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Solving equations for the exam. Irrational equations. An exhaustive guide. Scheme for solving complex equations

Equations, part $C$

An equality containing an unknown number, denoted 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 area of ​​​​admissible values ​​(ODV) for it.
  2. Solve the equation.
  3. Choose from the obtained roots of the equation those that satisfy the ODZ.

ODZ of various expressions (under the expression we will understand the alphanumeric record):

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

$(f(x))/(g(x)); g(x)≠0$

2. The root expression must not be negative.

$√(g(x)); g(x) ≥ 0$.

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

$(f(x))/(√(g(x))); g(x) > 0$

4. For the logarithm: the sublogarithmic expression must be positive; the base must be positive; the base cannot be equal to one.

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

Logarithmic Equations

Logarithmic equations are equations of the form $log_(a)f(x)=log_(a)g(x)$, where $a$ is a positive number different from $1$, and equations that reduce to this form.

To solve logarithmic equations, you need to know the properties of logarithms: we will consider all the properties of logarithms for $a > 0, a≠ 1, b> 0, c> 0, m$ - any real number.

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

$log_(a)b^m=mlog_(a)b;$

$log_(a^m)b=(1)/(m)log_(a)b.$

$log_(a^n)b^m=(m)/(n)log_(a)b$

$log_(3)3^(10)=10log_(3)3=10;$

$log_(5^3)7=(1)/(3)log_(5)7;$

$log_(3^7)4^5=(5)/(7)log_(3)4;$

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

$log_a(bc)=log_(a)b+log_(a)c$

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

$log_(a)(b)/(c)=log_(a)b-log_(a)c$

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

$log_(a)b∙log_(c)d=log_(c)b∙log_(a)d$ if $a, b, c$ and $d > 0, a≠1, b≠1.$

5. $c^(log_(a)b)=b^(log_(a)b)$, where $a, b, c > 0, a≠1$

6. Formula for moving to a new bottom

$log_(a)b=(log_(c)b)/(log_(c)a)$

7. In particular, if it is necessary to swap the base and the sublogarithmic expression

$log_(a)b=(1)/(log_(b)a)$

There are several main types of logarithmic equations:

The simplest logarithmic equations: $log_(a)x=b$. The solution of this type of equations follows from the definition of the logarithm, i.e. $x=a^b$ and $x > 0$

Let us represent both sides of the equation in the form of a logarithm in base $2$

$log_(2)x=log_(2)2^3$

If the logarithms are equal in the same base, then the sublogarithmic expressions are also equal.

Answer: $x = $8

Equations of the form: $log_(a)f(x)=log_(a)g(x)$. Because the bases are the same, then we equate the sublogarithmic expressions and take into account the ODZ:

$\table\(\ f(x)=g(x);\ f(x)>0;\ g(x) > 0, a > 0, a≠1;$

$log_(3)(x^2-3x-5)=log_(3)(7-2x)$

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

We transfer all the terms to the left side of the equation and give similar terms

Let's check the found roots according to the conditions $\table\(\ x^2-3x-5>0;\ 7-2x>0;$

When substituting into the second inequality, the root $x=4$ does not satisfy the condition, therefore, it is an extraneous root

Answer: $x=-3$

  • Variable replacement method.

In this method, you need:

  1. Write the ODZ equation.
  2. According to the properties of logarithms, ensure that the same logarithms are obtained 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 a value instead of a variable and get the simplest equation of the form: $log_(a)x=b$
  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 ODZ condition.

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

1. Let's write the ODZ equations:

$\table\(\ x>0,\text"because it is under the sign of the root and the logarithm";\ √x≠1→x≠1;$

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

$log_(2)√x+(2)/(log_(2)√x)-3=0$

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

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

$(t^2+2-3t)/(t)=0$

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

$t^2+2-3t=0$, $t≠0$

5. We solve the resulting quadratic equation using the Vieta theorem:

6. Let's go back to step 3, make the reverse substitution and get two simple logarithmic equations:

$log_(2)√x=1$, $log_(2)√x=2$

We take the logarithm of the right parts of the equations

$log_(2)√x=log_(2)2$, $log_(2)√x=log_(2)4$

Equate sublogarithmic expressions

$√x=2$, $√x=4$

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

$х_1=4$, $х_2= 16$

7. Let us substitute the roots of the logarithmic equation in item 1 and check the condition of the ODZ.

$\(\table\ 4 >0; \4≠1;$

The first root satisfies the ODZ.

$\(\table\ 16 >0; \16≠1;$ The second root also satisfies the DDE.

Answer: $4; 16$

  • Equations of the form $log_(a^2)x+log_(a)x+c=0$. 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 ODZ.

Fractionally rational equations

  • If the fraction is zero, then the numerator is zero and the denominator is not 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 ​​of the variable for which the equation does not make sense (ODV)
  2. Find the common denominator of the fractions included 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 ODZ condition.
  • If two fractions are involved in the equation and the numerators are their equal expressions, 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 ODZ of the entire original equation.

exponential equations

An exponential equation is an equation in which the unknown is contained in the exponent.

