Examples:

\(4^x=32\)
\(5^(2x-1)-5^(2x-3)=4,8\)
\((\sqrt(7))^(2x+2)-50\cdot(\sqrt(7))^(x)+7=0\)

How to solve exponential equations

When solving any exponential equation, we strive to bring it to the form \(a ^ (f (x)) \u003d a ^ (g (x)) \), and then make the transition to equality of indicators, that is:

\(a^(f(x))=a^(g(x))\) \(⇔\) \(f(x)=g(x)\)

For example:\(2^(x+1)=2^2\) \(⇔\) \(x+1=2\)

Important! From the same logic, two requirements follow for such a transition:
- number in left and right should be the same;
- degrees left and right must be "pure", that is, there should not be any, multiplications, divisions, etc.

For example:

To bring the equation to the form \(a^(f(x))=a^(g(x))\) and are used.

Example . Solve the exponential equation \(\sqrt(27) 3^(x-1)=((\frac(1)(3)))^(2x)\)
Solution:

\(\sqrt(27) 3^(x-1)=((\frac(1)(3)))^(2x)\)		We know that \(27 = 3^3\). With this in mind, we transform the equation.
\(\sqrt(3^3) 3^(x-1)=((\frac(1)(3)))^(2x)\)		By the property of the root \(\sqrt[n](a)=a^(\frac(1)(n))\) we get that \(\sqrt(3^3)=((3^3))^( \frac(1)(2))\). Further, using the degree property \((a^b)^c=a^(bc)\), we obtain \(((3^3))^(\frac(1)(2))=3^(3 \ cdot \frac(1)(2))=3^(\frac(3)(2))\).
\(3^(\frac(3)(2))\cdot 3^(x-1)=(\frac(1)(3))^(2x)\)		We also know that \(a^b a^c=a^(b+c)\). Applying this to the left side, we get: \(3^(\frac(3)(2)) 3^(x-1)=3^(\frac(3)(2)+ x-1)=3^ (1.5 + x-1)=3^(x+0.5)\).
\(3^(x+0,5)=(\frac(1)(3))^(2x)\)		Now remember that: \(a^(-n)=\frac(1)(a^n)\). This formula can also be used in reverse side: \(\frac(1)(a^n) =a^(-n)\). Then \(\frac(1)(3)=\frac(1)(3^1) =3^(-1)\).
\(3^(x+0.5)=(3^(-1))^(2x)\)		Applying the property \((a^b)^c=a^(bc)\) to the right side, we get: \((3^(-1))^(2x)=3^((-1) 2x) =3^(-2x)\).
\(3^(x+0.5)=3^(-2x)\)		And now we have the bases equal and there are no interfering coefficients, etc. So we can make the transition.
		Example . Solve the exponential equation \(4^(x+0.5)-5 2^x+2=0\) Solution: \(4^(x+0,5)-5 2^x+2=0\) We again use the degree property \(a^b \cdot a^c=a^(b+c)\) in reverse direction. \(4^x 4^(0,5)-5 2^x+2=0\) Now remember that \(4=2^2\). \((2^2)^x (2^2)^(0,5)-5 2^x+2=0\) Using the properties of the degree, we transform: \((2^2)^x=2^(2x)=2^(x 2)=(2^x)^2\) \((2^2)^(0.5)=2^(2 0.5)=2^1=2.\) \(2 (2^x)^2-5 2^x+2=0\) We look carefully at the equation, and we see that the replacement \(t=2^x\) suggests itself here. \(t_1=2\) \(t_2=\frac(1)(2)\) However, we found the values ​​\(t\), and we need \(x\). We return to the X, making the reverse substitution. \(2^x=2\) \(2^x=\frac(1)(2)\) We transform the second equation using the property negative degree… \(2^x=2^1\) \(2^x=2^(-1)\) ...and solve until the answer. \(x_1=1\) \(x_2=-1\) Answer : \(-1; 1\). The question remains - how to understand when to apply which method? It comes with experience. In the meantime, you have not earned it, use general recommendation for solutions challenging tasks“If you don’t know what to do, do what you can.” That is, look for how you can transform the equation in principle, and try to do it - what if it comes out? The main thing is to do only mathematically justified transformations. exponential equations without solutions Let's look at two more situations that often baffle students: - positive number equals zero to the power, for example, \(2^x=0\); - positive number to the power equals negative number, for example, \(2^x=-4\). Let's try to solve it by brute force. If x is a positive number, then as x grows, the entire power \(2^x\) will only