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 it takes strictly positive values ​​for any value of the argument, 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 the 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 is reduced to the 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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Solving most mathematical problems in one way or another involves transforming 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 successfully passing the Unified State Exam, but also for the reason that this skill will be useful when studying a mathematics course in high 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 the Unified State Examination in 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:

The discriminant of the resulting quadratic equation is 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: The equation has no restrictions on the range of permissible values, 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 is not equal to zero for any values 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. It reaches its smallest value at the same point as the parabola in the exponent, and this value is 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 the value , equal to 3 (the intersection of the ranges 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 teaching aids, problem books in elementary mathematics, collections of competitive problems, mathematics classes at school, as well as individual lessons with a professional tutor can help you in this difficult task. 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.

Lesson and presentation on the topic: "Exponential equations and exponential inequalities"

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Definition of Exponential Equations

Guys, we studied exponential functions, learned their properties and built graphs, analyzed examples of equations in which exponential functions were found. Today we will study exponential equations and inequalities.

Definition. Equations of the form: $a^(f(x))=a^(g(x))$, where $a>0$, $a≠1$ are called exponential equations.

Recalling the theorems that we studied in the topic "Exponential Function", we can introduce a new theorem:
Theorem. The exponential equation $a^(f(x))=a^(g(x))$, where $a>0$, $a≠1$ is equivalent to the equation $f(x)=g(x)$.

Examples of exponential equations

Example.
Solve equations:
a) $3^(3x-3)=27$.
b) $((\frac(2)(3)))^(2x+0.2)=\sqrt(\frac(2)(3))$.
c) $5^(x^2-6x)=5^(-3x+18)$.
Solution.
a) We know well that $27=3^3$.
Let's rewrite our equation: $3^(3x-3)=3^3$.
Using the theorem above, we find that our equation reduces to the equation $3x-3=3$; solving this equation, we get $x=2$.
Answer: $x=2$.

B) $\sqrt(\frac(2)(3))=((\frac(2)(3)))^(\frac(1)(5))$.
Then our equation can be rewritten: $((\frac(2)(3)))^(2x+0.2)=((\frac(2)(3)))^(\frac(1)(5) )=((\frac(2)(3)))^(0.2)$.
$2х+0.2=0.2$.
$x=0$.
Answer: $x=0$.

C) The original equation is equivalent to the equation: $x^2-6x=-3x+18$.
$x^2-3x-18=0$.
$(x-6)(x+3)=0$.
$x_1=6$ and $x_2=-3$.
Answer: $x_1=6$ and $x_2=-3$.

Example.
Solve the equation: $\frac(((0.25))^(x-0.5))(\sqrt(4))=16*((0.0625))^(x+1)$.
Solution:
Let's perform a series of actions sequentially and bring both sides of our equation to the same bases.
Let's perform a number of operations on the left side:
1) $((0.25))^(x-0.5)=((\frac(1)(4)))^(x-0.5)$.
2) $\sqrt(4)=4^(\frac(1)(2))$.
3) $\frac(((0.25))^(x-0.5))(\sqrt(4))=\frac(((\frac(1)(4)))^(x-0 ,5))(4^(\frac(1)(2)))= \frac(1)(4^(x-0.5+0.5))=\frac(1)(4^x) =((\frac(1)(4)))^x$.
Let's move on to the right side:
4) $16=4^2$.
5) $((0.0625))^(x+1)=\frac(1)((16)^(x+1))=\frac(1)(4^(2x+2))$.
6) $16*((0.0625))^(x+1)=\frac(4^2)(4^(2x+2))=4^(2-2x-2)=4^(-2x )=\frac(1)(4^(2x))=((\frac(1)(4)))^(2x)$.
The original equation is equivalent to the equation:
$((\frac(1)(4)))^x=((\frac(1)(4)))^(2x)$.
$x=2x$.
$x=0$.
Answer: $x=0$.

Example.
Solve the equation: $9^x+3^(x+2)-36=0$.
Solution:
Let's rewrite our equation: $((3^2))^x+9*3^x-36=0$.
$((3^x))^2+9*3^x-36=0$.
Let's make a change of variables, let $a=3^x$.
In the new variables, the equation will take the form: $a^2+9a-36=0$.
$(a+12)(a-3)=0$.
$a_1=-12$ and $a_2=3$.
Let's perform the reverse change of variables: $3^x=-12$ and $3^x=3$.
In the last lesson we learned that exponential expressions can only take positive values, remember the graph. This means that the first equation has no solutions, the second equation has one solution: $x=1$.
Answer: $x=1$.

Let's make a reminder of how to solve exponential equations:
1. Graphic method. We represent both sides of the equation in the form of functions and build their graphs, find the points of intersection 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 (exponents) of these bases are equal. $a^(f(x))=a^(g(x))$ $f(x)=g(x)$.
3. Variable replacement method. This method should be used if the equation, when replacing variables, simplifies its form and is much easier to solve.

