Equations, part $ C $

Equality containing an unknown number indicated by a letter is called an equation. The expression to the left of the equal sign is called the left side of the equation, and the expression to the right is called the right side of the equation.

Scheme for solving complex equations:

1. Before solving the equation, it is necessary to write down the range of permissible values \u200b\u200b(ODV) for it.
2. Solve the equation.
3. Choose from the obtained roots of the equation that which satisfy the ODZ.

ODZ of various expressions (by expression we mean an alphanumeric notation):

1. The expression in the denominator must not be zero.

$ (f (x)) / (g (x)); g (x) ≠ 0 $

2. The radical expression must not be negative.

$ √ (g (x)); g (x) ≥ 0 $.

3. The radical expression in the denominator must be positive.

$ (f (x)) / (√ (g (x))); g (x)\u003e 0 $

4. For a logarithm: sub-logarithmic expression must be positive; the basis must be positive; the base cannot be equal to one.

$ log_ (f (x)) g (x) \\ table \\ (\\ g (x)\u003e 0; \\ f (x)\u003e 0; \\ f (x) ≠ 1; $

Logarithmic Equations

Equations of the form $ log_ (a) f (x) \u003d log_ (a) g (x) $, where $ a $ is a positive number other than $ 1 $, and equations that reduce to this form are called logarithmic equations.

To solve logarithmic equations, it is necessary to know the properties of logarithms: we will consider all the properties of logarithms for $ a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0, m $ - any real number.

1. For any real numbers $ m $ and $ n $ the equalities are true:

$ log_ (a) b ^ m \u003d mlog_ (a) b; $

$ log_ (a ^ m) b \u003d (1) / (m) log_ (a) b. $

$ log_ (a ^ n) b ^ m \u003d (m) / (n) log_ (a) b $

$ log_ (3) 3 ^ (10) \u003d 10log_ (3) 3 \u003d 10; $

$ log_ (5 ^ 3) 7 \u003d (1) / (3) log_ (5) 7; $

$ log_ (3 ^ 7) 4 ^ 5 \u003d (5) / (7) log_ (3) 4; $

2. The logarithm of the product is equal to the sum of the logarithms in the same base of each factor.

$ log_a (bc) \u003d log_ (a) b + log_ (a) c $

3. The logarithm of the quotient is equal to the difference between the logarithms of the numerator and denominator for the same base

$ log_ (a) (b) / (c) \u003d log_ (a) b-log_ (a) c $

4. When multiplying two logarithms, you can swap their bases

$ log_ (a) b ∙ log_ (c) d \u003d log_ (c) b ∙ log_ (a) d $, if $ a, b, c $ and $ d\u003e 0, a ≠ 1, b ≠ 1. $

5. $ c ^ (log_ (a) b) \u003d b ^ (log_ (a) b) $, where $ a, b, c\u003e 0, a ≠ 1 $

6. The formula for the transition to a new base

$ log_ (a) b \u003d (log_ (c) b) / (log_ (c) a) $

7. In particular, if it is necessary to swap the base and the sub-logarithmic expression

$ log_ (a) b \u003d (1) / (log_ (b) a) $

There are several main types of logarithmic equations:

Simplest logarithmic equations: $ log_ (a) x \u003d b $. The solution of this type of equations follows from the definition of the logarithm, i.e. $ x \u003d a ^ b $ and $ x\u003e 0 $

We represent both sides of the equation as a logarithm to the base $ 2 $

$ log_ (2) x \u003d log_ (2) 2 ^ 3 $

If logarithms on the same base are equal, then sub-logarithmic expressions are also equal.

