How to solve exponential inequalities graphically. Exponential equations and inequalities. Isolating a stable expression and replacing a variable

We group degrees with the same bases. It is more convenient to separate them on opposite sides of the inequality:

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From each pair of powers we take out of brackets the common factor - the power with the smaller exponent. Taking the common factor out of brackets means dividing each term by this factor. When dividing degrees with the same bases, we leave the base the same and subtract the exponents:

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You can immediately divide by 20 (20=4∙5), but practice shows that dividing in two stages allows you to avoid possible errors:

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Since the base is 2/5<1, показательная функция

decreases, therefore the sign of inequality between exponents changes to the opposite:

Let's solve the quadratic inequality using the interval method. The zeros of the function on the left side of the inequality are x1=-1; x2=2. We mark them on the number line.

To check the sign, take a zero: 0²-0-2=-2, in the interval to which the zero belongs, put “-“. We arrange the remaining signs in a checkerboard pattern. Since we are solving an inequality in which the left side is less than zero, we choose the interval with the “-” sign.

Answer: x ∈ (-1; 2).

A variant of inequalities of this type is that all powers have the same bases, but differ in the coefficients of x in the exponents.

On the left side we put out of brackets the degree with the lowest exponent

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We arrived at an exponential inequality. Since the base 7>1, the function

increases, the sign of inequality between the indicators does not change:

To solve this inequality using the interval method, we move all terms to the left side and reduce the fractions to

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 of 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 the inequality sign does not need to be changed. 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.

Hello! My dear students, in this article we will learn how to solve exponential inequalities .

No matter how complicated the exponential inequality may seem to you, after some transformations (we'll talk about them a little later) all inequalities are reduced to solving the simplest exponential inequalities:

a x > b, a x< b And a x ≥ b, a x ≤ b.

Let's try to figure out how such inequalities are resolved.

We will look into a solution strict inequalities. The only difference when solving non-strict inequalities is that the resulting corresponding roots are included in the answer.

Suppose we need to solve an inequality of the form and f (x) > b, Where a>1 And b>0.

Look at the diagram for solving such inequalities (Figure 1):

Now let's look at a specific example. Solve inequality: 5 x – 1 > 125.

Since 5 > 1 and 125 > 0, then
x – 1 > log 5 125, that is
x – 1 > 3,
x > 4.

Answer: (4; +∞) .

What will be the solution to this same inequality? and f (x) >b, If 0 And b>0?