Let's consider the topic of transforming expressions with powers, but first let's dwell on a number of transformations that can be carried out with any expressions, including power ones. We will learn how to open parentheses, add similar terms, work with bases and exponents, and use the properties of powers.
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What are power expressions?
In school courses, few people use the phrase “powerful expressions,” but this term is constantly found in collections for preparing for the Unified State Exam. In most cases, a phrase denotes expressions that contain degrees in their entries. This is what we will reflect in our definition.
Definition 1
Power expression is an expression that contains degrees.
Let us give several examples of power expressions, starting with a power with a natural exponent and ending with a power with a real exponent.
The simplest power expressions can be considered powers of a number with a natural exponent: 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 . And also powers with zero exponent: 5 0, (a + 1) 0, 3 + 5 2 − 3, 2 0. And powers with negative integer powers: (0, 5) 2 + (0, 5) - 2 2.
It is a little more difficult to work with a degree that has rational and irrational exponents: 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 .
The indicator can be the variable 3 x - 54 - 7 3 x - 58 or the logarithm x 2 · l g x − 5 · x l g x.
We have dealt with the question of what power expressions are. Now let's start converting them.
Main types of transformations of power expressions
First of all, we will look at the basic identity transformations of expressions that can be performed with power expressions.
Example 1
Calculate the value of a power expression 2 3 (4 2 − 12).
Solution
We will carry out all transformations in compliance with the order of actions. In this case, we will start by performing the actions in brackets: we will replace the degree with a digital value and calculate the difference of two numbers. We have 2 3 (4 2 − 12) = 2 3 (16 − 12) = 2 3 4.
All we have to do is replace the degree 2 3 its meaning 8 and calculate the product 8 4 = 32. Here's our answer.
Answer: 2 3 · (4 2 − 12) = 32 .
Example 2
Simplify the expression with powers 3 a 4 b − 7 − 1 + 2 a 4 b − 7.
Solution
The expression given to us in the problem statement contains similar terms that we can give: 3 a 4 b − 7 − 1 + 2 a 4 b − 7 = 5 a 4 b − 7 − 1.
Answer: 3 · a 4 · b − 7 − 1 + 2 · a 4 · b − 7 = 5 · a 4 · b − 7 − 1 .
Example 3
Express the expression with powers 9 - b 3 · π - 1 2 as a product.
Solution
Let's imagine the number 9 as a power 3 2 and apply the abbreviated multiplication formula:
9 - b 3 π - 1 2 = 3 2 - b 3 π - 1 2 = = 3 - b 3 π - 1 3 + b 3 π - 1
Answer: 9 - b 3 · π - 1 2 = 3 - b 3 · π - 1 3 + b 3 · π - 1 .
Now let's move on to the analysis identity transformations, which can be applied specifically to power expressions.
Working with base and exponent
The degree in the base or exponent can have numbers, variables, and some expressions. For example, (2 + 0, 3 7) 5 − 3, 7 And . Working with such records is difficult. It is much easier to replace the expression in the base of the degree or the expression in the exponent with an identically equal expression.
Transformations of degree and exponent are carried out according to the rules known to us separately from each other. The most important thing is that the transformation results in an expression identical to the original one.
The purpose of transformations is to simplify the original expression or obtain a solution to the problem. For example, in the example we gave above, (2 + 0, 3 7) 5 − 3, 7 you can follow the steps to go to the degree 4 , 1 1 , 3 . By opening the parentheses, we can present similar terms to the base of the power (a · (a + 1) − a 2) 2 · (x + 1) and get a power expression of more simple type a 2 (x + 1).
Using Degree Properties
Properties of powers, written in the form of equalities, are one of the main tools for transforming expressions with powers. We present here the main ones, taking into account that a And b are any positive numbers, and r And s- arbitrary real numbers:
Definition 2
- a r · a s = a r + s ;
- a r: a s = a r − s ;
- (a · b) r = a r · b r ;
- (a: b) r = a r: b r ;
- (a r) s = a r · s .