Answer: $ x \u003d 8 $

Equations of the form: $ log_ (a) f (x) \u003d log_ (a) g (x) $. Because the grounds are the same, then we equate the sub-logarithmic expressions and take into account the ODZ:

$ \\ table \\ (\\ f (x) \u003d g (x); \\ f (x)\u003e 0; \\ g (x)\u003e 0, а\u003e 0, а ≠ 1; $

$ log_ (3) (x ^ 2-3x-5) \u003d log_ (3) (7-2x) $

Because the bases are the same, then we equate sublogarithmic expressions

We transfer all terms to the left-hand side of the equation and present similar terms

Let us check the found roots by the conditions $ \\ table \\ (\\ x ^ 2-3x-5\u003e 0; \\ 7-2x\u003e 0; $

When substituted into the second inequality, the root $ x \u003d 4 $ does not satisfy the condition, therefore, it is an extraneous root

Answer: $ x \u003d -3 $

• Variable replacement method.

In this method, you need:

1. Write down the ODZ equation.
2. According to the properties of logarithms, achieve the same logarithms in the equation.
3. Replace $ log_ (a) f (x) $ with any variable.
4. Solve the equation for the new variable.
5. Return to step 3, substitute the value for the variable and get the simplest equation of the form: $ log_ (a) x \u003d b $
6. Solve the simplest equation.
7. After finding the roots of the logarithmic equation, it is necessary to put them in item 1 and check the condition of the ODZ.

Solve the equation $ log_ (2) √x + 2log_ (√x) 2-3 \u003d 0 $

1. Let us write the ODZ equation:

$ \\ table \\ (\\ x\u003e 0, \\ text "since it stands under the sign of the root and the logarithm"; \\ √x ≠ 1 → x ≠ 1; $

2. Let's make the logarithms to the base $ 2 $, for this we use the rule of transition to a new base in the second term:

$ log_ (2) √x + (2) / (log_ (2) √x) -3 \u003d 0 $

4. We obtain a fractional - rational equation with respect to the variable t

Let us reduce all terms to a common denominator $ t $.

$ (t ^ 2 + 2-3t) / (t) \u003d 0 $

The fraction is zero when the numerator is zero and the denominator is not zero.

$ t ^ 2 + 2-3t \u003d 0 $, $ t ≠ 0 $

5. Solve the resulting quadratic equation by Vieta's theorem:

6. Return to step 3, make the reverse change and obtain two simple logarithmic equations:

$ log_ (2) √x \u003d 1 $, $ log_ (2) √x \u003d 2 $

Let us logarithm the right-hand sides of the equations

$ log_ (2) √x \u003d log_ (2) 2 $, $ log_ (2) √x \u003d log_ (2) 4 $

Equating sub-logarithmic expressions

$ √x \u003d 2 $, $ √x \u003d 4 $

To get rid of the root, square both sides of the equation

$ x_1 \u003d 4 $, $ x_2 \u003d 16 $

7. Substitute the roots of the logarithmic equation in item 1 and check the condition of the AED.

$ \\ (\\ table \\ 4\u003e 0; \\ 4 ≠ 1; $

The first root satisfies the ODV.

$ \\ (\\ table \\ 16\u003e 0; \\ 16 ≠ 1; $ The second root also satisfies ODV.

Answer: $ 4; 16 $

• Equations of the form $ log_ (a ^ 2) x + log_ (a) x + c \u003d 0 $. Such equations are solved by introducing a new variable and passing to the usual quadratic equation. After the roots of the equation are found, it is necessary to select them taking into account the ODV.

Fractional Rational Equations

• If the fraction is zero, then the numerator is zero and the denominator is non-zero.
• If at least one part of a rational equation contains a fraction, then the equation is called fractional rational.

To solve a fractionally rational equation, you need:

1. Find the values \u200b\u200bof the variable for which the equation does not make sense (ODV)
2. Find the common denominator of the fractions in the equation;
3. Multiply both sides of the equation by a common denominator;
4. Solve the resulting whole equation;
5. Exclude from its roots those that do not satisfy the DHS condition.

• If two fractions are involved in the equation and their numerators are equal, then the denominators can be equated to each other and the resulting equation can be solved without paying attention to the numerators. BUT given the ODV of the entire original equation.

