Complex exponential equation. exponential equations

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Preparation course for the exam for grades 10-11, as well as for teachers. Everything you need to solve part 1 of the exam in mathematics (the first 12 problems) and problem 13 (trigonometry). And this is more than 70 points on the Unified State Examination, and neither a hundred-point student nor a humanist can do without them.

All the necessary theory. Quick solutions, traps and secrets of the exam. All relevant tasks of part 1 from the Bank of FIPI tasks have been analyzed. The course fully complies with the requirements of the USE-2018.

The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.

Hundreds of exam tasks. Text problems and probability theory. Simple and easy to remember problem solving algorithms. Geometry. Theory, reference material, analysis of all types of USE tasks. Stereometry. Cunning tricks for solving, useful cheat sheets, development of spatial imagination. Trigonometry from scratch - to task 13. Understanding instead of cramming. Visual explanation of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Base for solving complex problems of the 2nd part of the exam.

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 equal 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 $t$.

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

The fraction is zero when the numerator zero, and the denominator is not equal to 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 deciding exponential equations properties of degrees are used, let's 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

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

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

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

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 expand this polynomial, we use the summand grouping method, for this we group the first two and the last two terms, while it is important to put the sign in front of the second grouping correctly, 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 available 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

Collection for solving exponential equations

Introduction

In the course of mathematics, one of the important places is given to the solution of exponential equations. For the first time, students meet with exponential equations in the NPO groups in the second year of study, and in the SVE groups in the first year of study. The exponential equations are also found in USE assignments. Accordingly, considerable attention should be paid to the study of methods for solving them. When solving exponential equations, difficulties often arise due to the following features: - bringing an algorithm for solving exponential equations; - when solving exponential equations, students make transformations that are equivalent to the original equations; - when solving an exponential equation, they introduce a new variable and forget to return to the reverse substitution. The proposed manual is the answers to the solution of exponential equations for independent work and successful completion of the exam.

The purpose of this collection: to study the theoretical material on the topic, to analyze this topic in algebra textbooks and the beginning of analysis, to systematize the USE tasks for solving exponential equations, to systematize and generalize methodological recommendations for solving exponential equations. To achieve this goal, it is necessary to solve the following tasks:

Explore Requirements state standards on the topic "Exponential Equations";

Analyze material on the topic in algebra textbooks and began analysis;

Systematize methods for solving exponential equations;

To systematize and summarize the methodological features of the study of this topic. The guide contains two sections. The first section defines an exponential equation, properties of powers, types of exponential equations and methods for solving them with solution samples. The second section presents a number of examples found in the tasks of the exam. Answers to these questions are provided at the end. This manual can be used both in the classroom and for individual learning, as well as for those who want to deepen their knowledge on the topic: "Exponential Equations".

Definition. An equation containing an unknown in the exponent, called exemplary.

Must remember!When solving exponential equations, it is often used:

1. Theorem: if a 0 ;, a≠ 1 and = , then = .

2. Properties of degrees : a x * a y = a x + y = = * ( x = , ( y = ,

a - x = ; a 0 = 1, a 1 = a.

Consider the main types of exponential equations and solution methods.

1. The simplest exponential equation of the form:

a x = b, wherea 0; b 0, a≠ 1, has a solutionx = .

Example 1 Solve the Equation 2 x = 3.

Solution : x =
Answer:

2. To solve equations of the form: a f ( x ) = b, wherea0; b0, a ≠ 1, need to provide reasons but as a power of the same number, and then compare the indicators.

Example 2 Solve Equation 5 2x+4 = 25.

3. An exponential equation of the form

a f ( x ) = a ȹ( x ) , a0, a ≠ 1

is solved by taking the logarithm of both sides of the equation to the base but. Its equivalent equation

f(x) = ȹ(x).

Example 3 Solve the Equation 6 2x - 8 = 216 x

Solution. 6 2x - 8 \u003d 6 3x, because 216 = 6 3 = 6 * 6 * 6

2x - 3x = 8

Example 4(USE) Indicate the interval to which the root belongs

equations 0.1x-1 = 16.

1). (-1;1]; 3). (-3; -1];

2). (1;10]; 4). (16; 20].

Solution. Let's represent the numbers and 16 as a power of 2:

2 -5 and 16 = 2 4

We get an equation equivalent to this one:

(2 -5) 0.1x-1 \u003d 2 4, i.e. 2 -5 (0.1x - 1) \u003d 2 4.

This equation is equivalent to the equation

5(0.1x - 1) = 4

0.5x \u003d 4 - 5

The number 2 is contained in the interval (1;10] ​​indicated as one of the answers. Therefore, the correct answer is 2.