grow: \(x=1\); \(2^1=2\) \(x=2\); \(2^2=4\) \(x=3\); \(2^3=8\). \(x=0\); \(2^0=1\) Also past. There are negative x's. Remembering the property \(a^(-n)=\frac(1)(a^n)\), we check: \(x=-1\); \(2^(-1)=\frac(1)(2^1) =\frac(1)(2)\) \(x=-2\); \(2^(-2)=\frac(1)(2^2) =\frac(1)(4)\) \(x=-3\); \(2^(-3)=\frac(1)(2^3) =\frac(1)(8)\) Despite the fact that the number becomes smaller with each step, it will never reach zero. So the negative degree did not save us either. We come to a logical conclusion: A positive number to any power will remain a positive number. Thus, both equations above have no solutions. exponential equations with different bases In practice, sometimes there are exponential equations with different bases that are not reducible to each other, and at the same time with the same exponents. They look like this: \(a^(f(x))=b^(f(x))\), where \(a\) and \(b\) are positive numbers. For example: \(7^(x)=11^(x)\) \(5^(x+2)=3^(x+2)\) \(15^(2x-1)=(\frac(1)(7))^(2x-1)\) Such equations can be easily solved by dividing by any of the parts of the equation (usually dividing by the right side, that is, by \ (b ^ (f (x)) \). You can divide in this way, because a positive number is positive to any degree (that is, we do not divide by zero.) We get: \(\frac(a^(f(x)))(b^(f(x)))\) \(=1\) Example . Solve the exponential equation \(5^(x+7)=3^(x+7)\) Solution: \(5^(x+7)=3^(x+7)\) Here we can’t turn a five into a three, or vice versa (according to at least, without use ). So we cannot come to the form \(a^(f(x))=a^(g(x))\). At the same time, the indicators are the same. Let's divide the equation by the right side, that is, by \(3^(x+7)\) (we can do this, because we know that the triple will not be zero in any degree). \(\frac(5^(x+7))(3^(x+7))\) \(=\)\(\frac(3^(x+7))(3^(x+7) )\) Now remember the property \((\frac(a)(b))^c=\frac(a^c)(b^c)\) and use it from the left in the opposite direction. On the right, we simply reduce the fraction. \((\frac(5)(3))^(x+7)\) \(=1\) It didn't seem to get any better. But remember another property of the degree: \(a^0=1\), in other words: "any number in zero degree equals \(1\)". The converse is also true: "a unit can be represented as any number raised to the power of zero." We use this by making the base on the right the same as the one on the left. \((\frac(5)(3))^(x+7)\) \(=\) \((\frac(5)(3))^0\) Voila! We get rid of the foundations. We write the answer. Answer : \(-7\). Sometimes the "sameness" of the exponents is not obvious, but the skillful use of the properties of the degree solves this issue. Example . Solve the exponential equation \(7^( 2x-4)=(\frac(1)(3))^(-x+2)\) Solution: \(7^( 2x-4)=(\frac(1)(3))^(-x+2)\) The equation looks very sad ... Not only that, the bases cannot be reduced to the same number(the seven will not be equal to \(\frac(1)(3)\)), so also the indicators are different ... However, let's have a deuce in the indicator of the left degree. \(7^( 2(x-2))=(\frac(1)(3))^(-x+2)\) Keeping in mind the property \((a^b)^c=a^(b c)\) , transform on the left: \(7^(2(x-2))=7^(2 (x-2))=(7^2)^(x-2)=49^(x-2)\). \(49^(x-2)=(\frac(1)(3))^(-x+2)\) Now, remembering the negative power property \(a^(-n)=\frac(1)(a)^n\), we transform on the right: \((\frac(1)(3))^(-x+2) =(3^(-1))^(-x+2)=3^(-1(-x+2))=3^(x-2)\) \(49^(x-2)=3^(x-2)\) Hallelujah! The scores are the same! Acting according to the scheme already familiar to us, we decide before the answer. Answer : \(2\).

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 exponential function, has no solution.

