Degree with different bases and indicators. The rules for multiplication of degrees with different bases. Properties of degrees with rational indicators
What is the degree?
Degree Call a work of several identical factors. For example:
2 × 2 × 2
The value of this expression is 8
2 × 2 × 2 \u003d 8
The left part of this equality can be shorter - first record a repetitive multiplier and indicate how many times it is repeated. A repeating multiplier in this case is 2. It repeats three times. Therefore, you write down the top two:
2 3 = 8
This expression is read like this: " two to the third degree is eight " or " the third degree of number 2 is 8 ".
The short form of multiplication recording of the same multiplers is used more often. Therefore, it should be remembered that if another number is inscribed over some number, then it is multiplying several identical factors.
For example, if expression 5 3 is given, then it should be borne in mind that this expression is equivalent to a 5 × 5 × 5 record.
The number that is repeated called the degree of degree. In expression 5 3, the foundation of the degree is the number 5.
And the number that is written above the number 5 is called indicator of degree. In expression 5 3, the indicator of the degree is the number 3. The indicator of the degree shows how many times the foundation is repeated. In our case, the base 5 is repeated three times.
The operation of multiplication of the same multipliers is called erend into degree.
For example, if you need to find a product of four identical multipliers, each of which is 2, then they say that number 2 erected into the fourth degree:
We see that the number 2 in the fourth degree is the number 16.
Note that in this lesson we consider degree with a natural indicator. This is a type of degree, the indicator of which is a natural number. Recall that naturally called integers that are more zero. For example, 1, 2, 3, and so on.
In general, the determination of the degree with a natural indicator is as follows:
Degree of a. With a natural indicator n. - this is an expression of the form a N.equal to the work n. multipliers, each of which is equal a.
Examples:
It should be attentive when the number is erected. Often by inattention, a person multiplies the foundation of the degree to the indicator.
For example, the number 5 to the second degree is the product of two factors each of which is 5. This product is equal to 25
Now imagine that we are inattentive multiplied the foundation 5 on the indicator 2
It turned out a mistake, since the number 5 to the second degree is not equal to 10.
Additionally, it should be mentioned that the degree of the number with the indicator 1, there is this number itself:
For example, the number 5 in the first degree is the number 5
Accordingly, if the number does not have an indicator, then it is necessary to assume that the indicator is equal to one.
For example, the numbers 1, 2, 3 are given without an indicator, so their indicators will be equal to one. Each of these numbers can be recorded with an indicator 1
And if you build 0 into any degree, then it will turn out 0. Indeed, no matter how many times did not multiply yourself, nothing happens. Examples:
And the expression 0 0 does not make sense. But in some sections of mathematics, in particular analysis and theory of sets, expression 0 0 may make sense.
For training, we solve several examples on the construction of numbers to the degree.
Example 1. Evaluate the number 3 in the second degree.
The number 3 of the second degree is the product of two factors, each of which is 3
3 2 \u003d 3 × 3 \u003d 9
Example 2. Evaluate the number 2 to the fourth degree.
The number 2 in the fourth degree is a product of four multipliers, each of which is 2
2 4 \u003d 2 × 2 × 2 × 2 \u003d 16
Example 3. Evaluate the number 2 to the third degree.
The number 2 to the third degree is the work of three factor, each of which is 2
2 3 \u003d 2 × 2 × 2 \u003d 8
Establishment into the degree of 10
To raise the number 10 to the degree, it is enough to add the number of zeros after the unit equal to the degree.
For example, erected the number 10 in the second degree. First, write the number 10 itself and specify the number 2 as an indicator
10 2
Now put the sign of equality, write down the unit and after this unit you write two zero, since the number of zeros should be equal to the degree
10 2 = 100
It means that the number 10 to the second degree is the number 100. This is due to the fact that the number 10 to the second degree is the product of two factors, each of which is 10
10 2 \u003d 10 × 10 \u003d 100
Example 2.. Erected the number 10 to the third degree.
In this case, after a unit, three zero will stand:
10 3 = 1000
Example 3.. Establish the number 10 in the fourth degree.
In this case, after a unit, four zero will stand:
10 4 = 10000
Example 4.. Establish the number 10 to the first degree.
In this case, after a unit, one zero will stand:
10 1 = 10
Pose of numbers 10, 100, 1000 in the form of a degree with a base of 10
To represent the numbers 10, 100, 1000 and 10,000 in the form of a degree with a base 10, you need to record the base 10, and as an indicator, indicate a number equal to the number of zeros of the source number.
Imagine the number 10 in the form of a degree with the base 10. We see that in it one zero. So, the number 10 in the form of a degree with the base 10 will be presented as 10 1
10 = 10 1
Example 2.. Imagine the number 100 in the form of a degree of base 10. We see that the number 100 contains two zero. So, the number 100 in the form of a degree with the base 10 will be presented as 10 2
100 = 10 2
Example 3.. Imagine the number 1,000 in the form of a degree with the base 10.
1 000 = 10 3
Example 4.. Imagine the number 10,000 in the form of a degree with the base 10.
10 000 = 10 4
Negative
When the degree of negative number is erected, it is necessary to enter into brackets.
For example, erect a negative number -2 to the second degree. The number -2 to the second degree is the product of two factors, each of which is equal to (-2)
(-2) 2 \u003d (-2) × (-2) \u003d 4
If we did not enter the number -2 brackets, it would have happened that we calculate the expression -2 2, which not equal four . Expression -2² will be -4. To understand why, we talked some moments.
When we put in front of a positive number of minus, we are thereby performing operation of the opposite value.
Suppose the number 2 is given, and it is necessary to find its opposite number. We know that the opposite of the number 2 is the number -2. In other words, to find the opposite number for 2, it is enough to put minus before this number. The inset of the minus in front of the number is already considered in mathematics with a full-fledged operation. This operation, as mentioned above, is called the transaction operation of the opposite value.
In the case of expression -2 2, there are two operations: the operation take the opposite value and the exercise. The exercise is a higher priority operation than taking the opposite value.
Therefore, the expression -2 2 is calculated in two stages. First, the exercise operation is performed. In this case, a positive number was erected into the second degree.
Then the opposite value was taken. This opposite value was found for value 4. And the opposite value for 4 is -4
−2 2 = −4
Brackets have the highest priority of execution. Therefore, in the case of calculating the expression (-2) 2, the opposite value is first completed, and then a negative number is erected into the second degree. The result is a positive answer 4, since the product of negative numbers is a positive number.
Example 2.. Build a number -2 to a third degree.
The number -2 to the third degree is the work of three multipliers, each of which is equal to (-2)
(-2) 3 \u003d (-2) × (-2) × (-2) \u003d -8
Example 3.. Build a number -2 in the fourth degree.
Number -2 in the fourth degree is a product of four multipliers, each of which is equal to (-2)
(-2) 4 \u003d (-2) × (-2) × (-2) × (-2) \u003d 16
It is easy to see that when the degree of negative number is erected, either a positive answer is either negative. The answer sign depends on the indicator of the initial degree.
If the indicator is clear, the answer will be positive. If the indicator is odd, the answer will be negative. Show it on the example of the number -3
In the first and in the third case, the indicator was odd number, so the answer became negative.
In the second and in the fourth case, the indicator was an even number, so the answer became positive.
Example 7. Evaluate the number -5 to the third degree.
The number -5 to the third degree is the work of three faults each of which is -5. The indicator 3 is an odd number, so we can say in advance that the answer will be negative:
(-5) 3 \u003d (-5) × (-5) × (-5) \u003d -125
Example 8. Evaluate the number -4 in the fourth degree.
The number -4 in the fourth degree is a product of four multipliers, each of which is -4. At the same time, the index 4 is even used, so we can say in advance that the answer will be positive:
(-4) 4 \u003d (-4) × (-4) × (-4) × (-4) \u003d 256
Finding values \u200b\u200bof expressions
When finding values \u200b\u200bof the expressions that do not contain brackets, the exercise will be carried out primarily, further multiplication and division in the order of their following, and then addition and subtraction in the order of their follows.
Example 1.. Find an expression value 2 + 5 2
First, the extermination is carried out. In this case, the second degree is erected by the number 5 - it turns out 25. Then this result is folded with Number 2
2 + 5 2 = 2 + 25 = 27
Example 10.. Find an expression value -6 2 × (-12)
First, the extermination is carried out. Note that the number -6 is not taken in the bracket, therefore the number 6 will be erected into the second degree, then the minus will be supplied before the result:
-6 2 × (-12) \u003d -36 × (-12)
Complete an example, multiplying -36 on (-12)
-6 2 × (-12) \u003d -36 × (-12) \u003d 432
Example 11.. Find an expression value -3 × 2 2
First, the extermination is carried out. The resulting result varies with the number -3
-3 × 2 2 \u003d -3 × 4 \u003d -12
If the expression contains brackets, you first need to perform actions in these brackets, then the extinguisure, then multiplication and division, and then addition and subtraction.
