Lecture: "Methods for solving exponential equations."
1 . exponential equations.
Equations containing unknowns in the exponent are called exponential equations. The simplest of these is the equation ax = b, where a > 0 and a ≠ 1.
1) For b< 0 и b = 0 это уравнение, согласно свойству 1 показательной функции, не имеет решения.
2) For b > 0, using the monotonicity of the function and the root theorem, the equation has a single root. In order to find it, b must be represented as b = aс, ax = bс ó x = c or x = logab.
The exponential equations, through algebraic transformations, lead to standard equations, which are solved using the following methods:
1) method of reduction to one base;
2) assessment method;
3) graphic method;
4) the method of introducing new variables;
5) factorization method;
6) exponential - power equations;
7) exponential with a parameter.
2 . Method of reduction to one basis.
The method is based on the following property of degrees: if two degrees are equal and their bases are equal, then their exponents are equal, i.e., the equation should be tried to be reduced to the form
Examples. Solve the equation:
1 . 3x=81;
Let's represent the right side of the equation in the form 81 = 34 and write the equation equivalent to the original 3 x = 34; x = 4. Answer: 4.
2. https://pandia.ru/text/80/142/images/image004_8.png" width="52" height="49"> and go to the equation for exponents 3x+1 = 3 – 5x; 8x = 4; x = 0.5 Answer: 0.5
3. https://pandia.ru/text/80/142/images/image006_8.png" width="105" height="47">
Note that the numbers 0.2, 0.04, √5, and 25 are powers of 5. Let's take advantage of this and transform the original equation as follows:
,
whence 5-x-1 = 5-2x-2 ó - x - 1 = - 2x - 2, from which we find the solution x = -1. Answer: -1.
5. 3x = 5. By definition of the logarithm, x = log35. Answer: log35.
6. 62x+4 = 33x. 2x+8.
Let's rewrite the equation as 32x+4.22x+4 = 32x.2x+8, i.e..png" width="181" height="49 src="> Hence x - 4 =0, x = 4. Answer: four.
7 . 2∙3x+1 - 6∙3x-2 - 3x = 9. Using the properties of powers, we write the equation in the form e. x+1 = 2, x =1. Answer: 1.
Bank of tasks No. 1.
Solve the equation:
Test number 1.
1) 0 2) 4 3) -2 4) -4 |
|
A2 32x-8 = √3. | 1)17/4 2) 17 3) 13/2 4) -17/4 |
A3 | 1) 3;1 2) -3;-1 3) 0;2 4) no roots |
1) 7;1 2) no roots 3) -7;1 4) -1;-7 |
|
A5 | 1) 0;2; 2) 0;2;3 3) 0 4) -2;-3;0 |
A6 | 1) -1 2) 0 3) 2 4) 1 |
Test #2
A1 | 1) 3 2) -1;3 3) -1;-3 4) 3;-1 |
A2 | 1) 14/3 2) -14/3 3) -17 4) 11 |
A3 | 1) 2;-1 2) no roots 3) 0 4) -2;1 |
A4 | 1) -4 2) 2 3) -2 4) -4;2 |
A5 | 1) 3 2) -3;1 3) -1 4) -1;3 |
3 Assessment method.
The root theorem: if the function f (x) increases (decreases) on the interval I, the number a is any value taken by f on this interval, then the equation f (x) = a has a single root on the interval I.
When solving equations by the estimation method, this theorem and the monotonicity properties of the function are used.
Examples. Solve Equations: 1. 4x = 5 - x.
Solution. Let's rewrite the equation as 4x + x = 5.
1. if x \u003d 1, then 41 + 1 \u003d 5, 5 \u003d 5 is true, then 1 is the root of the equation.
The function f(x) = 4x is increasing on R and g(x) = x is increasing on R => h(x)= f(x)+g(x) is increasing on R as the sum of increasing functions, so x = 1 is the only root of the equation 4x = 5 – x. Answer: 1.
2.
Solution. We rewrite the equation in the form 
.
1. if x = -1, then 
, 3 = 3-true, so x = -1 is the root of the equation.
2. prove that it is unique.
3. The function f(x) = - decreases on R, and g(x) = - x - decreases on R => h(x) = f(x) + g(x) - decreases on R, as the sum of decreasing functions . So by the root theorem, x = -1 is the only root of the equation. Answer: -1.
Bank of tasks No. 2. solve the equation
a) 4x + 1 = 6 - x;
b) 
c) 2x – 2 =1 – x;
4. Method for introducing new variables.
The method is described in section 2.1. The introduction of a new variable (substitution) is usually carried out after transformations (simplification) of the terms of the equation. Consider examples.
Examples.
R eat equation: 1.
.
Let's rewrite the equation differently: https://pandia.ru/text/80/142/images/image030_0.png" width="128" height="48 src="> i.e..png" width="210" height ="45">
Solution. Let's rewrite the equation differently: 
Denote https://pandia.ru/text/80/142/images/image035_0.png" width="245" height="57"> - not suitable.
t = 4 => https://pandia.ru/text/80/142/images/image037_0.png" width="268" height="51"> is an irrational equation. Note that 
The solution to the equation is x = 2.5 ≤ 4, so 2.5 is the root of the equation. Answer: 2.5.
Solution. Let's rewrite the equation in the form and divide both sides by 56x+6 ≠ 0. We get the equation
2x2-6x-7 = 2x2-6x-8 +1 = 2(x2-3x-4)+1, so..png" width="118" height="56">
The roots of the quadratic equation - t1 = 1 and t2<0, т. е..png" width="200" height="24">.
Solution . We rewrite the equation in the form
and note that it is a homogeneous equation of the second degree.
Divide the equation by 42x, we get 
Replace https://pandia.ru/text/80/142/images/image049_0.png" width="16" height="41 src="> .