In cases where we are dealing with natural, integer, positive exponents, the restrictions on the numbers a and b can be much less strict. So, for example, if we consider the equality a m · a n = a m + n, Where m And n are natural numbers, then it will be true for any values of a, both positive and negative, as well as for a = 0.
The properties of powers can be used without restrictions in cases where the bases of the powers are positive or contain variables whose range of permissible values is such that the bases take only positive values on it. In fact, within school curriculum in mathematics, the student's task is to choose an appropriate property and apply it correctly.
When preparing to enter universities, you may encounter problems in which inaccurate application of properties will lead to a narrowing of the DL and other difficulties in solving. In this section we will examine only two such cases. More information on the subject can be found in the topic “Converting expressions using properties of powers”.
Example 4
Imagine the expression a 2 , 5 (a 2) − 3: a − 5 , 5 in the form of a power with a base a.
Solution
First, we use the property of exponentiation and transform the second factor using it (a 2) − 3. Then we use the properties of multiplication and division of powers with the same base:
a 2 , 5 · a − 6: a − 5 , 5 = a 2 , 5 − 6: a − 5 , 5 = a − 3 , 5: a − 5 , 5 = a − 3 , 5 − (− 5 , 5) = a 2 .
Answer: a 2, 5 · (a 2) − 3: a − 5, 5 = a 2.
Transformation of power expressions according to the property of powers can be done 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 .
Solution
If we apply equality (a · b) r = a r · b r, from right to left, we get a product of the form 3 · 7 1 3 · 21 2 3 and then 21 1 3 · 21 2 3 . Let's add the exponents when multiplying powers with the same bases: 21 1 3 · 21 2 3 = 21 1 3 + 2 3 = 21 1 = 21.
There is another way to carry out the transformation:
3 1 3 · 7 1 3 · 21 2 3 = 3 1 3 · 7 1 3 · (3 · 7) 2 3 = 3 1 3 · 7 1 3 · 3 2 3 · 7 2 3 = = 3 1 3 · 3 2 3 7 1 3 7 2 3 = 3 1 3 + 2 3 7 1 3 + 2 3 = 3 1 7 1 = 21
Answer: 3 1 3 7 1 3 21 2 3 = 3 1 7 1 = 21
Example 6
Given a power expression a 1, 5 − a 0, 5 − 6, enter a new variable t = a 0.5.
Solution
Let's imagine the degree a 1, 5 How a 0.5 3. Using the property of degrees to degrees (a r) s = a r · s from right to left and we get (a 0, 5) 3: a 1, 5 − a 0, 5 − 6 = (a 0, 5) 3 − a 0, 5 − 6. You can easily introduce a new variable into the resulting expression t = a 0.5: we get t 3 − t − 6.
Answer: t 3 − t − 6 .
Converting fractions containing powers
We usually deal with two versions of power expressions with fractions: the expression represents a fraction with a power or contains such a fraction. All basic transformations of fractions are applicable to such expressions without restrictions. They can be reduced, brought to a new denominator, or worked separately with the numerator and denominator. Let's illustrate this with 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 .
Solution
We are dealing with a fraction, so we will carry out transformations in both the numerator and the denominator:
3 5 2 3 5 1 3 - 5 - 2 3 1 + 2 x 2 - 3 - 3 x 2 = 3 5 2 3 5 1 3 - 3 5 2 3 5 - 2 3 - 2 - x 2 = = 3 5 2 3 + 1 3 - 3 5 2 3 + - 2 3 - 2 - x 2 = 3 5 1 - 3 5 0 - 2 - x 2
Place a minus sign in front of the fraction to change the sign of the denominator: 12 - 2 - x 2 = - 12 2 + x 2
Answer: 3 5 2 3 5 1 3 - 5 - 2 3 1 + 2 x 2 - 3 - 3 x 2 = - 12 2 + x 2
Fractions containing powers are reduced to a new denominator in the same way as rational fractions. To do this, you need to find an additional factor and multiply the numerator and denominator of the fraction by it. It is necessary to select an additional factor in such a way that it does not go to zero for any values of variables from the ODZ variables for the original expression.
Example 8
Reduce the 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 the denominator x + 8 · y 1 2 .