Example.
Solve the system of equations: $\begin (cases) (27)^y*3^x=1, \\ 4^(x+y)-2^(x+y)=12. \end (cases)$.
Solution.
Let's consider both equations of the system separately:
$27^y*3^x=1$.
$3^(3y)*3^x=3^0$.
$3^(3y+x)=3^0$.
$x+3y=0$.
Consider the second equation:
$4^(x+y)-2^(x+y)=12$.
$2^(2(x+y))-2^(x+y)=12$.
Let's use the change of variables method, let $y=2^(x+y)$.
Then the equation will take the form:
$y^2-y-12=0$.
$(y-4)(y+3)=0$.
$y_1=4$ and $y_2=-3$.
Let's move on to the initial variables, from the first equation we get $x+y=2$. The second equation has no solutions. Then our initial system of equations is equivalent to the system: $\begin (cases) x+3y=0, \\ x+y=2. \end (cases)$.
Subtract the second from the first equation, we get: $\begin (cases) 2y=-2, \\ x+y=2. \end (cases)$.
$\begin (cases) y=-1, \\ x=3. \end (cases)$.
Answer: $(3;-1)$.

Exponential inequalities

Let's move on to inequalities. When solving inequalities, it is necessary to pay attention to the basis of the degree. There are two possible scenarios for the development of events when solving inequalities.

Theorem. If $a>1$, then the exponential inequality $a^(f(x))>a^(g(x))$ is equivalent to the inequality $f(x)>g(x)$.
If $0 a^(g(x))$ is equivalent to the inequality $f(x)

Example.
Solve inequalities:
a) $3^(2x+3)>81$.
b) $((\frac(1)(4)))^(2x-4) c) $(0.3)^(x^2+6x)≤(0.3)^(4x+15)$ .
Solution.
a) $3^(2x+3)>81$.
$3^(2x+3)>3^4$.
Our inequality is equivalent to inequality:
$2x+3>4$.
$2x>1$.
$x>0.5$.

B) $((\frac(1)(4)))^(2x-4) $((\frac(1)(4)))^(2x-4) In our equation, the base is when the degree is less than 1, then When replacing an inequality with an equivalent one, it is necessary to change the sign.
$2x-4>2$.
$x>3$.

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.

Let's look at the same inequality again and f (x) > b, If a>0 And b<0 .

So, the diagram in Figure 3:

An example of solving an inequality (1/3) x + 2 > –9. As we notice, no matter what number we substitute for x, (1/3) x + 2 is always greater than zero.

Answer: (–∞; +∞) .

How are inequalities of the form solved? and f(x)< b , Where a>1 And b>0?

Diagram in Figure 4:

And the following example: 3 3 – x ≥ 8.
Since 3 > 1 and 8 > 0, then
3 – x > log 3 8, that is
–x > log 3 8 – 3,
X< 3 – log 3 8.

Answer: (0; 3–log 3 8) .

How can the solution to the inequality change? and f(x)< b , at 0 And b>0?

Diagram in Figure 5:

And the following example: Solve inequality 0.6 2x – 3< 0,36 .

Following the diagram in Figure 5, we get
2x – 3 > log 0.6 0.36,
2х – 3 > 2,
2x > 5,
x > 2.5

Answer: (2,5; +∞) .

Let us consider the last scheme for solving an inequality of the form and f(x)< b , at a>0 And b<0 , presented in Figure 6:

For example, let's solve the inequality:

We note that no matter what number we substitute for x, the left side of the inequality is always greater than zero, and our expression is less than -8, i.e. and zero, which means there are no solutions.

Answer: no solutions.

Knowing how to solve the simplest exponential inequalities, you can proceed to solving exponential inequalities.

Example 1.

Find the largest integer value of x that satisfies the inequality

Since 6 x is greater than zero (at no x does the denominator go to zero), multiplying both sides of the inequality by 6 x, we get:

440 – 2 6 2x > 8, then
– 2 6 2x > 8 – 440,
– 2 6 2х > – 332,
6 2x< 216,
2x< 3,

x< 1,5. Наибольшее целое число из помежутка (–∞; 1,5) это число 1.

Answer: 1.

Example 2.

Solve inequality 2 2 x – 3 2 x + 2 ≤ 0

Let us denote 2 x by y, obtain the inequality y 2 – 3y + 2 ≤ 0, and solve this quadratic inequality.

y 2 – 3y +2 = 0,
y 1 = 1 and y 2 = 2.

The branches of the parabola are directed upward, let's draw a graph:

Then the solution to the inequality will be inequality 1< у < 2, вернемся к нашей переменной х и получим неравенство 1< 2 х < 2, решая которое и найдем ответ 0 < x < 1.

Answer: (0; 1) .

Example 3. Solve the inequality 5 x +1 – 3 x +2< 2·5 x – 2·3 x –1
Let's collect expressions with the same bases in one part of the inequality

5 x +1 – 2 5 x< 3 x +2 – 2·3 x –1

Let us take 5 x out of brackets on the left side of the inequality, and 3 x on the right side of the inequality and we get the inequality

5 x (5 – 2)< 3 х (9 – 2/3),
3·5 x< (25/3)·3 х

Divide both sides of the inequality by the expression 3 3 x, the sign of the inequality does not change, since 3 3 x is a positive number, we get the inequality:

X< 2 (так как 5/3 > 1).

Answer: (–∞; 2) .

If you have questions about solving exponential inequalities or would like to practice solving similar examples, sign up for my lessons. Tutor Valentina Galinevskaya.

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