Expressions, expression conversion

Power expressions (expressions with powers) and their transformation

In this article, we will talk about transforming expressions with powers. First, we will focus on transformations that are performed with expressions of any kind, including power expressions, such as opening brackets, reducing 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" is practically not found in school textbooks of mathematics, but it often appears in collections of problems, especially designed to prepare for the Unified State Examination and the OGE, for example,. After analyzing tasks in which it is required to perform any actions with power expressions, it becomes clear that power expressions are understood as expressions containing degrees in their entries. Therefore, for yourself, you can take the following definition:

Definition.

Power expressions are expressions containing powers.

Let's bring examples of power expressions. Moreover, we will represent them according to how the development of views on from a degree with a natural indicator to a degree with a real indicator takes place.

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

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

In the senior classes, they return to the degrees again. There, a degree with a rational exponent is introduced, which leads to the appearance of the corresponding power expressions: , , etc. 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 there are, for example, such expressions 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 figured out the question of what are power expressions. Next, we will learn how to transform them.

The main types of transformations of power expressions

With power expressions, you can perform any of the basic identity transformations of expressions. For example, you can expand brackets, replace numeric expressions with their values, add like terms, and so on. Naturally, in this case it is necessary to follow the accepted procedure for performing actions. Let's give examples.

Example.

Calculate the value of the power expression 2 3 ·(4 2 −12) .

Solution.

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

In the resulting expression, we replace the power of 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 Power Expressions 3 a 4 b −7 −1+2 a 4 b −7.

Solution.

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

To cope with the task allows the representation of the number 9 as a power of 3 2 and the subsequent use of the abbreviated multiplication formula, the difference of squares:

Answer:

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

Working with base and exponent

There are degrees, in the basis and / or indicator of which are not just numbers or variables, but some expressions. As an example, let's write (2+0.3 7) 5−3.7 and (a (a+1)−a 2) 2 (x+1) .

When working with such expressions, it is possible to replace both the expression in the base of the degree and the expression in the indicator with an identically equal expression on the DPV of its variables. In other words, according to the rules known to us, we can separately convert the base of the degree, and separately - the indicator. It is clear that as a result of this transformation, an expression is 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 (2+0.3 7) 5−3.7 mentioned above, you can perform operations with numbers in the base and exponent, which will allow you to go to the power of 4.1 1.3. And after opening the brackets and bringing like terms in the base of the degree (a·(a+1)−a 2) 2·(x+1) we get a power expression of a simpler form a 2·(x+1) .

Using Power Properties

One of the main tools for 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 power properties hold:

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

At school, the main attention in the transformation of power expressions is focused precisely on the ability to choose the appropriate property and apply it correctly. In this case, the bases of the degrees are usually positive, which allows you to use the properties of the degrees without restrictions. The same applies to the transformation of expressions containing variables in the bases of degrees - the range of acceptable values ​​​​of variables is usually such that the bases take only positive values ​​on it, which allows you to freely use the properties of degrees. In general, one must constantly ask the question, is it possible in this case apply any property of degrees, because inaccurate use of properties can lead to a narrowing of the ODZ and other troubles. These points are discussed in detail and with examples in the article transformation of expressions using the properties of degrees. Here we confine ourselves to 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 by the property of raising a power to a power: (a 2) −3 =a 2 (−3) =a −6. In this case, the initial 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 \u003d a 2.

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

Example.

Find the value of the power expression.

Solution.

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

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

Answer:

.

Example.

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

Solution.

The degree a 1.5 can be represented as a 0.5 3 and further on the basis of the property of the degree in the degree (a r) s =a r s applied from right to left, convert it to the form (a 0.5) 3 . In this way, a 1.5 -a 0.5 -6=(a 0.5) 3 -a 0.5 -6. Now it is easy to introduce a new variable t=a 0.5 , we get t 3 −t−6 .

Answer:

t 3 −t−6 .

Converting fractions containing powers

Power expressions can contain fractions with powers or represent such fractions. Any of the basic fraction transformations that are inherent in fractions of any kind are fully applicable to such fractions. That is, fractions that contain degrees can be reduced, reduced to a new denominator, work separately with their numerator and separately with the denominator, etc. To illustrate the above words, consider the solutions of 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 expression obtained after that using the properties of powers, and in the denominator we present similar terms:

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

Answer:

.

