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How to solve an exponential equation. Exponential equations

In this article you will get acquainted with all types exponential equations and algorithms for solving them, learn to recognize what type it belongs to exponential equation, which you need to solve, and apply the appropriate method to solve it. Detailed solution of examples exponential equations You can watch each type in the corresponding VIDEO LESSONS.

An exponential equation is an equation in which the unknown is contained in an exponent.

Before you start solving an exponential equation, it is useful to do a few preliminary actions , which can significantly facilitate the process of solving it. These are the steps:

1. Divide all bases of powers into prime factors.

2. Present the roots as a degree.

3. Decimals imagine as ordinary ones.

4. Mixed numbers write as improper fractions.

You will realize the benefits of these actions in the process of solving equations.

Let's look at the main types exponential equations and algorithms for solving them.

1. Equation of the form

This equation is equivalent to the equation

Watch the solution to the equation in this VIDEO TUTORIAL this type.

2. Equation of the form

In equations of this type:

b) the coefficients for the unknown in the exponent are equal.

To solve this equation, you need to factor out the smallest factor.

An example of solving an equation of this type:

watch the VIDEO TUTORIAL.

3. Equation of the form

Equations of this type differ in that

a) all degrees have the same bases

b) the coefficients for the unknown in the exponent are different.

Equations of this type are solved using changes of variables. Before introducing a replacement, it is advisable to get rid of free terms in the exponent. (, , etc)

Watch the VIDEO TUTORIAL to solve this type of equation:

4. Homogeneous equations kind

Distinctive features of homogeneous equations:

a) all monomials have the same degree,

b) the free term is zero,

c) the equation contains powers with two different bases.

Homogeneous equations are solved using a similar algorithm.

To solve this type of equation, we divide both sides of the equation by (can be divided by or by)

Attention! When dividing the right and left sides of an equation by an expression containing an unknown, you can lose roots. Therefore, it is necessary to check whether the roots of the expression by which we divide both sides of the equation are the roots of the original equation.

In our case, since the expression is not zero for any value of the unknown, we can divide by it without fear. Let's divide the left side of the equation by this expression term by term. We get:

Let's reduce the numerator and denominator of the second and third fractions:

Let's introduce the replacement:

Moreover title="t>0">при всех допустимых значениях неизвестного.!}

We get quadratic equation:

Let's solve the quadratic equation, find the values that satisfy the condition title="t>0">, а затем вернемся к исходному неизвестному.!}

Watch the VIDEO TUTORIAL detailed solution homogeneous equation:

5. Equation of the form

When solving this equation, we will proceed from the fact that title="f(x)>0">!}

The initial equality is satisfied in two cases:

1. If, since 1 to any power is equal to 1,

2. If two conditions are met:

Title="delim(lbrace)(matrix(2)(1)((f(x)>0) (g(x)=h(x)) (x-8y+9z=0))) ( )">!}

Watch the VIDEO TUTORIAL for a detailed solution to the equation

Lecture: “Methods for solving exponential equations.”

1 . Exponential equations.

Equations containing unknowns in exponents are called exponential equations. The simplest of them is the equation ax = b, where a > 0, a ≠ 1.

1) At b< 0 и b = 0 это уравнение, согласно свойству 1 exponential function, has no solution.

2) For b > 0, using the monotonicity of the function and the root theorem, the equation has a unique root. In order to find it, b must be represented in the form b = aс, аx = bс ó x = c or x = logab.

Exponential equations through algebraic transformations lead to standard equation which are solved using the following methods:

1) method of reduction to one base;

2) assessment method;

3) graphic method;

4) method of introducing new variables;

5) factorization method;

6) exponential – power equations;

7) demonstrative with a parameter.

2 . Method of reduction to one base.

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., one must try to reduce the equation 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 let's move on 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 represent 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 logarithm, x = log35. Answer: log35.

6. 62x+4 = 33x. 2x+8.

Let's rewrite the equation in the form 32x+4.22x+4 = 32x.2x+8, i.e..png" width="181" height="49 src="> Hence x – 4 =0, x = 4. Answer: 4.

7 . 2∙3x+1 - 6∙3x-2 - 3x = 9. Using the properties of powers, we write the equation in the form 6∙3x - 2∙3x – 3x = 9 then 3∙3x = 9, 3x+1 = 32, i.e. i.e. x+1 = 2, x =1. Answer: 1.

Problem bank No. 1.

Solve the equation:

Test No. 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 No. 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 Evaluation method.

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 using 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 = 1, then 41+1 = 5, 5 = 5 is true, which means 1 is the root of the equation.

Function f(x) = 4x – increases on R, and g(x) = x – increases on R => h(x)= f(x)+g(x) increases on R, as the sum of increasing functions, then x = 1 is the only root of the equation 4x = 5 – x. Answer: 1.