Answer: 0; 0.5.
Task Bank #3. solve the equation
b) 
G) 
Test #3 with a choice of answers. Minimum level.
A1 | 1) -0.2;2 2) log52 3) –log52 4) 2 |
А2 0.52x – 3 0.5x +2 = 0. | 1) 2;1 2) -1;0 3) no roots 4) 0 |
1) 0 2) 1; -1/3 3) 1 4) 5 |
|
A4 52x-5x - 600 = 0. | 1) -24;25 2) -24,5; 25,5 3) 25 4) 2 |
1) no roots 2) 2;4 3) 3 4) -1;2 |
Test #4 with a choice of answers. General level.
A1 | 1) 2;1 2) ½;0 3)2;0 4) 0 |
А2 2x – (0.5)2x – (0.5)x + 1 = 0 | 1) -1;1 2) 0 3) -1;0;1 4) 1 |
1) 64 2) -14 3) 3 4) 8 |
|
1)-1 2) 1 3) -1;1 4) 0 |
|
A5 | 1) 0 2) 1 3) 0;1 4) no roots |
5. Method of factorization.
1. Solve the equation: 5x+1 - 5x-1 = 24.
Solution..png" width="169" height="69"> , from where
2. 6x + 6x+1 = 2x + 2x+1 + 2x+2.
Solution. Let us take out 6x on the left side of the equation, and 2x on the right side. We get the equation 6x(1+6) = 2x(1+2+4) ó 6x = 2x.
Since 2x >0 for all x, we can divide both sides of this equation by 2x without fear of losing solutions. We get 3x = 1ó x = 0.
3. 
Solution. We solve the equation by factoring.
We select the square of the binomial 
4. https://pandia.ru/text/80/142/images/image067_0.png" width="500" height="181">
x = -2 is the root of the equation.
Equation x + 1 = 0 " style="border-collapse:collapse;border:none">
A1 5x-1 +5x -5x+1 = -19.
1) 1 2) 95/4 3) 0 4) -1
A2 3x+1 +3x-1 =270.
1) 2 2) -4 3) 0 4) 4
A3 32x + 32x+1 -108 = 0. x=1.5
1) 0,2 2) 1,5 3) -1,5 4) 3
1) 1 2) -3 3) -1 4) 0
A5 2x -2x-4 = 15.x=4
1) -4 2) 4 3) -4;4 4) 2
Test #6 General level.
A1 (22x-1)(24x+22x+1)=7. | 1) ½ 2) 2 3) -1;3 4) 0.2 |
A2 | 1) 2.5 2) 3;4 3) log43/2 4) 0 |
A3 2x-1-3x=3x-1-2x+2. | 1) 2 2) -1 3) 3 4) -3 |
A4 | 1) 1,5 2) 3 3) 1 4) -4 |
A5 | 1) 2 2) -2 3) 5 4) 0 |
6. Exponential - power equations.
The exponential equations are adjoined by the so-called exponential-power equations, i.e. equations of the form (f(x))g(x) = (f(x))h(x).
If it is known that f(x)>0 and f(x) ≠ 1, then the equation, like the exponential one, is solved by equating the exponents g(x) = f(x).
If the condition does not exclude the possibility of f(x)=0 and f(x)=1, then we have to consider these cases when solving the exponential power equation.
1..png" width="182" height="116 src=">
2. 
Solution. x2 +2x-8 - makes sense for any x, because a polynomial, so the equation is equivalent to the set
https://pandia.ru/text/80/142/images/image078_0.png" width="137" height="35">
b) 
7. Exponential equations with parameters.
1. For what values of the parameter p does the equation 4 (5 – 3)2 +4p2–3p = 0 (1) have a unique solution?
Solution. Let us introduce the change 2x = t, t > 0, then equation (1) will take the form t2 – (5p – 3)t + 4p2 – 3p = 0. (2)
The discriminant of equation (2) is D = (5p – 3)2 – 4(4p2 – 3p) = 9(p – 1)2.
Equation (1) has a unique solution if equation (2) has one positive root. This is possible in the following cases.
1. If D = 0, that is, p = 1, then equation (2) will take the form t2 – 2t + 1 = 0, hence t = 1, therefore, equation (1) has a unique solution x = 0.
2. If p1, then 9(p – 1)2 > 0, then equation (2) has two different roots t1 = p, t2 = 4p – 3. The set of systems satisfies the condition of the problem
Substituting t1 and t2 into the systems, we have
https://pandia.ru/text/80/142/images/image084_0.png" alt="(!LANG:no35_11" width="375" height="54"> в зависимости от параметра a?!}
Solution. Let 
then equation (3) will take the form t2 – 6t – a = 0. (4)
Let us find the values of the parameter a for which at least one root of equation (4) satisfies the condition t > 0.
Let us introduce the function f(t) = t2 – 6t – a. The following cases are possible.
https://pandia.ru/text/80/142/images/image087.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_14.gif" align="left" width="215" height="73 src=">где t0 - абсцисса вершины параболы и D - дискриминант квадратного трехчлена f(t);!}
https://pandia.ru/text/80/142/images/image089.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_16.gif" align="left" width="60" height="51 src=">!}
Case 2. Equation (4) has a unique positive solution if
D = 0, if a = – 9, then equation (4) will take the form (t – 3)2 = 0, t = 3, x = – 1.
