There is less and less time left before passing the exam in mathematics. The situation is heating up, the nerves of schoolchildren, parents, teachers and tutors are stretched more and more. Daily in-depth math classes will help you relieve nervous tension. After all, nothing, as you know, so charges with positive and does not help in passing exams, as confidence in one's abilities and knowledge. Today, a math tutor will tell you about solving systems of logarithmic and exponential inequalities, tasks that traditionally cause difficulties for many modern high school students.

In order to learn how to solve C3 problems from the Unified State Examination in mathematics, as a tutor in mathematics, I recommend that you pay attention to the following important points.

1. Before proceeding to solve systems of logarithmic and exponential inequalities, it is necessary to learn how to solve each of these types of inequalities separately. In particular, to understand how the area of ​​​​admissible values ​​is found, equivalent transformations of logarithmic and exponential expressions are carried out. You can comprehend some of the secrets related to this by studying the articles "" and "".

2. At the same time, it is necessary to realize that the solution of a system of inequalities does not always come down to solving each inequality separately and crossing the resulting gaps. Sometimes, knowing the solution of one inequality of the system, the solution of the second is greatly simplified. As a math tutor who prepares students for final exams in the USE format, I will reveal a couple of secrets related to this in this article.

3. It is necessary to clearly understand for yourself the difference between the intersection and the union of sets. This is one of the most important mathematical knowledge that an experienced professional tutor tries to give to his student from the very first lessons. A visual representation of the intersection and union of sets is given by the so-called "Euler circles".

Set intersection A set is called a set that contains only those elements that each of these sets has.

intersection

Image of the intersection of sets using "Euler circles"

Finger explanation. Diana has a “set” in her purse, consisting of ( pens, pencil, rulers, notebooks, combs). Alice has a "set" in her purse, consisting of ( notebook, pencil, mirrors, notebooks, the Kiev's cutlets). The intersection of these two "sets" will be the "set" consisting of ( pencil, notebooks), since both Diana and Alice have both of these “elements”.

Important to remember! If the solution of the inequality is the interval and the solution of the inequality is the interval, then the solution of the systems:

is the interval that is intersection original intervals. Here and belowany of the characters title="(!LANG:Rendered by QuickLaTeX.com" height="17" width="93" style="vertical-align: -4px;">!} and under is the opposite sign.

Union of sets is called the set that consists of all the elements of the original sets.

In other words, if two sets are given and then their association will be a set of the following form:

Image of union of sets using "Euler circles"

Finger explanation. The union of the "sets" taken in the previous example will be the "set" consisting of ( pens, pencil, rulers, notebooks, combs, notebook, mirrors, the Kiev's cutlets), since it consists of all the elements of the original "sets". One clarification that may not be superfluous. A bunch of can not contain the same elements.

Important to remember! If the solution to the inequality is the interval and the solution to the inequality is the interval, then the solution of the set is:

is the interval that is Union original intervals.

Let's go straight to the examples.

Example 1 Solve the system of inequalities:

Solution of problem C3.

1. We solve the first inequality first. Using the substitution, we pass to the inequality:

2. We now solve the second inequality. The range of its admissible values ​​is determined by the inequality:

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Within the acceptable range, given that the base of the logarithm title="(!LANG:Rendered by QuickLaTeX.com" height="18" width="52" style="vertical-align: -4px;"> переходим к равносильному неравенству:!}

Excluding solutions that are not within the range of admissible values, we obtain the interval

3. Answer to system inequalities will intersection

The resulting gaps on the number line. The solution is their intersection

Example 2 Solve the system of inequalities:

Solution of problem C3.

1. We solve the first inequality first. Multiply both parts by title="(!LANG:Rendered by QuickLaTeX.com" height="14" width="55" style="vertical-align: 0px;"> и делаем замену в результате чего приходим к неравенству:!}

Let's move on to the reverse substitution:

2.

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Graphical representation of the resulting span. Solution of the system - their intersection

Example 3 Solve the system of inequalities:

Solution of problem C3.