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.

exponential equations by algebraic transformations lead to standard equation, which are solved using the following methods:

1) method of reduction to one base;

2) evaluation method;

3) graphic method;

4) the method of introducing new variables;

5) factorization method;

6) indicative - 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"> - irrational equation. We 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 only decision?

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 p1, 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's find the values 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 !} square trinomial 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 decision, 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: 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="(!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 learning.

4. Guzeev and the practice of integral educational technology.

M. " public 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 skills 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 - training materials to prepare 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 learning at school.

26. Liimets work at the lesson. M. Knowledge, 1975

First level

exponential equations. Comprehensive guide (2019)

Hello! Today we will discuss with you how to solve equations that can be both elementary (and I hope that after reading this article, almost all of them will be so for you), and those that are usually given "backfill". Apparently, to fall asleep completely. But I will try to do my best so that now you do not get into trouble when faced with this type of equation. I will no longer beat around the bush, but I will immediately open little secret: today we will work exponential equations.

Before proceeding to an analysis of the ways to solve them, I will immediately outline for you a circle of questions (quite a small one) that you should repeat before you rush to storm this topic. So, to get best result, please, repeat:

1. properties and
2. Solution and Equations

Repeated? Wonderful! Then it will not be difficult for you to notice that the root of the equation is a number. Are you sure you understand how I did it? Truth? Then we continue. Now answer me the question, what is equal to the third power? You're absolutely right: . Eight is what power of two? That's right - the third! Because. Well, now let's try to solve the following problem: Let me multiply the number by itself once and get the result. The question is, how many times have I multiplied by itself? You can of course check this directly:

\begin(align) & 2=2 \\ & 2\cdot 2=4 \\ & 2\cdot 2\cdot 2=8 \\ & 2\cdot 2\cdot 2\cdot 2=16 \\ \end( align)

Then you can conclude that I multiplied times by itself. How else can this be verified? And here's how: directly by the definition of the degree: . But, you must admit, if I asked how many times two must be multiplied by itself in order to get, say, you would tell me: I will not fool myself and multiply by myself until I'm blue in the face. And he would be absolutely right. Because how can you write down all actions briefly(and brevity is the sister of talent)

where - this is the very "times" when you multiply by itself.

I think that you know (and if you don’t know, urgently, very urgently repeat the degrees!) that then my problem will be written in the form:

How can you reasonably conclude that:

So, quietly, I wrote down the simplest exponential equation:

And even found it root. Don't you think that everything is quite trivial? That's exactly what I think too. Here's another example for you:

But what to do? After all, it cannot be written as a degree of a (reasonable) number. Let's not despair and note that both of these numbers are perfectly expressed in terms of the power of the same number. What? Right: . Then the original equation is transformed to the form:

From where, as you already understood, . Let's not pull anymore and write down definition:

In our case with you: .

These equations are solved by reducing them to the form:

with subsequent solution of the equation

We, in fact, did this in the previous example: we got that. And we solved the simplest equation with you.

It seems to be nothing complicated, right? Let's practice on the simplest first. examples:

We again see that the right and left sides of the equation must be represented as a power of one number. True, this has already been done on the left, but on the right there is a number. But, it's okay, after all, and my equation miraculously will be transformed into this:

What did I have to do here? What rule? Power to Power Rule which reads:

What if:

Before answering this question, let's fill in the following table with you:

It is not difficult for us to notice that the less, the less value, but nonetheless, all of these values ​​are greater than zero. AND IT WILL ALWAYS BE SO!!! The same property is true FOR ANY BASE WITH ANY INDEX!! (for any and). Then what can we conclude about the equation? And here's one: it has no roots! Just like any equation has no roots. Now let's practice and Let's solve some simple examples:

Let's check:

1. Nothing is required of you here, except for knowing the properties of powers (which, by the way, I asked you to repeat!) As a rule, everything leads to the smallest base: , . Then the original equation will be equivalent to the following: All I need is to use the properties of powers: when multiplying numbers with the same base, the exponents are added, and when dividing, they are subtracted. Then I will get: Well, now with a clear conscience I will move from the exponential equation to the linear one: \begin(align)
& 2x+1+2(x+2)-3x=5 \\
& 2x+1+2x+4-3x=5 \\
&x=0. \\
\end(align)

2. In the second example, you need to be more careful: the trouble is that on the left side, we will not be able to represent the same number as a power. In this case it is sometimes useful represent numbers as a product of powers with different bases, but the same exponents:

The left side of the equation will take the form: What did this give us? And here's what: Numbers with different bases but the same exponent can be multiplied.In this case, the bases are multiplied, but the exponent does not change:

Applied to my situation, this will give:

\begin(align)
& 4\cdot ((64)^(x))((25)^(x))=6400, \\
& 4\cdot (((64\cdot 25))^(x))=6400, \\
& ((1600)^(x))=\frac(6400)(4), \\
& ((1600)^(x))=1600, \\
&x=1. \\
\end(align)

Not bad, right?