Solution
a) Let's select a factor that will allow us to reduce to a new denominator. a 0, 7 a 0, 3 = a 0, 7 + 0, 3 = a, therefore, as an additional factor we will take a 0 , 3. The range of permissible values of the variable a includes the set of all positive real numbers. Degree in this field a 0 , 3 does not go to zero.
Let's multiply the numerator and denominator of a fraction by a 0 , 3:
a + 1 a 0, 7 = a + 1 a 0, 3 a 0, 7 a 0, 3 = a + 1 a 0, 3 a
b) Let's pay attention to the denominator:
x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 2 - x 1 3 2 y 1 6 + 2 y 1 6 2
Let's multiply this expression by x 1 3 + 2 · y 1 6, we get the sum of the 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 reduce the original fraction.
This is how we found the additional factor x 1 3 + 2 · y 1 6 . On the range of permissible values of variables x And y the expression x 1 3 + 2 y 1 6 does not vanish, therefore, we can multiply the numerator and denominator of the fraction by it:
1 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = 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 = = x 1 3 + 2 y 1 6 x 1 3 3 + 2 y 1 6 3 = x 1 3 + 2 y 1 6 x + 8 y 1 2
Answer: a) a + 1 a 0, 7 = a + 1 a 0, 3 a, b) 1 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = 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.
Solution
a) We use the greatest common denominator (GCD), by which we can reduce the numerator and denominator. For numbers 30 and 45 it is 15. We can also make a reduction by x0.5+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 = 2 x 3 3 (x 0, 5 + 1)
b) Here the presence of identical factors is not obvious. You will have to perform some transformations in order to get the same factors in the numerator and denominator. To do this, we expand the denominator using the difference of squares formula:
a 1 4 - b 1 4 a 1 2 - b 1 2 = a 1 4 - b 1 4 a 1 4 2 - b 1 2 2 = = a 1 4 - b 1 4 a 1 4 + b 1 4 a 1 4 - b 1 4 = 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 = 2 · x 3 3 · (x 0 , 5 + 1) , b) a 1 4 - b 1 4 a 1 2 - b 1 2 = 1 a 1 4 + b 1 4 .
Basic operations with fractions include converting fractions to a new denominator and reducing fractions. Both actions are performed in compliance with a number of rules. When adding and subtracting fractions, the fractions are first reduced to common denominator, after which operations (addition or subtraction) are carried out with the numerators. The denominator remains the same. The result of our actions is a new fraction, the numerator of which is the product of the numerators, and the denominator is the product of the denominators.
Example 10
Do the steps x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 · 1 x 1 2 .
Solution
Let's start by subtracting the fractions that are in parentheses. Let's bring them to a common denominator:
x 1 2 - 1 x 1 2 + 1
Let's subtract the numerators:
x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = 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 = = x 1 2 + 1 2 - x 1 2 - 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = 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 = = 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 = = 4 x 1 2 x 1 2 - 1 x 1 2 + 1 x 1 2
Let's reduce by a power x 1 2, we get 4 x 1 2 - 1 · x 1 2 + 1 .
Additionally, you can simplify the power expression in the denominator using the difference of squares formula: squares: 4 x 1 2 - 1 x 1 2 + 1 = 4 x 1 2 2 - 1 2 = 4 x - 1 .
Answer: x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 1 x 1 2 = 4 x - 1
Example 11
Simplify the power-law expression x 3 4 x 2, 7 + 1 2 x - 5 8 x 2, 7 + 1 3.
Solution
We can reduce the fraction by (x 2 , 7 + 1) 2. We get the fraction x 3 4 x - 5 8 x 2, 7 + 1.
Let's continue transforming the powers of x x 3 4 x - 5 8 · 1 x 2, 7 + 1. Now you can use the property of dividing powers with the same bases: x 3 4 x - 5 8 1 x 2, 7 + 1 = x 3 4 - - 5 8 1 x 2, 7 + 1 = x 1 1 8 1 x 2 , 7 + 1 .
We move from 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 = x 1 3 8 x 2, 7 + 1.
