Types of exponential inequalities and methods for solving them. Solving exponential inequalities: basic methods. Definition of Exponential Equations

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 exponential function. It is on the properties that the solution of all exponential equations and inequalities.

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.

Base degree more 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:

For solutions quadratic inequality 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. When the argument is equal, the functions are equal, which means given 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 linear function, that is, to 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.

Majority decision mathematical problems is somehow related to the transformation of numerical, algebraic or functional expressions. The above applies especially to the decision. In the versions of the Unified State Exam in mathematics, this type of problem includes, in particular, task C3. Learning to solve C3 tasks is important not only for the purpose of successful completion Unified State Examination, but also for the reason that this skill will be useful when studying a mathematics course in higher school.

When completing C3 tasks, you have to solve various types of equations and inequalities. Among them are rational, irrational, exponential, logarithmic, trigonometric, containing modules ( absolute values), as well as combined ones. This article discusses the main types of exponential equations and inequalities, as well as various methods for solving them. Read about solving other types of equations and inequalities in the “” section in articles devoted to methods for solving C3 problems from Unified State Exam options mathematics.

Before we begin to analyze specific exponential equations and inequalities, as a math tutor, I suggest you brush up on some theoretical material that we will need.

Exponential function

What is an exponential function?

Function of the form y = a x, Where a> 0 and a≠ 1 is called exponential function.

Basic properties of exponential function y = a x:

Graph of an Exponential Function

The graph of the exponential function is exponent:

Graphs of exponential functions (exponents)

Solving exponential equations

Indicative are called equations in which the unknown variable is found only in exponents of some powers.

For solutions exponential equations you need to know and be able to use the following simple theorem:

Theorem 1. Exponential equation a f(x) = a g(x) (Where a > 0, a≠ 1) is equivalent to the equation f(x) = g(x).

In addition, it is useful to remember the basic formulas and operations with degrees:

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Example 1. Solve the equation:

Solution: We use the above formulas and substitution:

The equation then becomes:

Discriminant of the received quadratic equation positive:

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This means that this equation has two roots. We find them:

Moving on to reverse substitution, we get:

The second equation has no roots, since the exponential function is strictly positive throughout the entire domain of definition. Let's solve the second one:

Taking into account what was said in Theorem 1, we move on to the equivalent equation: x= 3. This will be the answer to the task.

Answer: x = 3.

Example 2. Solve the equation:

Solution: restrictions on the area acceptable values the equation does not, since the radical expression makes sense for any value x(exponential function y = 9 4 -x positive and not equal to zero).

We solve the equation by equivalent transformations using the rules of multiplication and division of powers:

The last transition was carried out in accordance with Theorem 1.

Answer:x= 6.

Example 3. Solve the equation:

Solution: both sides of the original equation can be divided by 0.2 x. This transition will be equivalent, since this expression is greater than zero for any value x(the exponential function is strictly positive in its domain of definition). Then the equation takes the form:

Answer: x = 0.

Example 4. Solve the equation:

Solution: we simplify the equation to an elementary one by means of equivalent transformations using the rules of division and multiplication of powers given at the beginning of the article:

Dividing both sides of the equation by 4 x, as in the previous example, is an equivalent transformation, since this expression not equal to zero for any value x.

Answer: x = 0.

Example 5. Solve the equation:

Solution: function y = 3x, standing on the left side of the equation, is increasing. Function y = —x The -2/3 on the right side of the equation is decreasing. This means that if the graphs of these functions intersect, then at most one point. In this case, it is easy to guess that the graphs intersect at the point x= -1. There will be no other roots.

Answer: x = -1.

Example 6. Solve the equation:

Solution: we simplify the equation by means of equivalent transformations, keeping in mind everywhere that the exponential function is strictly greater than zero for any value x and using the rules for calculating the product and quotient of powers given at the beginning of the article:

Answer: x = 2.

Solving exponential inequalities

Indicative are called inequalities in which the unknown variable is contained only in exponents of some powers.

For solutions exponential inequalities knowledge of the following theorem is required:

Theorem 2. If a> 1, then the inequality a f(x) > a g(x) is equivalent to an inequality of the same meaning: f(x) > g(x). If 0< a < 1, то показательное неравенство a f(x) > a g(x) is equivalent to an inequality with the opposite meaning: f(x) < g(x).