3. I don’t like it when I have two terms on one side of the equation, and none on the other (sometimes, of course, this is justified, but this is not the case now). Move the minus term to the right:

Now, as before, I will write everything through the powers of the triple:

I add the powers on the left and get an equivalent equation

You can easily find its root:

4. As in example three, the term with a minus - a place on the right side!

On the left, almost everything is fine with me, except for what? Yes, the “wrong degree” of the deuce bothers me. But I can easily fix this by writing: . Eureka - on the left, all bases are different, but all degrees are the same! We multiply quickly!

Here again, everything is clear: (if you didn’t understand how magically I got the last equality, take a break for a minute, take a break and read the properties of the degree again very carefully. Who said that you can skip the degree with negative indicator? Well, here I am about the same thing that no one). Now I will get:

\begin(align)
& ((2)^(4\left((x) -9 \right)))=((2)^(-1)) \\
&4((x) -9)=-1 \\
&x=\frac(35)(4). \\
\end(align)

Here are the tasks for you to practice, to which I will only give the answers (but in a “mixed” form). Solve them, check, and we will continue our research!

Ready? Answers like these ones:

1. any number

Okay, okay, I was joking! Here are the outlines of the solutions (some are quite brief!)

Don't you think it's no coincidence that one fraction on the left is an "inverted" other? It would be a sin not to use this:

This rule is very often used when solving exponential equations, remember it well!

Then the original equation becomes:

Solving it quadratic equation, you will get the following roots:

2. Another solution: dividing both parts of the equation by the expression on the left (or right). I will divide by what is on the right, then I will get:

Where (why?!)

3. I don’t even want to repeat myself, everything has already been “chewed” so much.

4. equivalent to a quadratic equation, the roots

5. You need to use the formula given in the first task, then you will get that:

The equation has turned into a trivial identity, which is true for any. Then the answer is any real number.

Well, here you are and practiced to decide the simplest exponential equations. Now I want to give you some life examples, which will help you understand why they are needed in principle. Here I will give two examples. One of them is quite everyday, but the other is more of scientific than practical interest.

Example 1 (mercantile) Let you have rubles, but you want to turn it into rubles. The bank offers you to take this money from you at an annual interest rate with a monthly capitalization of interest (monthly accrual). The question is, for how many months do you need to open a deposit in order to collect the desired final amount? Quite a mundane task, isn't it? Nevertheless, its solution is connected with the construction of the corresponding exponential equation: Let - the initial amount, - the final amount, - interest rate per period, - the number of periods. Then:

In our case (if the rate is per annum, then it is calculated per month). Why is it divided into? If you do not know the answer to this question, remember the topic ""! Then we get the following equation:

This exponential equation can already be solved only with a calculator (its appearance hints at this, and this requires knowledge of logarithms, which we will get acquainted with a little later), which I will do: ... Thus, in order to receive a million, we will need to make a deposit for a month (not very fast, right?).

Example 2 (rather scientific). Despite his, some "isolation", I recommend that you pay attention to him: he regularly "slips into the exam!! (task taken from the "real" version) During the collapse radioactive isotope its mass decreases according to the law, where (mg) is the initial mass of the isotope, (min) is the time elapsed from the initial moment, (min) is the half-life. AT initial moment time isotope mass mg. Its half-life is min. In how many minutes will the mass of the isotope be equal to mg? It's okay: we just take and substitute all the data in the formula proposed to us:

Let's divide both parts by, "in the hope" that on the left we get something digestible:

Well, we are very lucky! It stands on the left, then let's move on to the equivalent equation:

Where min.

