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Types of exponential equations. Solution of exponential equations. The basics. Classification of exponential equations

  • 14.09.2020

Lecture: "Methods for solving exponential equations."

1 . Exponential equations.

Equations containing unknowns in the exponent are called exponential equations. The simplest of them is the equation ax \u003d b, where a\u003e 0, and ≠ 1.

1) For b< 0 и b = 0 это уравнение, согласно свойству 1 показательной функции, не имеет решения.

2) For b\u003e 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 \u003d ac, ax \u003d bc ó x \u003d c or x \u003d logab.

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

1) method of reduction to one basis;

2) assessment method;

3) graphical method;

4) the method of introducing new variables;

5) the method of factorization;

6) exponential - power equations;

7) exponential with parameter.

2 . Method of coercion 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 indicators are also equal, that is, the equation must be tried to be reduced to the form

Examples. Solve the equation:

1 ... 3x \u003d 81;

Rewrite the right side of the equation as 81 \u003d 34 and rewrite the equation that is equivalent to the original 3 x \u003d 34; x \u003d 4. Answer: 4.

2. https://pandia.ru/text/80/142/images/image004_8.png "width \u003d" 52 "height \u003d" 49 "\u003e and turn to the equation for exponents 3x + 1 \u003d 3 - 5x; 8x \u003d 4; x \u003d 0.5 Answer: 0.5.

3. DIV_ADBLOCK217 "\u003e


Answer: 1 and 2.

4.

Note that the numbers 0.2, 0.04, √5, and 25 are powers of 5. Let's use this to transform the original equation as follows:

, whence 5-x-1 \u003d 5-2x-2 ó - x - 1 \u003d - 2x - 2, from which we find the solution x \u003d -1. Answer: -1.

5. 3x \u003d 5. By the definition of the logarithm x \u003d log35. Answer: log35.

6. 62x + 4 \u003d 33x. 2x + 8.

Let's rewrite the equation as 32x + 4.22x + 4 \u003d 32x.2x + 8, ie..png "width \u003d" 181 "height \u003d" 49 src \u003d "\u003e Hence x - 4 \u003d 0, x \u003d 4. Answer: 4.

7 ... 2 ∙ 3x + 1 - 6 ∙ 3x-2 - 3x \u003d 9. Using the properties of the degrees, we write the equation in the form 6 ∙ 3x - 2 ∙ 3x - 3x \u003d 9 then 3 ∙ 3x \u003d 9, 3x + 1 \u003d 32, i.e. i.e. x + 1 \u003d 2, x \u003d 1. Answer: 1.

Bank of tasks №1.

Solve the equation:

Test # 1.

1) 0 2) 4 3) -2 4) -4

A2 32x-8 \u003d √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 number 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) \u003d 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 \u003d 5 - x.

Decision. Rewrite the equation as 4x + x \u003d 5.

1.if x \u003d 1, then 41 + 1 \u003d 5, 5 \u003d 5 is true, so 1 is the root of the equation.

The function f (x) \u003d 4x - increases on R, and g (x) \u003d x - increases on R \u003d\u003e h (x) \u003d f (x) + g (x) increases on R, as the sum of increasing functions, so x \u003d 1 is the only root of the equation 4x \u003d 5 - x. Answer: 1.

2.

Decision. We rewrite the equation as .

1.if x \u003d -1, then , 3 \u003d 3-true, so x \u003d -1 is the root of the equation.

2. Let us prove that it is unique.

3. The function f (x) \u003d - decreases on R, and g (x) \u003d - x - decreases on R \u003d\u003e h (x) \u003d f (x) + g (x) - decreases on R, as the sum of decreasing functions ... Hence, by the root theorem, x \u003d -1 is the only root of the equation. Answer: -1.

Bank of tasks №2. Solve the equation

a) 4x + 1 \u003d 6 - x;

b)

c) 2x - 2 \u003d 1 - x;

4. Method for introducing new variables.

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

Examples. Rsolve the equation: 1. .

