Solution of exponential inequalities by introducing a new unknown. Solving exponential inequalities: basic methods. Lesson and presentation on the topic: "Exponential equations and exponential inequalities"
In this lesson, we will look at various exponential inequalities and learn how to solve them based on the methodology for solving the simplest exponential inequalities
1. Definition and properties of exponential function
Let us recall the definition and basic properties of the exponential function. It is on properties that the solution of all exponential equations and inequalities is based.
Exponential function is a function of the form, where the base of the degree and Here x is an independent variable, an argument; y - dependent variable, function.
Figure: 1. Exponential function graph
The graph shows increasing and decreasing exponents, illustrating the exponential function when the base is greater than one and less than one, but greater than zero, respectively.
Both curves pass through the point (0; 1)
Exponential function properties:
Domain: ;
Range of values:;
The function is monotonic, as it increases, as it decreases.
A monotone function takes each of its values \u200b\u200bfor a single argument value.
When, when the argument increases from minus to plus infinity, the function increases from zero, not inclusively, to plus infinity, that is, for given values \u200b\u200bof the argument, we have a monotonically increasing function (). For, on the contrary, when the argument increases from minus to plus infinity, the function decreases from infinity to zero, not inclusively, that is, for given values \u200b\u200bof the argument, we have a monotonically decreasing function ().
2. The simplest exponential inequalities, solution technique, example
Based on the above, we present a technique for solving the simplest exponential inequalities:
Methodology for solving inequalities:
Equalize the bases of the degrees;
Compare indicators, keeping or changing to the opposite sign of inequality.
The solution of complex exponential inequalities consists, as a rule, in their reduction to the simplest exponential inequalities.
The base of the degree is greater than one, which means that the inequality sign remains:
We transform the right side according to the properties of the degree:
The base of the degree is less than one, the inequality sign must be changed to the opposite:
To solve the quadratic inequality, we will solve the corresponding quadratic equation:
By Vieta's theorem, we find the roots:
The branches of the parabola are directed upwards.
Thus, we have a solution to the inequality:
It is easy to guess that the right-hand side can be represented as a power with a zero exponent:
The base of the degree is greater than one, the inequality sign does not change, we get:
Let us recall the technique for solving such inequalities.
Consider a fractional rational function:
Find the domain of definition:
Find the roots of the function:
The function has a single root, 
We select intervals of constant sign and determine the signs of the function on each interval:
Figure: 2. Intervals of constancy
So we got the answer.
Answer: 
3. Solution of typical exponential inequalities
Consider inequalities with the same indicators, but different bases.
One of the properties of an exponential function is that it takes strictly positive values \u200b\u200bfor any values \u200b\u200bof the argument, which means that it can be divided into an exponential function. Let's divide the given inequality by its right-hand side:
The base of the degree is greater than one, the inequality sign remains.
Let's illustrate the solution:
Figure 6.3 shows the graphs of the functions and. Obviously, when the argument is greater than zero, the graph of the function is higher, this function is larger. When the argument values \u200b\u200bare negative, the function goes lower, it is smaller. When the value of the argument, the functions are equal, which means that this point is also a solution to the given inequality.
Figure: 3. Illustration for example 4
We transform the given inequality according to the properties of the degree:
Here are similar terms:
Let's divide both parts into:
Now we continue to solve similarly to example 4, divide both parts into:
The base of the degree is greater than one, the inequality sign remains:
4. Graphical solution of exponential inequalities
Example 6 - Solve inequality graphically:
Let's consider the functions on the left and right sides and plot a graph of each of them.
The function is an exponential, increases over its entire domain of definition, that is, for all real values \u200b\u200bof the argument.
The function is linear, decreases over its entire domain of definition, that is, for all real values \u200b\u200bof the argument.
If these functions intersect, that is, the system has a solution, then such a solution is the only one and it can be easily guessed. To do this, we iterate over integers ()
It is easy to see that the root of this system is:
Thus, the graphs of the functions intersect at a point with an argument equal to one.
Now we need to get an answer. The meaning of the given inequality is that the exponent must be greater than or equal to the linear function, that is, be higher or coincide with it. The obvious answer is: (Figure 6.4)
Figure: 4. Illustration for example 6
So, we have considered the solution of various typical exponential inequalities. Next, we move on to considering more complex exponential inequalities.
Bibliography
Mordkovich A.G. Algebra and principles of mathematical analysis. - M .: Mnemosyne. Muravin G.K., Muravina O.V. Algebra and principles of mathematical analysis. - M .: Bustard. Kolmogorov A. N., Abramov A. M., Dudnitsyn Yu. P. et al. Algebra and principles of mathematical analysis. - M .: Education.