As you can see, exponential equations have a very real application in practice. Now I want to discuss with you another (simple) way to solve exponential equations, which is based on taking the common factor out of brackets and then grouping the terms. Do not be afraid of my words, you have already encountered this method in the 7th grade when you studied polynomials. For example, if you needed to factorize the expression:

Let's group: the first and third terms, as well as the second and fourth. It is clear that the first and third are the difference of the squares:

and the second and fourth have common factor top three:

Then the original expression is equivalent to this:

Where to take out the common factor is no longer difficult:

Consequently,

This is approximately how we will act when solving exponential equations: look for “commonality” among the terms and take it out of the brackets, and then - come what may, I believe that we will be lucky =)) For example:

On the right is far from the power of seven (I checked!) And on the left - a little better, you can, of course, “chop off” the factor a from the first term and from the second, and then deal with what you have received, but let's do more prudently with you. I don't want to deal with the fractions that are inevitably produced by "selection", so shouldn't I be better off enduring? Then I won’t have fractions: as they say, both the wolves are full and the sheep are safe:

Count the expression in brackets. Magically, magically, it turns out that (surprisingly, although what else can we expect?).

Then we reduce both sides of the equation by this factor. We get: where.

Here is a more complicated example (quite a bit, really):

Here's the trouble! We don't have one here common ground! It's not entirely clear what to do now. And let's do what we can: firstly, we will move the “fours” in one direction, and the “fives” in the other:

Now let's take out the "common" on the left and right:

So what now? What is the benefit of such a stupid grouping? At first glance, it is not visible at all, but let's look deeper:

Well, now let's make it so that on the left we have only the expression c, and on the right - everything else. How can we do it? And here's how: Divide both sides of the equation first by (so we get rid of the exponent on the right), and then divide both sides by (so we get rid of the numerical factor on the left). Finally we get:

Incredible! On the left we have an expression, and on the right - just. Then we immediately conclude that

Here's another example to reinforce:

I will bring him short solution(not really bothering to explain), try to figure out all the “subtleties” of the solution yourself.

Now the final consolidation of the material covered. Try to solve the following problems on your own. I will only bring brief recommendations and tips for solving them:

1. Let's take the common factor out of brackets:
2. We represent the first expression in the form: , divide both parts by and get that
3. , then the original equation is converted to the form: Well, now a hint - look for where you and I have already solved this equation!
4. Imagine how, how, ah, well, then divide both parts by, so you get the simplest exponential equation.
5. Take it out of brackets.
6. Take it out of brackets.

EXPOSITIONAL EQUATIONS. AVERAGE LEVEL

I assume that after reading the first article, which told what are exponential equations and how to solve them you have mastered necessary minimum knowledge needed to solve simple examples.

Now I will analyze another method for solving exponential equations, this is

"method of introducing a new variable" (or substitution). He solves most of the "difficult" problems, on the topic of exponential equations (and not only equations). This method is one of the most commonly used in practice. First, I recommend that you familiarize yourself with the topic.

As you already understood from the name, the essence of this method is to introduce such a change of variable that your exponential equation will miraculously transform into one that you can already easily solve. All that remains for you after solving this very “simplified equation” is to make a “reverse replacement”: that is, to return from the replaced to the replaced. Let's illustrate what we just said with a very simple example:

Example 1:

This equation is solved by a "simple substitution," as mathematicians disparagingly call it. Indeed, the substitution here is the most obvious. It just needs to be seen that

Then the original equation becomes:

If we additionally imagine how, then it is quite clear what needs to be replaced: of course, . What then becomes the original equation? And here's what:

You can easily find its roots on your own:. What should we do now? It's time to return to the original variable. What did I forget to include? Namely: when replacing a certain degree with a new variable (that is, when replacing a type), I will be interested in only positive roots! You yourself can easily answer why. Thus, we are not interested in you, but the second root is quite suitable for us:

Then where.

Answer:

As you can see, in the previous example, the replacement was asking for our hands. Unfortunately, this is not always the case. However, let's not go straight to the sad, but practice on one more example with a fairly simple replacement

Example 2

It is clear that most likely it will be necessary to replace (this is the smallest of the powers included in our equation), however, before introducing a replacement, our equation needs to be “prepared” for it, namely: , . Then you can replace, as a result I will get the following expression:

Oh God: cubic equation with absolutely terrible formulas for solving it (well, speaking in general terms). But let's not immediately despair, but think about what we should do. I'll suggest cheating: we know that in order to get a "beautiful" answer, we need to get some power of three (why would that be, huh?). And let's try to guess at least one root of our equation (I'll start guessing from the powers of three).