Example 12.. Find an expression value (3 2 + 1 × 3) - 15 + 5
First perform actions in brackets. Inside the brackets, we apply previously studied rules, namely, first erect the number 3 into the second degree, then we perform a multiplication of 1 × 3, then we add the results of the erection of the number 3 and the multiplication of 1 × 3. Next, subtraction and addition is performed in the order of them. Let us separate this procedure for performing action above the initial expression:
(3 2 + 1 × 3) - 15 + 5 \u003d 12 - 15 + 5 \u003d 2
Example 13.. Find an expression value 2 × 5 3 + 5 × 2 3
First erected a number to degree, then perform multiplication and lay down the results obtained:
2 × 5 3 + 5 × 2 3 \u003d 2 × 125 + 5 × 8 \u003d 250 + 40 \u003d 290
Identical transforms degrees
Over degrees, various identical transformations can be performed, thereby simplifying them.
Suppose it was necessary to calculate the expression (2 3) 2. In this example, two in the third degree is erected into a second degree. In other words, the degree is erected into another degree.
(2 3) 2 This is a product of two degrees, each of which is 2 3
At the same time, each of these degrees is a product of three factor, each of which is 2
Received a piece of 2 × 2 × 2 × 2 × 2 × 2, which is 64. Means the value of the expression (2 3) 2 or equal to 64
This example can be considerable to simplify. For this, the expression indicators (2 3) 2 can multiply and write this product above the base 2
Received 2 6. Two in sixth, this is a work of six multipliers, each of which is 2. This product is equal to 64
This property works due to the fact that 2 3 is a product of 2 × 2 × 2, which in turn is repeated twice. Then it turns out that the base 2 is repeated six times. From here you can write down that 2 × 2 × 2 × 2 × 2 × 2 is 2 6
In general, for any reason a. with indicators m. and n. The following equality is performed:
(a N.) m \u003d a n × m
This identical conversion is called erect. It can be found like this: "When erecting a degree to a degree, the foundation is left unchanged, and the indicators are prolonged" .
After multiplying the indicators, it will turn out another degree whose value can be found.
Example 2.. Find an expression value (3 2) 2
In this example, the base is 3, and the numbers 2 and 2 are indicators. We use the rules for the exercise degree into the degree. The base will be left unchanged, and indicators to change:
Received 3 4. And the number 3 in the fourth degree is 81
Consider the remaining transformations.
Multiplying degrees
To multiply degrees, you need to calculate each degree separately, and multiply the results.
For example, multiply 2 2 to 3 3.
2 2 This is the number 4, and 3 3 is the number 27. Alternate numbers 4 and 27, we get 108
2 2 × 3 3 \u003d 4 × 27 \u003d 108
In this example, the foundations of the degrees were different. In the event that the bases are the same, you can write a single base, and as an indicator to write down the sum of the indicators of the initial degrees.
For example, multiply 2 2 to 2 3
In this example, the foundations in degrees are the same. In this case, we can write one base 2 and write down the amount of the degrees 2 2 and 2 3 as an indicator. In other words, it is possible to leave no changes about again, and fold the initial degrees. It will look like this:
Received 2 5. Number 2 in the fifth degree is 32
This property works due to the fact that 2 2 is a product of 2 × 2, and 2 3 is a product of 2 × 2 × 2. Then the work of five identical multipliers is obtained, each of which is 2. This product is known as 2 5
In general, for any a. and indicators m. and n. The following equality is performed:
This identical conversion is called the main property of degree. It can be found like this: " Priotmate the degrees with the same bases, the base is left unchanged, and the indicators are folded " .
Note that this transformation can be applied with any number of degrees. The main thing is that the base is the same.
For example, find the value of the expression 2 1 × 2 2 × 2 3. Foundation 2.
Some tasks are sufficient to perform the appropriate transformation, without calculating the final degree. This is certainly very convenient because it is not so easy to calculate.
Example 1.. Represent an expression 5 8 × 25
This task needs to be done so that instead of expression 5 8 × 25 it turned out one degree.
Number 25 can be represented as 5 2. Then we get the following expression:
In this expression, it is possible to apply the main property of the degree - the base 5 should be left unchanged, and the indicators 8 and 2 fold:
Write the decision shorter:
Example 2.. Represent an expression in the form of a degree 2 9 × 32
The number 32 can be represented as 2 5. Then we obtain the expression 2 9 × 2 5. Next, you can apply the foundation of the degree property - the base 2 is left unchanged, and add indicators 9 and 5. As a result, the following solution will be:
Example 3.. Calculate the product 3 × 3 using the basic property of the degree.
Everyone knows well that three to multiply by three is equal to nine, but the task requires during the decision to take advantage of the main property of the degree. How to do it?
We remember that if the number is given without an indicator, the indicator should be considered an equal unit. There was a factory 3 and 3 in the form of 3 1 and 3 1
3 1 × 3 1
Now we use the main property of the degree. The base 3 is left unchanged, and the indicators 1 and 1 fold:
3 1 × 3 1 \u003d 3 2 \u003d 9
Example 4.. Calculate the product 2 × 2 × 3 2 × 3 3 using the basic property of the degree.
The product is 2 × 2 by replacing 2 1 × 2 1, then 2 1 + 1, and then 2 2. Production 3 2 × 3 3 Replace 3 2 + 3, and then 3 5
Example 5.. Perform multiplication x × x.
These are two identical letter factors with indicators 1. For clarity, we write these indicators. Next, the foundation x. Let's leave unchanged, and the indicators will lay down:
Being at the board, do not write multiplication of degrees with the same bases as detailed as it is done here. Such calculations must be performed in the mind. A detailed record will most likely be annoying the teacher and it will reduce the assessment for this. Here the detailed entry is given so that the material is as accessible as possible for understanding.
The solution of this example is preferably written as follows:
Example 6.. Perform multiplication x. 2 × X.
The second factory indicator is equal to one. For clarity to write it down. Next, the base will be left unchanged, and the indicators will lay down:
Example 7.. Perform multiplication y. 3 y. 2 y.
The figure of the third plant is equal to one. For clarity to write it down. Next, the base will be left unchanged, and the indicators will lay down:
Example 8.. Perform multiplication aA 3 A 2 A 5
The indicator of the first factor is equal to one. For clarity to write it down. Next, the base will be left unchanged, and the indicators will lay down:
Example 9.. Present a degree 3 8 as a product of degrees with the same bases.
In this task, you need to make a product of degrees, the foundations of which will be equal to 3, and the sum of which will be 8. You can use any indicators. Imagine the degree 3 8 in the form of the work of degrees 3 5 and 3 3
In this example, we again relied on the basic property of the degree. After all, the expression 3 5 × 3 3 can be written as 3 5 + 3, from where 3 8.
Of course, it was possible to present a degree 3 8 as a product of other degrees. For example, in the form of 3 7 × 3 1, since this product is also equal to 3 8
Representation of the degree in the form of a product of degrees with the same bases is for the most part creative work. Therefore, you do not need to be afraid to experiment.
Example 10.. Submit degree x. 12 in the form of various works of degrees with bases x. .
We use the main property of the degree. Imagine x. 12 in the form of works with bases x. , and the sum of which is equal to 12
Designs with the amounts of indicators were recorded for clarity. Most often, they can be skipped. Then the compact solution will turn out:
Exercise
To be taken into the degree of work, you need to build in the specified degree every multiplier of this work and multiply the results obtained.
For example, we erected the work of 2 × 3 in the second degree. We take this work in brackets and indicate as an indicator 2
Now erected in the second degree every multiplier of the work of 2 × 3 and variable the results obtained:
The principle of operation of this rule is based on determining the degree that was given at the very beginning.
Build a 2 × 3 product in a second degree means to repeat this product twice. And if you repeat it twice, then you can get the following:
2 × 3 × 2 × 3
From the permutation of places of the factors, the work does not change. This allows the same multipliers to group:
2 × 2 × 3 × 3
Repeating multipliers can be replaced by short records - grounds with indicators. The product of 2 × 2 can be replaced by 2 2, and the product 3 × 3 can be replaced by 3 2. Then the expression 2 × 2 × 3 × 3 refers to the expression 2 2 × 3 2.
Let be aB Source work. To raise this product to the degree n. , you need to build multipliers separately a. and b. in the specified degree n.
This property is valid for any number of multipliers. The following expressions are also valid:
Example 2.. Find an expression value (2 × 3 × 4) 2
In this example, we need to build 2 × 3 × 4 in the second degree. To do this, you need to build in the second degree every multiplier of this work and multiply the results obtained:
Example 3.. Build a third degree a × B × C
Let's enter into brackets this work, and we indicate the number 3 as an indicator
Example 4.. Evaluate the third degree of work 3 xyz.