Math. md. Mathematics-repetition. com. Diffur. kemsu. ru.
Homework
1. Algebra and the beginning of analysis, grade 10-11 (A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn) 1990, no. 472, 473;
2. Solve the inequality:
3. Solve inequality.
Lesson and presentation on the topic: "Exponential equations and exponential inequalities"
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Determination of exponential equations
Guys, we studied exponential functions, learned their properties and built graphs, analyzed examples of equations in which exponential functions were encountered. Today we will study exponential equations and inequalities.Definition. Equations of the form: $ a ^ (f (x)) \u003d a ^ (g (x)) $, where $ a\u003e 0 $, $ a ≠ 1 $ are called exponential equations.
Remembering the theorems that we studied in the topic "Exponential function", we can introduce a new theorem:
Theorem. The exponential equation $ a ^ (f (x)) \u003d a ^ (g (x)) $, where $ a\u003e 0 $, $ a ≠ 1 $, is equivalent to the equation $ f (x) \u003d g (x) $.
Examples of exponential equations
Example.Solve equations:
a) $ 3 ^ (3x-3) \u003d 27 $.
b) $ ((\\ frac (2) (3))) ^ (2x + 0.2) \u003d \\ sqrt (\\ frac (2) (3)) $.
c) $ 5 ^ (x ^ 2-6x) \u003d 5 ^ (- 3x + 18) $.
Decision.
a) We know well that $ 27 \u003d 3 ^ 3 $.
Let's rewrite our equation: $ 3 ^ (3x-3) \u003d 3 ^ 3 $.
Using the theorem above, we get that our equation is reduced to the equation $ 3x-3 \u003d 3 $, solving this equation, we get $ x \u003d 2 $.
Answer: $ x \u003d 2 $.
B) $ \\ sqrt (\\ frac (2) (3)) \u003d ((\\ frac (2) (3))) ^ (\\ frac (1) (5)) $.
Then our equation can be rewritten: $ ((\\ frac (2) (3))) ^ (2x + 0.2) \u003d ((\\ frac (2) (3))) ^ (\\ frac (1) (5) ) \u003d ((\\ frac (2) (3))) ^ (0,2) $.
$ 2x + 0.2 \u003d $ 0.2.
$ x \u003d 0 $.
Answer: $ x \u003d 0 $.
C) The original equation is equivalent to the equation: $ x ^ 2-6x \u003d -3x + 18 $.
$ x ^ 2-3x-18 \u003d 0 $.
$ (x-6) (x + 3) \u003d 0 $.
$ x_1 \u003d 6 $ and $ x_2 \u003d -3 $.
Answer: $ x_1 \u003d 6 $ and $ x_2 \u003d -3 $.
Example.
Solve the equation: $ \\ frac (((0.25)) ^ (x-0.5)) (\\ sqrt (4)) \u003d 16 * ((0.0625)) ^ (x + 1) $.
Decision:
We will sequentially perform a series of actions and bring both sides of our equation to the same bases.
Let's perform a series of operations on the left side:
1) $ ((0.25)) ^ (x-0.5) \u003d ((\\ frac (1) (4))) ^ (x-0.5) $.
2) $ \\ sqrt (4) \u003d 4 ^ (\\ frac (1) (2)) $.
3) $ \\ frac (((0.25)) ^ (x-0.5)) (\\ sqrt (4)) \u003d \\ frac (((\\ frac (1) (4))) ^ (x-0 , 5)) (4 ^ (\\ frac (1) (2))) \u003d \\ frac (1) (4 ^ (x-0.5 + 0.5)) \u003d \\ frac (1) (4 ^ x) \u003d ((\\ frac (1) (4))) ^ x $.
Let's move on to the right side:
4) $16=4^2$.
5) $ ((0.0625)) ^ (x + 1) \u003d \\ frac (1) ((16) ^ (x + 1)) \u003d \\ frac (1) (4 ^ (2x + 2)) $.
6) $ 16 * ((0.0625)) ^ (x + 1) \u003d \\ frac (4 ^ 2) (4 ^ (2x + 2)) \u003d 4 ^ (2-2x-2) \u003d 4 ^ (- 2x ) \u003d \\ frac (1) (4 ^ (2x)) \u003d ((\\ frac (1) (4))) ^ (2x) $.
The original equation is equivalent to the equation:
$ ((\\ frac (1) (4))) ^ x \u003d ((\\ frac (1) (4))) ^ (2x) $.
$ x \u003d 2x $.
$ x \u003d 0 $.
Answer: $ x \u003d 0 $.
Example.
Solve the equation: $ 9 ^ x + 3 ^ (x + 2) -36 \u003d 0 $.