Solution. Let's rewrite the equation in the form .

1. if x = -1, then , 3 = 3 is true, which means x = -1 is the root of the equation.

2. prove that he is the only one.

3. 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 . This means, according to the root theorem, x = -1 is the only root of the equation. Answer: -1.

Problem bank No. 2. Solve the equation

a) 4x + 1 =6 – x;

c) 2x – 2 =1 – x;

4. Method of introducing new variables.

The method is described in paragraph 2.1. The introduction of a new variable (substitution) is usually carried out after transformations (simplification) of the terms of the equation. Let's look at examples.

Examples. R Solve the 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:

Let's designate 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"> - irrational equation. We note that

The solution to the equation is x = 2.5 ≤ 4, which means 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, t..png" width="118" height="56">

The roots of the quadratic equation are t1 = 1 and t2<0, т. е..png" width="200" height="24">.

Solution . Let's 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

Let's replace https://pandia.ru/text/80/142/images/image049_0.png" width="16" height="41 src="> .

Answer: 0; 0.5.

Problem bank No. 3. Solve the equation

Test No. 3 with a choice of answers. Minimum level.

A1	1) -0.2;2 2) log52 3) –log52 4) 2
A2 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 No. 4 with a choice of answers. General level.

A1	1) 2;1 2) ½;0 3)2;0 4) 0
A2 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. Factorization method.

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's put 6x out of brackets 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.

Solution. Let's solve the equation using the factorization method.

Let us 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 No. 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.

Adjacent to exponential equations are 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 an exponential equation.

1..png" width="182" height="116 src=">

Solution. x2 +2x-8 – makes sense for any x, since it is a polynomial, which means the equation is equivalent to the totality

https://pandia.ru/text/80/142/images/image078_0.png" width="137" height="35">

7. Exponential equations with parameters.

1. For what values of the parameter p does equation 4 (5 – 3)2 +4p2–3p = 0 (1) have a unique solution?

Solution. Let us introduce the replacement 2x = t, t > 0, then equation (1) will take the form t2 – (5p – 3)t + 4p2 – 3p = 0. (2)

Discriminant of equation (2) 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 p1, then 9(p – 1)2 > 0, then equation (2) has two different roots t1 = p, t2 = 4p – 3. The conditions of the problem are satisfied by a set of systems

Substituting t1 and t2 into the systems, we have

https://pandia.ru/text/80/142/images/image084_0.png" alt="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="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="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="no35_17" width="267" height="63">!}

Thus, for a 0, equation (4) has a single positive root . Then equation (3) has a unique solution

When 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 perfect square; Thus, the roots of equation (2) were immediately calculated using the formula for the roots of a quadratic equation, and then conclusions were drawn regarding these roots. Equation (3) has been 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 quadratic trinomial and a graphical model. Note that equation (4) can be solved using Vieta's theorem.

Let's solve more complex equations.

Problem 3: Solve the equation

Solution. ODZ: x1, x2.

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="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="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="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.

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2. Guzeev technology: from reception to philosophy.

M. “School Director” No. 4, 1996

3. Guzeev and organizational forms of training.

4. Guzeev and the practice of integral educational technology.

M. " Public education", 2001

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Mathematics at school No. 2, 1987 pp. 9 – 11.

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M. “Public Education”, 1998

7. Episheva schoolchildren to study mathematics.

M. "Enlightenment", 1990

8. Ivanova prepare lessons - workshops.

Mathematics at school No. 6, 1990 p. 37 – 40.

9. Smirnov’s 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 skills schoolchildren.

Mathematics at school No. 2, 1989 p. 10.

13. Scanavi. Publisher, 1997

14. and others. Algebra and the beginnings of analysis. Didactic materials For

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entering universities. “A S T - press school”, 2002

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Minsk and Russian Federation “Review”, 1996

18. Written D. We are preparing for the exam in mathematics. M. Rolf, 1999

19. etc. Learning to solve equations and inequalities.

M. "Intellect - Center", 2003

20. etc. Educational and training materials for preparing for the EGE.

M. "Intelligence - Center", 2003 and 2004.

21 and others. CMM options. 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.

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Go to the youtube channel of our website to stay up to date with all the new video lessons.

First, let's remember the basic formulas of powers and their properties.

Product of a number a occurs on itself n times, we can write this expression as a a … a=a n

1. a 0 = 1 (a ≠ 0)

3. a n a m = a n + m

4. (a n) m = a nm

5. a n b n = (ab) n

7. a n / a m = a n - m

Power or exponential equations– these are equations in which the variables are in powers (or exponents), and the base is a number.

Examples of exponential equations:

IN in this example the number 6 is the base, it is always at the bottom, and the variable x degree or indicator.