First guess. Is not a root. Alas and ah...

.
The left side is equal.
Right part: !
There is! Guessed the first root. Now things will get easier!

Do you know about the "corner" division scheme? Of course you know, you use it when you divide one number by another. But few people know that the same can be done with polynomials. There is one wonderful theorem:

Applicable to my situation it tells me what is divisible without a remainder by. How is division carried out? That's how:

I look at which monomial I should multiply to get Clear on, then:

I subtract the resulting expression from, I get:

Now, what do I need to multiply to get? It is clear that on, then I will get:

and again subtract the resulting expression from the remaining one:

Well, the last step, I multiply by, and subtract from the remaining expression:

Hooray, the division is over! What have we accumulated in private? By itself: .

Then we got the following expansion of the original polynomial:

Let's solve the second equation:

It has roots:

Then the original equation:

has three roots:

We, of course, discard the last root, since it less than zero. And the first two after the reverse replacement will give us two roots:

Answer: ..

By this example, I did not at all want to frighten you; rather, I set out to show that at least we had enough simple replacement, nevertheless it led to quite complex equation, the solution of which required some special skills from us. Well, no one is immune from this. But the replacement in this case was pretty obvious.

Here's an example with a slightly less obvious substitution:

It is not at all clear what we should do: the problem is that in our equation there are two different bases and one foundation is not obtained from another by raising it to any (reasonable, naturally) degree. However, what do we see? Both bases differ only in sign, and their product is the difference of squares equal to one:

Definition:

Thus, the numbers that are bases in our example are conjugate.

In that case, the smart move would be multiply both sides of the equation by the conjugate number.

For example, on, then the left side of the equation will become equal, and the right side. If we make a replacement, then our original equation with you will become like this:

its roots, then, but remembering that, we get that.

Answer: , .

As a rule, the replacement method is enough to solve most of the "school" exponential equations. The following tasks are taken from the USE C1 ( elevated level difficulties). You are already literate enough to solve these examples on your own. I will only give the required replacement.

1. Solve the equation:
2. Find the roots of the equation:
3. Solve the equation: . Find all the roots of this equation that belong to the segment:

Now for some quick explanations and answers:

1. Here it suffices to note that and. Then the original equation will be equivalent to this: This equation solved by replacement Do further calculations yourself. In the end, your task will be reduced to solving the simplest trigonometric (depending on the sine or cosine). Solution similar examples we will explore in other sections.
2. Here you can even do without a replacement: just move the subtrahend to the right and represent both bases through powers of two: and then immediately go to the quadratic equation.
3. The third equation is also solved in a rather standard way: imagine how. Then, replacing we get a quadratic equation: then,

   Do you already know what a logarithm is? Not? Then urgently read the topic!

   The first root, obviously, does not belong to the segment, and the second is incomprehensible! But we will find out very soon! Since, then (this is a property of the logarithm!) Let's compare:

   Subtract from both parts, then we get:

   The left side can be represented as:

   multiply both sides by:

   can be multiplied by, then

   Then let's compare:

   since then:

   Then the second root belongs to the desired interval

   Answer:

As you see, the selection of the roots of exponential equations requires sufficient profound knowledge properties of logarithms, so I advise you to be as careful as possible when solving exponential equations. As you know, in mathematics everything is interconnected! As my math teacher used to say: "You can't read math like history overnight."

As a rule, all the difficulty in solving problems C1 is precisely the selection of the roots of the equation. Let's practice with another example:

It is clear that the equation itself is solved quite simply. Having made the substitution, we reduce our original equation to the following:

Let's look at the first root first. Compare and: since, then. (property of the logarithmic function, at). Then it is clear that the first root does not belong to our interval either. Now the second root: . It is clear that (since the function is increasing). It remains to compare and

since, then, at the same time. Thus, I can "drive a peg" between and. This peg is a number. The first expression is less than and the second is greater than. Then the second expression is greater than the first and the root belongs to the interval.

Answer: .