In most cases, it is more convenient to transfer factors with negative exponents from the numerator to the denominator and back, changing the sign of the exponent. This action allows you to simplify the further decision. Let's give an example: the power expression (x + 1) - 0, 2 3 · x - 1 can be replaced by x 3 · (x + 1) 0, 2.
Converting expressions with roots and powers
In problems there are power expressions that contain not only powers with fractional indicators, but also roots. It is advisable to reduce such expressions only to roots or only to powers. Going for degrees is preferable as they are easier to work with. This transition is especially preferable when the ODZ of variables for the original expression allows you to replace the roots with powers without the need to access the modulus or split the ODZ into several intervals.
Example 12
Express the expression x 1 9 · x · x 3 6 as a power.
Solution
Range of permissible variable values x is defined by two inequalities x ≥ 0 and x x 3 ≥ 0, which define the set [ 0 , + ∞) .
On this set we have the right to move from roots to powers:
x 1 9 · x · x 3 6 = x 1 9 · x · x 1 3 1 6
Using the properties of powers, we simplify the resulting power expression.
x 1 9 · x · x 1 3 1 6 = x 1 9 · x 1 6 · x 1 3 1 6 = x 1 9 · x 1 6 · x 1 · 1 3 · 6 = = x 1 9 · x 1 6 x 1 18 = x 1 9 + 1 6 + 1 18 = x 1 3
Answer: x 1 9 · x · x 3 6 = x 1 3 .
Converting powers with variables in the exponent
These transformations are quite easy to make if you use the properties of the degree correctly. For example, 5 2 x + 1 − 3 5 x 7 x − 14 7 2 x − 1 = 0.
We can replace by the product of powers, the exponents of which are the sum of some variable and a number. On the left side, this can be done with the first and last terms of the left side of the expression:
5 2 x 5 1 − 3 5 x 7 x − 14 7 2 x 7 − 1 = 0, 5 5 2 x − 3 5 x 7 x − 2 7 2 x = 0 .
Now let's divide both sides of the equality by 7 2 x. This expression for the variable x takes only positive values:
5 5 - 3 5 x 7 x - 2 7 2 x 7 2 x = 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 = 0 , 5 5 2 x 7 2 x - 3 5 x 7 x 7 x 7 x - 2 7 2 x 7 2 x = 0
Let's reduce fractions with powers, we get: 5 · 5 2 · x 7 2 · x - 3 · 5 x 7 x - 2 = 0.
Finally, the ratio of powers with the same exponents is replaced by powers of ratios, resulting in the equation 5 5 7 2 x - 3 5 7 x - 2 = 0, which is equivalent to 5 5 7 x 2 - 3 5 7 x - 2 = 0 .
Let us introduce a new variable t = 5 7 x, which reduces the solution of the original exponential equation to the solution quadratic equation 5 · t 2 − 3 · t − 2 = 0 .
Converting expressions with powers and logarithms
Expressions containing powers and logarithms are also found in problems. An example of such expressions is: 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 approaches and properties of logarithms discussed above, which we discussed in detail in the topic “Transformation of logarithmic expressions”.
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Expressions, expression conversion
Power expressions(expressions with powers) and their transformation
In this article we will talk about converting expressions with powers. First, we will focus on transformations that are performed with expressions of any kind, including power expressions, such as opening parentheses and bringing similar terms. And then we will analyze the transformations inherent specifically in expressions with degrees: working with the base and exponent, using the properties of degrees, etc.
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What are power expressions?
The term “power expressions” practically does not appear in school mathematics textbooks, but it appears quite often in collections of problems, especially those intended for preparation for the Unified State Exam and the Unified State Exam, for example. After analyzing the tasks in which it is necessary to perform any actions with power expressions, it becomes clear that power expressions are understood as expressions containing powers in their entries. Therefore, you can accept the following definition for yourself:
Definition.
Power expressions are expressions containing degrees.
Let's give examples of power expressions. Moreover, we will present them according to how the development of views on from a degree with a natural exponent to a degree with a real exponent occurs.