Let us give more examples of exponential equations.
2 x *5=10
16 x - 4 x - 6=0

Now let's look at how exponential equations are solved?

Let's take a simple equation:

2 x = 2 3

This example can be solved even in your head. It can be seen that x=3. After all, in order for the left and right sides to be equal, you need to put the number 3 instead of x.
Now let’s see how to formalize this decision:

2 x = 2 3
x = 3

In order to solve such an equation, we removed identical grounds(that is, twos) and wrote down what was left, these are degrees. We got the answer we were looking for.

Now let's summarize our decision.

Algorithm for solving the exponential equation:
1. Need to check the same whether the equation has bases on the right and left. If the reasons are not the same, we are looking for options to solve this example.
2. After the bases become the same, equate degrees and solve the resulting new equation.

Now let's look at a few examples:

Let's start with something simple.

The bases on the left and right sides are equal to the number 2, which means we can discard the base and equate their degrees.

x+2=4 The simplest equation is obtained.
x=4 – 2
x=2
Answer: x=2

IN following example It can be seen that the bases are different: 3 and 9.

3 3x - 9 x+8 = 0

First, move the nine to the right side, we get:

Now you need to make the same bases. We know that 9=3 2. Let's use the power formula (a n) m = a nm.

3 3x = (3 2) x+8

We get 9 x+8 =(3 2) x+8 =3 2x+16

3 3x = 3 2x+16 Now it is clear that on the left and right sides the bases are the same and equal to three, which means we can discard them and equate the degrees.

3x=2x+16 we get the simplest equation
3x - 2x=16
x=16
Answer: x=16.

Let's look at the following example:

2 2x+4 - 10 4 x = 2 4

First of all, we look at the bases, bases two and four. And we need them to be the same. We transform the four using the formula (a n) m = a nm.

4 x = (2 2) x = 2 2x

And we also use one formula a n a m = a n + m:

2 2x+4 = 2 2x 2 4

Add to the equation:

2 2x 2 4 - 10 2 2x = 24

We gave an example for the same reasons. But other numbers 10 and 24 bother us. What to do with them? If you look closely you can see that on the left side we have 2 2x repeated, here is the answer - we can put 2 2x out of brackets:

2 2x (2 4 - 10) = 24

Let's calculate the expression in brackets:

2 4 — 10 = 16 — 10 = 6

We divide the entire equation by 6:

Let's imagine 4=2 2:

2 2x = 2 2 bases are the same, we discard them and equate the degrees.
2x = 2 is the simplest equation. Divide it by 2 and we get
x = 1
Answer: x = 1.

Let's solve the equation:

9 x – 12*3 x +27= 0

Let's convert:
9 x = (3 2) x = 3 2x

We get the equation:
3 2x - 12 3 x +27 = 0

Our bases are the same, equal to three. In this example, you can see that the first three has a degree twice (2x) than the second (just x). In this case, you can solve replacement method. We replace the number with the smallest degree:

Then 3 2x = (3 x) 2 = t 2

We replace all x powers in the equation with t:

t 2 - 12t+27 = 0
We get a quadratic equation. Solving through the discriminant, we get:
D=144-108=36
t 1 = 9
t2 = 3

Returning to the variable x.

Take t 1:
t 1 = 9 = 3 x

That is,

3 x = 9
3 x = 3 2
x 1 = 2

One root was found. We are looking for the second one from t 2:
t 2 = 3 = 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 = 2; x 2 = 1.

On the website you can ask any questions you may have in the HELP DECIDE section, we will definitely answer you.

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Exponential equations. As you know, the Unified State Examination includes simple equations. We have already considered some - these are logarithmic, trigonometric, rational. Here are the exponential equations.

In a recent article we worked with exponential expressions, it will be useful. The equations themselves are solved simply and quickly. You just need to know the properties of exponents and... About thisFurther.

Let us list the properties of exponents:

The zero power of any number is equal to one.

A corollary from this property:

A little more theory.

An exponential equation is an equation containing a variable in the exponent, that is, it is an equation of the form:

f(x) expression that contains a variable

Methods for solving exponential equations

1. As a result of transformations, the equation can be reduced to the form:

Then we apply the property:

2. Upon obtaining an equation of the form a f (x) = b using the definition of logarithm, we get:

3. As a result of transformations, you can obtain an equation of the form:

Logarithm applied:

Express and find x.

In tasks Unified State Exam options It will be enough to use the first method.

That is, it is necessary to represent the left and right sides in the form of powers with the same base, and then we equate the exponents and solve the usual linear equation.

Consider the equations:

Find the root of equation 4 1–2x = 64.

