is written only once, and to the right and a little higher next to it, but in a slightly smaller font it is written how many times the root acts as a factor. This number or letter is called exponent or degree numbers. So, a 2 is equal to a.a or aa, because the root of a must be multiplied by itself twice to get the power of aa. Also, a 3 means aaa, that is, here a is repeated three times as a multiplier.

The exponent of the first power is 1, but it is usually not written down. So, a 1 is written as a.

You should not confuse degrees with coefficients. The coefficient shows how often the value is taken as part whole. The exponent indicates how often the value is taken as factor in the work.
So, 4a = a + a + a + a. But a 4 = a.a.a.a

The exponential notation has the peculiar advantage of allowing us to express unknown degree. For this purpose, instead of a number, the exponent is written letter. In the process of solving the problem, we can get a value that, as we know, is some degree of another magnitude. But so far we do not know if it is a square, a cube, or another, higher degree. So, in the expression a x , the exponent means that this expression has some degree, although not defined what degree. So, b m and d n are raised to the powers of m and n. When the exponent is found, number substituted for a letter. So, if m=3, then b m = b 3 ; but if m = 5 then b m =b 5 .

The method of writing values ​​with exponents is also a great advantage when using expressions. Thus, (a + b + d) 3 is (a + b + d).(a + b + d).(a + b + d), that is, the cube of the trinomial (a + b + d). But if we write this expression after cubed, it will look like
a 3 + 3a 2 b + 3a 2 d + 3ab 2 + 6abd + 3ad 2 + b 3 + d 3 .

If we take a series of powers whose exponents increase or decrease by 1, we find that the product increases by common factor or reduced by common divisor, and this factor or divisor is the original number that is raised to a power.

So, in the series aaaaa, aaaa, aaa, aa, a;
or a 5 , a 4 , a 3 , a 2 , a 1 ;
indicators, if counted from right to left, are 1, 2, 3, 4, 5; and the difference between their values ​​is 1. If we start on right multiply on a, we will successfully get multiple values.

So a.a = a 2 , the second term. And a 3 .a = a 4
a 2 .a = a 3 , the third term. a 4 .a = a 5 .

If we start left divide on a,
we get a 5:a = a 4 and a 3:a = a 2 .
a 4:a = a 3 a 2:a = a 1

But such a division process can be continued further, and we get a new set of values.

So, a:a = a/a = 1. (1/a):a = 1/aa
1:a = 1/a (1/aa):a = 1/aaa.

The full row will be: aaaaa, aaaa, aaa, aa, a, 1, 1/a, 1/aa, 1/aaa.

Or a 5 , a 4 , a 3 , a 2 , a, 1, 1/a, 1/a 2 , 1/a 3 .

Here values on right from unit is reverse values ​​to the left of one. Therefore, these degrees can be called inverse powers a. One can also say that the powers on the left are the inverse of the powers on the right.

So, 1:(1/a) = 1.(a/1) = a. And 1:(1/a 3) = a 3 .

The same recording plan can be applied to polynomials. So, for a + b, we get a set,
(a + b) 3 , (a + b) 2 , (a + b), 1, 1/(a + b), 1/(a + b) 2 , 1/(a + b) 3 .

For convenience, another form of writing inverse powers is used.

According to this form, 1/a or 1/a 1 = a -1 . And 1/aaa or 1/a 3 = a -3 .
1/aa or 1/a 2 = a -2 . 1/aaaa or 1/a 4 = a -4 .

And to make the exponents a complete series with 1 as the total difference, a/a or 1 is considered as such that has no degree and is written as a 0 .

Then, taking into account the direct and inverse powers
instead of aaaa, aaa, aa, a, a/a, 1/a, 1/aa, 1/aaa, 1/aaaa
you can write a 4 , a 3 , a 2 , a 1 , a 0 , a -1 , a -2 , a -3 , a -4 .
Or a +4 , a +3 , a +2 , a +1 , a 0 , a -1 , a -2 , a -3 , a -4 .

And a series of only separately taken degrees will have the form:
+4,+3,+2,+1,0,-1,-2,-3,-4.

The root of the degree can be expressed by more than one letter.

Thus, aa.aa or (aa) 2 is the second power of aa.
And aa.aa.aa or (aa) 3 is the third power of aa.

All degrees of the number 1 are the same: 1.1 or 1.1.1. will be equal to 1.

Exponentiation is finding the value of any number by multiplying that number by itself. Exponentiation rule:

Multiply the value by itself as many times as indicated in the power of the number.

This rule is common to all examples that may arise in the process of exponentiation. But it will be correct to explain how it applies to particular cases.

If only one term is raised to a power, then it is multiplied by itself as many times as the exponent indicates.

The fourth power a is a 4 or aaaa. (Art. 195.)
The sixth power of y is y 6 or yyyyyy.
The nth power of x is x n or xxx..... n times repeated.

If it is necessary to raise an expression of several terms to a power, the principle that the degree of the product of several factors is equal to the product of these factors raised to a power.

So (ay) 2 =a 2 y 2 ; (ay) 2 = ay.ay.
But ay.ay = ayay = aayy = a 2 y 2 .
So, (bmx) 3 = bmx.bmx.bmx = bbbmmmxxx = b 3 m 3 x 3 .

Therefore, in finding the degree of a product, we can either operate on the entire product at once, or we can operate on each factor separately, and then multiply their values ​​with degrees.

Example 1. The fourth power of dhy is (dhy) 4 , or d 4 h 4 y 4 .

Example 2. The third power of 4b is (4b) 3 , or 4 3 b 3 , or 64b 3 .

Example 3. The nth power of 6ad is (6ad) n or 6 n a n d n .

Example 4. The third power of 3m.2y is (3m.2y) 3 , or 27m 3 .8y 3 .

The degree of a binomial, consisting of terms connected by + and -, is calculated by multiplying its terms. Yes,

(a + b) 1 = a + b, the first power.
(a + b) 1 = a 2 + 2ab + b 2 , second power (a + b).
(a + b) 3 \u003d a 3 + 3a 2 b + 3ab 2 + b 3, third degree.
(a + b) 4 \u003d a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4, fourth degree.

Square a - b, there is a 2 - 2ab + b 2 .