Example 7. Solve the inequality:

Solution: Let's present the original inequality in the form:

Let's divide both sides of this inequality by 3 2 x, in this case (due to the positivity of the function y= 3 2x) the inequality sign will not change:

Let's use the substitution:

Then the inequality will take the form:

So, the solution to the inequality is the interval:

moving to the reverse substitution, we get:

Due to the positivity of the exponential function, the left inequality is satisfied automatically. Using the well-known property of the logarithm, we move on to the equivalent inequality:

Since the base of the degree is a number greater than one, equivalent (by Theorem 2) is the transition to the following inequality:

So, we finally get answer:

Example 8. Solve the inequality:

Solution: Using the properties of multiplication and division of powers, we rewrite the inequality in the form:

Let's introduce a new variable:

Taking this substitution into account, the inequality takes the form:

Multiplying the numerator and denominator of the fraction by 7, we obtain the following equivalent inequality:

So, the following values of the variable satisfy the inequality t:

Then, moving to the reverse substitution, we get:

Since the base of the degree here is greater than one, the transition to the inequality will be equivalent (by Theorem 2):

Finally we get answer:

Example 9. Solve the inequality:

Solution:

We divide both sides of the inequality by the expression:

It is always greater than zero (due to the positivity of the exponential function), so there is no need to change the inequality sign. We get:

t located in the interval:

Moving on to the reverse substitution, we find that the original inequality splits into two cases:

The first inequality has no solutions due to the positivity of the exponential function. Let's solve the second one:

Example 10. Solve the inequality:

Solution:

Parabola branches y = 2x+2-x 2 are directed downwards, therefore it is limited from above by the value that it reaches at its vertex:

Parabola branches y = x 2 -2x The +2 in the indicator are directed upward, which means it is limited from below by the value that it reaches at its vertex:

At the same time, the function also turns out to be bounded from below y = 3 x 2 -2x+2, which is on the right side of the equation. She achieves her goal lowest value at the same point as the parabola in the exponent, and this value is equal to 3 1 = 3. So, the original inequality can only be true if the function on the left and the function on the right take on a value equal to 3 at the same point (by the intersection The range of values of these functions is only this number). This condition is satisfied at a single point x = 1.

Answer: x= 1.

In order to learn to decide exponential equations and inequalities, it is necessary to constantly train in solving them. Various things can help you with this difficult task. methodological manuals, problem books in elementary mathematics, collections of competitive problems, mathematics classes at school, as well as individual sessions with a professional tutor. I sincerely wish you success in your preparation and excellent results in the exam.

Sergey Valerievich

P.S. Dear guests! Please do not write requests to solve your equations in the comments. Unfortunately, I have absolutely no time for this. Such messages will be deleted. Please read the article. Perhaps in it you will find answers to questions that did not allow you to solve your task on your own.

That is compulsory when solving a system of exponential equations? Certainly, transformation this system into a system of simple equations.

Examples.

Solve systems of equations:

Let's express at through X from (2) the system equation and substitute this value into the (1) system equation.

We solve (2) the th equation of the resulting system:

2 x +2 x +2 =10, apply the formula: a x + y=a x∙ a y.

2 x +2 x ∙2 2 =10, let’s take the common factor 2 x out of brackets:

2 x (1+2 2)=10 or 2 x ∙5=10, hence 2 x =2.

2 x =2 1, from here x=1. Let's return to the system of equations.

Answer: (1; 2).

Solution.

We represent the left and right sides of (1) equation in the form of powers with a base 2 , and the right side of (2) the equation as the zero power of the number 5 .

If two powers with the same bases are equal, then the exponents of these powers are equal - we equate the exponents with the bases 2 and exponents with bases 5 .

The resulting system linear equations with two variables we solve using the addition method.

We find x=2 and we substitute this value instead X into the second equation of the system.

We find at.

Answer: (2; 1.5).

Solution.

If in the previous two examples we moved to a simpler system by equating the indicators of two degrees with the same bases, then in the 3rd example this operation is impossible. It is convenient to solve such systems by introducing new variables. We will introduce variables u And v, and then express the variable u through v and we get an equation for the variable v.

We solve (2) the th equation of the system.

v 2 +63v-64=0. Let's select the roots using Vieta's theorem, knowing that: v 1 +v 2 = -63; v 1 ∙v 2 =-64.

We get: v 1 =-64, v 2 =1. We return to the system and find u.

Since the values of the exponential function are always positive, the equations 4 x = -1 and 4 y = -64 have no solutions.

Methods for solving systems of equations

To begin with, let us briefly recall what methods generally exist for solving systems of equations.

Exist four main ways solutions to systems of equations:

Substitution method: take any of the given equations and express $y$ in terms of $x$, then $y$ is substituted into the system equation, from where the variable $x.$ is found. After this, we can easily calculate the variable $y.$

Addition method: In this method, you need to multiply one or both equations by such numbers that when you add both together, one of the variables “disappears.”

Graphical method: both equations of the system are depicted on coordinate plane and the point of their intersection is found.

Method of introducing new variables: in this method we replace some expressions to simplify the system, and then use one of the above methods.