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Consider an example: solve inequality (0.5) (7 - 3*x)< 4.

Note that 4 = (0.5) 2 . Then the inequality will take the form (0.5)(7 - 3*x)< (0.5) (-2) . Основание показательной функции 0.5 меньше единицы, следовательно, она убывает. В этом случае надо поменять знак неравенства и не записывать только показатели.

We get: 7 - 3*x>-2.

Hence: x<3.

Answer: x<3.

If the base in the inequality was greater than one, then when getting rid of the base, there would be no need to change the sign of the inequality.

In this lesson we will look at various exponential inequalities and learn how to solve them, based on the technique for solving the simplest exponential inequalities

1. Definition and properties of an exponential function

Let us recall the definition and basic properties of the exponential function. The solution of all exponential equations and inequalities is based on these properties.

Exponential function is a function of the form , where the base is the degree and Here x is the independent variable, argument; y is the dependent variable, function.

Rice. 1. Graph of exponential function

The graph shows increasing and decreasing exponents, illustrating the exponential function with a base greater than one and less than one but greater than zero, respectively.

Both curves pass through the point (0;1)

Properties of the Exponential Function:

Domain: ;

Range of values: ;

The function is monotonic, increases with, decreases with.

A monotonic function takes each of its values ​​given a single argument value.

When , when the argument increases from minus to plus infinity, the function increases from zero inclusive to plus infinity, i.e., for given values ​​of the argument we have a monotonically increasing function (). On the contrary, when the argument increases from minus to plus infinity, the function decreases from infinity to zero inclusive, i.e., for given values ​​of the argument we have a monotonically decreasing function ().

2. The simplest exponential inequalities, solution method, example

Based on the above, we present a method for solving simple exponential inequalities:

Technique for solving inequalities:

Equalize the bases of degrees;

Compare indicators by maintaining or changing the inequality sign to the opposite one.

The solution to complex exponential inequalities usually consists in reducing them to the simplest exponential inequalities.

The base of the degree is greater than one, which means the inequality sign is preserved:

Let's transform the right-hand side according to the properties of the degree:

The base of the degree is less than one, the inequality sign must be reversed:

To solve the quadratic inequality, we solve the corresponding quadratic equation:

Using Vieta's theorem we find the roots:

The branches of the parabola are directed upward.

Thus, we have a solution to the inequality:

It’s easy to guess that the right side can be represented as a power with an exponent of zero:

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 the fractional-rational function:

We find the domain of definition:

Finding 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:

Rice. 2. Intervals of constancy of sign

Thus, we received the answer.

Answer:

3. Solving standard exponential inequalities

Let's consider inequalities with the same indicators, but different bases.

One of the properties of the exponential function is that for any value of the argument it takes strictly positive values, which means that it can be divided into an exponential function. Let us divide the given inequality by its right side:

The base of the degree is greater than one, the inequality sign is preserved.

Let's illustrate the solution:

Figure 6.3 shows graphs of 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 ​​are negative, the function goes lower, it is smaller. If the argument is equal, the functions are equal, which means that this point is also a solution to the given inequality.

Rice. 3. Illustration for example 4

Let us transform the given inequality according to the properties of the degree:

Here are some similar terms:

Let's divide both parts into:

Now we continue to solve similarly to example 4, divide both parts by:

The base of the degree is greater than one, the inequality sign remains:

4. Graphical solution of exponential inequalities

Example 6 - Solve the inequality graphically:

Let's look at the functions on the left and right sides and build a graph for each of them.

The function is exponential and increases over its entire domain of definition, i.e., for all real values ​​of the argument.

The function is linear and decreases over its entire domain of definition, i.e., for all real values ​​of the argument.

If these functions intersect, that is, the system has a solution, then such a solution is unique and 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 answer is obvious: (Figure 6.4)

Rice. 4. Illustration for example 6

So, we looked at solving various standard exponential inequalities. Next we move on to consider more complex exponential inequalities.

Bibliography

Mordkovich A. G. Algebra and the beginnings of mathematical analysis. - M.: Mnemosyne. Muravin G. K., Muravin O. V. Algebra and the beginnings of mathematical analysis. - M.: Bustard. Kolmogorov A. N., Abramov A. M., Dudnitsyn Yu. P. et al. Algebra and the beginnings of mathematical analysis. - M.: Enlightenment.

Math. md. Mathematics-repetition. com. Diffur. kemsu. ru.

Homework

1. Algebra and the beginnings of analysis, grades 10-11 (A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn) 1990, No. 472, 473;

2. Solve the inequality:

3. Solve inequality.

Exponential equations and inequalities are those in which the unknown is contained in the exponent.