Example 4(USE) Find the sum of the squares of the roots of the equation -5 = 9 -2x .

1) 26 2) 25 3) 17 4)13.

Solution. Using the properties of degrees, we transform the right side of the equation: 9 -2x \u003d (3 2) -2x \u003d 3 -4x

This equation will take the form: -5 = 3 -4 .

It follows from the monotonicity properties of an exponential function that the exponential equation is equivalent to the equation

x 2 - 5 \u003d -4x.

Solve the quadratic equation x 2 + 4x -5 = 0

D = b 2 – 4ac

D \u003d 4 2 - 4 * 1 * (-5) \u003d 16 + 20 \u003d 36 0, the equation has two roots:

Since the quadratic equation is equivalent to the original equation, the resulting roots are the horses of this equation. In other matters, you can check by direct substitution that the numbers -5 and 1 are the roots of this equation. Thus, the sum of the squares of the roots of the equation -5 = 9 -2x equals (-5) 2 + 1 2 = 25 +1 = 26.

Correct answer number - 1

4. Type equation a 0 a 2x +a 1 a x +a 2 = 0.

This equation is called a three-term exponential equation. Stand a x = y converts it to the usual quadratic equation a 0 y 2 x + a 1 y + a 2 = 0 . Having solved it, we will find the roots y 1 And y 2 . After that, the solution of the original equation is reduced to the solution of two equations a x = y 1 , a x = y 2 . The last equations have a solution for y 1 0 And y 2 0 .

Example 5. Solve Equation 2 2 x - 2 x - 2=0.

Solution. Let 2 x = y, then the equation will take the form

y 2 – y – 2 = 0

D \u003d (-1) 2 - 41 (-2) \u003d 9 0, 2 roots

a) 2 x = 2; b) 2 x = -1, no solution, because -one

Example 6. Solve Equation 9 x – 3 x – 6 = 0

Solution. The first term of the equation can be represented as 9 x = 3 2 x = (3 x) 2 . Then the original equation will take the form (3 x) 2 - 3 x - 6 = 0. Denote 3 x = y, then we have y 2 - y - 6 = 0

y 1 = 3; y 2 \u003d -2.

a) 3 x = 3 b) 3 x = -2 – no solution, because -2

5. Type equation

This equation is solved by taking the common factor out of brackets.

Example 7. Solve the equation

2 x +1 + 32 x -1 – 52 x + 6 = 0

Solution. Let's take the common factor 2 x -1 out of brackets, we get

2 x -1 (2 2 + 3 - 52) = -6

2 x -1 (-3) = -6

2 x -1 = -6: (-3)

6. Equation of the form, where f(x) is an expression containing an unknown number; a0; a ≠ 1.

To solve these equations, you need:

1. replace 1 = a 0 ; a f (x) = a 0 ;

2. solve the equation f (x) = 0

Example 8. Solve the equation

By definition of a degree with a zero exponent, we have:

x 2 - 7x + 12 = 0, (because 1 = 2 0)

D = b 2 – 4ac

Solving the quadratic equation, we get: x 1 \u003d 3, x 2 \u003d 4.

Answer: 3; 4.

7. Equation view

This equation is reduced to a three-term exponential equation by dividing both parts by a x or b x .

Example 9 Solve the Equation 9 x + 6 x = 2 2 x +1

Solution. Let's rewrite the equation as 3 2 x + 2 x 3 x – 22 2 x = 0.

Dividing both sides of the equation by 2 2 x ≠ 0, we get

Let the equation then take the form

y 2 + y -2 = 0 . Solving the quadratic equation, we get = -2, = 1.

a) - there is no solution, because -2

Examples.

I. Solve equations:

31. 0.5 x +7 0.5 1-2 x = 2

32.0.6 x 0.6 3 =

34. 3 2 x -1 + 3 2 x = 108

35. 2x +1 + 2x -1 + 2x = 28

36. 2 3 x +2 - 2 3 x -2 = 30

37. 3x -1 - 3x + 3x +1 = 63

40. 7x - 7x-1 = 6

41.5 3x += 140

42. 3 2y-1 +3 2y-2 -3 2y-4 = 315

43. 2x+1 + 32x-1 -52x+6=0

44.9x - 43x +3 =0

45.16 x -174 x +16 =0

46. ​​25x - 65x + 5 = 0

47.64x-8x-56=0

48. 84x - 62x + 1 = 0

50. 13 2 x +1 - 13 x - 12 = 0

II. (USE) Indicate to which interval the root of the equation belongs:

1. 3 4 x +5 = 81

1) (-1;0] 2) (0;3] 3) (3;4] 4) (4;+∞]

2.45 x -8 = 64

1) (-∞; -3] 2) (-3; -2] 3) (-2;0] 4) (0; 3]