Reducing fractions containing powers to a new denominator is carried out similarly to reducing rational fractions to a new denominator. At the same time, 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 DPV. To prevent this from happening, it is necessary that the additional factor does not vanish for any values ​​of the variables from the ODZ variables for the original expression.

Example.

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

Solution.

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

b) Looking more closely at the denominator, we 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 bring the original fraction.

So we found an additional factor . The expression does not vanish on the range of acceptable values ​​of the variables x and y, therefore, we can multiply the numerator and denominator of the fraction by it:

Answer:

a) , b) .

There is also nothing new in the reduction of fractions containing degrees: the numerator and denominator are represented as a certain number of factors, and the same factors of the numerator and denominator are reduced.

Example.

Reduce the fraction: a) , b).

Solution.

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

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

Answer:

a)

b) .

Reducing fractions to a new denominator and reducing fractions is mainly used to perform operations on 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), and 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 reciprocal.

Example.

Follow the steps .

Solution.

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

Now we multiply fractions:

Obviously, a reduction by the power x 1/2 is possible, after which we have .

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

Answer:

Example.

Simplify 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 convert the resulting fraction into a product. This gives us the opportunity to use the property of dividing powers with the same bases: . And at the end of the process, we pass from the last product to the fraction.

Answer:

.

And we 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 by 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, along with degrees with fractional exponents, there are also roots. To convert 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 degrees, they usually move from roots to degrees. 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 degrees without the need to access the module or split the ODZ into several intervals (we discussed this in detail in the article, the transition from roots to powers and vice versa After getting acquainted with the degree with a rational exponent a degree with an irrational indicator is introduced, which makes it possible to speak of a degree with an arbitrary real indicator.At this stage, the school begins to study exponential function, which is analytically given by the degree, in the basis of which there is a number, and in the indicator - a variable. So we are faced with power expressions containing numbers in the base of the degree, 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 transformations are quite simple. In the vast majority of cases, they are based on the properties of the degree and are aimed mostly 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.

First, the exponents, in whose exponents the sum of some variable (or expression with variables) and a number, is found, are replaced by products. This applies to the first and last terms of the expression on the left 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 takes only positive values ​​on the ODZ variable x for the original equation (this is a standard technique for solving equations of this kind, we are not talking about it now, so focus on subsequent transformations of expressions with powers ):

Now fractions with powers are cancelled, which gives .

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

I. V. Boikov, L. D. Romanova Collection of tasks for preparing for the exam. Part 1. Penza 2003.

Section 5 EXPRESSIONS AND EQUATIONS

In the section you will learn:

ü o expressions and their simplifications;

ü what are the properties of equalities;

ü how to solve equations based on the properties of equalities;

ü what types of problems are solved with the help of equations; what are perpendicular lines and how to build them;

ü what lines are called parallel and how to build them;

ü what is a coordinate plane;

ü how to determine the coordinates of a point on a plane;

ü what is a dependency graph between quantities and how to build it;

ü how to apply the learned material in practice

§ 30. EXPRESSIONS AND THEIR SIMPLIFICATION

You already know what literal expressions are and know how to simplify them using the laws of addition and multiplication. For example, 2a ∙ (-4 b) = -8 ab . In the resulting expression, the number -8 is called the coefficient of the expression.

Does the expression cd coefficient? So. It is equal to 1 because cd - 1 ∙ cd .

Recall that converting an expression with parentheses to an expression without parentheses is called parenthesis expansion. For example: 5(2x + 4) = 10x + 20.

The reverse action in this example is to put the common factor out of brackets.

Terms containing the same literal factors are called similar terms. By taking the common factor out of brackets, similar terms are erected:

5x + y + 4 - 2x + 6 y - 9 =

= (5x - 2x) + (y + 6y )+ (4 - 9) = = (5-2)* + (1 + 6)* y-5=

B x + 7y - 5.

Bracket expansion rules

1. If there is a “+” sign in front of the brackets, then when opening the brackets, the signs of the terms in brackets are preserved;

2. If there is a “-” sign in front of the brackets, then when the brackets are opened, the signs of the terms in brackets are reversed.