Exponential Equations

Equations in which the unknown is contained in the exponent are called exponential.

When solving exponential equations, the properties of degrees are used, recall some of them:

1. When multiplying degrees with the same bases, the base remains the same, and the indicators are added.

$ a ^ n a ^ m \u003d a ^ (n + m) $

2. When dividing degrees with the same bases, the base remains the same, and the indicators are subtracted

$ a ^ n: a ^ m \u003d a ^ (n-m) $

3. When raising a power to a power, the base remains the same, and the indicators are multiplied

$ (a ^ n) ^ m \u003d a ^ (n ∙ m) $

4. When raising to the power of a product, each factor is raised to this power

$ (a b) ^ n \u003d a ^ n b ^ n $

5. When raising to a power of a fraction, the numerator and denominator are raised to this power

$ ((a) / (b)) ^ n \u003d (a ^ n) / (b ^ n) $

6. When raising any base to a zero exponent, the result is equal to one

7. The base in any negative exponent can be represented as a base in the same positive exponent by changing the position of the base relative to the fraction line

$ a ^ (- n) \u003d (1) / (a \u200b\u200b^ n) $

$ (a ^ (- n)) / (b ^ (- k)) \u003d (b ^ k) / (a \u200b\u200b^ n) $

8. The radical (root) can be represented as a power with a fractional exponent

$ √ ^ n (a ^ k) \u003d a ^ ((k) / (n)) $

Types of exponential equations:

1. Simple exponential equations:

a) The form $ a ^ (f (x)) \u003d a ^ (g (x)) $, where $ a\u003e 0, a ≠ 1, x $ is unknown. To solve such equations, we will use the property of degrees: degrees with the same base ($ a\u003e 0, a ≠ 1 $) are equal only if their exponents are equal.

b) Equation of the form $ a ^ (f (x)) \u003d b, b\u003e 0 $

To solve such equations, both sides must be logarithm with the base $ a $, it turns out

$ log_ (a) a ^ (f (x)) \u003d log_ (a) b $

2. Method of equalizing the bases.

3. Method of factorization and change of variable.

• For this method, in the entire equation by the property of degrees, it is necessary to transform the degrees to one form $ a ^ (f (x)) $.
• Change the variable $ a ^ (f (x)) \u003d t, t\u003e 0 $.
• We get a rational equation, which must be solved by factoring the expression.
• We make the reverse substitution, taking into account that $ t\u003e

Solve the equation $ 2 ^ (3x) -7 2 \u200b\u200b^ (2x-1) + 7 2 ^ (x-1) -1 \u003d 0 $

By the property of degrees, we transform the expression so that we get the degree 2 ^ x.