As is known, first one gets acquainted with the power of a number with a natural exponent; at this stage, the first simplest power expressions of the type 3 2, 7 5 +1, (2+1) 5, (−0.1) 4, 3 a 2 appear −a+a 2 , x 3−1 , (a 2) 3 etc.
A little later, the power of a number with an integer exponent is studied, which leads to the appearance of power expressions with negative integer powers, like the following: 3 −2, 
, a −2 +2 b −3 +c 2 .
In high school they return to degrees. There a degree with a rational exponent is introduced, which entails the appearance of the corresponding power expressions: 
, , 
and so on. Finally, degrees with irrational exponents and expressions containing them are considered: , .
The matter is not limited to the listed power expressions: further the variable penetrates into the exponent, and, for example, the following expressions arise: 2 x 2 +1 or 
. And after getting acquainted with , expressions with powers and logarithms begin to appear, for example, x 2·lgx −5·x lgx.
So, we have dealt with the question of what power expressions represent. Next we will learn to transform them.
Main types of transformations of power expressions
With power expressions you can do any of the basic identity transformations of expressions. For example, you can open parentheses, replace numerical expressions with their values, add similar terms, etc. Naturally, it is necessary to comply with the accepted order of actions. Let's give examples.
Example.
Calculate the value of the power expression 2 3 ·(4 2 −12) .
Solution.
According to the order of execution of actions, first perform the actions in brackets. There, firstly, we replace the power 4 2 with its value 16 (if necessary, see), and secondly, we calculate the difference 16−12=4. We have 2 3 ·(4 2 −12)=2 3 ·(16−12)=2 3 ·4.
In the resulting expression, we replace the power 2 3 with its value 8, after which we calculate the product 8·4=32. This is the desired value.
So, 2 3 ·(4 2 −12)=2 3 ·(16−12)=2 3 ·4=8·4=32.
Answer:
2 3 ·(4 2 −12)=32.
Example.
Simplify expressions with powers 3 a 4 b −7 −1+2 a 4 b −7.
Solution.
Obviously, this expression contains similar terms 3·a 4 ·b −7 and 2·a 4 ·b −7 , and we can give them: .
Answer:
3 a 4 b −7 −1+2 a 4 b −7 =5 a 4 b −7 −1.
Example.
Express an expression with powers as a product.
Solution.
You can cope with the task by representing the number 9 as a power of 3 2 and then using abbreviated multiplication formulas difference of squares: 
Answer:
There are also a number of identical transformations inherent specifically in power expressions. We will analyze them further.
Working with base and exponent
There are degrees whose base and/or exponent are not just numbers or variables, but some expressions. As an example, we give the entries (2+0.3·7) 5−3.7 and (a·(a+1)−a 2) 2·(x+1) .
When working with similar expressions, you can replace both the expression in the base of the degree and the expression in the exponent with an identically equal expression on ODZ its variables. In other words, according to the rules known to us, we can separately transform the base of the degree and separately the exponent. It is clear that as a result of this transformation, an expression will be obtained that is identically equal to the original one.
Such transformations allow us to simplify expressions with powers or achieve other goals we need. For example, in the power expression mentioned above (2+0.3 7) 5−3.7, you can perform operations with the numbers in the base and exponent, which will allow you to move to the power 4.1 1.3. And after opening the brackets and bringing similar terms to the base of the degree (a·(a+1)−a 2) 2·(x+1), we obtain a power expression of a simpler form a 2·(x+1) .
Using Degree Properties
One of the main tools for transforming expressions with powers is equalities that reflect . Let us recall the main ones. For any positive numbers a and b and arbitrary real numbers r and s, the following properties of powers are true:
- a r ·a s =a r+s ;
- a r:a s =a r−s ;
- (a·b) r =a r ·b r ;
- (a:b) r =a r:b r ;
- (a r) s =a r·s .
Note that for natural, integer, and positive exponents, the restrictions on the numbers a and b may not be so strict. For example, for natural numbers m and n the equality a m ·a n =a m+n is true not only for positive a, but also for negative a, and for a=0.