Let's enter into brackets this work, and we indicate 3 as an indicator 3
(3xyz.) 3
Erend to the third degree every multiplier of this work:
(3xyz.) 3 = 3 3 x. 3 y. 3 z. 3
The number 3 to the third degree is equal to the number 27. The rest will leave unchanged:
(3xyz.) 3 = 3 3 x. 3 y. 3 z. 3 = 27x. 3 y. 3 z. 3
In some examples, multiplication of degrees with the same indicators can be replaced by the basis of the bases with one indicator.
For example, we calculate the value of expression 5 2 × 3 2. Establish each number in the second degree and variables the results obtained:
5 2 × 3 2 \u003d 25 × 9 \u003d 225
But you can not calculate each degree separately. Instead, this product of degrees can be replaced with a product with one indicator (5 × 3) 2. Next, calculate the value in brackets and build the result obtained in the second degree:
5 2 × 3 2 \u003d (5 × 3) 2 \u003d (15) 2 \u003d 225
In this case, again, the rule of construction is used to the degree of work. After all, if (a × B.) N. = a n × b n T. a n × b n \u003d (A × b) n . That is, the left and right part of the equality changed in places.
Erect
This transformation we considered as an example when they tried to understand the essence of identical transformations of degrees.
If the degree is raised into the degree, the base is left unchanged, and the indicators are prolonged:
(a N.) m \u003d a n × m
For example, the expression (2 3) 2 is the erection of a degree into a degree in the third degree being built into a second degree. To find the meaning of this expression, the base can be left unchanged, and multiply indicators:
(2 3) 2 \u003d 2 3 × 2 \u003d 2 6
(2 3) 2 \u003d 2 3 × 2 \u003d 2 6 \u003d 64
This rule is based on previous rules: erection of the work and the main property of the degree.
Let's return to expression (2 3) 2. The expression in brackets 2 3 is a product of three identical factors, each of which is equal to 2. Then in the expression (2 3) 2, the degree inside the brackets can be replaced with a piece of 2 × 2 × 2.
(2 × 2 × 2) 2
And this is the construction of the work we studied earlier. Recall that for the construction of a piece of work, you need to build in the specified degree every multiplier of this product and multiply the results:
(2 × 2 × 2) 2 \u003d 2 2 × 2 2 × 2 2
Now we are dealing with the basic degree property. The base is left unchanged, and we fold the indicators:
(2 × 2 × 2) 2 \u003d 2 2 × 2 2 × 2 2 \u003d 2 2 + 2 + 2 \u003d 2 6
As used to receive 2 6. The value of this degree is 64
(2 × 2 × 2) 2 \u003d 2 2 × 2 2 × 2 2 \u003d 2 2 + 2 + 2 \u003d 2 6 \u003d 64
The degree may also be built a work, the factors of which are also degrees.
For example, find the value of the expression (2 2 × 3 2) 3. Here, the indicators of each factor must be multiplied by the overall indicator 3. Next to find the value of each degree and calculate the work:
(2 2 × 3 2) 3 \u003d 2 2 × 3 × 3 2 × 3 \u003d 2 6 × 3 6 \u003d 64 × 729 \u003d 46656
Approximately the same thing happens when the work is erected. We said that when the work is erected into the degree of work, each multiplier of this work is being built into the specified degree.
For example, to build a product of 2 × 4 to a third degree, you need to record the following expression:
But it was previously said that if the number is given without an indicator, the indicator should be considered an equal unit. It turns out that the multipliers of the work of 2 × 4 initially have indicators equal to 1. So, the expression 2 1 × 4 1 \u200b\u200bwas erected in the third degree. And this is the erection of degree.
I will rewrite the solution using the degree rates. We must have the same result:
Example 2.. Find an expression value (3 3) 2
The base is left unchanged, and the indicators are alternating:
Received 3 6. Number 3 in the sixth degree is number 729
Example 3.xY.)³
Example 4.. Perform the exercise in the expression ( aBC)⁵
Erend to the fifth degree every multiplier of the work:
Example 5.aX.) 3
Erend to the third degree every multiplier of the work:
Since a negative number of -2 was erected into the third degree, it was taken in brackets.
Example 6.. Perform the exercise to the degree in expression (10 xY.) 2
Example 7.. Perform a degree in expression (-5 x.) 3
Example 8.. Perform the exercise in the expression (-3 y.) 4
Example 9.. Perform the exercise in the expression (-2 aBX.)⁴
Example 10.. Simplify expression x. 5 × ( x. 2) 3
Power x. 5 until we leave unchanged, and in the expression ( x. 2) 3 Perform the exercise to the degree to the degree:
x. 5 × (x. 2) 3 \u003d X. 5 × X. 2 × 3. \u003d X. 5 × X. 6
Now perform multiplication x. 5 × X. 6. To do this, we use the main property of the degree - the foundation x. Let's leave unchanged, and the indicators will lay down:
x. 5 × (x. 2) 3 \u003d X. 5 × X. 2 × 3. \u003d X. 5 × X. 6 = x. 5 + 6 = x. 11
Example 9.. Find the value of expression 4 3 × 2 2 using the basic property of the degree.
The main property of the degree can be used if the bases of the initial degrees are the same. In this example, the base is different, therefore, for the beginning, the initial expression must be changed a little, namely, to make the bases of degrees become the same.
Let's look carefully at 4 3. The basis of this degree is the number 4, which can be represented as 2 2. Then the initial expression takes the form (2 2) 3 × 2 2. By exposing the exercise to the degree in expression (2 2) 3, we will receive 2 6. Then the initial expression takes the form 2 6 × 2 2, calculated which can be used using the main property of the degree.
Write the solution to this example:
Decision degrees
To make the division of degrees, you need to find the value of each degree, then make the division of ordinary numbers.
For example, we divide 4 3 to 2 2.
Calculate 4 3, we get 64. Calculate 2 2, we get 4. Now we divide 64 to 4, we get 16
If, when dividing degrees of bases, they will be the same, then the base can be left unchanged, and from the indicator of the degree of divide, the indicator of the degree of divider.
For example, find the value of expression 2 3: 2 2
The base 2 will be left unchanged, and from the degree of divide, the degree of divider is submitted:
So, the value of expression 2 3: 2 2 is 2.
This property is based on multiplying degrees with the same bases, or as we used to speak on the main property of the degree.
Let's return to the previous example 2 3: 2 2. Here divide it is 2 3, and divider 2 2.
Divide the same number to another means to find such a number that, when multiplying the divider, will give a divider as a result.
In our case, divided 2 3 to 2 2 means to find such a degree that, when multiplying the divider 2 2, will give as a result of 2 3. What degree can be multiplied by 2 2 to get 2 3? Obviously, only a degree 2 1. From the main property of the degree we have:
Ensure that the value of expression 2 3: 2 2 is 2 1 can be directly calculated by the expression 2 3: 2 2. To do this, first find the value of the degree 2 3, we get 8. Then we will find the value of the degree 2 2, we get 4. We divide 8 to 4, we obtain 2 or 2 1, since 2 \u003d 2 1.
2 3: 2 2 = 8: 4 = 2
Thus, when dividing degrees with the same bases, the following equality is performed:
It may happen and so that not only the foundations can be the same, but also indicators. In this case, the answer will be able to answer.
For example, find the value of expression 2 2: 2 2. Calculate the value of each degree and execute the division of the resulting numbers:
When solving an example 2 2: 2 2, you can also apply the degree separation rule with the same bases. The result is a number to zero degree, since the difference in degrees 2 2 and 2 2 is zero:
Why is the number 2 to zero degree equal to one we found out higher. If you calculate 2 2: 2 2 by the usual method, without using the degree division rule, the unit will be.
Example 2.. Find an expression value 4 12: 4 10
4 We will leave unchanged, and from the indicator of the dividend submitter the degree of the divider:
4 12: 4 10 = 4 12 − 10 = 4 2 = 16
Example 3.. Present private x. 3: x. in the form of a degree x.
We use the rule of dividing degrees. Base x. Let us leave unchanged, and from the indicator of the dividend submount the degree of the divider. The indicator of the divider is equal to one. For clarity, write it:
Example 4.. Present private x. 3: x. 2 in the form of a degree with the base x.
We use the rule of dividing degrees. Base x.
Decision degrees can be recorded in the form of a fraction. So, the previous example can be written as follows:
The numerator and denominator of the fraci is allowed to record in an expanded form, namely in the form of works of the same multipliers. Power x. 3 can write as x × x × x and degree x. 2 as x × x. . Then design x. 3 - 2 can be skipped and take advantage of the cutting of the fraction. In the numerator and in the denominator it will be possible to cut two factors x. . As a result, one multiplier will remain x.