$ (2 ^ x) ^ 3- (7 (2 ^ x) ^ 2) / (2) + (7 2 ^ x) / (2-1) \u003d 0 $

Let's change the variable $ 2 ^ x \u003d t; t\u003e 0 $

We obtain a cubic equation of the form

$ t ^ 3- (7 t ^ 2) / (2) + (7 t) / (2) -1 \u003d 0 $

Multiply the whole equation by $ 2 $ to get rid of the denominators

$ 2t ^ 3-7 t ^ 2 + 7 t-2 \u003d 0 $

Expand the left side of the equation by the grouping method

$ (2t ^ 3-2) - (7 t ^ 2-7 t) \u003d 0 $

Take out from the first bracket the common factor $ 2 $, from the second $ 7t $

$ 2 (t ^ 3-1) -7t (t-1) \u003d 0 $

Additionally, in the first parenthesis we see the formula difference of cubes

$ (t-1) (2t ^ 2 + 2t + 2-7t) \u003d 0 $

The product is zero when at least one of the factors is zero

1) $ (t-1) \u003d 0; $ 2) $ 2t ^ 2 + 2t + 2-7t \u003d 0 $

Let's solve the first equation

Let's solve the second equation through the discriminant

$ D \u003d 25-4 2 2 \u003d 9 \u003d 3 ^ 2 $

$ t_2 \u003d (5-3) / (4) \u003d (1) / (2) $

$ t_3 \u003d (5 + 3) / (4) \u003d 2 $

$ 2 ^ x \u003d 1; 2 ^ x \u003d (1) / (2); 2 ^ x \u003d 2 $

$ 2 ^ x \u003d 2 ^ 0; 2 ^ x \u003d 2 ^ (- 1); 2 ^ x \u003d 2 ^ 1 $

$ x_1 \u003d 0; x_2 \u003d -1; x_3 \u003d 1 $

Answer: $ -1; 0; 1 $

4. Quadratic conversion method

• We have an equation of the form $ A a ^ (2f (x)) + B a ^ (f (x)) + C \u003d 0 $, where $ A, B $ and $ C $ are coefficients.
• We make the substitution $ a ^ (f (x)) \u003d t, t\u003e 0 $.
• A quadratic equation of the form $ A t ^ 2 + B t + C \u003d 0 $ is obtained. We solve the resulting equation.
• We make the reverse replacement, taking into account the fact that $ t\u003e 0 $. We get the simplest exponential equation $ a ^ (f (x)) \u003d t $, solve it and write the result in the answer.

Factoring methods:

• Factor out the common factor.

To factor a polynomial by factoring out the common factor outside the brackets, you need:

1. Determine the common factor.
2. Divide the given polynomial by it.
3. Write down the product of the common factor and the resulting quotient (enclosing this quotient in brackets).

Factor the polynomial: $ 10a ^ (3) b-8a ^ (2) b ^ 2 + 2a $.

The common factor for this polynomial is $ 2a $, since all terms are divisible by $ 2 $ and by "a". Next, we find the quotient of dividing the original polynomial by "2a", we get:

$ 10a ^ (3) b-8a ^ (2) b ^ 2 + 2a \u003d 2a ((10a ^ (3) b) / (2a) - (8a ^ (2) b ^ 2) / (2a) + ( 2a) / (2a)) \u003d 2a (5a ^ (2) b-4ab ^ 2 + 1) $

This is the end result of the factorization.

Applying abbreviated multiplication formulas

1. The square of the sum is decomposed into the square of the first number plus twice the product of the first number by the second number and plus the square of the second number.

$ (a + b) ^ 2 \u003d a ^ 2 + 2ab + b ^ 2 $

2. The square of the difference is decomposed into the square of the first number minus twice the product of the first number by the second and plus the square of the second number.

$ (a-b) ^ 2 \u003d a ^ 2-2ab + b ^ 2 $

3. The difference of the squares is decomposed into the product of the difference of numbers and their sum.

$ a ^ 2-b ^ 2 \u003d (a + b) (a-b) $

4. The cube of the sum is equal to the cube of the first number plus three times the square of the first number by the second number plus three times the product of the first and the square of the second number plus the cube of the second number.

$ (a + b) ^ 3 \u003d a ^ 3 + 3a ^ 2b + 3ab ^ 2 + b ^ 3 $

5. The cube of the difference is equal to the cube of the first number minus the triple product of the square of the first number by the second number, plus the triple product of the first and the square of the second number, and minus the cube of the second number.

$ (a-b) ^ 3 \u003d a ^ 3-3a ^ 2b + 3ab ^ 2-b ^ 3 $

6. The sum of the cubes is equal to the product of the sum of numbers by the incomplete square of the difference.

$ a ^ 3 + b ^ 3 \u003d (a + b) (a ^ 2-ab + b ^ 2) $

7. The difference of the cubes is equal to the product of the difference of numbers by the incomplete square of the sum.

$ a ^ 3-b ^ 3 \u003d (a-b) (a ^ 2 + ab + b ^ 2) $

Grouping method

The grouping method is convenient to use when it is necessary to factorize a polynomial with an even number of terms. In this method, it is necessary to collect the terms in groups and take out the common factor from each group outside the bracket. Several groups, after placing them in parentheses, should have the same expressions, then move this parenthesis as a common factor forward and multiply by the parenthesis of the resulting quotient.