Task 1 . Simplify the expression:

1) 4x+(-7x + 5);

2) 15 y -(-8 + 7 y ).

Solutions. 1. There is a “+” sign before the brackets, therefore, when opening the brackets, the signs of all terms are preserved:

4x + (-7x + 5) \u003d 4x - 7x + 5 \u003d -3x + 5.

2. There is a “-” sign in front of the brackets, therefore, during the opening of the brackets: the signs of all terms are reversed:

15 - (- 8 + 7y) \u003d 15y + 8 - 7y \u003d 8y +8.

To open brackets, use the distributive property of multiplication: a( b + c) = ab + ac. If a > 0, then the signs of the terms b and with do not change. If a< 0, то знаки слагаемых b and from are reversed.

Task 2. Simplify the expression:

1) 2(6y -8) + 7y;

2) -5 (2-5x) + 12.

Solutions. 1. The factor 2 in front of the brackets e is positive, therefore, when opening the brackets, we keep the signs of all terms: 2(6 y - 8) + 7 y = 12 y - 16 + 7 y =19 y -16.

2. The factor -5 in front of the brackets e is negative, therefore, when opening the brackets, we change the signs of all terms to the opposite ones:

5(2 - 5x) + 12 = -10 + 25x +12 = 2 + 25x.

Find out more

1. The word "sum" comes from the Latin summa , which means "total", "total".

2. The word "plus" comes from the Latin plus , which means "more", and the word "minus" - from the Latin minus , which means "less". The signs "+" and "-" are used to indicate the operations of addition and subtraction. These signs were introduced by the Czech scientist J. Vidman in 1489 in the book "A quick and pleasant account for all merchants"(Fig. 138).

Rice. 138

REMEMBER THE MAIN THINGS

1. What terms are called similar? How are like terms constructed?

2. How do you open brackets preceded by a “+” sign?

3. How do you open brackets preceded by a "-" sign?

4. How do you open brackets that are preceded by a positive factor?

5. How do you open brackets that are preceded by a negative factor?

1374". Name the coefficient of the expression:

1) 12 a; 3) -5.6 xy;

2)4 6; 4)-s.

1375". Name the terms that differ only by the coefficient:

1) 10a + 76-26 + a; 3) 5n + 5m -4n + 4;

2) bc -4d - bc + 4d; 4) 5x + 4y-x + y.

What are these terms called?

1376". Are there similar terms in the expression:

1) 11a + 10a; 3)6n + 15n; 5) 25r - 10r + 15r;

2) 14s-12; 4)12 m + m; 6) 8k +10k - n?

1377". Is it necessary to change the signs of the terms in brackets, opening the brackets in the expression:

1)4 + (a + 3b); 2)-c +(5-d ); 3) 16-(5m-8n)?

1378°. Simplify the expression and underline the coefficient:

1379°. Simplify the expression and underline the coefficient:

1380°. Reduce like terms:

1) 4a - Po + 6a - 2a; 4) 10 - 4 d - 12 + 4d;

2) 4b - 5b + 4 + 5b; 5) 5a - 12b - 7a + 5b;

3)-7ang="EN-US">c+ 5-3 c + 2; 6) 14 n - 12 m -4 n -3 m.

1381°. Reduce like terms:

1) 6a - 5a + 8a -7a; 3) 5s + 4-2s-3s;

2)9 b +12-8-46; 4) -7n + 8m - 13n - 3m.

1382°. Take the common factor out of brackets:

1) 1.2 a +1.2 b; 3) -3 n - 1.8 m; 5) -5p + 2.5k -0.5t;

2) 0.5 s + 5d; 4) 1.2 n - 1.8 m; 6) -8p - 10k - 6t.

1383°. Take the common factor out of brackets:

1) 6a-12b; 3) -1.8 n -3.6 m;

2) -0.2 s + 1 4 d; A) 3p - 0.9k + 2.7t.

1384°. Open brackets and reduce like terms;

1) 5 + (4a -4); 4) -(5 c - d) + (4 d + 5c);

2) 17x-(4x-5); 5) (n - m) - (-2 m - 3 n);

3) (76 - 4) - (46 + 2); 6) 7 (-5x + y) - (-2y + 4x) + (x - 3y).