1. We solve the first inequality first. Multiply both parts of it by title="(!LANG:Rendered by QuickLaTeX.com" height="18" width="61" style="vertical-align: -4px;"> после чего получаем неравенство:!}

Using substitution, we pass to the following inequality:

Let's move on to the reverse substitution:

2. We now solve the second inequality. Let us first determine the range of admissible values ​​of this inequality:

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Please note that

Then, taking into account the range of permissible values, we obtain:

3. We find the general solution of inequalities. Comparing the obtained irrational values ​​of the nodal points is by no means a trivial task in this example. This can be done in the following way. As

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then and the final response to the system is:

Example 4 Solve the system of inequalities:

Solution of problem С3.

1. Let's solve the second inequality first:

2. The first inequality of the original system is a logarithmic variable-base inequality. A convenient way to solve such inequalities is described in the article "Complex logarithmic inequalities", it is based on a simple formula:

Instead of a sign, any inequality sign can be substituted, the main thing is that it be the same in both cases. Using this formula greatly simplifies the solution of the inequality:

Let us now determine the range of admissible values ​​of this inequality. It is given by the following system:

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It is easy to see that at the same time this interval will also be the solution of our inequality.

3. The final answer to the original systems inequalities will intersection obtained intervals, that is

Example 5 Solve the system of inequalities:

Problem solution C3.

1. We solve the first inequality first. We use substitution We pass to the following quadratic inequality:

2. We now solve the second inequality. The range of its allowable values ​​is determined by the system:

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This inequality is equivalent to the following mixed system:

In the range of valid values, that is, with title="(!LANG:Rendered by QuickLaTeX.com" height="18" width="53" style="vertical-align: -4px;"> используя равносильные преобразования переходим к следующей смешанной системе:!}

Taking into account the range of permissible values, we obtain:

3. The final decision of the original systems is an

Solution of problem C3.

1. We solve the first inequality first. By equivalent transformations we bring it to the form:

2. We now solve the second inequality. The range of its valid values ​​is determined by the span: title="(!LANG:Rendered by QuickLaTeX.com" height="14" width="68" style="vertical-align: 0px;"> Используя замену переменной переходим к следующему квадратичному неравенству:!}

This answer belongs entirely to the range of acceptable values ​​of inequality.

3. By crossing the intervals obtained in the previous paragraphs, we obtain the final answer to the system of inequalities:

Today we have solved systems of logarithmic and exponential inequalities. Tasks of this kind were offered in trial versions of the USE in mathematics throughout the current academic year. However, as a math tutor with experience preparing for the USE, I can say that this does not mean at all that similar tasks will be in the real versions of the USE in mathematics in June.

Let me express one warning, addressed primarily to tutors and school teachers involved in preparing high school students for the USE in mathematics. It is very dangerous to prepare schoolchildren for an exam strictly on given topics, because in this case there is a risk of completely “filling up” it even with a slight change in the previously stated task format. Mathematics education must be complete. Dear colleagues, please do not liken your students to robots by the so-called "training" to solve a certain type of problem. After all, there is nothing worse than the formalization of human thinking.

Good luck to everyone and creative success!

Sergey Valerievich

If you try, then there are two options: it will work or it will not work. If you don't try, there's only one.
© Folk wisdom

The solution of most mathematical problems is somehow connected with the transformation of numerical, algebraic or functional expressions. This applies especially to the solution. In the USE variants in mathematics, this type of task includes, in particular, task C3. Learning how to solve C3 tasks is important not only for the successful passing of the exam, but also for the reason that this skill will come in handy when studying a mathematics course in higher education.

Performing tasks C3, 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 heading "" in articles devoted to methods for solving C3 problems from the USE variants in mathematics.

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

Exponential function

What is an exponential function?

View function y = a x, where a> 0 and a≠ 1, called exponential function.

Main exponential function properties y = a x:

Graph of an exponential function

The graph of the exponential function is exhibitor:

Graphs of exponential functions (exponents)

Solution of exponential equations

indicative called equations in which the unknown variable is found only in exponents of any 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 actions with degrees:

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Example 1 Solve the equation:

Decision: use the above formulas and substitution:

The equation then becomes:

The discriminant of the obtained quadratic equation is positive:

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

Going back to substitution, we get:

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

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

Answer: x = 3.