Factor polynomial $ 2a ^ 3-a ^ 2 + 4a-2 $

To decompose this polynomial, we use the method of grouping the terms, for this we group the first two and the last two terms, while it is important to put the right sign in front of the second grouping, we put the + sign and therefore write the terms with their signs in brackets.

$ (2a ^ 3-a ^ 2) + (4a-2) \u003d a ^ 2 (2a-1) +2 (2a-1) $

After taking out common factors, we got a pair of identical brackets. Now we take out this bracket as a common factor.

$ a ^ 2 (2a-1) +2 (2a-1) \u003d (2a-1) (a ^ 2 + 2) $

The product of these parentheses is the end result of the factorization.

Using the square trinomial formula.

If there is a square trinomial of the form $ ax ^ 2 + bx + c $, then it can be expanded by the formula

$ ax ^ 2 + bx + c \u003d a (x-x_1) (x-x_2) $, where $ x_1 $ and $ x_2 $ are the roots of the square trinomial

Today we will train the skill of solving the problem of the 5th exam - find the root of the equation. We will look for the root of the equation. Let's consider examples of solving such tasks. But first, let's remember - what does it mean to find the root of the equation?

This means finding such a number encrypted under x, which we substitute for x and our equation will be the correct equality.

For example, 3x \u003d 9 is an equation, and 3. 3 \u003d 9 - this is already a true equality. That is, in this case, we substituted the number 3 instead of x - we got the correct expression or equality, this means that we solved the equation, that is, we found the given number x \u003d 3, which turns the equation into the correct equality.

This is what we will do - we will find the root of the equation.

Task 1 - Find the root of equation 2 1-4x \u003d 32

This is a revealing equation. It is solved as follows - it is necessary that both the left and the right of the "equal" sign have a degree with the same base.

On the left we have a base of degree 2, and on the right there is no degree at all. But we know that 32 is 2 to the fifth power. That is, 32 \u003d 2 5

So our equation will look like this: 2 1-4x \u003d 2 5

On the left and right, our bases of the degree are the same, which means that in order for us to have equality, the exponents must also be equal:

We get an ordinary equation. We solve in the usual way - we leave all unknowns on the left, and transfer the known ones to the right, we get:

Checking: 2 1-4 (-1) \u003d 32

We found the root of the equation. Answer: x \u003d -1.

Find the root of the equation yourself in the following tasks:

b) 2 1-3x \u003d 128

Task 2 - find the root of the equation

We solve the equation in the same way - by reducing the left and right sides of the equation to the same base of the degree. In our case, to the base of degree 2.

We use the following property of the degree:

By this property, we get for the right side of our equation:

If the bases of the degree are equal, then the exponents are also equal:

Answer: x \u003d 9.

Let's check - we substitute the found value of x into the original equation - if we get the correct equality, then we solved the equation correctly.

We found the root of the equation correctly.

Task 3 - Find the root of the equation

Note that on the right we have 1/8, and 1/8 is

Then our equation will be written as:

If the bases of the degree are equal, then the exponents are also equal, we get a simple equation:

Answer: x \u003d 5. Check it yourself.

Task 4 - find the root of the equation log 3 (15-x) \u003d log 3 2

This equation is solved in the same way as the exponential one. We want the bases of the logarithms to the left and right of the equal sign to be the same. Now they are the same, which means we equate those expressions that stand under the sign of logarithms:

Answer: x \u003d 13

Task 5 - Find the root of the equation log 3 (3-x) \u003d 3

The number 3 is log 3 27. To make it clear at the bottom, the subscript under the sign of the logarithm is the number that is raised to a power, in our case 3, under the sign of the logarithm is the number that was obtained when raising to the power - this is 27, and the logarithm itself is the exponent to which 3 must be raised to get 27.