Case 3. Equation (4) has two roots, but one of them does not satisfy the inequality t > 0. This is possible if
https://pandia.ru/text/80/142/images/image092.png" alt="(!LANG:no35_17" width="267" height="63">!}
Thus, at a 0 equation (4) has a single positive root 
. Then equation (3) has a unique solution 
For a< – 9 уравнение (3) корней не имеет.
if a< – 9, то корней нет; если – 9 < a < 0, то
if a = – 9, then x = – 1;
if a 0, then
Let us compare the methods for solving equations (1) and (3). Note that when solving equation (1) was reduced to a quadratic equation, the discriminant of which is a full square; thus, the roots of equation (2) were immediately calculated by the formula of the roots of the quadratic equation, and then conclusions were drawn regarding these roots. Equation (3) was reduced to a quadratic equation (4), the discriminant of which is not a perfect square, therefore, when solving equation (3), it is advisable to use theorems on the location of the roots of a square trinomial and a graphical model. Note that equation (4) can be solved using the Vieta theorem.
Let's solve more complex equations.
Task 3. Solve the equation 
Solution. ODZ: x1, x2.
Let's introduce a replacement. Let 2x = t, t > 0, then as a result of transformations the equation will take the form t2 + 2t – 13 – a = 0. (*) Let us find the values of a for which at least one root of the equation (*) satisfies the condition t > 0.
https://pandia.ru/text/80/142/images/image098.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_23.gif" align="left" width="71" height="68 src=">где t0 - абсцисса вершины f(t) = t2 + 2t – 13 – a, D - дискриминант квадратного трехчлена f(t).!}
https://pandia.ru/text/80/142/images/image100.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_25.gif" align="left" width="360" height="32 src=">!}
https://pandia.ru/text/80/142/images/image102.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_27.gif" align="left" width="218" height="42 src=">!}
Answer: if a > - 13, a 11, a 5, then if a - 13,
a = 11, a = 5, then there are no roots.
Bibliography.
1. Guzeev foundations of educational technology.
2. Guzeev technology: from reception to philosophy.
M. "Headmaster" No. 4, 1996
3. Guzeev and organizational forms of education.
4. Guzeev and the practice of integral educational technology.
M. "People's education", 2001
5. Guzeev from the forms of the lesson - seminar.
Mathematics at school No. 2, 1987, pp. 9 - 11.
6. Selevko educational technologies.
M. "People's education", 1998
7. Episheva schoolchildren learn mathematics.
M. "Enlightenment", 1990
8. Ivanov to prepare lessons - workshops.
Mathematics at School No. 6, 1990, p. 37-40.
9. Smirnov model of teaching mathematics.
Mathematics at School No. 1, 1997, p. 32-36.
10. Tarasenko ways of organizing practical work.
Mathematics at School No. 1, 1993, p. 27 - 28.
11. About one of the types of individual work.
Mathematics at School No. 2, 1994, pp. 63 - 64.
12. Khazankin creative abilities of schoolchildren.
Mathematics at School No. 2, 1989, p. ten.
13. Scanavi. Publisher, 1997
14. et al. Algebra and the beginnings of analysis. Didactic materials for
15. Krivonogov tasks in mathematics.
M. "First of September", 2002
16. Cherkasov. Handbook for high school students and
entering universities. "A S T - press school", 2002
17. Zhevnyak for applicants to universities.
Minsk and RF "Review", 1996
18. Written D. Preparing for the exam in mathematics. M. Rolf, 1999
19. and others. Learning to solve equations and inequalities.
M. "Intellect - Center", 2003
20. and others. Educational and training materials for preparing for the E G E.
M. "Intellect - Center", 2003 and 2004
21 and others. Variants of CMM. Testing Center of the Ministry of Defense of the Russian Federation, 2002, 2003
22. Goldberg equations. "Quantum" No. 3, 1971
23. Volovich M. How to successfully teach mathematics.
Mathematics, 1997 No. 3.
24 Okunev for the lesson, children! M. Enlightenment, 1988
25. Yakimanskaya - oriented education at school.
26. Liimets work at the lesson. M. Knowledge, 1975
At the stage of preparation for the final testing, high school students need to improve their knowledge on the topic "Exponential Equations". The experience of past years indicates that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of preparation, need to carefully master the theory, memorize the formulas and understand the principle of solving such equations. Having learned to cope with this type of tasks, graduates will be able to count on high scores when passing the exam in mathematics.
Get ready for the exam testing together with Shkolkovo!
When repeating the materials covered, many students are faced with the problem of finding the formulas needed to solve the equations. A school textbook is not always at hand, and the selection of the necessary information on a topic on the Internet takes a long time.
The Shkolkovo educational portal invites students to use our knowledge base. We are implementing a completely new method of preparing for the final test. Studying on our site, you will be able to identify gaps in knowledge and pay attention to precisely those tasks that cause the greatest difficulties.
Teachers of "Shkolkovo" collected, systematized and presented all the material necessary for the successful passing of the exam in the simplest and most accessible form.
The main definitions and formulas are presented in the "Theoretical Reference" section.
For a better assimilation of the material, we recommend that you practice the assignments. Carefully review the examples of exponential equations with solutions presented on this page in order to understand the calculation algorithm. After that, proceed with the tasks in the "Catalogs" section. You can start with the easiest tasks or go straight to solving complex exponential equations with several unknowns or . The database of exercises on our website is constantly supplemented and updated.
Those examples with indicators that caused you difficulties can be added to the "Favorites". So you can quickly find them and discuss the solution with the teacher.
To successfully pass the exam, study on the Shkolkovo portal every day!
Belgorod State University
CHAIR algebra, number theory and geometry
Work theme: Exponential-power equations and inequalities.
Graduate work student of the Faculty of Physics and Mathematics
Scientific adviser:
______________________________
Reviewer: _______________________________
________________________
Belgorod. 2006
| Introduction | 3 | ||
| Topic I. | Analysis of the literature on the research topic. | ||
| Topic II. | Functions and their properties used in solving exponential-power equations and inequalities. | ||
| I.1. | Power function and its properties. | ||
| I.2. | The exponential function and its properties. | ||
| Topic III. | Solution of exponential-power equations, algorithm and examples. | ||
| Topic IV. | Solving exponential-power inequalities, solution plan and examples. | ||
| Topic v. | Experience in conducting classes with schoolchildren on the topic: "Solution of exponential-power equations and inequalities." | ||
| v. 1. | Teaching material. | ||
| v. 2. | Tasks for independent solution. | ||
| Conclusion. | Conclusions and offers. | ||
| Bibliography. | |||
| Applications | |||
Introduction.