Example 2 Solve the equation:

Decision: the equation has no restrictions on the area of ​​​​admissible 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:

Decision: 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 on its domain). Then the equation takes the form:

Answer: x = 0.

Example 4 Solve the equation:

Decision: we simplify the equation to an elementary one by 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:

Decision: function y = 3x, standing on the left side of the equation, is increasing. Function y = —x-2/3, standing on the right side of the equation, is decreasing. This means that if the graphs of these functions intersect, then at most at 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:

Decision: we simplify the equation by equivalent transformations, bearing 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 partial powers given at the beginning of the article:

Answer: x = 2.

Solving exponential inequalities

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

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

Theorem 2. If a 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:

Decision: represent the original inequality in the form:

Divide both parts of this inequality by 3 2 x, and (due to the positiveness of the function y= 3 2x) the inequality sign will not change:

Let's use a substitution:

Then the inequality takes the form:

So, the solution to the inequality is the interval:

passing to the reverse substitution, we get:

The left inequality, due to the positiveness of the exponential function, is fulfilled automatically. Using the well-known property of the logarithm, we pass to the equivalent inequality:

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

So we finally get answer:

Example 8 Solve the inequality:

Decision: using the properties of multiplication and division of powers, we rewrite the inequality in the form:

Let's introduce a new variable:

With this substitution, the inequality takes the form:

Multiply the numerator and denominator of the fraction by 7, we get the following equivalent inequality:

So, the inequality is satisfied by the following values ​​of the variable t:

Then, going back to substitution, we get:

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

Finally we get answer:

Example 9 Solve the inequality:

Decision:

We divide both sides of the inequality by the expression:

It is always greater than zero (because the exponential function is positive), so the inequality sign does not need to be changed. We get:

t , which are in the interval:

Passing 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:

Decision:

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

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

At the same time, the function turns out to be bounded from below y = 3 x 2 -2x+2 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 the value , equal to 3 (the intersection of the ranges of these functions is only this number). This condition is satisfied at a single point x = 1.

Answer: x= 1.

To learn how to solve exponential equations and inequalities, you need to constantly train in their solution. Various methodological manuals, elementary mathematics problem books, 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 brilliant results in the exam.

Sergey Valerievich

P.S. Dear guests! Please do not write requests for solving your equations in the comments. Unfortunately, I don't have time for this at all. 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.

Irrational inequalities

An irrational inequality is understood as an inequality in which the unknown quantities are under the sign of the radical. The solution of such inequalities usually consists in the fact that, with the help of some transformations, they are replaced with equivalent rational equations, inequalities, or systems of equations and inequalities (often mixed systems, i.e., those that include both equations and inequalities), and further the solution can follow the steps outlined above. These transformations are, in addition to the change of variables (introduction of new variables) and factorization, also the elevation of both parts of the inequality to the same degree. However, in this case it is necessary to monitor the equivalence of transitions from one inequality to another. With thoughtless exponentiation, the roots of inequality can be both lost and gained at the same time. For example, squaring the correct inequality -1<2, мы получим верное неравенство 1<4; из верного неравенства -5<2 получается уже неверное неравенство 25<4;из неверного неравенства 1<-2 получим верное неравенство 1<4; наконец, из неверного неравенства 5<2 получим неверное неравенство 25<4. Вы видите, что возможны все комбинации верных и неверных неравенств!

However, the main assertion used here is true: if both sides of an inequality are nonnegative, then it is equivalent to the inequality obtained from it by termwise exponentiation.

When solving inequalities in this way, care must be taken not to acquire extraneous solutions. Therefore, it is useful, where possible, to find the domain of definition of the inequality, as well as the domain of possible values ​​of the solutions.