Let's rewrite the equation differently: https://pandia.ru/text/80/142/images/image030_0.png "width \u003d" 128 "height \u003d" 48 src \u003d "\u003e ie..png" width \u003d "210" height \u003d "45"\u003e

Decision. Let's rewrite the equation differently:

Let's designate https://pandia.ru/text/80/142/images/image035_0.png "width \u003d" 245 "height \u003d" 57 "\u003e - does not fit.

t \u003d 4 \u003d\u003e https://pandia.ru/text/80/142/images/image037_0.png "width \u003d" 268 "height \u003d" 51 "\u003e is an irrational equation. Note that

The solution to the equation is x \u003d 2.5 ≤ 4, which means 2.5 is the root of the equation. Answer: 2.5.


Decision. Rewrite the equation in the form and divide both sides by 56x + 6 ≠ 0. We obtain the equation

2x2-6x-7 \u003d 2x2-6x-8 +1 \u003d 2 (x2-3x-4) +1, t..png "width \u003d" 118 "height \u003d" 56 "\u003e

Quadratic roots - t1 \u003d 1 and t2<0, т. е..png" width="200" height="24">.

Decision . We rewrite the equation as

and note that it is a homogeneous second-degree equation.

Divide the equation by 42x, we get

Let's replace https://pandia.ru/text/80/142/images/image049_0.png "width \u003d" 16 "height \u003d" 41 src \u003d "\u003e.

Answer: 0; 0.5.

Bank of tasks number 3. Solve the equation

b)

d)

Test number 3 with a choice of answer. The minimum level.

A1

1) -0.2; 2 2) log52 3) –log52 4) 2

A2 0.52x - 3 0.5x +2 \u003d 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 \u003d 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 number 4 with a choice of answer. General level.

A1

1) 2; 1 2) ½; 0 3) 2; 0 4) 0

A2 2x - (0.5) 2x - (0.5) x + 1 \u003d 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 \u003d 24.

Solution..png "width \u003d" 169 "height \u003d" 69 "\u003e, from where

2. 6x + 6x + 1 \u003d 2x + 2x + 1 + 2x + 2.

Decision. Factor out 6x on the left and 2x on the right. We get the equation 6x (1 + 6) \u003d 2x (1 + 2 + 4) ó 6x \u003d 2x.

Since 2x\u003e 0 for all x, both sides of this equation can be divided by 2x without fear of losing solutions. We get 3x \u003d 1ó x \u003d 0.

3.

Decision. Solve the equation by factoring.

Select the square of the binomial

4. https://pandia.ru/text/80/142/images/image067_0.png "width \u003d" 500 "height \u003d" 181 "\u003e

x \u003d -2 is the root of the equation.

Equation x + 1 \u003d 0 "style \u003d" border-collapse: collapse; border: none "\u003e

A1 5x-1 + 5x -5x + 1 \u003d -19.

1) 1 2) 95/4 3) 0 4) -1

A2 3x + 1 + 3x-1 \u003d 270.

1) 2 2) -4 3) 0 4) 4

A3 32x + 32x + 1 -108 \u003d 0.x \u003d 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 \u003d 15.x \u003d 4

1) -4 2) 4 3) -4;4 4) 2

Test number 6 General level.

A1 (22x-1) (24x + 22x + 1) \u003d 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 \u003d 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. Indicative - power equations.

The so-called exponential - power equations are adjacent to the exponential equations, i.e., equations of the form (f (x)) g (x) \u003d (f (x)) h (x).

If it is known that f (x)\u003e 0 and f (x) ≠ 1, then the equation, like the exponential one, is solved by equating the exponents g (x) \u003d f (x).

If the condition does not exclude the possibility of f (x) \u003d 0 and f (x) \u003d 1, then we have to consider these cases when solving the exponential - power equation.

1..png "width \u003d" 182 "height \u003d" 116 src \u003d "\u003e

2.

Decision. x2 + 2x-8 - makes sense for any x, since it is a polynomial, so the equation is equivalent to a set

https://pandia.ru/text/80/142/images/image078_0.png "width \u003d" 137 "height \u003d" 35 "\u003e

b)

7. Exponential equations with parameters.

1. For what values \u200b\u200bof the parameter p does equation 4 (5 - 3)  2 + 4p2–3p \u003d 0 (1) have a unique solution?