At school, the main focus when transforming power expressions is on the ability to choose the appropriate property and apply it correctly. In this case, the bases of degrees are usually positive, which allows the properties of degrees to be used without restrictions. The same applies to the transformation of expressions containing variables in the bases of powers - the range of permissible values of variables is usually such that the bases take only positive values on it, which allows you to freely use the properties of powers. In general, you need to constantly ask yourself whether it is possible to use any property of degrees in this case, because inaccurate use of properties can lead to a narrowing of the educational value and other troubles. These points are discussed in detail and with examples in the article. converting expressions using properties of powers. Here we will limit ourselves to considering a few simple examples.
Example.
Express the expression a 2.5 ·(a 2) −3:a −5.5 as a power with base a.
Solution.
First, we transform the second factor (a 2) −3 using the property of raising a power to a power: (a 2) −3 =a 2·(−3) =a −6. The original power expression will take the form a 2.5 ·a −6:a −5.5. Obviously, it remains to use the properties of multiplication and division of powers with the same base, we have
a 2.5 ·a −6:a −5.5 =
a 2.5−6:a −5.5 =a −3.5:a −5.5 =
a −3.5−(−5.5) =a 2 .
Answer:
a 2.5 ·(a 2) −3:a −5.5 =a 2.
Properties of powers when transforming power expressions are used both from left to right and from right to left.
Example.
Find the value of the power expression.
Solution.
The equality (a·b) r =a r ·b r, applied from right to left, allows us to move from the original expression to a product of the form and further. And when multiplying powers with the same bases, the exponents add up: 
.
It was possible to transform the original expression in another way: 
Answer:
.
Example.
Given the power expression a 1.5 −a 0.5 −6, introduce a new variable t=a 0.5.
Solution.
The degree a 1.5 can be represented as a 0.5 3 and then, based on the property of the degree to the degree (a r) s =a r s, applied from right to left, transform it to the form (a 0.5) 3. Thus, a 1.5 −a 0.5 −6=(a 0.5) 3 −a 0.5 −6. Now it’s easy to introduce a new variable t=a 0.5, we get t 3 −t−6.
Answer:
t 3 −t−6 .
Converting fractions containing powers
Power expressions can contain or represent fractions with powers. Any of the basic ones are fully applicable to such fractions fraction conversions, which are inherent in fractions of any kind. That is, fractions that contain powers can be reduced, reduced to a new denominator, worked separately with their numerator and separately with the denominator, etc. To illustrate these words, consider solutions to several examples.
Example.
Simplify power expression 
.
Solution.
This power expression is a fraction. Let's work with its numerator and denominator. In the numerator we open the brackets and simplify the resulting expression using the properties of powers, and in the denominator we present similar terms: 
And let’s also change the sign of the denominator by placing a minus in front of the fraction: 
.
Answer:
.
Reduction of fractions containing powers to a new denominator is carried out in the same way as reduction to a new denominator rational fractions. In this case, an additional factor is also found and the numerator and denominator of the fraction are multiplied by it. When performing this action, it is worth remembering that reduction to a new denominator can lead to a narrowing of the VA. To prevent this from happening, it is necessary that the additional factor does not go to zero for any values of the variables from the ODZ variables for the original expression.
Example.
Reduce the fractions to a new denominator: a) to denominator a, b) 
to the denominator.
Solution.
a) In this case, it is quite easy to figure out which additional multiplier helps to achieve the desired result. This is a multiplier of a 0.3, since a 0.7 ·a 0.3 =a 0.7+0.3 =a. Note that in the range of permissible values of the variable a (this is the set of all positive real numbers), the power of a 0.3 does not vanish, therefore, we have the right to multiply the numerator and denominator of a given fraction by this additional factor: 
b) Taking a closer look at the denominator, you will find that 
and multiplying this expression by will give the sum of cubes and , that is, . And this is the new denominator to which we need to reduce the original fraction.
This is how we found an additional factor. In the range of acceptable values of the variables x and y, the expression does not vanish, therefore, we can multiply the numerator and denominator of the fraction by it: 
Answer:
A) 
, b) 
.