"... the joy of seeing and understanding ..."
A. Einstein.
In this work, I tried to convey my experience as a teacher of mathematics, to convey, at least to some extent, my attitude to teaching mathematics - a human matter in which mathematical science, pedagogy, didactics, psychology, and even philosophy are surprisingly intertwined.
I had a chance to work with kids and graduates, with children standing at the poles of intellectual development: those who were registered with a psychiatrist and who were really interested in mathematics
I had to solve many methodological problems. I will try to talk about those that I managed to solve. But even more - it was not possible, and in those that seem to be resolved, new questions appear.
But even more important than the experience itself are the teacher's reflections and doubts: why is it exactly like this, this experience?
And the summer is different now, and the turn of education has become more interesting. “Under the Jupiters” today is not the search for a mythical optimal system of teaching “everyone and everything”, but the child himself. But then - with necessity - and the teacher.
In the school course of algebra and the beginning of analysis, grades 10 - 11, when passing the exam for a high school course and at entrance exams to universities, there are equations and inequalities containing an unknown at the base and exponents - these are exponential-power equations and inequalities.
Little attention is paid to them at school, there are practically no tasks on this topic in textbooks. However, mastering the methodology for solving them, it seems to me, is very useful: it increases the mental and creative abilities of students, completely new horizons open up before us. When solving problems, students acquire the first skills of research work, their mathematical culture is enriched, and the ability to think logically develops. Schoolchildren develop such personality traits as purposefulness, goal-setting, independence, which will be useful to them in later life. And also there is a repetition, expansion and deep assimilation of educational material.
I started working on this topic of my thesis research with writing a term paper. In the course of which I studied and analyzed the mathematical literature on this topic in more depth, I identified the most appropriate method for solving exponential-power equations and inequalities.
It lies in the fact that in addition to the generally accepted approach when solving exponential-power equations (the base is taken greater than 0) and when solving the same inequalities (the base is taken greater than 1 or greater than 0, but less than 1), cases are also considered when the bases are negative, are 0 and 1.
An analysis of the written examination papers of students shows that the lack of coverage of the issue of the negative value of the argument of the exponential-power function in school textbooks causes a number of difficulties for them and leads to errors. And they also have problems at the stage of systematization of the results obtained, where, due to the transition to an equation - a consequence or an inequality - a consequence, extraneous roots may appear. In order to eliminate errors, we use a check by the original equation or inequality and an algorithm for solving exponential-power equations, or a plan for solving exponential-power inequalities.
In order for students to be able to successfully pass the final and entrance exams, I think it is necessary to pay more attention to solving exponential-power equations and inequalities in the classroom, or additionally in electives and circles.
In this way topic , my thesis is defined as follows: "Exponential-power equations and inequalities."
Goals of this work are:
1. Analyze the literature on this topic.
2. Give a complete analysis of the solution of exponential-power equations and inequalities.
3. Give a sufficient number of examples on this topic of various types.
4. Check in class, optional and circle classes how the proposed methods for solving exponential-power equations and inequalities will be perceived. Give appropriate recommendations for the study of this topic.
Subject our research is to develop a technique for solving exponential-power equations and inequalities.
The purpose and subject of the study required the solution of the following tasks:
1. Study the literature on the topic: "Exponential-power equations and inequalities."
2. Master the methods of solving exponential-power equations and inequalities.
3. Select training material and develop a system of exercises at different levels on the topic: "Solving exponential-power equations and inequalities."
In the course of the thesis research, more than 20 papers devoted to the application of various methods for solving exponential-power equations and inequalities were analyzed. From here we get.
Thesis plan:
Introduction.
Chapter I. Analysis of the literature on the research topic.
Chapter II. Functions and their properties used in solving exponential-power equations and inequalities.
II.1. Power function and its properties.
II.2. The exponential function and its properties.
Chapter III. Solution of exponential-power equations, algorithm and examples.
Chapter IV. Solving exponential-power inequalities, solution plan and examples.
Chapter V. Experience in conducting classes with schoolchildren on this topic.
1. Educational material.
2. Tasks for independent solution.
Conclusion. Conclusions and offers.
List of used literature.
Literature analyzed in Chapter I
This lesson is intended for those who are just starting to learn exponential equations. As always, let's start with a definition and simple examples.
If you are reading this lesson, then I suspect that you already have at least a minimal understanding of the simplest equations - linear and square: $56x-11=0$; $((x)^(2))+5x+4=0$; $((x)^(2))-12x+32=0$ etc. To be able to solve such constructions is absolutely necessary in order not to “hang” in the topic that will be discussed now.
So, exponential equations. Let me give you a couple of examples:
\[((2)^(x))=4;\quad ((5)^(2x-3))=\frac(1)(25);\quad ((9)^(x))=- 3\]
Some of them may seem more complicated to you, some of them, on the contrary, are too simple. But all of them are united by one important feature: they contain an exponential function $f\left(x \right)=((a)^(x))$. Thus, we introduce the definition:
An exponential equation is any equation that contains an exponential function, i.e. an expression of the form $((a)^(x))$. In addition to the specified function, such equations can contain any other algebraic constructions - polynomials, roots, trigonometry, logarithms, etc.
OK then. Understood the definition. Now the question is: how to solve all this crap? The answer is both simple and complex at the same time.
Let's start with the good news: from my experience with many students, I can say that for most of them, exponential equations are much easier than the same logarithms, and even more so trigonometry.