1385°. Open the brackets and reduce like terms:

1) 10a + (4 - 4a); 3) (s - 5 d) - (- d + 5s);

2) -(46-10) + (4-56); 4) - (5 n + m) + (-4 n + 8 m) - (2 m -5 n).

1386°. Expand the brackets and find the meaning of the expression:

1)15+(-12+ 4,5); 3) (14,2-5)-(12,2-5);

2) 23-(5,3-4,7); 4) (-2,8 + 13)-(-5,6 + 2,8) + (2,8-13).

1387°. Expand the brackets and find the meaning of the expression:

1) (14- 15,8)- (5,8 + 4);

2)-(18+22,2)+ (-12+ 22,2)-(5- 12).

1388°. Open parenthesis:

1) 0.5 ∙ (a + 4); 4) (n - m) ∙ (-2.4 p);

2)-s ∙ (2.7-1.2 d ); 5) 3 ∙ (-1.5 p + k - 0.2 t);

3) 1.6 ∙ (2n + m); 6) (4.2 p - 3.5 k -6 t) ∙ (-2a).

1389°. Open parenthesis:

1) 2.2 ∙ (x-4); 3)(4 c - d )∙(-0.5 y );

2) -2 ∙ (1.2 n - m); 4) 6- (-p + 0.3 k - 1.2 t).

1390. Simplify the expression:

1391. Simplify the expression:

1392. Reduce like terms:

1393. Reduce like terms:

1394. Simplify the expression:

1) 2.8 - (0.5 a + 4) - 2.5 ∙ (2a - 6);

2) -12 ∙ (8 - 2, by) + 4.5 ∙ (-6 y - 3.2);

4) (-12.8 m + 24.8 n) ∙ (-0.5)-(3.5 m -4.05 m) ∙ 2.

1395. Simplify the expression:

1396. Find the meaning of the expression;

1) 4-(0.2 a-3) - (5.8 a-16), if a \u003d -5;

2) 2-(7-56)+ 156-3∙(26+ 5), if = -0.8;

m = 0.25, n = 5.7.

1397. Find the value of the expression:

1) -4∙ (i-2) + 2∙(6x - 1), if x = -0.25;

1398*. Find the error in the solution:

1) 5- (a-2.4) -7 ∙ (-a + 1.2) \u003d 5a - 12-7a + 8.4 \u003d -2a-3.6;

2) -4 ∙ (2.3 a - 6) + 4.2 ∙ (-6 - 3.5 a) \u003d -9.2 a + 46 + 4.26 - 14.7 a \u003d -5.5 a + 8.26.

1399*. Expand the brackets and simplify the expression:

1) 2ab - 3(6(4a - 1) - 6(6 - 10a)) + 76;

1400*. Arrange the parentheses to get the correct equality:

1) a-6-a + 6 \u003d 2a; 2) a -2 b -2 a + b \u003d 3 a -3 b.

1401*. Prove that for any numbers a and b if a > b , then the following equality holds:

1) (a + b) + (a-b) \u003d 2a; 2) (a + b) - (a - b) \u003d 2 b.

Will this equality be correct if: a) a< b; b) a = 6?

1402*. Prove that for any natural number a, the arithmetic mean of the preceding and following numbers is equal to a.

APPLY IN PRACTICE

1403. To prepare a fruit dessert for three people, you need: 2 apples, 1 orange, 2 bananas and 1 kiwi. How to make a literal expression to determine the amount of fruit needed to prepare a dessert for guests? Help Marin to calculate how many fruits she needs to buy if she comes to visit: 1) 5 friends; 2) 8 friends.

1404. Make a literal expression to determine the time required to complete homework in mathematics, if:

1) a min was spent on solving problems; 2) simplification of expressions is 2 times more than for solving problems. How much time did Vasilko do his homework if he spent 15 minutes solving problems?

1405. Lunch in the school canteen consists of salad, borscht, cabbage rolls and compote. The cost of salad is 20%, borscht - 30%, cabbage rolls - 45%, compote - 5% of the total cost of the entire meal. Write an expression to find the cost of lunch at the school cafeteria. How much does lunch cost if the price of a salad is 2 UAH?