3.6 3 x +5 = 36

1) (-∞;-8] 2) (-8;0] 3) (0;20) 4) 4) (1;3)

6.6 10 x -1 = 36

1) (-4;-1) 2) [-1;0) 3) (0;1) 4) 2) (0;1) 3) 4)

1) [-1;1] 2) (1;2) 3)

10.5 2 x +1 = 125

1) [-2;0] 2) (0;2) 3) 4)

11.25 x +1 = 4

1) [-4;-2] 2) [-2;-1] 3) [-1;1] 4)

1) [-6;-4] 2) [-4;-3] 3) [-3;1] 4)

13.6 2 x +2 = 216

1) 2) 3) [-2;0] 4)

14.72 x +2 = 343

1) [-4;-3] 2) [-3;-2] 3) [-2;0] 4)

15. 3 3 x +3 = 9

1) [-1;1] 2) 3) 4)

16.2 3 x +1 = 8

1) [-6;-4] 2) [-4;-2] 3) [-2;2] 4)

1) [-7;-5] 2) [-5;-3] 3) [-3;0] 4)

18. 0.1 2 x = 100 3 x +1

1) [-] 2) [; 1] 3) (-1;-0.5) 4) (0.5;1)

19. 0.2 x -0.5 = 0.04 x -1

1) [-1] 2) 3) (-1;0) 4) (1.5; 3)

20.008 x = 5 1-2 x

1) [-1; 1.5] 2) 3) (-1; -0.5) 4) (0.5;1)

III. Find the sum of the squares of the roots of the equation

1) 9 2) 0 3) 4 4)

1) 9 2) 1 3) 8 4)

1) 10 2) 4 3) 8 4) 0.04

1) 10 2) 13 3) 37 4) 0.25

1) 0 2) 2 3) 1 4) 0.25

1) 26 2) 25 3) 17 4) 13

1) 9 2) 0 3) 4 4)

1) 9 2) 1 3) 8 4)

Answers

I.Solve Equations

II. (USE) Indicate to which interval the root of the equation belongs

III. Find the sum of the squares of the roots of the equation

Additional examples:

1. 4 3-2x = 4 2-x

2. 2 5 x +1 = 4 2 x

3.5 3 = 25 x +0.5

8.5 x -4 = 25 2

11.4 x +2 x -24 = 0

12. 9 x - 4 * 3 x - 45 = 0

13. 4x - 3 * 2x = 40

14. 2 4 x - 50 * 2 2 x \u003d 896

15. 7 2 x - 6 * 7 x - 7 = 0

16. 9 x - 8 * 3 x - 9 = 0

17. 16 x + 4 * 4 x - 5 = 0

18. 4x -9 * 2x + 8 = 0

19. 36 x - 4 * 6 x - 12 = 0

20. 64 x - 8 x - 56 = 0

21. 7 x +2 + 4 * 7 x +1 = 539

22. 2x +1 + 3 * 2x -1 - 5 * 2x + 6 = 0

23. 7x + 7x +2 = 350

24. 7 * 5 x - 5 x +1 = 2 * 5 3

25. 3 x +2 + 4 * 3 x +1 = 21

26.5 1+2 x + 5 2 x +3 = 650

27. 6 x +1 + 35 * 6 x -1 = 71

28. 4x +1 +4x = 320

29. 3 x +1 - 2 * 3 x -2 = 25

30. 2 3 x +2 - 2 3 x -2 = 30

33.4 x = 5 - x

35. 2-3 x = 2x - 3

36. 3 * 2 2 x + 6 x -2 * 3 2 x = 0

37. 2 * 2 2 x - 5 * 2 x * 3 x + 3 * 3 2 x \u003d 0

38. 3 * 16 x + 2 * 81 x = 5 * 36 x

39. 3 * 4 2 x - 4 x * 9 x + 2 * 9 2 x = 0

40. 6 * 4 x - 13 * 6 x + 6 * 9 x = 0

41. 3 * 2 2 x + * 9 x +1 - 6 * 4 x +1 = - * 9 x +2

42. 4x + 3x -1 = 4x -1 + 3x +2

44. 7 x -5 * - 49 * + 3 * 7 x -5 = 147

45. 3 * 2 x +1 +2 * 5 x -2 = 5 x + 2 x -2

47. 0.125 * 2 -4x-16 \u003d

51. (0.2) x + 0.5 = (0.04) x

53. 32 (x + 8) (x-4) \u003d 0.25 *

54. 5x+1 = 5x-1

55. 7 x + 1 - 7 x + 2 * 7 x-1 - 14 * 7 x-2 \u003d 48

56. 3 2x-1 - 9 x + = 675

57. 5 2x-1 + 5 x + 1 = 250

58. – 5 * + 4 = 0

59. 2 2+x + 2 2-x = 17

60. 2 x + 1 * 5 x \u003d 10 x + 1 * 5 x + 2

61. 2 x * 5 x-1 = 200

64. 7 x + 1 + 3 * 7 x \u003d 3 x + 2 + 3 x

65. 9 x - 5 x - 3 2x * 15 + 5 x + 1 * 3 = 0

66. 25 x - 7 x - 7 * 5 2x + 1 + 5 * 7 x + 1 \u003d 0

67. 9 x + 6 x - 2 * 4 x \u003d 0

68. 4 * 2 2x - 6 x \u003d 18 * 9 x

69. 4 x \u003d 2 * 10 x + 3 * 25 x

70. 64 * 9 -x - 84 * 12 -x + 27 * 16 -x \u003d 0

72. 8 x + 8 = 3 * 4 x + 3 * 2 x + 1

73. 3 -12x-1 - 9 -6x-1 - 27 -4x-1 + 81 1-3x \u003d 2192

Conclusion

Summing up, the following conclusions can be drawn:

1, The exponential equations are of interest to students. When solving exponential equations, the skills of systematization, logical thinking are developed when choosing right method solutions, increases creativity and mental abilities.

2. For each type of equations, difficulties may arise in determining the solution method.

In the course of algebra and the beginning of analysis, exponential equations are often found in USE assignments. In the lessons, little time is devoted to studying this topic, not all methods for solving exponential equations are shown in textbooks, few examples are given for independent solution. Therefore, this manual will help students to delve deeper into the solution, to assimilate the program material of this topic for the successful passing of the written exam for the course of a comprehensive school, as well as for those who wish to pass the exam.

Literature

    Mathematics in tables and diagrams. For schoolchildren and entrants. St. Petersburg, Victoria Plus LLC, 2004, 224 p.

    Maths. Control measuring materials of the unified state exam in 2004. M .: Testing Center of the Ministry of Education of Russia, 2004.

    The system of training tasks and exercises in mathematics / A.Ya. Simonov, D.S. Bakaev, A.G. Epelman and others - M .: Education, 1991. -208 p.

    Getting ready for the unified state exam. Mathematics / J1.0. Denishcheva, E. M. Boychenko, Yu.A. Glazkov and others - 2nd ed., stereotype. - M.: Bustard, 2004, - 120 p.

    Lappo L.D., Popov M.A. Maths. Typical test tasks: Educational and practical guide / L.D. Lappo, M.A. Popov. - M.: Publishing house "Exam", 2004 - 48 p.

    Unified state exam: mathematics: 2004 - 2005: Control. will measure, materials / L. O. Denishcheva, G.K. Bezrukova, E.M. Boychenko and others; ed. G.S. Kovaleva; M - in education and science Ros. Federation. Federal. service for supervision in the field of education and science. - M. : Education, 2005. - 80 p.

    Maths. Training USE tests 2004 - 2005 / T.A. Koreshkova, V.V. Miroshin, N.V. Shevelev. - M.6 Ed - in Eksmo, 2005. - 80 p. (Preparation for the exam)

a) Solve the equation: .

b) Indicate the roots of this equation that belong to the segment.

The solution of the problem

This lesson shows how to correctly use the replacement in an exponential equation, how to solve the simplest trigonometric equation and determine its roots that belong to a certain interval. The first part of the problem is the solution of the exponential equation. To do this, a replacement is performed and a fractional rational equation is obtained, the solution of which is possible in several ways: reduction to a quadratic equation or selection. IN this case both ways are acceptable as the equation is not very complicated. After obtaining the roots, we perform the reverse substitution and obtain two simple trigonometric equations of the form sina=t. The roots of this equation are found by standard formulas. In order to determine the extra roots in the solution, the most optimal is to use a unit circle, with the roots of the equation marked on it. Thus we get common decision equations - the answer to point a) of the problem. To answer point b), it is necessary to correctly take into account the gap and calculate the roots. In this case, this is very easy to do, since it is easy to mark all the roots on a unit circle and find their value using the periodicity of the sine and cosine (we should not forget that the period of the sine and cosine is 2π). Decision received.

The solution to this problem is recommended for students in grade 10 when studying the topic "Trigonometric Equations" ("Arcsine", "Arcsine and the solution of the equation sina = t"); for students in grades 11 when studying the topic "Exponential and logarithmic functions" (" Exponential function, its properties. The simplest exponential equations”, “Exponential equations”). When preparing for the exam, the lesson is recommended when repeating the topics “Trigonometric Equations”, “Exponential and Logarithmic Functions”.

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