But there is also bad news: sometimes the compilers of problems for all kinds of textbooks and exams are visited by "inspiration", and their drug-inflamed brain begins to produce such brutal equations that it becomes problematic not only for students to solve them - even many teachers get stuck on such problems.
However, let's not talk about sad things. And let's return to those three equations that were given at the very beginning of the story. Let's try to solve each of them.
First equation: $((2)^(x))=4$. Well, to what power must the number 2 be raised to get the number 4? Perhaps the second? After all, $((2)^(2))=2\cdot 2=4$ — and we have obtained the correct numerical equality, i.e. indeed $x=2$. Well, thanks, cap, but this equation was so simple that even my cat could solve it. :)
Let's look at the following equation:
\[((5)^(2x-3))=\frac(1)(25)\]
But here it is a little more difficult. Many students know that $((5)^(2))=25$ is the multiplication table. Some also suspect that $((5)^(-1))=\frac(1)(5)$ is essentially the definition of negative exponents (similar to the formula $((a)^(-n))= \frac(1)(((a)^(n)))$).
Finally, only a select few guess that these facts can be combined and the output is the following result:
\[\frac(1)(25)=\frac(1)(((5)^(2)))=((5)^(-2))\]
Thus, our original equation will be rewritten as follows:
\[((5)^(2x-3))=\frac(1)(25)\Rightarrow ((5)^(2x-3))=((5)^(-2))\]
And now this is already completely solved! On the left side of the equation there is an exponential function, on the right side of the equation there is an exponential function, there is nothing but them anywhere else. Therefore, it is possible to “discard” the bases and stupidly equate the indicators:
We got the simplest linear equation that any student can solve in just a couple of lines. Okay, in four lines:
\[\begin(align)& 2x-3=-2 \\& 2x=3-2 \\& 2x=1 \\& x=\frac(1)(2) \\\end(align)\]
If you don’t understand what happened in the last four lines, be sure to return to the topic “linear equations” and repeat it. Because without a clear assimilation of this topic, it is too early for you to take on exponential equations.
\[((9)^(x))=-3\]
Well, how do you decide? First thought: $9=3\cdot 3=((3)^(2))$, so the original equation can be rewritten like this:
\[((\left(((3)^(2)) \right))^(x))=-3\]
Then we recall that when raising a degree to a power, the indicators are multiplied:
\[((\left(((3)^(2)) \right))^(x))=((3)^(2x))\Rightarrow ((3)^(2x))=-(( 3)^(1))\]
\[\begin(align)& 2x=-1 \\& x=-\frac(1)(2) \\\end(align)\]
And for such a decision, we get an honestly deserved deuce. For we, with the equanimity of a Pokémon, sent the minus sign in front of the three to the power of this very three. And you can't do that. And that's why. Take a look at the different powers of the triple:
\[\begin(matrix) ((3)^(1))=3& ((3)^(-1))=\frac(1)(3)& ((3)^(\frac(1)( 2)))=\sqrt(3) \\ ((3)^(2))=9& ((3)^(-2))=\frac(1)(9)& ((3)^(\ frac(1)(3)))=\sqrt(3) \\ ((3)^(3))=27& ((3)^(-3))=\frac(1)(27)& (( 3)^(-\frac(1)(2)))=\frac(1)(\sqrt(3)) \\\end(matrix)\]
When compiling this tablet, I didn’t pervert as soon as I did: I considered positive degrees, and negative ones, and even fractional ones ... well, where is at least one negative number here? He is not! And it cannot be, because the exponential function $y=((a)^(x))$, firstly, always takes only positive values (no matter how much you multiply one or divide by two, it will still be a positive number), and secondly, the base of such a function, the number $a$, is by definition a positive number!
Well, how then to solve the equation $((9)^(x))=-3$? No, there are no roots. And in this sense, exponential equations are very similar to quadratic ones - there may also be no roots. But if in quadratic equations the number of roots is determined by the discriminant (the discriminant is positive - 2 roots, negative - no roots), then in exponential equations it all depends on what is to the right of the equal sign.
Thus, we formulate the key conclusion: the simplest exponential equation of the form $((a)^(x))=b$ has a root if and only if $b>0$. Knowing this simple fact, you can easily determine whether the equation proposed to you has roots or not. Those. is it worth solving it at all or immediately write down that there are no roots.
This knowledge will help us many times over when we have to solve more complex problems. In the meantime, enough lyrics - it's time to study the basic algorithm for solving exponential equations.
How to solve exponential equations
So, let's formulate the problem. It is necessary to solve the exponential equation:
\[((a)^(x))=b,\quad a,b>0\]
According to the "naive" algorithm that we used earlier, it is necessary to represent the number $b$ as a power of the number $a$:
In addition, if instead of the variable $x$ there is any expression, we will get a new equation, which can already be solved. For example:
\[\begin(align)& ((2)^(x))=8\Rightarrow ((2)^(x))=((2)^(3))\Rightarrow x=3; \\& ((3)^(-x))=81\Rightarrow ((3)^(-x))=((3)^(4))\Rightarrow -x=4\Rightarrow x=-4; \\& ((5)^(2x))=125\Rightarrow ((5)^(2x))=((5)^(3))\Rightarrow 2x=3\Rightarrow x=\frac(3)( 2). \\\end(align)\]
And oddly enough, this scheme works in about 90% of cases. What about the other 10% then? The remaining 10% are slightly "schizophrenic" exponential equations of the form:
\[((2)^(x))=3;\quad ((5)^(x))=15;\quad ((4)^(2x))=11\]
To what power do you need to raise 2 to get 3? In the first? But no: $((2)^(1))=2$ is not enough. In the second? Neither: $((2)^(2))=4$ is too much. What then?