There is also nothing new in reducing fractions containing powers: the numerator and denominator are represented as a number of factors, and the same factors of the numerator and denominator are reduced.
Example.
Reduce the fraction: a) 
, b) .
Solution.
a) Firstly, the numerator and denominator can be reduced by the numbers 30 and 45, which is equal to 15. It is also obviously possible to perform a reduction by x 0.5 +1 and by 
. Here's what we have: 
b) In this case, identical factors in the numerator and denominator are not immediately visible. To obtain them, you will have to perform preliminary transformations. In this case, they consist in factoring the denominator using the difference of squares formula: 
Answer:
A) 
b) 
.
Converting fractions to a new denominator and reducing fractions are mainly used to do things with fractions. Actions are performed according to known rules. When adding (subtracting) fractions, they are reduced to a common denominator, after which the numerators are added (subtracted), but the denominator remains the same. The result is a fraction whose numerator is the product of the numerators, and the denominator is the product of the denominators. Division by a fraction is multiplication by its inverse.
Example.
Follow the steps 
.
Solution.
First, we subtract the fractions in parentheses. To do this, we bring them to a common denominator, which is 
, after which we subtract the numerators: 
Now we multiply the fractions: 
Obviously, it is possible to reduce by a power of x 1/2, after which we have 
.
You can also simplify the power expression in the denominator by using the difference of squares formula: 
.
Answer:
Example.
Simplify the Power Expression 
.
Solution.
Obviously, this fraction can be reduced by (x 2.7 +1) 2, this gives the fraction 
. It is clear that something else needs to be done with the powers of X. To do this, we transform the resulting fraction into a product. This gives us the opportunity to take advantage of the property of dividing powers with the same bases: 
. And at the end of the process we move from the last product to the fraction.
Answer:
.
And let us also add that it is possible, and in many cases desirable, to transfer factors with negative exponents from the numerator to the denominator or from the denominator to the numerator, changing the sign of the exponent. Such transformations often simplify further actions. For example, a power expression can be replaced by .
Converting expressions with roots and powers
Often, in expressions in which some transformations are required, roots with fractional exponents are also present along with powers. To transform such an expression to the desired form, in most cases it is enough to go only to roots or only to powers. But since it is more convenient to work with powers, they usually move from roots to powers. However, it is advisable to carry out such a transition when the ODZ of variables for the original expression allows you to replace the roots with powers without the need to refer to the module or split the ODZ into several intervals (we discussed this in detail in the article transition from roots to powers and back After getting acquainted with the degree with a rational exponent a degree with an irrational exponent is introduced, which allows us to talk about a degree with an arbitrary real exponent. At this stage, the school begins to study exponential function , which is analytically given by a power, the base of which is a number, and the exponent is a variable. So we are faced with power expressions containing numbers in the base of the power, and in the exponent - expressions with variables, and naturally the need arises to perform transformations of such expressions.
It should be said that the transformation of expressions of the indicated type usually has to be performed when solving exponential equations And exponential inequalities, and these conversions are quite simple. In the overwhelming majority of cases, they are based on the properties of the degree and are aimed, for the most part, at introducing a new variable in the future. The equation will allow us to demonstrate them 5 2 x+1 −3 5 x 7 x −14 7 2 x−1 =0.
Firstly, powers, in the exponents of which is the sum of a certain variable (or expression with variables) and a number, are replaced by products. This applies to the first and last terms of the expression on the left side:
5 2 x 5 1 −3 5 x 7 x −14 7 2 x 7 −1 =0,
5 5 2 x −3 5 x 7 x −2 7 2 x =0.
Next, both sides of the equality are divided by the expression 7 2 x, which on the ODZ of the variable x for the original equation takes only positive values (this is a standard technique for solving equations of this type, we are not talking about it now, so focus on subsequent transformations of expressions with powers ): 
Now we can cancel fractions with powers, which gives 
.
Finally, the ratio of powers with the same exponents is replaced by powers of relations, resulting in the equation 
, which is equivalent 
. The transformations made allow us to introduce a new variable, which reduces the solution of the original exponential equation to the solution of a quadratic equation