Methods for solving indicative equations. Lecture: "Methods for solving indicative equations

Indicatively called equations in which the unknown is contained in an indicator. The simplest indicative equation is: a x \u003d a b, where a\u003e 0, and 1, x - unknown.

The main properties of degrees, with which the indicative equations are converted: a\u003e 0, b\u003e 0.

When solving indicatory equations also use the following properties of the indicative function: Y \u003d A x, A\u003e 0, A1:

To represent the number of degree, the main logarithmic identity is used: B \u003d, A\u003e 0, A1, B\u003e \u200b\u200b0.

Tasks and tests on the topic "Indicative equations"

Indicative equations
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For successful solution Indicative equations you need to know the basic properties of degrees, the properties of the indicative function, the main logarithmic identity.

When solving indicative equations, two main methods are used:

transition from equation a f (x) \u003d a g (x) to equation f (x) \u003d g (x);
introduction new straight lines.

Examples.

1. Equations reduced to the simplest. Resolved by bringing both parts of the equation to the degree with the same base.

3 x \u003d 9 x - 2.

Decision:

3 x \u003d (3 2) x - 2;
3 x \u003d 3 2x - 4;
x \u003d 2x -4;
x \u003d 4.

Answer: 4.

2. Equations solved by making a general multiplier for parentheses.

Decision:

3 x - 3 x - 2 \u003d 24
3 x - 2 (3 2 - 1) \u003d 24
3 x - 2 × 8 \u003d 24
3 x - 2 \u003d 3
x - 2 \u003d 1
x \u003d 3.

Answer: 3.

3. Equations solved by replacing the variable.

Decision:

2 2x + 2 x - 12 \u003d 0
We indicate 2 x \u003d y.
Y 2 + Y - 12 \u003d 0
y 1 \u003d - 4; Y 2 \u003d 3.
a) 2 x \u003d - 4.The does not have solutions, because 2 x\u003e 0.
b) 2 x \u003d 3; 2 x \u003d 2 log 2 3; x \u003d log 2 3.

Answer: Log 2 3.

4. Equations containing degrees with two different (not reduced to each other) grounds.

3 × 2 x + 1 - 2 × 5 x - 2 \u003d 5 x + 2 x - 2.

3 × 2 x + 1 - 2 x - 2 \u003d 5 x - 2 × 5 x - 2
2 x - 2 × 23 \u003d 5 x - 2
× 23.
2 x - 2 \u003d 5 x - 2
(5/2) x- 2 \u003d 1
x - 2 \u003d 0
x \u003d 2.

Answer:2.

5. Equations are homogeneous relative to the x and b x.

General form: .

9 x + 4 x \u003d 2.5 × 6 x.

Decision:

3 2x - 2.5 × 2 x × 3 x +2 2x \u003d 0 |: 2 2x\u003e 0
(3/2) 2x - 2.5 × (3/2) x + 1 \u003d 0.
Denote (3/2) x \u003d y.
y 2 - 2.5Y + 1 \u003d 0,
y 1 \u003d 2; Y 2 \u003d ½.

Answer: log 3/2 2; - Log 3/2 2.

1º. Accurate equationscalled equations containing a variable in an indicator.

The solution of the indicative equations is based on the degree property: two degrees with the same base are equal then and only if their indicators are equal.

2º Main ways to solve indicative equations:

1) the simplest equation has a solution;

2) equation of type of logarithming based on a. bring to the form;

3) equation of species is equivalent to equation;

4) equation of type equivalent equation.

5) the equation of the form through the replacement is reduced to the equation, and then solve the set of simpleness of the demonstration equations;

6) equation with mutually reverse values replacement is reduced to the equation, and then solve the set of equations;

7) equations homogeneous relatively a G (X) and b G (x) given that View Through the replacement, it is reduced to the equation, and then the set of equations solve.

Classification of indicative equations.

1. Equations solved by the transition to one base.

Example 18. Solve equation .

Solution: take advantage of the fact that all the foundations of the degrees are degrees of the number 5 :.

2. Equations solved by the transition to one indicator.

These equations are solved by the transformation of the initial equation to the form which use the proportion properties is driven to the simplest.

Example 19. Solve equation:

3. Equations solved by the issuance of a common factor for brackets.

If in the equation each indicator of the extent differs from the other for a certain number, the equations are solved by the degree with the smallest indicator for the bracket.

Example 20. Solve equation.