Knowledgeable students have probably already guessed: in such cases, when it is impossible to solve “beautifully”, “heavy artillery” is connected to the case - logarithms. Let me remind you that using logarithms, any positive number can be represented as a power of any other positive number (with the exception of one):
Remember this formula? When I tell my students about logarithms, I always warn you: this formula (it is also the basic logarithmic identity or, if you like, the definition of the logarithm) will haunt you for a very long time and “emerge” in the most unexpected places. Well, she surfaced. Let's look at our equation and this formula:
\[\begin(align)& ((2)^(x))=3 \\& a=((b)^(((\log )_(b))a)) \\\end(align) \]
If we assume that $a=3$ is our original number on the right, and $b=2$ is the very base of the exponential function to which we so want to reduce the right side, we get the following:
\[\begin(align)& a=((b)^(((\log )_(b))a))\Rightarrow 3=((2)^(((\log )_(2))3 )); \\& ((2)^(x))=3\Rightarrow ((2)^(x))=((2)^(((\log )_(2))3))\Rightarrow x=( (\log )_(2))3. \\\end(align)\]
We got a slightly strange answer: $x=((\log )_(2))3$. In some other task, with such an answer, many would doubt and begin to double-check their solution: what if there was a mistake somewhere? I hasten to please you: there is no error here, and logarithms in the roots of exponential equations are quite a typical situation. So get used to it. :)
Now we solve by analogy the remaining two equations:
\[\begin(align)& ((5)^(x))=15\Rightarrow ((5)^(x))=((5)^(((\log )_(5))15)) \Rightarrow x=((\log )_(5))15; \\& ((4)^(2x))=11\Rightarrow ((4)^(2x))=((4)^(((\log )_(4))11))\Rightarrow 2x=( (\log )_(4))11\Rightarrow x=\frac(1)(2)((\log )_(4))11. \\\end(align)\]
That's all! By the way, the last answer can be written differently:
It was we who introduced the multiplier into the argument of the logarithm. But no one prevents us from adding this factor to the base:
Moreover, all three options are correct - they are just different forms of writing the same number. Which one to choose and write down in this decision is up to you.
Thus, we have learned to solve any exponential equations of the form $((a)^(x))=b$, where the numbers $a$ and $b$ are strictly positive. However, the harsh reality of our world is that such simple tasks will meet you very, very rarely. More often you will come across something like this:
\[\begin(align)& ((4)^(x))+((4)^(x-1))=((4)^(x+1))-11; \\& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& ((100)^(x-1))\cdot ((2,7)^(1-x))=0.09. \\\end(align)\]
Well, how do you decide? Can this be resolved at all? And if so, how?
No panic. All these equations are quickly and simply reduced to those simple formulas that we have already considered. You just need to know to remember a couple of tricks from the algebra course. And of course, there are no rules for working with degrees here. I'll talk about all this now. :)
Transformation of exponential equations
The first thing to remember is that any exponential equation, no matter how complex it may be, one way or another must be reduced to the simplest equations - the very ones that we have already considered and which we know how to solve. In other words, the scheme for solving any exponential equation looks like this:
- Write down the original equation. For example: $((4)^(x))+((4)^(x-1))=((4)^(x+1))-11$;
- Do some stupid shit. Or even some crap called "transform the equation";
- At the output, get the simplest expressions like $((4)^(x))=4$ or something else like that. Moreover, one initial equation can give several such expressions at once.
With the first point, everything is clear - even my cat can write the equation on a leaf. With the third point, too, it seems, it is more or less clear - we have already solved a whole bunch of such equations above.
But what about the second point? What are the transformations? What to convert to what? And How?
Well, let's figure it out. First of all, I would like to point out the following. All exponential equations are divided into two types:
- The equation is composed of exponential functions with the same base. Example: $((4)^(x))+((4)^(x-1))=((4)^(x+1))-11$;
- The formula contains exponential functions with different bases. Examples: $((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x))$ and $((100)^(x-1) )\cdot ((2,7)^(1-x))=0.09$.
Let's start with equations of the first type - they are the easiest to solve. And in their solution we will be helped by such a technique as the selection of stable expressions.
Highlighting a stable expression
Let's look at this equation again:
\[((4)^(x))+((4)^(x-1))=((4)^(x+1))-11\]
What do we see? The four are raised to different degrees. But all these powers are simple sums of the variable $x$ with other numbers. Therefore, it is necessary to remember the rules for working with degrees:
\[\begin(align)& ((a)^(x+y))=((a)^(x))\cdot ((a)^(y)); \\& ((a)^(x-y))=((a)^(x)):((a)^(y))=\frac(((a)^(x)))(((a )^(y))). \\\end(align)\]
Simply put, addition of exponents can be converted to a product of powers, and subtraction is easily converted to division. Let's try to apply these formulas to the powers from our equation:
\[\begin(align)& ((4)^(x-1))=\frac(((4)^(x)))(((4)^(1)))=((4)^ (x))\cdot \frac(1)(4); \\& ((4)^(x+1))=((4)^(x))\cdot ((4)^(1))=((4)^(x))\cdot 4. \ \\end(align)\]
We rewrite the original equation taking this fact into account, and then we collect all the terms on the left:
\[\begin(align)& ((4)^(x))+((4)^(x))\cdot \frac(1)(4)=((4)^(x))\cdot 4 -eleven; \\& ((4)^(x))+((4)^(x))\cdot \frac(1)(4)-((4)^(x))\cdot 4+11=0. \\\end(align)\]
The first four terms contain the element $((4)^(x))$ — let's take it out of the bracket:
\[\begin(align)& ((4)^(x))\cdot \left(1+\frac(1)(4)-4 \right)+11=0; \\& ((4)^(x))\cdot \frac(4+1-16)(4)+11=0; \\& ((4)^(x))\cdot \left(-\frac(11)(4) \right)=-11. \\\end(align)\]
It remains to divide both parts of the equation by the fraction $-\frac(11)(4)$, i.e. essentially multiply by the inverted fraction - $-\frac(4)(11)$. We get:
\[\begin(align)& ((4)^(x))\cdot \left(-\frac(11)(4) \right)\cdot \left(-\frac(4)(11) \right )=-11\cdot \left(-\frac(4)(11) \right); \\& ((4)^(x))=4; \\& ((4)^(x))=((4)^(1)); \\&x=1. \\\end(align)\]
That's all! We reduced the original equation to the simplest and got the final answer.