When solving exponential equations, the properties of powers are used, let us recall some of them:

1. When multiplying powers with the same bases, the base remains the same, and the exponents are added.

$a^n a^m=a^(n+m)$

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

$a^n:a^m=a^(n-m)$

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

$(a^n)^m=a^(n∙m)$

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

$(a b)^n=a^n b^n$

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

$((a)/(b))^n=(a^n)/(b^n)$

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 line of the fraction

$a^(-n)=(1)/(a^n)$

$(a^(-n))/(b^(-k))=(b^k)/(a^n)$

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

$√^n(a^k)=a^((k)/(n))$

Types of exponential equations:

1. Simple exponential equations:

a) The form $a^(f(x))=a^(g(x))$, where $a >0, a≠1, x$ is unknown. To solve such equations, we use the property of powers: powers with the same base ($а >0, a≠1$) are equal only when their exponents are equal.

b) An equation of the form $a^(f(x))=b, b>0$

To solve such equations, it is necessary to take both parts of the logarithm in the base $a$, it turns out

$log_(a)a^(f(x))=log_(a)b$

2. Base adjustment method.

3. Method of factorization and change of variable.

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

Solve the equation $2^(3x)-7 2^(2x-1)+7 2^(x-1)-1=0$

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

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

Let's change the variable $2^x=t; t>0$

We get a cubic equation of the form

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

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

$2t^3-7 t^2+7 t-2=0$

Let us expand the left side of the equation by the grouping method

$(2t^3-2)-(7 t^2-7 t)=0$

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

$2(t^3-1)-7t(t-1)=0$

Additionally, in the first bracket we see the formula for the difference of cubes

$(t-1)(2t^2+2t+2-7t)=0$

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

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

Let's solve the first equation

We solve the second equation through the discriminant

$D=25-4 2 2=9=3^2$

$t_2=(5-3)/(4)=(1)/(2)$

$t_3=(5+3)/(4)=2$

$2^x=1; 2^x=(1)/(2); 2^x=2$

$2^x=2^0; 2^x=2^(-1); 2^x=2^1$

$x_1=0; x_2=-1; x_3=1$

Answer: $-1; 0; 1$

4. Method for converting to a quadratic equation

  • We have an equation of the form $A·a^(2f(x))+В·a^(f(x))+С=0$, where $A, B$ and $C$ are coefficients.
  • We make the change $a^(f(x))=t, t > 0$.
  • It turns out a quadratic equation of the form $A·t^2+B·t+С=0$. We solve the resulting equation.
  • We make the reverse substitution, taking into account that $t > 0$. We get the simplest exponential equation $a^(f(x))=t$, solve it and write the result in response.

Factoring methods:

  • Taking the common factor out of brackets.

To factorize a polynomial by taking the common factor out of 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).

Factorize 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 "a". Next, we find the quotient of dividing the original polynomial by "2a", we get:

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

This is the end result of the factorization.

Application of 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.

$(a+b)^2=a^2+2ab+b^2$

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.

$(a-b)^2=a^2-2ab+b^2$

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

$a^2-b^2=(a+b)(a-b)$

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

$(a+b)^3=a^3+3a^2b+3ab^2+b^3$

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

$(a-b)^3=a^3-3a^2b+3ab^2-b^3$

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

$a^3+b^3=(a+b)(a^2-ab+b^2)$

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

$a^3-b^3=(a-b)(a^2+ab+b^2)$

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 the common factor out of the bracket from each group. Several groups, after being placed in brackets, should get the same expressions, then we take this bracket forward as a common factor and multiply it by the bracket of the resulting quotient.

Factorize the polynomial $2a^3-a^2+4a-2$

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

$(2a^3-a^2)+(4a-2)=a^2(2a-1)+2(2a-1)$

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

$a^2(2a-1)+2(2a-1)=(2a-1)(a^2+2)$

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

Using the formula of a square trinomial.

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

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

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Today we will train the skill of solving task 5 of the USE - find the root of the equation. Let's look for the root of the equation. 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 a number encrypted under x, which we will substitute for x and our equation will be a true equality.

For example, 3x=9 is an equation and 3 . 3=9 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, which means that we solved the equation, that is, we found the given number x=3, which turns the equation into a true 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 =32

This is an exponential equation. It is solved as follows - it is necessary that both to the left and to the right of the “equal” sign there is 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=2 5

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

On the left and on the 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 the unknowns on the left, and transfer the known ones to the right, we get:

Checking: 2 1-4(-1) =32

We have found the root of the equation. Answer: x=-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 a similar way - by bringing 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 degree property:

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

If the bases of the exponent are equal, then the exponents are equal:

Answer: x=9.

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

We have found the root of the equation correctly.

Task 3 - find the root of the equation

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

Then our equation will be written as:

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

Answer: x=5. Do the check yourself.

Task 4 - find the root of the equation log 3 (15's) = 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 equals sign to be the same. Now they are the same, so we equate those expressions that are under the sign of logarithms:

Answer: x=13

Task 5 - find the root of the equation log 3 (3-x)=3

The number 3 is log 3 27. To make it clear below, the subscript below the logarithm sign is the number that is raised to the power, in our case 3, the logarithm sign is the number that turned out when raised to the power is 27, and the logarithm itself is exponent to which you need to raise 3 to get 27.

Look at 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)=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))=log 3 27

log 3 (3+24)= log 3 27

log 3 27=log 3 27

Answer: x=-24.

Find the root of the equation. Task 6.

log 2 (x+3)=log 2 (3x-15)

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

log 2 12=log 2 12

Answer: x=9.

Find the root of the equation. Task 7.

log 2 (14-2x)=2log 2 3

log 2 (14-2x)=log 2 3 2

Check: log 2 (14-5)=2log 2 3

log29=2log23

log 2 3 2 =2log 2 3

2log 2 3=2log 2 3

Answer: x=2.5

Prepare for the exam and for the OGE - see the previous topics and.