Decision. We introduce the replacement 2x \u003d t, t\u003e 0, then equation (1) takes the form t2 - (5p - 3) t + 4p2 - 3p \u003d 0. (2)

The discriminant of equation (2) D \u003d (5p - 3) 2 - 4 (4p2 - 3p) \u003d 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 \u003d 0, that is, p \u003d 1, then equation (2) takes the form t2 - 2t + 1 \u003d 0, hence t \u003d 1, therefore, equation (1) has a unique solution x \u003d 0.

2. If p1, then 9 (p - 1) 2\u003e 0, then equation (2) has two different roots t1 \u003d p, t2 \u003d 4p - 3. The condition of the problem is satisfied by the set of systems

Substituting t1 and t2 into the systems, we have

https://pandia.ru/text/80/142/images/image084_0.png "alt \u003d" (! LANG: no35_11" width="375" height="54"> в зависимости от параметра a?!}

Decision. Let be then equation (3) takes the form t2 - 6t - a \u003d 0. (4)

Let us find the values \u200b\u200bof the parameter a for which at least one root of Eq. (4) satisfies the condition t\u003e 0.

Let us introduce the function f (t) \u003d t2 - 6t - a. The following cases are possible.

https://pandia.ru/text/80/142/images/image087.png "alt \u003d" (! 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 \u003d" (! 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 \u003d 0, if a \u003d - 9, then equation (4) takes the form (t - 3) 2 \u003d 0, t \u003d 3, x \u003d - 1.

Case 3. Equation (4) has two roots, but one of them does not satisfy the inequality t\u003e 0. This is possible if

https://pandia.ru/text/80/142/images/image092.png "alt \u003d" (! LANG: no35_17" width="267" height="63">!}

Thus, for a 0, Eq. (4) has a unique 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 \u003d - 9, then x \u003d - 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 using the formula for the roots of the quadratic equation, and then conclusions were drawn about 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 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

Decision. ODZ: x1, x2.

Let's introduce a replacement. Let 2x \u003d t, t\u003e 0, then, as a result of transformations, the equation takes the form t2 + 2t - 13 - a \u003d 0. (*) Find the values \u200b\u200bof a for which at least one root of the equation (*) satisfies the condition t\u003e 0.

https://pandia.ru/text/80/142/images/image098.png "alt \u003d" (! 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 \u003d" (! 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 \u003d" (! LANG: http: //1september.ru/ru/mat/2002/35/no35_27.gif" align="left" width="218" height="42 src=">!}

Answer: if a\u003e - 13, a  11, a  5, then if a - 13,

a \u003d 11, a \u003d 5, then there are no roots.

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entering universities. "AS T - press school", 2002

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This is the name of equations of the form where the unknown is located both in the exponent and in the base of the degree.

You can specify a completely clear algorithm for solving an equation of the form. For this, it is necessary to pay attention to the fact that for oh) not equal to zero, one and minus one, equality of degrees with the same bases (be it positive or negative) is possible only if the indicators are equal That is, all the roots of the equation will be the roots of the equation f (x) \u003d g (x) The converse statement is not true, for oh)< 0 and fractional values f (x) and g (x)expressions oh) f (x) and

oh) g (x) lose their meaning. That is, when going from to f (x) \u003d g (x) (for and, extraneous roots may appear, which must be eliminated by checking against the original equation. a \u003d 0, a \u003d 1, a \u003d -1must be considered separately.

So, for a complete solution of the equation, we consider the cases:

a (x) \u003d O f (x) and g (x) are positive numbers, then this is the solution. Otherwise, no

a (x) \u003d 1... The roots of this equation are also the roots of the original equation.

a (x) \u003d -1... If for a value x satisfying this equation, f (x) and g (x) are integers of the same parity (either both are even or both are odd), then this is the solution. Otherwise, no

For and, we solve the equation f (x) \u003d g (x) and substituting the obtained results into the original equation, we cut off extraneous roots.

Examples of solving exponential equations.

Example # 1.

1) x - 3 \u003d 0, x \u003d 3. since 3\u003e 0, and 3 2\u003e 0, then x 1 \u003d 3 is the solution.

2) x - 3 \u003d 1, x 2 \u003d 4.