Solving exponential equations often comes down to solving the equation a x = a b, where a > 0, a ≠ 1, x is an unknown. This equation has a single root x = b, since the following theorem is true:

Theorem. If a > 0, a ≠ 1 and a x 1 = a x 2, then x 1 = x 2.

Let us substantiate the considered statement.

Let us assume that the equality x 1 = x 2 does not hold, i.e. x 1< х 2 или х 1 = х 2 . Пусть, например, х 1 < х 2 . Тогда если а >1, then exponential function y = a x increases and therefore the inequality a x 1 must be satisfied< а х 2 ; если 0 < а < 1, то функция убывает и должно выполняться неравенство а х 1 >a x 2. In both cases we received a contradiction to the condition a x 1 = a x 2.

Let's consider several problems.

Solve the equation 4 ∙ 2 x = 1.

Solution.

Let's write the equation in the form 2 2 ∙ 2 x = 2 0 – 2 x+2 = 2 0, from which we get x + 2 = 0, i.e. x = -2.

Answer. x = -2.

Solve equation 2 3x ∙ 3 x = 576.

Solution.

Since 2 3x = (2 3) x = 8 x, 576 = 24 2, the equation can be written as 8 x ∙ 3 x = 24 2 or as 24 x = 24 2.

From here we get x = 2.

Answer. x = 2.

Solve the equation 3 x+1 – 2∙3 x - 2 = 25.

Solution.

Taking the common factor 3 x - 2 out of brackets on the left side, we get 3 x - 2 ∙ (3 3 – 2) = 25 – 3 x - 2 ∙ 25 = 25,

whence 3 x - 2 = 1, i.e. x – 2 = 0, x = 2.

Answer. x = 2.

Solve the equation 3 x = 7 x.

Solution.

Since 7 x ≠ 0, the equation can be written as 3 x /7 x = 1, whence (3/7) x = 1, x = 0.

Answer. x = 0.

Solve the equation 9 x – 4 ∙ 3 x – 45 = 0.

Solution.

By replacing 3 x = a this equation reduces to quadratic equation a 2 – 4a – 45 = 0.

Solving this equation, we find its roots: a 1 = 9, and 2 = -5, whence 3 x = 9, 3 x = -5.

The equation 3 x = 9 has root 2, and the equation 3 x = -5 has no roots, since the exponential function cannot take negative values.

Answer. x = 2.

Solving exponential inequalities often comes down to solving the inequalities a x > a b or a x< а b . Эти неравенства решаются с помощью свойства возрастания или убывания показательной функции.

Let's look at some problems.

Solve inequality 3 x< 81.

Solution.

Let's write the inequality in the form 3 x< 3 4 . Так как 3 >1, then the function y = 3 x is increasing.

Therefore, for x< 4 выполняется неравенство 3 х < 3 4 , а при х ≥ 4 выполняется неравенство 3 х ≥ 3 4 .

Thus, at x< 4 неравенство 3 х < 3 4 является верным, а при х ≥ 4 – неверным, т.е. неравенство
3 x< 81 выполняется тогда и только тогда, когда х < 4.

Answer. X< 4.

Solve the inequality 16 x +4 x – 2 > 0.

Solution.

Let us denote 4 x = t, then we obtain the quadratic inequality t2 + t – 2 > 0.

This inequality holds for t< -2 и при t > 1.

Since t = 4 x, we get two inequalities 4 x< -2, 4 х > 1.

The first inequality has no solutions, since 4 x > 0 for all x € R.

We write the second inequality in the form 4 x > 4 0, whence x > 0.

Answer. x > 0.

Graphically solve the equation (1/3) x = x – 2/3.

Solution.

1) Let's build graphs of the functions y = (1/3) x and y = x – 2/3.

2) Based on our figure, we can conclude that the graphs of the considered functions intersect at the point with the abscissa x ≈ 1. Checking proves that

x = 1 is the root of this equation:

(1/3) 1 = 1/3 and 1 – 2/3 = 1/3.

In other words, we have found one of the roots of the equation.

3) Let's find other roots or prove that there are none. The function (1/3) x is decreasing, and the function y = x – 2/3 is increasing. Therefore, for x > 1, the values ​​of the first function are less than 1/3, and the second – more than 1/3; at x< 1, наоборот, значения первой функции больше 1/3, а второй – меньше 1/3. Геометрически это означает, что графики этих функций при х >1 and x< 1 «расходятся» и потому не могут иметь точек пересечения при х ≠ 1.

Answer. x = 1.

Note that from the solution of this problem, in particular, it follows that the inequality (1/3) x > x – 2/3 is satisfied for x< 1, а неравенство (1/3) х < х – 2/3 – при х > 1.

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