Exponential and logarithmic inequalities

The solution of exponential and logarithmic inequalities is preceded by the study of the properties of the corresponding functions; performing many tasks on the transformation of exponential and logarithmic expressions; solution of equations containing logarithms and variables in the exponent. The solution of the simplest inequalities, which are considered

where means one of the inequalities<,>,.

The point is that this topic is usually introduced as an absolutely new one, based only on the previously studied properties of these functions. It is expedient, in my opinion, to connect it with the solution of inequalities in general (that is, with the already known algorithm). It should be noted that the interval method cannot be used directly. But the solution of various exponential and logarithmic inequalities is based on the following rules:

If a>1, then

If 0

If a>1, then

If 0

Where the sign means the opposite in meaning to the sign.

Using which exponential and logarithmic inequalities are usually reduced to rational ones, which can already be solved by the method of intervals described above.

Inequalities containing trigonometric functions

This topic is poorly covered in the educational literature, and in some textbooks it is generally taken out of the scope of the course being studied (as already mentioned in Chapter I of this work). Of the trigonometric inequalities, as a rule, only the simplest types are considered.

Whereas the tasks presented in the practical part related to this paragraph are found in collections of competitive problems, in collections for applicants and materials for entrance exams to technical faculties of universities. Those. this material is not included in the required study in elementary and high school, but is useful.

The interval method is especially effective in solving inequalities containing trigonometric functions. When solving purely trigonometric inequalities by this method, instead of the number axis, it is convenient to use a number circle, which is divided by the roots of the corresponding trigonometric equations (numerator and denominator) into arcs that play the same role as the intervals on the number axis. On these arcs, the trigonometric expression corresponding to the inequality being solved has constant signs, which can be determined using the rule of a separate “convenient” point and the property of multiplicity of roots. Often, to determine the arcs themselves, it is not at all necessary to find the entire (infinite) set of roots of the corresponding equations; it is enough from these equations to find the values ​​​​of the main trigonometric functions (sine, cosine, tangent, cotangent) and mark the points on the number circle corresponding to these values.

You can use the number circle directly to solve the original trigonometric inequality using the interval method if all the functions through which the inequality is written have the main (smallest positive) period or, where m is some positive integer. If the main period of these functions is greater than or, then you should first change the variables, and then use the number circle.

If the inequality contains both trigonometric and other functions, then the numerical axis should be used to solve it by the interval method.

All the B7 problems I've seen have been formulated in much the same way: solve an equation. In this case, the equations themselves belong to one of three types:

1. logarithmic;
2. Demonstrative;
3. Irrational.

Generally speaking, a complete guide to each type of equation will take more than a dozen pages, going far beyond the scope of the exam. Therefore, we will consider only the simplest cases that require unpretentious reasoning and calculations. This knowledge will be quite enough to solve any B7 problem.

In mathematics, the term "solve an equation" means to find the set of all roots of a given equation, or to prove that this set is empty. But only numbers can be entered in the USE form - no sets. Therefore, if there was more than one root in task B7 (or, conversely, none) - an error was made in the solution.

Logarithmic Equations

A logarithmic equation is any equation that reduces to the form log a f(x) = k, where a > 0, a≠ 1 is the base of the logarithm, f(x) is an arbitrary function, k is some constant.

Such an equation is solved by introducing the constant k under the sign of the logarithm: k= log a a k. The base of the new logarithm is equal to the base of the original. We get the equation log a f(x) = log a a k, which is solved by discarding the logarithm.

Note that, by the condition a> 0, so f(x) = a k> 0, i.e. the original logarithm exists.

Task. Solve the equation: log 7 (8 − x) = 2.

Decision. log 7 (8 − x) = 2 ⇔ log 7 (8 − x) = log 7 7 2 ⇔ 8 − x = 49 ⇔ x = −41.

Task. Solve the equation: log 0.5 (6 − x) = −2.

Decision. log 0.5 (6 − x) = −2 ⇔ log 0.5 (6 − x) = log 0.5 0.5 −2 ⇔ 6 − x = 4 ⇔ x = 2.

But what if the original equation turns out to be more complicated than the standard log a f(x) = k? Then we reduce it to the standard one, collecting all the logarithms in one direction, and the numbers in the other.