Decision:
Let's rewrite our equation: $ ((3 ^ 2)) ^ x + 9 * 3 ^ x-36 \u003d 0 $.
$ ((3 ^ x)) ^ 2 + 9 * 3 ^ x-36 \u003d 0 $.
Let's do the change of variables, let $ a \u003d 3 ^ x $.
In new variables, the equation will take the form: $ a ^ 2 + 9a-36 \u003d 0 $.
$ (a + 12) (a-3) \u003d 0 $.
$ a_1 \u003d -12 $ and $ a_2 \u003d 3 $.
Let's perform the reverse change of variables: $ 3 ^ x \u003d -12 $ and $ 3 ^ x \u003d 3 $.
In the last lesson, we learned that exponential expressions can only take positive values, remember the graph. Hence, the first equation has no solutions, the second equation has one solution: $ x \u003d 1 $.
Answer: $ x \u003d 1 $.
Let's put together a checklist for solving exponential equations:
1. Graphic method. We represent both sides of the equation in the form of functions and build their graphs, find the intersection points of the graphs. (We used this method in the last lesson).
2. The principle of equality of indicators. The principle is based on the fact that two expressions with the same bases are equal if and only if the degrees (indicators) of these bases are equal. $ a ^ (f (x)) \u003d a ^ (g (x)) $ $ f (x) \u003d g (x) $.
3. Variable replacement method. This method should be used if the equation, when changing variables, simplifies its form and is much easier to solve.
Example.
Solve the system of equations: $ \\ begin (cases) (27) ^ y * 3 ^ x \u003d 1, \\\\ 4 ^ (x + y) -2 ^ (x + y) \u003d 12. \\ end (cases) $.
Decision.
Consider both equations of the system separately:
$ 27 ^ y * 3 ^ x \u003d 1 $.
$ 3 ^ (3y) * 3 ^ x \u003d 3 ^ 0 $.
$ 3 ^ (3y + x) \u003d 3 ^ 0 $.
$ x + 3y \u003d 0 $.
Consider the second equation:
$ 4 ^ (x + y) -2 ^ (x + y) \u003d 12 $.
$ 2 ^ (2 (x + y)) - 2 ^ (x + y) \u003d 12 $.
Let's use the change of variables method, let $ y \u003d 2 ^ (x + y) $.
Then the equation will take the form:
$ y ^ 2-y-12 \u003d 0 $.
$ (y-4) (y + 3) \u003d 0 $.
$ y_1 \u003d 4 $ and $ y_2 \u003d -3 $.
Moving on to the initial variables, from the first equation we get $ x + y \u003d 2 $. The second equation has no solutions. Then our initial system of equations is equivalent to the system: $ \\ begin (cases) x + 3y \u003d 0, \\\\ x + y \u003d 2. \\ end (cases) $.
Subtracting the second from the first equation, we get: $ \\ begin (cases) 2y \u003d -2, \\\\ x + y \u003d 2. \\ end (cases) $.
$ \\ begin (cases) y \u003d -1, \\\\ x \u003d 3. \\ end (cases) $.
Answer: $ (3; -1) $.
Exponential inequalities
Let's move on to inequalities. When solving inequalities, it is necessary to pay attention to the base of the degree. There are two possible scenarios for solving inequalities.Theorem. If $ a\u003e 1 $, then the exponential inequality $ a ^ (f (x))\u003e a ^ (g (x)) $ is equivalent to the inequality $ f (x)\u003e g (x) $.
If $ 0
Example.
Solve inequalities:
a) $ 3 ^ (2x + 3)\u003e 81 $.
b) $ ((\\ frac (1) (4))) ^ (2x-4) c) $ (0.3) ^ (x ^ 2 + 6x) ≤ (0.3) ^ (4x + 15) $ ...
Decision.
a) $ 3 ^ (2x + 3)\u003e 81 $.
$ 3 ^ (2x + 3)\u003e 3 ^ 4 $.
Our inequality is equivalent to the inequality:
$ 2x + 3\u003e 4 $.
$ 2x\u003e 1 $.
$ x\u003e 0.5 $.
B) $ ((\\ frac (1) (4))) ^ (2x-4) $ ((\\ frac (1) (4))) ^ (2x-4) In our equation, the base at degree less than 1, then when replacing an inequality with an equivalent one, the sign must be changed.
$ 2x-4\u003e 2 $.
$ x\u003e 3 $.
C) Our inequality is equivalent to the inequality:
$ x ^ 2 + 6x≥4x + 15 $.
$ x ^ 2 + 2x-15≥0 $.
$ (x-3) (x + 5) ≥0 $.
Let's use the interval solution method:
Answer: $ (- ∞; -5] U)