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

Equations are called exponential if the unknown is contained in the exponent. The simplest exponential equation has the form: a x \u003d a b, where a> 0, and 1, x is an unknown.

The main properties of the degrees, with the help of which the exponential equations are transformed: a>0, b>0.

When deciding exponential equations they also use the following properties of the exponential function: y = a x , a > 0, a1:

To represent a number as a power, the basic logarithmic identity is used: b = , a > 0, a1, b > 0.

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For successful solution exponential equations You must know the basic properties of powers, the properties of the exponential function, the basic logarithmic identity.

When solving exponential equations, two main methods are used:

transition from the equation a f(x) = a g(x) to the equation f(x) = g(x);
introduction of new lines.

Examples.

1. Equations Reducing to the Simplest. They are solved by bringing both sides of the equation to a power with the same base.

3x \u003d 9x - 2.

Decision:

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

Answer: 4.

2. Equations solved by bracketing the common factor.

Decision:

3x - 3x - 2 = 24
3 x - 2 (3 2 - 1) = 24
3 x - 2 x 8 = 24
3 x - 2 = 3
x - 2 = 1
x=3.

Answer: 3.

3. Equations Solved by Change of Variable.

Decision:

2 2x + 2 x - 12 = 0
We denote 2 x \u003d y.
y 2 + y - 12 = 0
y 1 = - 4; y 2 = 3.
a) 2 x = - 4. The equation has no solutions, because 2 x > 0.
b) 2 x = 3; 2 x = 2 log 2 3 ; x = log 2 3.

Answer: log 2 3.

4. Equations containing powers with two different (not reducible to each other) bases.

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

3 x 2 x + 1 - 2 x - 2 = 5 x - 2 x 5 x - 2
2 x - 2 × 23 = 5 x - 2
×23
2 x - 2 = 5 x - 2
(5/2) x– 2 = 1
x - 2 = 0
x = 2.

Answer: 2.

5. Equations that are homogeneous with respect to a x and b x .

General form: .

9 x + 4 x = 2.5 x 6 x .

Decision:

3 2x – 2.5 × 2x × 3x +2 2x = 0 |: 2 2x > 0
(3/2) 2x - 2.5 × (3/2) x + 1 = 0.
Denote (3/2) x = y.
y 2 - 2.5y + 1 \u003d 0,
y 1 = 2; y2 = ½.

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

1º. exponential equations name equations containing a variable in the exponent.

The solution of exponential equations is based on the power property: two powers with the same base are equal if and only if their exponents are equal.

2º. Basic ways to solve exponential equations:

1) the simplest equation has a solution;

2) an equation of the form by logarithm to the base a bring to mind;

3) the equation of the form is equivalent to the equation ;

4) an equation of the form is equivalent to the equation.

5) an equation of the form through a replacement is reduced to an equation, and then a set of simplest exponential equations is solved;

6) equation with reciprocal quantities by replacement reduce to the equation , and then solve the set of equations ;

7) equations homogeneous with respect to a g(x) and b g (x) given that kind through the substitution reduce to the equation , and then solve the set of equations .

Classification of exponential equations.

1. Equations Solved by Transition to One Base.

Example 18. Solve the equation .

Solution: Let's take advantage of the fact that all bases of powers are powers of 5: .

2. Equations solved by passing to one exponent.

These equations are solved by transforming the original equation to the form , which is reduced to its simplest using the proportion property.

Example 19. Solve the equation:

3. Equations Solved by Bracketing the Common Factor.

If in the equation each exponent differs from the other by some number, then the equations are solved by bracketing the degree with the smallest exponent.

Example 20. Solve the equation.

Solution: Let's put the degree with the smallest exponent out of brackets on the left side of the equation:

Example 21. Solve the equation

Solution: We group separately on the left side of the equation the terms containing degrees with base 4, on the right side - with base 3, then put the degrees with the smallest exponent out of brackets:

4. Equations Reducing to Quadratic (or Cubic) Equations.

The following equations are reduced to a quadratic equation with respect to the new variable y:

a) the type of substitution , while ;

b) the type of substitution , while .

Example 22. Solve the equation .

Solution: Let's make a change of variable and solve quadratic equation:

Answer: 0; one.

5. Homogeneous equations with respect to exponential functions.

The view equation is homogeneous equation second degree relative to unknown a x and b x. Such equations are reduced by preliminary division of both parts by and subsequent substitution to quadratic equations.

Example 23. Solve the equation.

Solution: Divide both sides of the equation by:

Putting , we get a quadratic equation with roots .

Now the problem is reduced to solving the set of equations . From the first equation, we find that . The second equation has no roots, since for any value x.

Answer: -1/2.

6. Equations rational with respect to exponential functions.

Example 24. Solve the equation.

Solution: Divide the numerator and denominator of the fraction by 3 x and instead of two we get one exponential function:

7. Equations of the form .

Such equations with a set of admissible values (ODV) determined by the condition , by taking the logarithm of both parts of the equation, are reduced to an equivalent equation , which in turn are equivalent to the combination of two equations or .

Example 25. Solve the equation:.

didactic material.

Solve the 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 sum of the roots of the equation .

Find the value of the expression:

28. , where x0- root of the equation ;

29. , where x0 is the root of the 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.

exponential inequalities.

1º. An inequality containing a variable in the exponent is called exemplary inequality.

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

if , then the inequality is equivalent to ;

if , then the inequality is equivalent to .

When solving exponential inequalities, the same techniques are used as when solving exponential equations.

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

Solution: Because , then the given inequality can be written as: . Since , this inequality is equivalent to the inequality .

Solving the last inequality, we get .

Example 27. Solve the inequality: ( the method of taking the common factor out of brackets).

Solution: We take out the brackets on the left side of the inequality, on the right side of the inequality and divide both sides of the inequality by (-2), changing the sign of the inequality to the opposite:

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

Example 28. Solve the inequality ( method of introducing a new variable).

Solution: Let . Then this inequality takes the form: or , whose solution is the interval .

From here. Since the function is increasing, then .

didactic material.

Specify the set of solutions to the inequality:

1. ; 2. ; 3. ;

6. At what values x do the points of the graph of the function lie below the line?

7. At what values x do the points of the graph of the function lie not below the line?

Solve the inequality:

8. ; 9. ; 10. ;

13. Indicate the largest integer solution of the inequality .

14. Find the product of the largest integer and the smallest integer solutions of the inequality .

Solve the inequality:

15. ; 16. ; 17. ;

18. ; 19. ; 20. ;

21. ; 22. ; 23. ;

24. ; 25. ; 26. .

Find the scope of the function:

27. ; 28. .

29. Find the set of argument values for which the values of each of the functions are greater than 3:

and .

Answers: 11.3; 12.3; 13.-3; 14.1; 15. (0; 0.5); 16. ; 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. )