REPETITION TASKS

1406. Solve the equation:

1407. Tanya spent on ice creamall available money, and for sweets -the rest. How much money does Tanya have?

if sweets cost 12 UAH?

Among the various expressions that are considered in algebra, sums of monomials occupy an important place. Here are examples of such expressions:
\(5a^4 - 2a^3 + 0.3a^2 - 4.6a + 8 \)
\(xy^3 - 5x^2y + 9x^3 - 7y^2 + 6x + 5y - 2 \)

The sum of monomials is called a polynomial. The terms in a polynomial are called members of the polynomial. Mononomials are also referred to as polynomials, considering a monomial as a polynomial consisting of one member.

For example, polynomial
\(8b^5 - 2b \cdot 7b^4 + 3b^2 - 8b + 0.25b \cdot (-12)b + 16 \)
can be simplified.

We represent all the terms as monomials of the standard form:
\(8b^5 - 2b \cdot 7b^4 + 3b^2 - 8b + 0.25b \cdot (-12)b + 16 = \)
\(= 8b^5 - 14b^5 + 3b^2 -8b -3b^2 + 16 \)

We give similar terms in the resulting polynomial:
\(8b^5 -14b^5 +3b^2 -8b -3b^2 + 16 = -6b^5 -8b + 16 \)
The result is a polynomial, all members of which are monomials of the standard form, and among them there are no similar ones. Such polynomials are called polynomials of standard form.

Per polynomial degree standard form take the largest of the powers of its members. So, the binomial \(12a^2b - 7b \) has the third degree, and the trinomial \(2b^2 -7b + 6 \) has the second.

Usually, the terms of standard form polynomials containing one variable are arranged in descending order of its exponents. For example:
\(5x - 18x^3 + 1 + x^5 = x^5 - 18x^3 + 5x + 1 \)

The sum of several polynomials can be converted (simplified) into a standard form polynomial.

Sometimes the members of a polynomial need to be divided into groups, enclosing each group in parentheses. Since parentheses are the opposite of parentheses, it is easy to formulate parentheses opening rules:

If the + sign is placed before the brackets, then the terms enclosed in brackets are written with the same signs.

If a "-" sign is placed in front of the brackets, then the terms enclosed in brackets are written with opposite signs.

Transformation (simplification) of the product of a monomial and a polynomial

Using the distributive property of multiplication, one can transform (simplify) the product of a monomial and a polynomial into a polynomial. For example:
\(9a^2b(7a^2 - 5ab - 4b^2) = \)
\(= 9a^2b \cdot 7a^2 + 9a^2b \cdot (-5ab) + 9a^2b \cdot (-4b^2) = \)
\(= 63a^4b - 45a^3b^2 - 36a^2b^3 \)

The product of a monomial and a polynomial is identically equal to the sum of the products of this monomial and each of the terms of the polynomial.

This result is usually formulated as a rule.

To multiply a monomial by a polynomial, one must multiply this monomial by each of the terms of the polynomial.

We have repeatedly used this rule for multiplying by a sum.

The product of polynomials. Transformation (simplification) of the product of two polynomials

In general, the product of two polynomials is identically equal to the sum of the product of each term of one polynomial and each term of the other.

Usually use the following rule.

To multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other and add the resulting products.

Abbreviated multiplication formulas. Sum, Difference, and Difference Squares

Some expressions in algebraic transformations have to be dealt with more often than others. Perhaps the most common expressions are \((a + b)^2, \; (a - b)^2 \) and \(a^2 - b^2 \), that is, the square of the sum, the square of the difference, and square difference. You have noticed that the names of these expressions seem to be incomplete, so, for example, \((a + b)^2 \) is, of course, not just the square of the sum, but the square of the sum of a and b. However, the square of the sum of a and b is not so common, as a rule, instead of the letters a and b, it contains various, sometimes quite complex expressions.

Expressions \((a + b)^2, \; (a - b)^2 \) are easy to convert (simplify) into polynomials of the standard form, in fact, you have already met with such a task when multiplying polynomials:
\((a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = \)
\(= a^2 + 2ab + b^2 \)

The resulting identities are useful to remember and apply without intermediate calculations. Short verbal formulations help this.