Or even shorter:
It is also useful to be able to quickly cut the fractions consisting of degrees. For example, the fraction can be reduced by x. 2. To reduce the fraction on x. 2 need a numerator and denominator of a fraction to divide on x. 2
Decision degrees can not be painted in detail. The reduced reduction can be in short:
Or even shorter:
Example 5.. Perform division x. 12 : X. 3
We use the rule of dividing degrees. Base x. Let us leave unchanged, and from the degree of dividing the degree of divider:
We write down the decision by reducing the fraction. Decision degrees x. 12 : X. 3 Write in the form. Next will reduce this fraction on x. 3 .
Example 6.. Find an expression value
In the numerator, you will perform multiplication of degrees with the same bases:
Now apply the degree detection rule with the same bases. The base 7 is left unchanged, and from the degree of dividing the degree of divider:
Complete an example, calculating degree 7 2
Example 7.. Find an expression value
Perform a degree in the extent to the degree. Make it needed with expression (2 3) 4
Now performed in the numerator multiplication of degrees with the same bases.
How to multiply the degree? What degrees can multiply, and which is not? How to multiply the degree?
In the algebra to find a product of degrees in two cases:
1) if the degrees have the same bases;
2) If the degrees have the same indicators.
When multiplying degrees with the same bases, it is necessary to leave the base for the same, and the indicators are folded:
When multiplying degrees with the same indicators, the overall indicator can be reached by braces:
Consider how to multiply the degrees on specific examples.
The unit is not written in an indicator, but when multiplying degrees - take into account:
When multiplying, the number of degrees can be any. It should be remembered that in front of the lettering sign of multiplication can not write:
In expressions, the construction of the extent is performed first.
If the number is needed to multiply to the degree, you must first be raised to the degree, and only later - multiplication:
www.algebraclass.ru.
Addition, subtraction, multiplication, and dividing degrees
Addition and subtraction of degrees
Obviously, numbers with degrees can be accurate as other values , by adding them one after another with their signs.
So, the sum A 3 and B 2 is a 3 + b 2.
The sum A 3 - B n and H 5 -D 4 is A 3 - B N + H 5 - D 4.
Factors identical degrees of the same variables May be designed or deducted.
Thus, the amount 2a 2 and 3a 2 is 5a 2.
It is also obvious that if you take two squares a, or three squares a, or five squares a.
But degrees different variables and various degrees identical variablesmust be made by their addition with their signs.
So, the sum A 2 and A 3 is the sum A 2 + A 3.
This is obvious that the square of the number A, and the Cube of the number A, not equal to a double square A, but a double Cuba a.
Amount A 3 B n and 3a 5 B 6 is A 3 B N + 3A 5 B 6.
Subtraction The degrees are carried out in the same way as addition, except that the signs of subtractable must be changed accordingly.
Or:
2a 4 - (-6a 4) \u003d 8A 4
3H 2 B 6 - 4H 2 B 6 \u003d -H 2 B 6
5 (a - h) 6 - 2 (a - h) 6 \u003d 3 (a - h) 6
Multiplying degrees
Numbers with degrees can be multiplied by other values \u200b\u200bby writing them one after another, with a sign of multiplication or without it between them.
Thus, the result of multiplication A 3 on B 2 is a 3 b 2 or aaabb.
Or:
x -3 ⋅ a m \u003d a m x -3
3A 6 Y 2 ⋅ (-2x) \u003d -6A 6 XY 2
a 2 B 3 Y 2 ⋅ A 3 B 2 Y \u003d A 2 B 3 Y 2 A 3 B 2 y
The result in the latter example can be ordered by the addition of the same variables.
The expression will take the form: A 5 B 5 Y 3.
Comparing several numbers (variables) with degrees, we can see that if any two of them are multiplied, the result is the number (variable) with a degree equal to sum Degrees of the terms.
So, a 2 .a 3 \u003d aa.aaa \u003d aaaaa \u003d a 5.
Here 5 is the degree of multiplication result, equal to 2 + 3, the sum of the degrees of the components.
So, a n .a m \u003d a m + n.
For a n, A is taken as the multiplier as many times as the degree n;
And a m, takes as a multiplier as many times as the degree m;
Therefore, the degrees with the same bases can be multiplied by the addition of degrees.
So, a 2 .a 6 \u003d a 2 + 6 \u003d a 8. And x 3 .x 2 .x \u003d x 3 + 2 + 1 \u003d x 6.
Or:
4a N ⋅ 2a n \u003d 8a 2n
b 2 y 3 ⋅ b 4 y \u003d b 6 y 4
(B + H - Y) N ⋅ (B + H - Y) \u003d (B + H - Y) n + 1
Multiply (x 3 + x 2 y + xy 2 + y 3) ⋅ (x - y).
Answer: x 4 - y 4.
Multiply (x 3 + x - 5) ⋅ (2x 3 + x + 1).
This rule is valid for numbers, the degree of which - negative.
1. So, A -2 .a -3 \u003d a -5. This can be written in the form (1 / AA). (1 / AAA) \u003d 1 / AAAAA.
2. Y -n .y -m \u003d y -n-m.
3. A -N .A M \u003d A M-N.
If A + B is multiplied by A - B, the result will be equal to A 2 - B 2: That is
The result of multiplication of the amount or difference of two numbers is equal to the sum or difference of their squares.
If the sum is multiplied and the difference of two numbers erected into square, the result will be equal to the amount or difference of these numbers in fourth degree.
So, (a - y). (A + Y) \u003d A 2 - Y 2.
(A 2 - Y 2) ⋅ (A 2 + Y 2) \u003d A 4 \u200b\u200b- Y 4.
(A 4 - Y 4) ⋅ (a 4 + y 4) \u003d a 8 - y 8.
Decision degrees
The numbers with degrees can be divided, like other numbers, by taking a divide divider, or the placement of them in the form of a fraction.
Thus, a 3 b 2 divided by b 2, equal to A 3.
Recording A 5 divided by A 3 looks like $ \\ FRAC $. But this is equal to A 2. In a number of numbers
a +4, A +3, A +2, A +1, A 0, A -1, A -2, A -3, A -4.
any number can be divided into another, and the degree will be equal to difference Indicators of divisible numbers.
When dividing degrees with the same base, their indicators are deducted..
So, y 3: y 2 \u003d y 3-2 \u003d y 1. That is, $ \\ frac \u003d y $.
And a n + 1: a \u003d a n + 1-1 \u003d a n. That is, $ \\ frac \u003d a ^ n $.
Or:
y 2m: y m \u003d y m
8A n + m: 4a m \u003d 2a n
12 (B + Y) N: 3 (B + Y) 3 \u003d 4 (B + Y) N-3
The rule is also fair and for numbers with negative values \u200b\u200bof degrees.
The result of division A -5 on A -3 is equal to A -2.
Also, $ \\ FRAC: \\ FRAC \u003d \\ FRAC. \\ FRAC \u003d \\ FRAC \u003d \\ FRAC $.
h 2: H -1 \u003d H 2 + 1 \u003d H 3 or $ H ^ 2: \\ FRAC \u003d H ^ 2. \\ FRAC \u003d H ^ $ 3
It is necessary to very well assimilate the multiplication and division of degrees, as such operations are very widely used in algebra.
Examples of solving examples with fractions containing numbers with degrees
1. Reduce degrees in $ \\ FRAC $ reply: $ \\ FRAC $.
2. Reduce degrees in $ \\ FRAC $. Answer: $ \\ FRAC $ or 2x.
3. Reduce the degrees of A 2 / A 3 and A -3 / A -4 and bring to a common denominator.
a 2 .A -4 is a -2 first numerator.
a 3 .A -3 is a 0 \u003d 1, the second numerator.
a 3 .A -4 is a -1, a common numerator.
After simplification: A -2 / A -1 and 1 / A -1.
4. Reduce the indicators of the degrees 2a 4 / 5a 3 and 2 / a 4 and bring to the common denominator.
Answer: 2A 3 / 5A 7 and 5A 5 / 5A 7 or 2A 3 / 5A 2 and 5 / 5A 2.
5. Multiply (A 3 + B) / B 4 on (A - B) / 3.
6. Multiply (A 5 + 1) / x 2 on (B 2 - 1) / (X + A).
7. Multiply B 4 / A -2 on H -3 / X and A N / Y -3.
8. Divide A 4 / Y 3 on A 3 / Y 2. Answer: A / Y.
Properties of degree
We remind you that in this lesson you understand properties of degrees with natural indicators and zero. The degrees with rational indicators and their properties will be considered in lessons for 8 classes.
The ratio with a natural indicator has several important properties that allow you to simplify calculations in examples with degrees.
Property number 1.
The work of degrees
When multiplying degrees with the same bases, the base remains unchanged, and the indicators of degrees are folded.
a m · a n \u003d a m + n, where "A" is any number, and "M", "N" - any natural numbers.