In conclusion, let's look at another example of an equation where the replacement is rather non-standard:

Let's start right away with what you can do, and what - in principle, you can, but it's better not to do it. It is possible - to represent everything through the powers of three, two and six. Where it leads? Yes, and will not lead to anything: a hodgepodge of degrees, some of which will be quite difficult to get rid of. What then is needed? Let's note that a And what will it give us? And the fact that we can reduce the solution of this example to the solution of a fairly simple exponential equation! First, let's rewrite our equation as:

Now we divide both sides of the resulting equation into:

Eureka! Now we can replace, we get:

Well, now it's your turn to solve problems for demonstration, and I will give them only brief comments so that you do not go astray! Good luck!

1. The most difficult! Seeing a replacement here is oh, how ugly! Nevertheless, this example can be completely solved using allocation full square . To solve it, it suffices to note that:

So here's your replacement:

(Note that here, in our substitution, we cannot discard negative root!!! Why do you think?)

Now, to solve the example, you have to solve two equations:

Both of them are solved by the "standard replacement" (but the second one in one example!)

2. Notice that and make a substitution.

3. Expand the number into coprime factors and simplify the resulting expression.

4. Divide the numerator and denominator of the fraction by (or if you prefer) and make the substitution or.

5. Note that the numbers and are conjugate.

EXPOSITIONAL EQUATIONS. ADVANCED LEVEL

In addition, let's look at another way - solution of exponential equations by the logarithm method. I can’t say that the solution of exponential equations by this method is very popular, but in some cases only it can lead us to right decision our equation. Especially often it is used to solve the so-called " mixed equations ': that is, those where there are functions of different types.

For example, an equation like:

in general case can only be solved by taking the logarithm of both parts (for example, by base), in which the original equation turns into the following:

Let's consider the following example:

It is clear that we are only interested in the ODZ of the logarithmic function. However, this follows not only from the ODZ of the logarithm, but for another reason. I think that it will not be difficult for you to guess which one.

Let's take the logarithm of both sides of our equation to the base:

As you can see, taking the logarithm of our original equation quickly led us to the correct (and beautiful!) answer. Let's practice with another example:

Here, too, there is nothing to worry about: we take the logarithm of both sides of the equation in terms of the base, then we get:

Let's make a replacement:

However, we missed something! Did you notice where I made a mistake? After all, then:

which does not satisfy the requirement (think where it came from!)

Answer:

Try to write down the solution of the exponential equations below:

Now check your solution with this:

1. We logarithm both parts to the base, given that:

(the second root does not suit us due to the replacement)

2. Logarithm to the base:

Let's transform the resulting expression to the following form:

EXPOSITIONAL EQUATIONS. BRIEF DESCRIPTION AND BASIC FORMULA

exponential equation

Type equation:

called the simplest exponential equation.

Degree properties

Solution Approaches

• Reduction to the same base
• Casting to the same indicator degree
• Variable substitution
• Simplify the expression and apply one of the above.

Equations are called exponential if the unknown is contained in the exponent. The simplest exponential equation has the form: a x \u003d a b, where a> 0, and 1, x is an unknown.

The main properties of the degrees, with the help of which the exponential equations are transformed: a>0, b>0.

When solving exponential equations, the following properties of the exponential function are also used: y = a x , a > 0, a1:

To represent a number as a power, use the base logarithmic identity: b = , a > 0, a1, b > 0.

Tasks and tests on the topic "Exponential Equations"

• exponential equations

  Lessons: 4 Assignments: 21 Tests: 1

• exponential equations - Important Topics to repeat the exam in mathematics

  Tasks: 14

• Systems of exponential and logarithmic equations - Demonstrative and logarithmic function Grade 11

  Lessons: 1 Assignments: 15 Tests: 1

• §2.1. Solution of exponential equations

  Lessons: 1 Assignments: 27

• §7 Exponential and logarithmic equations and inequalities - Section 5. Exponential and logarithmic functions Grade 10

  Lessons: 1 Assignments: 17

For successful solution exponential equations You must know the basic properties of powers, the properties of the exponential function, the basic logarithmic identity.

When solving exponential equations, two main methods are used:

1. transition from the equation a f(x) = a g(x) to the equation f(x) = g(x);
2. introduction of new lines.

Examples.

1. Equations Reducing to the Simplest. They are solved by bringing both sides of the equation to a power with the same base.