\((a + b)^2 = a^2 + b^2 + 2ab \) - the square of the sum is equal to the sum of the squares and the double product.

\((a - b)^2 = a^2 + b^2 - 2ab \) - the square of the difference is the sum of the squares without doubling the product.

\(a^2 - b^2 = (a - b)(a + b) \) - the difference of squares is equal to the product of the difference and the sum.

These three identities allow in transformations to replace their left parts with right ones and vice versa - right parts with left ones. The most difficult thing in this case is to see the corresponding expressions and understand what the variables a and b are replaced in them. Let's look at a few examples of using abbreviated multiplication formulas.

Some algebraic examples of one kind are capable of terrifying schoolchildren. Long expressions are not only intimidating, but also very difficult to calculate. Trying to immediately understand what follows and what follows, not to get confused for long. It is for this reason that mathematicians always try to simplify the “terrible” task as much as possible and only then proceed to solve it. Oddly enough, such a trick greatly speeds up the process.

Simplification is one of the fundamental points in algebra. If in simple tasks it is still possible to do without it, then more difficult to calculate examples may be “too tough”. This is where these skills come in handy! Moreover, complex mathematical knowledge is not required: it will be enough just to remember and learn how to put into practice a few basic techniques and formulas.

Regardless of the complexity of the calculations, when solving any expression, it is important follow the order of operations with numbers:

1. parentheses;
2. exponentiation;
3. multiplication;
4. division;
5. addition;
6. subtraction.

The last two points can be safely swapped and this will not affect the result in any way. But adding two neighboring numbers, when next to one of them there is a multiplication sign, is absolutely impossible! The answer, if any, is wrong. Therefore, you need to remember the sequence.

The use of such

Such elements include numbers with a variable of the same order or the same degree. There are also so-called free members that do not have next to them the letter designation of the unknown.

The bottom line is that in the absence of parentheses You can simplify the expression by adding or subtracting like.

A few illustrative examples:

• 8x 2 and 3x 2 - both numbers have the same second order variable, so they are similar and when added, they are simplified to (8+3)x 2 =11x 2, while when subtracted, it turns out (8-3)x 2 =5x 2;
• 4x 3 and 6x - and here "x" has a different degree;
• 2y 7 and 33x 7 - contain different variables, therefore, as in the previous case, they do not belong to similar ones.

Factoring a Number

This little mathematical trick, if you learn how to use it correctly, will help you to cope with a tricky problem more than once in the future. And it’s easy to understand how the “system” works: a decomposition is a product of several elements, the calculation of which gives the original value. Thus, 20 can be represented as 20x1, 2x10, 5x4, 2x5x2, or some other way.

On a note: multipliers are always the same as divisors. So you need to look for a working “pair” for expansion among the numbers by which the original is divisible without a remainder.

You can perform such an operation both with free members and with digits attached to a variable. The main thing is not to lose the latter during calculations - even after decomposition, the unknown cannot take and "go nowhere." It remains at one of the factors:

• 15x=3(5x);
• 60y 2 \u003d (15y 2) 4.

Prime numbers that can only be divided by themselves or 1 never factor - it makes no sense..

Basic Simplification Methods

The first thing that catches the eye:

• the presence of brackets;
• fractions;
• roots.

Algebraic examples in the school curriculum are often compiled with the assumption that they can be beautifully simplified.

Bracket Calculations

Pay close attention to the sign in front of the brackets! Multiplication or division is applied to each element inside, and minus - reverses the existing "+" or "-" signs.

Parentheses are calculated according to the rules or according to the formulas of abbreviated multiplication, after which similar ones are given.

Fraction reduction

Reduce fractions is also easy. They themselves “willingly run away” once in a while, it is worth making operations with bringing such members. But you can simplify the example even before this: pay attention to the numerator and denominator. They often contain explicit or hidden elements that can be mutually reduced. True, if in the first case you just need to delete the superfluous, in the second you will have to think, bringing part of the expression to the form for simplification. Methods used:

• search and bracketing of the greatest common divisor of the numerator and denominator;
• dividing each top element by the denominator.