This property of degrees also acts on the work of three and more degrees.
b · b 2 · b 3 · b 4 · b 5 \u003d b 1 + 2 + 3 + 4 + 5 \u003d b 15
6 15 · 36 \u003d 6 15 · 6 2 \u003d 6 15 · 6 2 \u003d 6 17
(0.8) 3 · (0.8) 12 \u003d (0.8) 3 + 12 \u003d (0.8) 15
Note that in the specified property it was only about multiplying degrees with the same bases. . It does not apply to their addition.
It is impossible to replace the amount (3 3 + 3 2) by 3 5. This is understandable if
calculate (3 3 + 3 2) \u003d (27 + 9) \u003d 36, a 3 5 \u003d 243
Property number 2.
Private degree
When dividing degrees with the same bases, the base remains unchanged, and from the indicator of the division deductible the degree of divider.
(2b) 5: (2b) 3 \u003d (2b) 5 - 3 \u003d (2b) 2
11 3 - 2 · 4 2 - 1 \u003d 11 · 4 \u003d 44
Example. Solve equation. We use the property of private degrees.
3 8: T \u003d 3 4
Answer: T \u003d 3 4 \u003d 81
Using properties No. 1 and No. 2, you can easily simplify expressions and make calculations.
- Example. Simplify the expression.
4 5m + 6 · 4 m + 2: 4 4m + 3 \u003d 4 5m + 6 + m + 2: 4 4m + 3 \u003d 4 6m + 8 - 4m - 3 \u003d 4 2m + 5
Example. Find the value of the expression using the degree properties.
2 11 − 5 = 2 6 = 64
Please note that in the property 2 it was only about dividing degrees with the same bases.
It is impossible to replace the difference (4 3 -4 2) by 4 1. This is understandable if you calculate (4 3 -4 2) \u003d (64 - 16) \u003d 48, a 4 1 \u003d 4
Property number 3.
Erect
When erecting the degree to the degree, the foundation remains unchanged, and the indicators of degrees are variable.
(a n) m \u003d a n · m, where "A" is any number, and "M", "N" - any natural numbers.
Please note that property number 4, as well as other properties of degrees, apply in reverse order.
(a n · b n) \u003d (a · b) n
That is, in order to multiply the degrees with the same indicators, it is possible to multiply the bases, and the degree indicator is unchanged.
2 4 · 5 4 \u003d (2 · 5) 4 \u003d 10 4 \u003d 10 000
0.5 16 · 2 16 \u003d (0,5 · 2) 16 \u003d 1
In more complex examples, there may be cases when multiplication and division must be performed above degrees with different bases and different indicators. In this case, we recommend to act as follows.
For example, 4 5 · 3 2 \u003d 4 3 · 4 2 · 3 2 \u003d 4 3 · (4 · 3) 2 \u003d 64 · 12 2 \u003d 64 · 144 \u003d 9216
Example of decimal fraction.
4 21 · (-0.25) 20 \u003d 4 · 4 20 · (-0.25) 20 \u003d 4 · (4 · (-0.25)) 20 \u003d 4 · (-1) 20 \u003d 4 · 1 \u003d four
Properties 5.
Private degree (fraction)
To invite the degree in private, you can build a separate and divider into this degree, and the first result is divided into the second.
(A: b) n \u003d a n: b n, where "a", "b" - any rational numbers, b ≠ 0, n - any natural number.
(5: 3) 12 = 5 12: 3 12
We remind you that private can be represented as a fraction. Therefore, on the topic, we will focus more in more detail on the next page.
Degrees and roots
Operations with degrees and roots. Degree with negative ,
zero and fractional indicator. About expressions that do not make sense.
Operations with degrees.
1. When multiplying degrees with the same base, their indicators fold:
a M. · a n \u003d a m + n.
2. When dividing degrees with the same basis, their indicators remove .
3. The degree of work of two or several woggles is equal to the work of the degrees of these factors.
4. The degree of relation (fracted) is equal to the ratio of degrees of the divide (numerator) and the divider (denominator):
(a / B.) n \u003d a n / b n.
5. When erecting a degree to the degree, their indicators are multiplied:
All of the above formulas are read and are performed in both directions from left to right and vice versa.
PRI MERS (2 · 3 · 5/15) ² = 2 ² · 3 ² · 5 ² / 15 ² \u003d 900/225 \u003d 4 .
Root operations. In all the following formulas, the symbol means arithmetic root (Courted expression positively).
1. The root of the work of several womb is equal to the product of the roots of these factors:
2. The root from the relationship is equal to the attitude of the roots of the divide and divider:
3. When the root is erected, it is enough to build this degree subject:
4. If you increase the degree of root in M \u200b\u200btimes and at the same time build a feed number into a m-degree, the root value will not change:
5. If you reduce the degree of root in M \u200b\u200btimes and at the same time remove the root of the M-degree from the feed number, the root value will not change:
Expansion of the concept of degree. So far, we have considered degrees only with a natural indicator; But the actions with degrees and roots can also lead to negative, zero and fractional Indicators. All these indicators of degrees require additional definition.
Degree with a negative indicator. The degree of a certain number with a negative (whole) indicator is defined as a unit divided by the degree of the same number with an indicator equal to the absolute veliver of the negative indicator:
T Heathe formula a M. : a N. = a M - N can be used not only at m. more than n. but also m. less than n. .
PRI MERS a. 4: a. 7 \u003d A. 4 — 7 \u003d A. — 3 .
If we want the formula a M. : a N. = a M. — n. It was fair for m \u003d N. We need to determine zero degree.
The degree with the zero indicator. The degree of any nonzero number with zero is equal to 1.
PRI MERS. 2 0 \u003d 1, ( – 5) 0 = 1, (– 3 / 5) 0 = 1.
Degree with fractional indicator. In order to build a valid number A into the degree M / N, it is necessary to extract the root of the N-degree from M-degree of this number A:
About expressions that do not make sense. There are several such expressions.
where a. ≠ 0 , does not exist.
In fact, assuming that x. - Some number, then in accordance with the definition of the division operation, we have: a. = 0· x.. a. \u003d 0, which contradicts the condition: a. ≠ 0
— any number.
In fact, assuming that this expression is equal to some number x., according to the definition of the division operation, we have: 0 \u003d 0 · x. . But this equality takes place when any number X.As required to prove.
0 0 — any number.
Consider three fundamental cases:
1) x. = 0 – This value does not satisfy this equation.
2) for x. \u003e 0 We get: x / X. \u003d 1, i.e. 1 \u003d 1, from where it follows
what x. - any number; But taking into account that in
oUR Case x. \u003e 0, answer is x. > 0 ;
Decision multiplication rules with different bases
A degree with a rational indicator
Power Function IV.
§ 69. Multiplication and division of degrees with the same grounds
Theorem 1. In order to multiply the degrees with the same bases, it is sufficient to add degrees, and the reason to leave the former, that is
Evidence. By definition of degree
2 2 2 3 = 2 5 = 32; (-3) (-3) 3 = (-3) 4 = 81.
We looked at the work of two degrees. In fact, the proven property is true for any number of degrees with the same bases.
Theorem 2. To divide degrees with the same bases when the divisory indicator is greater than the divider indicator, it is sufficient from the indicator of the dividend deductory indicator, and the base is left for the same, that is for t\u003e P.
(a. =/= 0)
Evidence. Recall that private from dividing one number to another is called the number that, when multiplying the divider, gives divisible. Therefore, to prove the formula where a. \u003d / \u003d 0, it does not care what to prove the formula
If a t\u003e P. , then t - P. will be natural; Consequently, by Theorem 1
Theorem 2 is proved.
It should be paid to the fact that the formula
we are proved only in the assumption that t\u003e P. . Therefore, of the proven can not be done yet, for example, such conclusions:
In addition, with negative indicators, we have not yet been considered and we still do not know what sense can I give expression 3 - 2 .
Theorem 3. To build a degree to a degree, rather multiply the indicators, leaving the reason for the degree of the degree, i.e
Evidence. Using the degree of degree and theorem of 1 of this paragraph, we get:
q.E.D.
For example, (2 3) 2 \u003d 2 6 \u003d 64;
518 (orally.) Determine h. From equations:
1) 2 2 2 2 3 2 4 2 5 2 6 = 2 x. ; 3) 4 2 4 4 4 6 4 8 4 10 = 2 x. ;
2) 3 3 3 3 5 3 7 3 9 = 3 x. ; 4) 1 / 5 1 / 25 1 / 125 1 / 625 = 1 / 5 x. .
519. (S T N O.) Simplify:
520. (U with tn about.) Simplify:
521. These expressions are submitted in the form of degrees with the same bases:
1) 32 and 64; 3) 8 5 and 16 3; 5) 4 100 and 32 50;
2) -1000 and 100; 4) -27 and -243; 6) 81 75 8 200 and 3,600 4 150.
Lesson on the topic: "Rules for multiplication and division of degrees with the same and different indicators. Examples"
Additional materials
Dear users, do not forget to leave your comments, reviews, wishes. All materials are checked by antivirus program.