Solution: I will bring the degree in the left part of the equation with the smallest indicator for brackets:

Example 21. Solve equation

Solution: grouped separately in the left part of the equation, the terms containing degrees with the base 4, in the right part - with the base 3, then we will result in the lowest indicator for brackets:

4. Equations reduced to square (or cubic) equations.

The equation is reduced to a square equation relative to the new variable y

a) the type of substitution, while;

b) the type of substitution, while.

Example 22. Solve equation .

Solution: We will replace the variable and solve quadratic equation:

Answer: 0; one.

5. Uniform relatively indicative functions of the equation.

The equation of the species is uniform equation second degree relative to unknown a X. and b X. . Such equations are reduced by pre-division of both parts on and subsequent substitution to square equations.

Example 23. Solve equation.

Solution: We divide both parts of the equation on:

Putting, we get a square equation with roots.

Now the task is reduced to the solution of the totality of equations . From the first equation we find that. The second equation does not have roots, since with any meaning x..

Answer: -1/2.

6. Rational relatively indicative functions of the equation.

Example 24. Solve equation.

Solution: Divide the numerator and denomoter of the fraction on 3 X. and get instead of two - one indicative function:

7. View equations .

Such equations with many permissible values \u200b\u200b(OTZ), determined by the condition, the logarithming of both parts of the equation are given to an equivalent equation, which in turn are equivalent to the totality of two equations or.

Example 25. Solve equation :.

Didactic material.

Decide equations:

1. ; 2. ; 3. ;

4. ; 5. ; 6. ;

9. ; 10. ; 11. ;

14. ; 15. ;

16. ; 17. ;

18. ; 19. ;

20. ; 21. ;

22. ; 23. ;

24. ; 25. .

26. Find the product of the roots of the equation .

27. Find the amount of the roots of the equation .

Find the value of the expression:

28., where x 0 - root of the equation ;

29., Where x 0 - whole root equation .

Solve the equation:

31. ; 32. .

Answers:10; 2. -2/9; 3. 1/36; 4. 0, 0.5; fifty; 6. 0; 7. -2; 8. 2; 9. 1, 3; 10. 8; 11. 5; 12. 1; 13. ¼; 14. 2; 15. -2, -1; 16. -2, 1; 17. 0; 18. 1; 19. 0; 20. -1, 0; 21. -2, 2; 22. -2, 2; 23. 4; 24. -1, 2; 25. -2, -1, 3; 26. -0.3; 27. 3; 28. 11; 29. 54; 30. -1, 0, 2, 3; 31.; 32.

Topic number 8.

Indicative inequalities.

1º. Inequality containing a variable in an indicator is called indicative inequality.

2º The solution of demonstration inequalities of the form is based on the following statements:

if, the inequality is equivalent;

if, the inequality is equivalent.

When solving indicative inequalities, the same techniques are used as in solving indicative equations.

Example 26. Solve inequality (method of transition to one base).

Solution: since The specified inequality can be written in the form: . Since, this inequality is equivalent to inequality .

Deciding the last inequality, we get.

Example 27. Solve inequality: ( method of making a common factor for brackets).

Solution: I will bring for brackets in the left part of inequality, in the right part of the inequality and we divide both part of inequality on (-2), changing the sign of inequality to the opposite:

Since, when transition to inequality of indicators, the sign of inequality changes again to the opposite. We get. Thus, the set of all solutions of this inequality is the interval.

Example 28. Solve inequality ( by introducing a new variable).

Solution: Let. Then this inequality will take the form: or , the solution of which is the interval.

From here. Since the function increases, then.

Didactic material.

Indicate many solutions inequality:

1. ; 2. ; 3. ;

6. Under what values x. Play points function lie below direct?

7. Under what values x. Function graphics points are not lower than straight?

Solve inequality:

8. ; 9. ; 10. ;

13. Specify the greatest whole solution of inequality. .

14. Find the work of the greatest whole and the smallest whole of inequality solutions .

Solve inequality:

15. ; 16. ; 17. ;

18. ; 19. ; 20. ;

21. ; 22. ; 23. ;

24. ; 25. ; 26. .

Find the function definition area:

27. ; 28. .

29. Find a plurality of argument values \u200b\u200bat which the values \u200b\u200bof each of the functions are greater than 3:

and .

Answers: 11. 3; 12. 3; 13. -3; 14. 1; 15. (0; 0.5); sixteen. ; 17. (-1; 0) U (3; 4); 18. [-2; 2]; 19. (0; + ∞); 20. (0; 1); 21. (3; + ∞); 22. (-∞; 0) U (0.5; + ∞); 23. (0; 1); 24. (-1; 1); 25. (0; 2]; 26. (3; 3.5) U (4; + ∞); 27. (-∞; 3) U (5); 28.)