At the same time, in the process of solving, we discovered (and even took out of the bracket) the common factor $((4)^(x))$ - this is the stable expression. It can be designated as a new variable, or you can simply accurately express it and get an answer. In any case, the key principle of the solution is as follows:
Find in the original equation a stable expression containing a variable that is easily distinguished from all exponential functions.
The good news is that almost every exponential equation admits such a stable expression.
But there is also bad news: such expressions can be very tricky, and it can be quite difficult to distinguish them. So let's look at another problem:
\[((5)^(x+2))+((0,2)^(-x-1))+4\cdot ((5)^(x+1))=2\]
Perhaps someone will now have a question: “Pasha, are you stoned? Here are different bases - 5 and 0.2. But let's try to convert a power with base 0.2. For example, let's get rid of the decimal fraction, bringing it to the usual:
\[((0,2)^(-x-1))=((0,2)^(-\left(x+1 \right)))=((\left(\frac(2)(10 ) \right))^(-\left(x+1 \right)))=((\left(\frac(1)(5) \right))^(-\left(x+1 \right)) )\]
As you can see, the number 5 still appeared, albeit in the denominator. At the same time, the indicator was rewritten as negative. And now we recall one of the most important rules for working with degrees:
\[((a)^(-n))=\frac(1)(((a)^(n)))\Rightarrow ((\left(\frac(1)(5) \right))^( -\left(x+1 \right)))=((\left(\frac(5)(1) \right))^(x+1))=((5)^(x+1))\ ]
Here, of course, I cheated a little. Because for a complete understanding, the formula for getting rid of negative indicators had to be written as follows:
\[((a)^(-n))=\frac(1)(((a)^(n)))=((\left(\frac(1)(a) \right))^(n ))\Rightarrow ((\left(\frac(1)(5) \right))^(-\left(x+1 \right)))=((\left(\frac(5)(1) \ right))^(x+1))=((5)^(x+1))\]
On the other hand, nothing prevented us from working with only one fraction:
\[((\left(\frac(1)(5) \right))^(-\left(x+1 \right)))=((\left(((5)^(-1)) \ right))^(-\left(x+1 \right)))=((5)^(\left(-1 \right)\cdot \left(-\left(x+1 \right) \right) ))=((5)^(x+1))\]
But in this case, you need to be able to raise a degree to another degree (I remind you: in this case, the indicators are added up). But I didn’t have to “flip” the fractions - perhaps for someone it will be easier. :)
In any case, the original exponential equation will be rewritten as:
\[\begin(align)& ((5)^(x+2))+((5)^(x+1))+4\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+5\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+((5)^(1))\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+((5)^(x+2))=2; \\& 2\cdot ((5)^(x+2))=2; \\& ((5)^(x+2))=1. \\\end(align)\]
So it turns out that the original equation is even easier to solve than the previously considered one: here you don’t even need to single out a stable expression - everything has been reduced by itself. It remains only to remember that $1=((5)^(0))$, whence we get:
\[\begin(align)& ((5)^(x+2))=((5)^(0)); \\&x+2=0; \\&x=-2. \\\end(align)\]
That's the whole solution! We got the final answer: $x=-2$. At the same time, I would like to note one trick that greatly simplified all the calculations for us:
In exponential equations, be sure to get rid of decimal fractions, translate them into ordinary ones. This will allow you to see the same bases of the degrees and greatly simplify the solution.
Now let's move on to more complex equations in which there are different bases, which are generally not reducible to each other using powers.
Using the exponent property
Let me remind you that we have two more particularly harsh equations:
\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& ((100)^(x-1))\cdot ((2,7)^(1-x))=0.09. \\\end(align)\]
The main difficulty here is that it is not clear what and to what basis to lead. Where are the fixed expressions? Where are the common grounds? There is none of this.
But let's try to go the other way. If there are no ready-made identical bases, you can try to find them by factoring the available bases.
Let's start with the first equation:
\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& 21=7\cdot 3\Rightarrow ((21)^(3x))=((\left(7\cdot 3 \right))^(3x))=((7)^(3x))\ cdot ((3)^(3x)). \\\end(align)\]
But after all, you can do the opposite - make up the number 21 from the numbers 7 and 3. It is especially easy to do this on the left, since the indicators of both degrees are the same:
\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((\left(7\cdot 3 \right))^(x+ 6))=((21)^(x+6)); \\& ((21)^(x+6))=((21)^(3x)); \\&x+6=3x; \\& 2x=6; \\&x=3. \\\end(align)\]
That's all! You took the exponent out of the product and immediately got a beautiful equation that can be solved in a couple of lines.
Now let's deal with the second equation. Here everything is much more complicated:
\[((100)^(x-1))\cdot ((2,7)^(1-x))=0.09\]
\[((100)^(x-1))\cdot ((\left(\frac(27)(10) \right))^(1-x))=\frac(9)(100)\]
In this case, the fractions turned out to be irreducible, but if something could be reduced, be sure to reduce it. This will often result in interesting grounds that you can already work with.