Training manuals and simulators in the online store "Integral" for grade 7
Manual for textbook Yu.N. Makarychev benefit to the textbook A.G. Mordkovich
The purpose of the lesson: learns to perform actions with the degrees of the number.
To begin with, remember the concept of "degree of number". The expression of the type $ \\ underbrace (A * A * \\ ldots * a) _ (n) $ can be represented as $ a ^ n $.
It is also true inverse: $ a ^ n \u003d \\ underbrace (A * A * \\ ldots * a) _ (n) $.
This equality is called "a degree record in the form of a work." It will help us determine how to multiply and share degrees.
Remember:
a. - The foundation of the degree.
n. - Indicator.
If a n \u003d 1., So, the number but They took once and, accordingly: $ a ^ n \u003d a $.
If a n \u003d 0., then $ a ^ 0 \u003d 1 $.
Why this happens, we will be able to find out when we will get acquainted with the rules of multiplication and division of degrees.
Multiplication rules
a) If degrees are multiplied with the same base.To $ a ^ n * a ^ m $, write down the degree in the form of a work: $ \\ underbrace (a * a * \\ ldots * a) _ (n) * \\ underbrace (a * a * \\ ldots * a) _ (m ) $.
Figure shows that the number but have taken n + M. Once, then $ a ^ n * a ^ m \u003d a ^ (n + m) $.
Example.
$2^3 * 2^2 = 2^5 = 32$.
This property is convenient to use what to simplify work when erecting a number to a greater degree.
Example.
$2^7= 2^3 * 2^4 = 8 * 16 = 128$.
b) If degrees are multiplied with different bases, but the same indicator.
To $ a ^ n * b ^ n $, write down the degree in the form of a work: $ \\ underbrace (a * a * \\ ldots * a) _ (n) * \\ underbrace (b * b * \\ ldots * b) _ (m ) $.
If you change the multiplier places and calculate the resulting pairs, we get: $ \\ underbrace ((a * b) * (A * B) * \\ ldots * (A * B)) _ (n) $.
So, $ a ^ n * b ^ n \u003d (a * b) ^ n $.
Example.
$3^2 * 2^2 = (3 * 2)^2 = 6^2= 36$.
Rules of division
a) The base of the degree is the same, different indicators.Consider dividing the degree with a high figure for dividing the degree with a smaller indicator.
So, it is necessary $ \\ FRAC (a ^ n) (a ^ m) $where n\u003e M..
We write a degree in the form of a fraction:
$ \\ FRAC (\\ underbrace (A * A * \\ ldots * a) _ (n)) (\\ underbrace (a * a * \\ ldots * a) _ (m)) $.
For convenience, the division will write in the form of a simple fraction.Now will reduce the fraction.
It turns out: $ \\ underbrace (a * a * \\ ldots * a) _ (n-m) \u003d a ^ (n-m) $.
It means $ \\ FRAC (a ^ n) (a ^ m) \u003d a ^ (n-M) $.
This property will help explain the situation with the erection of the number to the zero degree. Suppose that n \u003d M., then $ a ^ 0 \u003d a ^ (n - n) \u003d \\ FRAC (A ^ n) (a ^ n) \u003d 1 $.
Examples.
$ \\ FRAC (3 ^ 3) (3 ^ 2) \u003d 3 ^ (3-2) \u003d 3 ^ 1 \u003d 3 $.
$ \\ FRAC (2 ^ 2) (2 ^ 2) \u003d 2 ^ (2-2) \u003d 2 ^ 0 \u003d 1 $.
b) the foundation of the degree is different, the indicators are the same.
Suppose it is necessary $ \\ FRAC (A ^ n) (B ^ n) $. We write the degree of numbers in the form of a fraction:
$ \\ FRAC (\\ underbrace (A * A * \\ ldots * a) _ (n)) (\\ underbrace (b * b * \\ ldots * b) _ (n)) $.
For convenience imagine.
Using the fraction property, we break a big fraction on the work of small, we get.
$ \\ underbrace (\\ FRAC (A) (B) * \\ FRAC (A) (B) * \\ ldots * \\ FRAC (A) (B)) _ (n) $.
Accordingly, $ \\ FRAC (A ^ n) (B ^ n) \u003d (\\ FRAC (A) (B)) ^ n $.
Example.
$ \\ FRAC (4 ^ 3) (2 ^ 3) \u003d (\\ FRAC (4) (2)) ^ 3 \u003d 2 ^ 3 \u003d $ 8.
Consider the topic of transforming expressions with degrees, but first let's stop at a number of transformations that can be carried out with any expressions, including with power. We will learn to reveal brackets, bring similar terms, to work with the basis and indicator of the degree, use the properties of degrees.
What are powerful expressions?
In the school year, few people use the phrase "powerful expressions", but this term is constantly meeting in collections to prepare for the exam. In most cases, phrases are indicated by expressions that contain in their degree records. This is we reflect in our definition.
Definition 1.
Power expression - This is an expression that contains degrees.
Let us give a few examples of power expressions, starting with the degree with a natural indicator and ending with the real indicator.
The most simple power expressions can be considered the degree of the number with the natural indicator: 3 2, 7 5 + 1, (2 + 1) 5, (- 0, 1) 4, 2 2 3 3, 3 · A 2 - A + A 2, x 3 - 1, (a 2) 3. As well as degrees with zero indicator: 5 0, (A + 1) 0, 3 + 5 2 - 3, 2 0. And degrees with whole negative degrees: (0, 5) 2 + (0, 5) - 2 2.
Easily more difficult to work with a degree having rational and irrational indicators: 264 1 4 - 3 · 3 · 3 1 2, 2 3, 5 · 2 - 2 2 - 1, 5, 1 A 1 4 · A 1 2 - 2 · A - 1 6 · b 1 2, x π · x 1 - π, 2 3 3 + 5.
As an indicator, a variable 3 x - 54 - 7 · 3 x - 58 or logarithm can be x 2 · L g x - 5 · x L g x.
With the question of what power expressions are, we figured out. Now we will deal with their conversion.
The main types of transformations of power expressions
First of all, we will consider the basic identity transformations of expressions that can be performed with power expressions.
Example 1.
Calculate the value of the power expression 2 3 · (4 2 - 12).
Decision
All transformations we will be carried out in compliance with the procedure for performing actions. In this case, we begin with the implementation of actions in brackets: replace the degree of digital value and calculate the difference of two numbers. Have 2 3 · (4 2 - 12) \u003d 2 3 · (16 - 12) \u003d 2 3 · 4.
We still have to replace the degree 2 3 His meaning 8 and calculate the work 8 · 4 \u003d 32. Here is our answer.
Answer: 2 3 · (4 2 - 12) \u003d 32.
Example 2.
Simplify expression with degrees 3 · a 4 · b - 7 - 1 + 2 · a 4 · b - 7.
Decision
An expression given to us in terms of the task contains similar terms that we can lead: 3 · a 4 · b - 7 - 1 + 2 · a 4 · b - 7 \u003d 5 · a 4 · b - 7 - 1.
Answer: 3 · a 4 · b - 7 - 1 + 2 · a 4 · b - 7 \u003d 5 · a 4 · b - 7 - 1.
Example 3.
Prepare an expression with degrees 9 - b 3 · π - 1 2 as a piece.
Decision
Imagine number 9 as a degree 3 2 and apply the formula of abbreviated multiplication:
9 - B 3 · π - 1 2 \u003d 3 2 - B 3 · π - 1 2 \u003d \u003d 3 - B 3 · π - 1 3 + B 3 · π - 1
Answer: 9 - B 3 · π - 1 2 \u003d 3 - B 3 · π - 1 3 + B 3 · π - 1.
And now we turn to the analysis of identical transformations that can be applied precisely in relation to power expressions.
Work with the basis and indicator of the degree
The degree in the base or indicator may also have numbers, variables, and some expressions. For example, (2 + 0, 3 · 7) 5 - 3, 7 and . Work with such entries is difficult. It is much easier to replace the expression at the base of the degree or expression in the indicator identically equal to an expression.
Degree and indicator transformations are carried out according to the rules known to us separately from each other. The most important thing is that as a result of the transformation, an expression is identical to the initial one.
The purpose of the transformations is to simplify the initial expression or get the solution to the problem. For example, in the example, which we led above, (2 + 0, 3 · 7) 5 - 3, 7, you can perform actions to transition to degree 4 , 1 1 , 3 . Open brackets, we can lead similar terms at the bottom (A · (A + 1) - a 2) 2 · (x + 1) and get a powerful expression of a simpler type a 2 · (x + 1).
Use the properties of degrees
The properties of degrees recorded in the form of equalities are one of the main tools for transforming expressions with degrees. Here are the main of them, given that A. and B. - these are any positive numbers, and R. and S. - arbitrary valid numbers:
Definition 2.