When an expression or part of it is under the root, the primary simplification problem is almost the same as the case with fractions. It is necessary to look for ways to completely get rid of it or, if this is not possible, to minimize the sign interfering with calculations. For example, to unobtrusive √(3) or √(7).

A sure way to simplify the radical expression is to try to factor it out, some of which are outside the sign. An illustrative example: √(90)=√(9×10) =√(9)×√(10)=3√(10).

Other little tricks and nuances:

• this simplification operation can be carried out with fractions, taking it out of the sign both as a whole and separately as a numerator or denominator;
• it is impossible to decompose and take out a part of the sum or difference beyond the root;
• when working with variables, be sure to take into account its degree, it must be equal to or a multiple of the root for the possibility of rendering: √(x 2 y)=x√(y), √(x 3)=√(x 2 ×x)=x√( x);
• sometimes it is allowed to get rid of the radical variable by raising it to a fractional power: √ (y 3)=y 3/2.

Power Expression Simplification

If in the case of simple calculations by minus or plus, examples are simplified by bringing similar ones, then what about when multiplying or dividing variables with different powers? They can be easily simplified by remembering two main points:

1. If there is a multiplication sign between the variables, the exponents are added.
2. When they are divided by each other, the same denominator is subtracted from the degree of the numerator.

The only condition for such a simplification is that both terms have the same basis. Examples for clarity:

• 5x 2 × 4x 7 + (y 13 / y 11) \u003d (5 × 4)x 2+7 + y 13- 11 \u003d 20x 9 + y 2;
• 2z 3 +z×z 2 -(3×z 8 /z 5)=2z 3 +z 1+2 -(3×z 8-5)=2z 3 +z 3 -3z 3 =3z 3 -3z 3 = 0.

We note that operations with numerical values ​​in front of variables occur according to the usual mathematical rules. And if you look closely, it becomes clear that the power elements of the expression "work" in a similar way:

• raising a member to a power means multiplying it by itself a certain number of times, i.e. x 2 \u003d x × x;
• division is similar: if you expand the degree of the numerator and denominator, then some of the variables will be reduced, while the rest are “gathered”, which is equivalent to subtraction.

As in any business, when simplifying algebraic expressions, not only knowledge of the basics is necessary, but also practice. After just a few lessons, examples that once seemed complicated will be reduced without much difficulty, turning into short and easily solved ones.

Video

This video will help you understand and remember how expressions are simplified.

Didn't get an answer to your question? Suggest a topic to the authors.

Let's consider the topic of transforming expressions with powers, but first we will dwell on a number of transformations that can be carried out with any expressions, including power ones. We will learn how to open brackets, give like terms, work with the base and exponent, use the properties of degrees.

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What are Power Expressions?

In the school course, few people use the phrase "power expressions", but this term is constantly found in collections for preparing for the exam. In most cases, the 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.

We give several examples of power expressions, starting with a degree with a natural exponent and ending with a degree 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 . As well as 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 a variable 3 x - 54 - 7 3 x - 58 or a 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 take a look at their transformation.

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 Power Expression Value 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 between the two numbers. We have 2 3 (4 2 − 12) = 2 3 (16 − 12) = 2 3 4.

It remains for us to replace the degree 2 3 its meaning 8 and calculate the product 8 4 = 32. Here is our answer.

Answer: 2 3 (4 2 − 12) = 32 .

Example 2

Simplify expression with powers 3 a 4 b − 7 − 1 + 2 a 4 b − 7.

Solution

The expression given to us in the condition of the problem contains similar terms, which we can bring: 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 an expression with powers of 9 - b 3 · π - 1 2 as a product.

Solution

Let's represent 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 .

And now let's move on to the analysis of identical transformations that 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 . It is difficult to work with such records. 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.

The transformations of the degree and the indicator 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 transformations, an expression is obtained that is identical to the original one.

The purpose of transformations is to simplify the original expression or to obtain a solution to the problem. For example, in the example we gave above, (2 + 0 , 3 7) 5 − 3 , 7 you can perform operations to go to the degree 4 , 1 1 , 3 . Opening the brackets, we can bring like terms in the base of the degree (a (a + 1) − a 2) 2 (x + 1) and get a power expression of a simpler form a 2 (x + 1).