Unfortunately, we haven't come up with anything. But we see that the exponents on the left in the product are opposite:
Let me remind you: to get rid of the minus sign in the exponent, you just need to “flip” the fraction. So let's rewrite the original equation:
\[\begin(align)& ((100)^(x-1))\cdot ((\left(\frac(10)(27) \right))^(x-1))=\frac(9 )(100); \\& ((\left(100\cdot \frac(10)(27) \right))^(x-1))=\frac(9)(100); \\& ((\left(\frac(1000)(27) \right))^(x-1))=\frac(9)(100). \\\end(align)\]
In the second line, we just bracketed the total from the product according to the rule $((a)^(x))\cdot ((b)^(x))=((\left(a\cdot b \right))^ (x))$, and in the latter they simply multiplied the number 100 by a fraction.
Now note that the numbers on the left (at the base) and on the right are somewhat similar. How? Yes, obviously: they are powers of the same number! We have:
\[\begin(align)& \frac(1000)(27)=\frac(((10)^(3)))(((3)^(3)))=((\left(\frac( 10)(3) \right))^(3)); \\& \frac(9)(100)=\frac(((3)^(2)))(((10)^(3)))=((\left(\frac(3)(10) \right))^(2)). \\\end(align)\]
Thus, our equation will be rewritten as follows:
\[((\left(((\left(\frac(10)(3) \right))^(3)) \right))^(x-1))=((\left(\frac(3 )(10) \right))^(2))\]
\[((\left(((\left(\frac(10)(3) \right))^(3)) \right))^(x-1))=((\left(\frac(10 )(3) \right))^(3\left(x-1 \right)))=((\left(\frac(10)(3) \right))^(3x-3))\]
At the same time, on the right, you can also get a degree with the same base, for which it is enough just to “flip” the fraction:
\[((\left(\frac(3)(10) \right))^(2))=((\left(\frac(10)(3) \right))^(-2))\]
Finally, our equation will take the form:
\[\begin(align)& ((\left(\frac(10)(3) \right))^(3x-3))=((\left(\frac(10)(3) \right)) ^(-2)); \\& 3x-3=-2; \\& 3x=1; \\& x=\frac(1)(3). \\\end(align)\]
That's the whole solution. Its main idea boils down to the fact that even with different reasons, we try by hook or by crook to reduce these reasons to the same one. In this we are helped by elementary transformations of equations and the rules for working with powers.
But what rules and when to use? How to understand that in one equation you need to divide both sides by something, and in another - to decompose the base of the exponential function into factors?
The answer to this question will come with experience. Try your hand at first on simple equations, and then gradually complicate the tasks - and very soon your skills will be enough to solve any exponential equation from the same USE or any independent / test work.
And to help you in this difficult task, I suggest downloading a set of equations on my website for an independent solution. All equations have answers, so you can always check yourself.
In this lesson, we will consider the solution of more complex exponential equations, recall the main theoretical provisions regarding the exponential function.
1. Definition and properties of an exponential function, a technique for solving the simplest exponential equations
Recall the definition and main properties of an exponential function. It is on the properties that the solution of all exponential equations and inequalities is based.
Exponential function is a function of the form , where the base is the degree and Here x is an independent variable, an argument; y - dependent variable, function.
Rice. 1. Graph of the exponential function
The graph shows an increasing and decreasing exponent, illustrating the exponential function at 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 as , decreases as .
A monotonic function takes each of its values with a single value of the argument.
When the argument increases from minus to plus infinity, the function increases from zero, inclusive, to plus infinity. On the contrary, when the argument increases from minus to plus infinity, the function decreases from infinity to zero, inclusive.
2. Solution of typical exponential equations
Recall how to solve the simplest exponential equations. Their solution is based on the monotonicity of the exponential function. Almost all complex exponential equations are reduced to such equations.
The equality of exponents with equal bases is due to the property of the exponential function, namely its monotonicity.
Solution Method:
Equalize the bases of the degrees;
Equate exponents.
Let's move on to more complex exponential equations, our goal is to reduce each of them to the simplest.
Let's get rid of the root on the left side and reduce the degrees to the same base:
In order to reduce a complex exponential equation to a simple one, a change of variables is often used.
Let's use the degree property:
We introduce a replacement. Let then 
We multiply the resulting equation by two and transfer all the terms to the left side:
The first root does not satisfy the interval of y values, we discard it. We get:
Let's bring the degrees to the same indicator:
We introduce a replacement:
Let then 
. With this replacement, it is obvious that y takes strictly positive values. We get:
We know how to solve similar quadratic equations, we write out the answer:
To make sure that the roots are found correctly, you can check according to the Vieta theorem, that is, find the sum of the roots and their product and check with the corresponding coefficients of the equation.
We get:
3. Technique for solving homogeneous exponential equations of the second degree
Let us study the following important type of exponential equations:
Equations of this type are called homogeneous of the second degree with respect to the functions f and g. On its left side there is a square trinomial with respect to f with parameter g or a square trinomial with respect to g with parameter f.
Solution Method:
This equation can be solved as a quadratic one, but it is easier to do it the other way around. Two cases should be considered:
In the first case, we get
In the second case, we have the right to divide by the highest degree and we get:
You should introduce a change of variables , we get a quadratic equation for y:
Note that the functions f and g can be arbitrary, but we are interested in the case when these are exponential functions.
4. Examples of solving homogeneous equations
Let's move all the terms to the left side of the equation:
Since the exponential functions acquire strictly positive values, we have the right to immediately divide the equation by , without considering the case when:
We get:
We introduce a replacement: 
(according to the properties of the exponential function)
We got a quadratic equation:
We determine the roots according to the Vieta theorem:
The first root does not satisfy the interval of y values, we discard it, we get:
Let's use the properties of the degree and reduce all degrees to simple bases:
It is easy to notice the functions f and g:
Since the exponential functions acquire strictly positive values, we have the right to immediately divide the equation by , without considering the case when .