If there is more than one logarithm in the original equation, you will have to look for the area of ​​​​admissible values ​​(ODV) of each function standing under the logarithm. Otherwise, extra roots may appear.

Task. Solve the equation: log 5 ( x+ 1) + log 5 ( x + 5) = 1.

Since there are two logarithms in the equation, we find the ODZ:

1. x + 1 > 0 ⇔ x > −1
2. x + 5 > 0 ⇔ x > −5

We get that the ODZ is the interval (−1, +∞). Now we solve the equation:

log 5 ( x+ 1) + log 5 ( x+ 5) = 1 ⇒ log 5 ( x + 1)(x+ 5) = 1 ⇔ log 5 ( x + 1)(x+ 5) = log 5 5 1 ⇔ ( x + 1)(x + 5) = 5 ⇔ x 2 + 6x + 5 = 5 ⇔ x (x + 6) = 0 ⇔ x 1 = 0, x 2 = −6.

But x 2 = -6 does not qualify for ODZ. Remains the root x 1 = 0.

exponential equations

An exponential equation is any equation that reduces to the form a f(x) = k, where a > 0, a≠ 1 - base of degree, f(x) is an arbitrary function, k is some constant.

This definition almost verbatim repeats the definition of a logarithmic equation. The exponential equations are solved even easier than the logarithmic ones, because here it is not required that the function f(x) was positive.

To solve this, we make the substitution k = a t, where t Generally speaking, the logarithm ( t= log a k), but in the USE the numbers a and k will be chosen so that to find t will be easy. In the resulting equation a f(x) = a t the bases are equal, which means that the exponents are equal, i.e. f(x) = t. The solution of the last equation, as a rule, does not cause problems.

Task. Solve Equation: 7 x − 2 = 49.

Decision. 7 x − 2 = 49 ⇔ 7 x − 2 = 7 2 ⇔ x − 2 = 2 ⇔ x = 4.

Task. Solve the equation: 6 16 − x = 1/36.

Decision. 6 16 - x = 1/36 ⇔ 6 16 − x = 6 −2 ⇔ 16 − x = −2 ⇔ x = 18.

A little about the transformation of exponential equations. If the original equation is different from a f(x) = k , we apply the rules for working with degrees:

1. a n · a m = a n + m ,
2. a n / a m = a n − m ,
3. (a n) m = a n · m .

In addition, you need to know the rules for replacing roots and fractions with degrees with a rational exponent:

Such equations are extremely rare in the USE, but without them, the analysis of problem B7 would be incomplete.

Task. Solve Equation: (5/7) x− 2 (7/5) 2 x − 1 = 125/343

Notice, that:

1. (7/5) 2x − 1 = ((5/7) −1) 2x − 1 = (5/7) 1 − 2x ,
2. 125/343 = (5 3) /(7 3) = (5/7) 3 .

We have: (5/7) x− 2 (7/5) 2 x − 1 = 125/343 ⇔ (5/7) x− 2 · (5/7) 1 − 2 x = (5/7) 3 ⇔ (5/7) x − 2 + 1 − 2x = (5/7) 3 ⇔ (5/7) −x − 1 = (5/7) 3 ⇔ −x − 1 = 3 ⇔ x = −4.

Irrational equations

Irrational is understood as any equation containing the sign of the root. Of the whole variety of irrational equations, we will consider only the simplest case, when the equation has the form:

To solve this equation, we square both sides. We get the equation f(x) = a 2. In this case, the requirement of the ODZ is automatically fulfilled: f(x) ≥ 0, because a 2 ≥ 0. It remains to solve a simple equation f(x) = a 2 .

Task. Solve the equation:

We square both sides and get: 5 x − 6 = 8 2 ⇔ 5x − 6 = 64 ⇔ 5x = 70 ⇔ x = 14.

Task. Solve the equation:

First, like last time, we square both sides. And then we will add a minus sign to the numerator. We have:

Note that when x= −4 there will be a positive number under the root, i.e. the requirement of the ODZ has been met.