Using Power Properties

The properties of degrees, written as equalities, are one of the main tools for transforming expressions with degrees. We present here the main ones, considering 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 stringent. 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.

You can apply the properties of degrees without restrictions in cases where the bases of the degrees are positive or contain variables whose range of acceptable values ​​is such that the bases take only positive values ​​on it. In fact, within the framework of the school curriculum in mathematics, the task of the student is to choose the appropriate property and apply it correctly.

When preparing for admission to universities, there may be tasks in which inaccurate application of properties will lead to a narrowing of the ODZ and other difficulties with the solution. In this section, we will consider only two such cases. More information on the subject can be found in the topic "Transforming expressions using exponent properties".

Example 4

Represent the expression a 2 , 5 (a 2) - 3: a - 5 , 5 as a degree with a base a.

Solution

To begin with, we use the exponentiation property 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 .

The transformation of power expressions according to the property of degrees 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 the equality (a b) r = a r b r, from right to left, then 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 \u003d 21 1 3 + 2 3 \u003d 21 1 \u003d 21.

There is another way to make transformations:

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

Imagine the degree a 1 , 5 how a 0 , 5 3. Using the degree property in a degree (a r) s = a r s from right to left and get (a 0 , 5) 3: a 1 , 5 - a 0 , 5 - 6 = (a 0 , 5) 3 - a 0 , 5 - 6 . In the resulting expression, you can easily introduce a new variable t = a 0 , 5: get t 3 − t − 6.

Answer: t 3 − t − 6 .

Converting fractions containing powers

We usually deal with two variants of power expressions with fractions: the expression is a fraction with a degree or contains such a fraction. All basic fraction transformations are applicable to such expressions without restrictions. They can be reduced, brought to a new denominator, work 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

Put a minus 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 vanish for any values ​​of the variables from the ODZ variables for the original expression.

Example 8

Bring 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) We choose 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 take a 0 , 3. The range of admissible values ​​of the variable a includes the set of all positive real numbers. In this area, the degree 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) 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

Multiply this expression by x 1 3 + 2 · y 1 6 , we get 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 factor x 1 3 + 2 · y 1 6 . On the range of acceptable values ​​of variables x and y the expression x 1 3 + 2 y 1 6 does not vanish, so 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) Use the greatest common denominator (GCD) by which the numerator and denominator can be reduced. For the numbers 30 and 45, this is 15 . We can also reduce 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 = 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 .

The main operations with fractions include reduction to a new denominator and reduction of fractions. Both actions are performed in compliance with a number of rules. When adding and subtracting fractions, the fractions are first reduced to a common denominator, after which actions (addition or subtraction) are performed with 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 brackets. 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 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 degree 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 formula for the difference of squares: 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 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 a fraction x 3 4 x - 5 8 x 2, 7 + 1.

Let's continue transformations of x powers x 3 4 x - 5 8 · 1 x 2 , 7 + 1 . Now you can use the power division property 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 pass from the last product 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 multipliers with negative exponents from the numerator to the denominator and vice versa by changing the sign of the exponent. This action simplifies 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 tasks, there are power expressions that contain not only degrees with fractional exponents, but also roots. It is desirable to reduce such expressions only to roots or only to powers. The transition to degrees is preferable, since they are easier to work with. Such a transition is especially advantageous when the DPV of the variables for the original expression allows you to replace the roots with powers without having to access the modulus or split the DPV into several intervals.

Example 12

Express the expression x 1 9 x x 3 6 as a power.

Solution

Valid range of a variable x is determined 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 degrees, 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 simple to make if you correctly use the properties of the degree. For example, 5 2 x + 1 − 3 5 x 7 x − 14 7 2 x − 1 = 0.

We can replace the product of the degree, in terms of which the sum of some variable and a number is found. On the left side, this can be done with the first and last terms on 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 equation by 7 2 x. This expression on the ODZ of 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 the 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, which leads to 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 .

We introduce a new variable t = 5 7 x , which reduces the solution of the original exponential equation to the solution of the 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. Examples of such expressions are: 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 discussed above and the properties of logarithms, which we have analyzed in detail in the topic “Transformation of logarithmic expressions”.

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