- a r · a s \u003d a r + s;
- a r: a s \u003d a r - s;
- (a · b) r \u003d a r · b r;
- (A: B) R \u003d A R: B R;
- (A R) S \u003d A R · s.
In cases where we are dealing with natural, integer, positive indicators of the degree, the limitations on the number A and B may be much less strict. So, for example, if we consider equality a m · a n \u003d a m + nwhere M. and N. - natural numbers, it will be true for any values \u200b\u200bof A, both positive and negative, as well as for a \u003d 0..
It is possible to apply the properties of degrees without restrictions in cases where the bases of degrees are positive or contain variables, the area of \u200b\u200bpermissible values \u200b\u200bof which is such that only positive values \u200b\u200bare taken on it. In fact, in the framework of the school program on mathematics, the student's task is to choose the appropriate property and its correct application.
When preparing for admission to universities, tasks may occur in which the inaccient use of properties will lead to a narrowing of OTZ and other difficulties with the solution. In this section, we will analyze only two such cases. More information on the issue can be found in the topic "Transformation of expressions using the properties of degrees".
Example 4.
Imagine an expression A 2, 5 · (A 2) - 3: A - 5, 5 in the form of a degree A..
Decision
To begin with, we use the exercise property and we transform the second factor on it. (A 2) - 3 . Then use the properties of multiplication and division of degrees with the same base:
a 2, 5 · a - 6: a - 5, 5 \u003d a 2, 5 - 6: a - 5, 5 \u003d a - 3, 5: a - 5, 5 \u003d a - 3, 5 - (- 5, 5) \u003d a 2.
Answer: A 2, 5 · (A 2) - 3: A - 5, 5 \u003d A 2.
The transformation of power expressions according to the property of degrees can be made both from left to right and in the opposite direction.
Example 5.
Find the value of the power expression 3 1 3 · 7 1 3 · 21 2 3.
Decision
If we apply equality (A · b) r \u003d a r · b r, right to left, then we get a product of the form 3 · 7 1 3 · 21 2 3 and further 21 1 3 · 21 2 3. Moving the indicators when multiplying degrees with the same bases: 21 1 3 · 21 2 3 \u003d 21 1 3 + 2 3 \u003d 21 1 \u003d 21.
There is another way to carry out conversion:
3 1 3 · 7 1 3 · 21 2 3 \u003d 3 1 3 · 7 1 3 · (3 · 7) 2 3 \u003d 3 1 3 · 7 1 3 · 3 2 3 · 7 2 3 \u003d 3 1 3 · 3 2 3 · 7 1 3 · 7 2 3 \u003d 3 1 3 + 2 3 · 7 1 3 + 2 3 \u003d 3 1 · 7 1 \u003d 21
Answer: 3 1 3 · 7 1 3 · 21 2 3 \u003d 3 1 · 7 1 \u003d 21
Example 6.
Power expression is given A 1, 5 - A 0, 5 - 6Enter a new variable T \u003d A 0, 5.
Decision
Imagine a degree A 1, 5 as a 0, 5 · 3 . Use the degree property to the degree (a r) s \u003d a r · s On the right left and obtain (a 0, 5) 3: a 1, 5 - a 0, 5 - 6 \u003d (a 0, 5) 3 - a 0, 5 - 6. In the resulting expression, you can easily enter a new variable. T \u003d A 0, 5: Receive T 3 - T - 6.
Answer: T 3 - T - 6.
Transformation of fractions containing degrees
We usually deal with two variants of power expressions with fractions: the expression is a fraction with a degree or contains such a fraction. These expressions apply all major transformations of fractions without restrictions. They can be reduced, lead to a new denominator, work separately with a numerator and denominator. We illustrate this by examples.
Example 7.
Simplify the power expression 3 · 5 2 3 · 5 1 3 - 5 - 2 3 1 + 2 · x 2 - 3 - 3 · x 2.
Decision
We are dealing with a fraction, so we carry out transformations in the numerator, and in the denominator:
3 · 5 2 3 · 5 1 3 - 5 - 2 3 1 + 2 · x 2 - 3 - 3 · x 2 \u003d 3 · 5 2 3 · 5 1 3 - 3 · 5 2 3 · 5 - 2 3 - 2 - x 2 \u003d \u003d 3 · 5 2 3 + 1 3 - 3 · 5 2 3 + - 2 3 - 2 - x 2 \u003d 3 · 5 1 - 3 · 5 0 - 2 - x 2
Position minus before the fraction in order to change the sign of the denominator: 12 - 2 - x 2 \u003d - 12 2 + x 2
Answer: 3 · 5 2 3 · 5 1 3 - 5 - 2 3 1 + 2 · x 2 - 3 - 3 · x 2 \u003d - 12 2 + x 2
The fractions containing degrees are given to the new denominator in exactly as well as rational fractions. To do this, you need to find an additional multiplier and multiply the numerator and denominator of the fraction. It is necessary to select an additional factor in such a way that it does not apply to zero under any values \u200b\u200bof the variables from the odd variables for the initial expression.
Example 8.
Give fractions to a new denominator: a) A + 1 A 0, 7 to the denominator A., b) 1 x 2 3 - 2 · x 1 3 · y 1 6 + 4 · y 1 3 to denominator X + 8 · Y 1 2.
Decision
a) We will select a multiplier who will allow us to bring to a new denominator. a 0, 7 · a 0, 3 \u003d a 0, 7 + 0, 3 \u003d a,therefore, as an additional multiplier we will take A 0, 3. The area of \u200b\u200bpermissible values \u200b\u200bof the variable A includes many of all positive valid numbers. In this area A 0, 3 Not accessed to zero.
Perform multiplication of the numerator and denominator of the fraction on A 0, 3:
a + 1 a 0, 7 \u003d a + 1 · a 0, 3 a 0, 7 · a 0, 3 \u003d a + 1 · a 0, 3 a
b) pay attention to the denominator:
x 2 3 - 2 · x 1 3 · y 1 6 + 4 · y 1 3 \u003d x 1 3 2 - x 1 3 · 2 · y 1 6 + 2 · y 1 6 2
Multiply this expression on x 1 3 + 2 · Y 1 6, we obtain the sum of cubes x 1 3 and 2 · y 1 6, i.e. X + 8 · Y 1 2. This is our new denominator to which we need to bring the original fraction.
So we found an additional multiplier X 1 3 + 2 · Y 1 6. On the area of \u200b\u200bpermissible values \u200b\u200bof variables X. and Y. The expression x 1 3 + 2 · Y 1 6 does not turn to zero, so we can multiply the numerator and denominator of the fraction:
1 x 2 3 - 2 · x 1 3 · y 1 6 + 4 · y 1 3 \u003d x 1 3 + 2 · y 1 6 x 1 3 + 2 · y 1 6 x 2 3 - 2 · x 1 3 · y 1 6 + 4 · y 1 3 \u003d x 1 3 + 2 · y 1 6 x 1 3 3 + 2 · y 1 6 3 \u003d x 1 3 + 2 · y 1 6 x + 8 · y 1 2
Answer: a) a + 1 a 0, 7 \u003d a + 1 · a 0, 3 a, b) 1 x 2 3 - 2 · x 1 3 · y 1 6 + 4 · y 1 3 \u003d x 1 3 + 2 · y 1 6 x + 8 · y 1 2.
Example 9.
Reduce the fraction: a) 30 · x 3 · (x 0, 5 + 1) · x + 2 · x 1 1 3 - 5 3 45 · x 0, 5 + 1 2 · x + 2 · x 1 1 3 - 5 3, b) a 1 4 - b 1 4 A 1 2 - B 1 2.
Decision
a) We use the largest common denominator (node) to which the numerator and denominator can be reduced. For numbers 30 and 45, this is 15. We can also reduce on x 0, 5 + 1 and on x + 2 · x 1 1 3 - 5 3.
We get:
30 · x 3 · (x 0, 5 + 1) · x + 2 · x 1 1 3 - 5 3 45 · x 0, 5 + 1 2 · x + 2 · x 1 1 3 - 5 3 \u003d 2 · x 3 3 · (x 0, 5 + 1)
b) Here the presence of the same multipliers is not obvious. You will have to perform some conversions in order to obtain the same multipliers in a numerator and denominator. To do this, lay an denominator using the square difference formula:
a 1 4 - B 1 4 A 1 2 - B 1 2 \u003d A 1 4 - B 1 4 A 1 4 2 - B 1 2 2 \u003d A 1 4 - B 1 4 A 1 4 + B 1 4 · A 1 4 - B 1 4 \u003d 1 A 1 4 + B 1 4
Answer:a) 30 · x 3 · (x 0, 5 + 1) · x + 2 · x 1 1 3 - 5 3 45 · x 0, 5 + 1 2 · x + 2 · x 1 1 3 - 5 3 \u003d 2 · X 3 3 · (x 0, 5 + 1), b) a 1 4 - b 1 4 A 1 2 - b 1 2 \u003d 1 A 1 4 + B 1 4.