See the picture:

Thus, any number can be written as a logarithm. In this case, it is very convenient to write the number 3 as a logarithm with base 3. We get:

log 3 (3-x) \u003d log 3 27

The bases of the logarithms are equal, which means that the numbers under the sign of the logarithm are also equal:

Let's check:

log 3 (3 - (- 24)) \u003d log 3 27

log 3 (3 + 24) \u003d log 3 27

log 3 27 \u003d log 3 27

Answer: x \u003d -24.

Find the root of the equation. Task 6.

log 2 (x + 3) \u003d log 2 (3x-15)

Check: log 2 (9 + 3) \u003d log 2 (27-15)

log 2 12 \u003d log 2 12

Answer: x \u003d 9.

Find the root of the equation. Task 7.

log 2 (14-2x) \u003d 2log 2 3

log 2 (14-2x) \u003d log 2 3 2

Check: log 2 (14-5) \u003d 2log 2 3

log 2 9 \u003d 2log 2 3

log 2 3 2 \u003d 2log 2 3

2log 2 3 \u003d 2log 2 3

Answer: x \u003d 2.5

Prepare for the Unified State Exam and for the OGE -See the previous topics and.

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EQUATIONS IN THE USE IN MATH EXAMPLES AND SOLUTIONS Kravchenko N.А. Mathematics teacher GBOU SOSH №891, Moscow Educational presentation to prepare for the Unified State Exam

CONTENTS Task Annotation Example 1 (irrational equation) Example 2 (exponential equation) Example 3 (irrational equation) Example 4 (fractional rational equation) Example 5 (logarithmic equation) Example 6 (logarithmic equation) Example 7 (trigonometric equation) Example 8 ( exponential equation) Example 9 (irrational equation) Example 10 (logarithmic equation)

REFERENCE TYPE: Equation. CHARACTERISTIC OF THE PROBLEM: Simple exponential, logarithmic, trigonometric or irrational equation. COMMENT: The equation is reduced in one action to linear or square (in this case, the answer needs to indicate only one of the roots - greater or less). Incorrect answers are mainly due to arithmetic errors.

Solve the equation. EXAMPLE 1 Solution. Let's square: Then we get where the Answer is: -2

EXAMPLE 2 Solve the equation. Decision. Let's move on to one base of the degree: From equality of bases to equality of degrees: Whence Answer: 3

EXAMPLE 3 Solve the equation. Decision. Let's raise both sides of the equation to the third power: After elementary transformations we get: Answer: 23

EXAMPLE 4 Solve the equation. If your equation has more than one root, indicate the smaller one in your answer. Decision. Range of valid values: x ≠ 10. On this area, we multiply by the denominator: Both roots lie in the ODZ. The lesser of these is −3. Answer: -3

EXAMPLE 5 Solve the equation. Decision. Using the formula we get: Answer: 6

EXAMPLE 6 Solve the equation. Decision. The logarithms of two expressions are equal if the expressions themselves are equal and at the same time positive: From where we get the Answer: 6

EXAMPLE 7 Solve the equation. Indicate the smallest positive root in your answer. Decision. Let's solve the equation:

The values \u200b\u200bcorrespond to large positive roots. If k \u003d 1, then x 1 \u003d 6.5 and x 2 \u003d 8.5. If k \u003d 0, then x 3 \u003d 0.5 and x 4 \u003d 2.5. The values \u200b\u200bcorrespond to the smaller values \u200b\u200bof the roots. The smallest positive decision is 0.5. Answer: 0, 5

EXAMPLE 8 Solve the equation. Decision. Reducing the left and right sides of the equation to the powers of 6, we get: Whence it means, Answer: 2

EXAMPLE 9 Solve the equation. Decision. Squaring both sides of the equation, we get: Obviously from where Answer: 5

EXAMPLE 10 Solve the equation. Decision. Let's rewrite the equation so that on both sides there is a logarithm to the base 4: Further, it is obvious where the Answer is: -11

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