It is necessary to make sure that in the left and right parts there were demonstrative expressions with one base. We can represent 64 as 4 to the power of 3. We get:

4 1–2x = 4 3

1 – 2x = 3

– 2x = 2

x = – 1

Examination:

4 1–2 (–1) = 64

4 1 + 2 = 64

4 3 = 64

64 = 64

Answer: –1

Find the root of equation 3 x–18 = 1/9.

It is known that

So 3 x-18 = 3 -2

The bases are equal, we can equate the indicators:

x – 18 = – 2

x = 16

Examination:

3 16–18 = 1/9

3 –2 = 1/9

1/9 = 1/9

Answer: 16

Find the root of the equation:

Let's represent the fraction 1/64 as one-fourth to the third power:

2x – 19 = 3

2x = 22

x = 11

Examination:

Answer: 11

Find the root of the equation:

Let's imagine 1/3 as 3 –1, and 9 as 3 squared, we get:

(3 –1) 8–2x = 3 2

3 –1∙(8–2x) = 3 2

3 –8+2x = 3 2

Now we can equate the indicators:

– 8+2x = 2

2x = 10

x = 5

Examination:

Answer: 5

26654. Find the root of the equation:

Solution:

Answer: 8.75

Indeed, no matter to what degree we raise positive number a, we can't get a negative number in any way.

Any exponential equation after appropriate transformations is reduced to solving one or more simple ones.In this section we will also look at solving some equations, don’t miss it!That's all. Good luck to you!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell me about the site on social networks.

Equipment:

computer,
multimedia projector,
screen,
Annex 1(PowerPoint slide presentation) “Methods for solving exponential equations”
Appendix 2(Solving an equation like “Three different bases degrees” in Word)
Appendix 3(handout in Word for practical work).
Appendix 4(handout in Word for homework).

During the classes

1. Organizational stage

message of the lesson topic (written on the board),
the need for a general lesson in grades 10-11:

The stage of preparing students for active learning

Repetition

Definition.

An exponential equation is an equation containing a variable with an exponent (student answers).

Teacher's note. Exponential equations belong to the class of transcendental equations. This unpronounceable name suggests that such equations, generally speaking, cannot be solved in the form of formulas.

They can only be solved approximately by numerical methods on computers. But what about exam tasks? The trick is that the examiner frames the problem in such a way that it allows for an analytical solution. In other words, you can (and should!) do the following identity transformations, which reduce this exponential equation to the simplest exponential equation. This simplest equation is called: the simplest exponential equation. It's being resolved by logarithm.

The situation with solving an exponential equation is reminiscent of traveling through a labyrinth, which was specially invented by the author of the problem. Of these very general reasoning Very specific recommendations follow.

To successfully solve exponential equations you must:

1. Not only actively know all the exponential identities, but also find the sets of variable values on which these identities are defined, so that when using these identities you do not acquire unnecessary roots, and even more so, do not lose solutions to the equation.

2. Actively know all exponential identities.

3. Clearly, in detail and without errors, carry out mathematical transformations of equations (transfer terms from one part of the equation to another, not forgetting to change the sign, bring fractions to a common denominator, etc.). This is called mathematical culture. At the same time, the calculations themselves should be done automatically by hand, and the head should think about the general guiding thread of the solution. Transformations must be made as carefully and in detail as possible. Only this will guarantee a correct, error-free decision. And remember: small arithmetic error may simply create a transcendental equation, which in principle cannot be solved analytically. It turns out that you have lost your way and have hit the wall of the labyrinth.

4. Know methods for solving problems (that is, know all the paths through the solution maze). To navigate correctly at each stage, you will have to (consciously or intuitively!):

define equation type;
remember the corresponding type solution method tasks.

The stage of generalization and systematization of the studied material.

The teacher, together with students using a computer, conducts a review of all types of exponential equations and methods for solving them, compiles general scheme. (Used training computer program L.Ya. Borevsky "Mathematics Course - 2000", the author of the PowerPoint presentation is T.N. Kuptsova.)

Rice. 1. The figure shows a general diagram of all types of exponential equations.

As can be seen from this diagram, the strategy for solving exponential equations is to reduce the given exponential equation to the equation, first of all, with the same bases of degrees , and then – and with the same degree indicators.

Having received an equation with the same bases and exponents, you replace this exponent with a new variable and get a simple algebraic equation (usually fractional-rational or quadratic) with respect to this new variable.

Having solved this equation and made the reverse substitution, you end up with a set of simple exponential equations that can be solved in general view using logarithm.

Equations in which only products of (partial) powers are found stand out. Using exponential identities, it is possible to reduce these equations immediately to one base, in particular, to the simplest exponential equation.

Let's look at how to solve an exponential equation with three different bases.

(If the teacher has the educational computer program by L.Ya. Borevsky “Course of Mathematics - 2000”, then naturally we work with the disk, if not, you can make a printout of this type of equation from it for each desk, presented below.)