3x \u003d 9x - 2.

Solution:

3 x \u003d (3 2) x - 2;
3x = 3 2x - 4;
x = 2x -4;
x=4.

Answer: 4.

2. Equations solved by bracketing the common factor.

Solution:

3x - 3x - 2 = 24
3 x - 2 (3 2 - 1) = 24
3 x - 2 x 8 = 24
3 x - 2 = 3
x - 2 = 1
x=3.

Answer: 3.

3. Equations Solved by Change of Variable.

Solution:

2 2x + 2 x - 12 = 0
We denote 2 x \u003d y.
y 2 + y - 12 = 0
y 1 = - 4; y 2 = 3.
a) 2 x = - 4. The equation has no solutions, because 2 x > 0.
b) 2 x = 3; 2 x = 2 log 2 3 ; x = log 2 3.

Answer: log 2 3.

4. Equations containing powers with two different (not reducible to each other) bases.

3 × 2 x + 1 - 2 × 5 x - 2 \u003d 5 x + 2 x - 2.

3 x 2 x + 1 - 2 x - 2 = 5 x - 2 x 5 x - 2
2 x - 2 × 23 = 5 x - 2
×23
2 x - 2 = 5 x - 2
(5/2) x– 2 = 1
x - 2 = 0
x = 2.

Answer: 2.

5. Equations that are homogeneous with respect to a x and b x .

General form: .

9 x + 4 x = 2.5 x 6 x .

Solution:

3 2x – 2.5 × 2x × 3x +2 2x = 0 |: 2 2x > 0
(3/2) 2x - 2.5 × (3/2) x + 1 = 0.
Denote (3/2) x = y.
y 2 - 2.5y + 1 \u003d 0,
y 1 = 2; y2 = ½.

Answer: log 3/2 2; - log 3/2 2.

To the youtube channel of our site site to be aware of all new video lessons.

To begin with, let's remember basic formulas degrees and their properties.

Product of a number a happens 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 \u003d 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:

AT this example the number 6 is the base, it is always at the bottom, and the variable x degree or measure.

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

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

Let's take a simple equation:

2 x = 2 3

Such an example can be solved even in the mind. It can be seen that x=3. After all, so that the left and right part were equal, you need to put the number 3 instead of x.
Now let's see how this decision should be made:

2 x = 2 3
x = 3

To solve this equation, we removed same grounds(that is, deuces) and wrote down what was left, these are degrees. We got the answer we were looking for.

Now let's summarize our solution.

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

Now let's solve some examples:

Let's start 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 has turned out.
x=4 - 2
x=2
Answer: x=2

AT following example It can be seen that the bases are different - 3 and 9.

3 3x - 9 x + 8 = 0

To begin with, we transfer 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 \u003d (3 2) x + 8

We get 9 x + 8 \u003d (3 2) x + 8 \u003d 3 2 x + 16

3 3x \u003d 3 2x + 16 now you can see that in the left and right side the bases are the same and equal to three, which means we can discard them and equate the degrees.

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

Let's look at the following example:

2 2x + 4 - 10 4 x \u003d 2 4

First of all, we look at the bases, the bases are different two and four. And we need to be the same. We transform the quadruple according to 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 interfere with us. What to do with them? If you look closely, you can see that on the left side we repeat 2 2x, 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 whole equation by 6:

Imagine 4=2 2:

2 2x \u003d 2 2 bases are the same, discard them and equate the degrees.
2x \u003d 2 turned out to be the simplest equation. We divide it by 2, we get
x = 1
Answer: x = 1.

Let's solve the equation:

9 x - 12*3 x +27= 0

Let's transform:
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, it is clear that the first triple has a degree twice (2x) than the second (just x). In this case, you can decide substitution method. Number with least degree replace:

Then 3 2x \u003d (3 x) 2 \u003d t 2

We replace all degrees with x's in the equation with t:

t 2 - 12t + 27 \u003d 0
We get a quadratic equation. We solve through the discriminant, we get:
D=144-108=36
t1 = 9
t2 = 3

Back to Variable x.

We take t 1:
t 1 \u003d 9 \u003d 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 \u003d 3 \u003d 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 \u003d 2; x 2 = 1.

On the site you can in the section HELP DECIDE to ask questions of interest, we will definitely answer you.

Join a group