The essential actions with fractions include bringing to a new denominator and cutting fractions. Both actions are performed in compliance with a number of rules. When adding and subtracting fractions first, the fractions are given to a common denominator, after which actions (addition or subtraction) are carried out with numerals. The denominator remains the same. The result of our actions is a new fraction, the numerator of which is the product of numerators, and the denominator is a product of denominators.
Example 10.
Perform actions x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 · 1 x 1 2.
Decision
Let's start with the subtraction of fractions that are located in brackets. We give them to the general denominator:
x 1 2 - 1 · x 1 2 + 1
Subscribe numbers:
x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 · 1 x 1 2 \u003d x 1 2 + 1 · x 1 2 + 1 x 1 2 - 1 · x 1 2 + 1 - x 1 2 - 1 · x 1 2 - 1 x 1 2 + 1 · x 1 2 - 1 · 1 x 1 2 \u003d x 1 2 + 1 2 - x 1 2 - 1 2 x 1 2 - 1 · x 1 2 + 1 · 1 x 1 2 \u003d x 1 2 2 + 2 · x 1 2 + 1 - x 1 2 2 - 2 · x 1 2 + 1 x 1 2 - 1 · x 1 2 + 1 · 1 x 1 2 \u003d \u003d 4 · x 1 2 x 1 2 - 1 · x 1 2 + 1 · 1 x 1 2
Now we multiply the fractions:
4 · x 1 2 x 1 2 - 1 · x 1 2 + 1 · 1 x 1 2 \u003d 4 · x 1 2 x 1 2 - 1 · x 1 2 + 1 · x 1 2
We will reduce to the degree x 1 2., we obtain 4 x 1 2 - 1 · x 1 2 + 1.
Additionally, it is possible to simplify the power expression in the denominator, using the square difference formula: Squares: 4 x 1 2 - 1 · x 1 2 + 1 \u003d 4 x 1 2 2 - 1 2 \u003d 4 x - 1.
Answer: x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 · 1 x 1 2 \u003d 4 x - 1
Example 11.
Simplify the power expression x 3 4 · x 2, 7 + 1 2 x - 5 8 · x 2, 7 + 1 3.
Decision
We can reduce the fraction on (x 2, 7 + 1) 2. We obtain the fraction x 3 4 x - 5 8 · x 2, 7 + 1.
We continue to transform the degrees of X 3 4 x - 5 8 · 1 x 2, 7 + 1. Now you can use the defenses of degrees with the same bases: x 3 4 x - 5 8 · 1 x 2, 7 + 1 \u003d x 3 4 - - 5 8 · 1 x 2, 7 + 1 \u003d x 1 1 8 · 1 x 2 , 7 + 1.
Go from the last work to the fraction x 1 3 8 x 2, 7 + 1.
Answer: x 3 4 · x 2, 7 + 1 2 x - 5 8 · x 2, 7 + 1 3 \u003d x 1 3 8 x 2, 7 + 1.
Multiplers with negative indicators in most cases are more convenient to transfer from the numerator to the denominator and back, changing the indicator sign. This action allows you to simplify the further solution. Let us give an example: a power expression (X + 1) - 0, 2 3 · X - 1 can be replaced by x 3 · (x + 1) 0, 2.
Transformation of expressions with roots and degrees
There are significant expressions in the tasks that contain not only degrees with fractional indicators, but also roots. Such expressions are desirable to bring only to roots or only to degrees. The transition to degrees is preferable, since they are easier to work with them. Such a transition is particularly preferable when the OTZ variables for the original expression makes it possible to replace the roots by degrees without the need to turn to the module or split OTZ into several gaps.
Example 12.
Prepare the expression x 1 9 · x · x 3 6 as a degree.
Decision
Area of \u200b\u200bpermissible variable values X. Determined by two inequalities x ≥ 0. and x · x 3 ≥ 0, which set many [ 0 , + ∞) .
On this set we have the right to move from the roots to the degrees:
x 1 9 · x · x 3 6 \u003d x 1 9 · x · x 1 3 1 6
Using the properties of degrees, simplifies the resulting power expression.
x 1 9 · x · x 1 3 1 6 \u003d x 1 9 · x 1 6 · x 1 3 1 6 \u003d x 1 9 · x 1 6 · x 1 · 1 3 · 6 \u003d x 1 9 · x 1 6 · X 1 18 \u003d x 1 9 + 1 6 + 1 18 \u003d x 1 3
Answer: x 1 9 · x · x 3 6 \u003d x 1 3.
Transformation of degrees with variables in the indicator
The conversion data simply simply produce if competently use the degree properties. For example, 5 2 · x + 1 - 3 · 5 x · 7 x - 14 · 7 2 · x - 1 \u003d 0.
We can replace the degree in the indicators of which there is a sum of some variable and the number. In the left side, this can be done with the first and last term of the left part of the expression:
5 2 · x · 5 1 - 3 · 5 x · 7 x - 14 · 7 2 · x · 7 - 1 \u003d 0, 5 · 5 2 · x - 3 · 5 x · 7 x - 2 · 7 2 · x \u003d 0.
Now share both parts of equality on 7 2 · X. This expression on the OTZ variable X receives only positive values:
5 · 5 - 3 · 5 x · 7 x - 2 · 7 2 · x 7 2 · x \u003d 0 7 2 · x, 5 · 5 2 · x 7 2 · x - 3 · 5 x · 7 x 7 2 · x - 2 · 7 2 · x 7 2 · x \u003d 0, 5 · 5 2 · x 7 2 · x - 3 · 5 x · 7 x 7 x · 7 x - 2 · 7 2 · x 7 2 · x \u003d 0.
We will reduce the fractions with degrees, we get: 5 · 5 2 · x 7 2 · x - 3 · 5 x 7 x - 2 \u003d 0.
Finally, the ratio of degrees with the same indicators is replaced by degrees of relations, which leads to an equation 5 · 5 7 2 · x - 3 · 5 7 x - 2 \u003d 0, which is equivalent to 5 · 5 7 x 2 - 3 · 5 7 x - 2 \u003d 0.
We introduce a new variable T \u003d 5 7 x, which reduces the solution of the initial indicative equation to the solution of the square equation 5 · T 2 - 3 · T - 2 \u003d 0.
Transformation of expressions with degrees and logarithms
Expressions containing the degree and logarithm recording are also found in tasks. An example of such expressions can be: 1 4 1 - 5 · LOG 2 3 or LOG 3 27 9 + 5 (1 - log 3 5) · Log 5 3. The transformation of such expressions is carried out using the above approaches and the properties of the logarithms, which we dismantled in detail in the topic "Transformation of logarithmic expressions".
If you notice a mistake in the text, please select it and press Ctrl + Enter
Formulas degrees Used in the process of abbreviation and simplify complex expressions, in solving equations and inequalities.
Number c. is an n.Little degree a. when:
Operations with degrees.
1. Multiplying the degree with the same basis, their indicators fold:
a M.· A n \u003d a m + n.
2. In dividing degrees with the same basis, their indicators are deducted:
3. The degree of work of 2 or more multipliers is equal to the product of these factors:
(ABC ...) n \u003d a n · b n · C n ...
4. The degree of fraction is equal to the ratio of degrees of the divide and divider:
(A / B) n \u003d a n / b n.
5. Earring the degree to the degree, the indicators of degrees are prolonged:
(a m) n \u003d a m n.
Each above formula is true in directions from left to right and vice versa.
for example. (2 · 3 · 5/15) ² \u003d 2² · 3² · 5² / 15 ² \u003d 900/225 \u003d 4.
Root operations.
1. The root of the work of several factors is equal to the product of the roots of these factors:
2. The root of the relationship is equal to the attitude of the divide and divider of the roots:
3. When the root is erected, it is fairly built into this degree.
4. If you increase the degree of root in n. once and at the same time build in n.The degree of the feed number, the value of the root will not change:
5. If you reduce the root degree in n. once and at the same time extract the root n.degree from an undercurned number, the value of the root will not change:
Degree with a negative indicator.The degree of a certain number with an indisputable (whole) indicator is determined as a unit divided by the degree of the same number with an indicator equal to the absolute value of the non-positive indicator:
Formula a M.: a n \u003d a m - n can be used not only when m.> n. but also m.< n..
for example. a. 4: A 7 \u003d A 4 \u200b\u200b- 7 \u003d A -3.
To formula a M.: a n \u003d a m - n became fair as m \u003d N.The presence of a zero degree is needed.
The degree with the zero indicator.The degree of any number that is not equal to zero, with the zero indicator equals one.
for example. 2 0 = 1,(-5) 0 = 1,(-3/5) 0 = 1.
Degree with fractional indicator.To build a valid number but in degree m / N., it is necessary to extract the root n.degree from m.degree of this number but.