3) x - 3 \u003d -1, x \u003d 2. Both exponents are even. This solution is x 3 \u003d 1.

4) x - 3? 0 and x? ± 1.x \u003d x 2, x \u003d 0 or x \u003d 1. For x \u003d 0, (-3) 0 \u003d (-3) 0 -this solution is correct x 4 \u003d 0. For x \u003d 1, (-2) 1 \u003d (-2) 1 - this solution is correct x 5 \u003d 1.

Answer: 0, 1, 2, 3, 4.

Example # 2.

By definition of the arithmetic square root: x - 1? 0, x? 1.

1) x - 1 \u003d 0 or x \u003d 1, \u003d 0, 0 0 is not a solution.

2) x - 1 \u003d 1 x 1 \u003d 2.

3) x - 1 \u003d -1 x 2 \u003d 0 does not fit in the ODZ.

D \u003d (-2) - 4 * 1 * 5 \u003d 4 - 20 \u003d -16 - no roots.

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

Performing tasks C3, you have to solve various types of equations and inequalities. Among them are rational, irrational, exponential, logarithmic, trigonometric, containing modules (absolute values), as well as combined ones. This article discusses the main types of exponential equations and inequalities, as well as various methods for solving them. Read about the solution of other types of equations and inequalities in the heading "" in articles devoted to methods for solving problems C3 from the USE versions in mathematics.

Before proceeding with the analysis of specific exponential equations and inequalitiesAs a math tutor, I suggest you brush up on some of the theoretical material we need.

Exponential function

What is an exponential function?

View function y = a xwhere a \u003e 0 and a ≠ 1 is called exponential function.

The main exponential properties y = a x:

Exponential function graph

The exponential function graph is exhibitor:

Exponential function plots (exponentials)

Solving exponential equations

Illustrative are called equations in which the unknown variable is only in terms of some powers.

For solutions exponential equations you need to know and be able to use the following simple theorem:

Theorem 1. Exponential equation a f(x) = a g(x) (where a > 0, a ≠ 1) is equivalent to the equation f(x) = g(x).

In addition, it is useful to remember the basic formulas and actions with degrees:

Title \u003d "(! LANG: Rendered by QuickLaTeX.com">!}

Example 1. Solve the equation:

Decision: we use the above formulas and substitution:

The equation then takes the form:

The discriminant of the resulting quadratic equation is positive:

Title \u003d "(! LANG: Rendered by QuickLaTeX.com">!}

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

Passing to the reverse substitution, we get:

The second equation has no roots, since the exponential function is strictly positive over the entire domain. We solve the second:

Taking into account what has been said in Theorem 1, we pass to the equivalent equation: x \u003d 3. This will be the answer to the task.

Answer: x = 3.

Example 2. Solve the equation:

Decision: the equation has no restrictions on the range of admissible values, since the radical expression makes sense for any value x (exponential function y = 9 4 -x is positive and nonzero).

We solve the equation by equivalent transformations using the rules of multiplication and division of powers:

The last transition was carried out in accordance with Theorem 1.

Answer:x= 6.

Example 3. Solve the equation:

Decision: both sides of the original equation can be divided by 0.2 x ... This transition will be equivalent, since this expression is greater than zero for any value x (the exponential function is strictly positive in its domain of definition). Then the equation takes the form:

Answer: x = 0.

Example 4. Solve the equation:

Decision:we simplify the equation to an elementary one by equivalent transformations using the rules for division and multiplication of powers given at the beginning of the article:

Dividing both sides of the equation by 4 x , as in the previous example, is an equivalent conversion, since the given expression is not equal to zero for any values x.

Answer: x = 0.

Example 5. Solve the equation:

Decision: function y = 3 xon the left side of the equation is increasing. Function y = —x-2/3 on the right side of the equation is decreasing. This means that if the graphs of these functions intersect, then at no more than one point. In this case, it is easy to guess that the graphs intersect at the point x \u003d -1. There will be no other roots.

Answer: x = -1.

Example 6. Solve the equation:

Decision: we simplify the equation by means of equivalent transformations, keeping in mind everywhere that the exponential function is strictly greater than zero for any value xand using the rules for calculating the product and the quotient degrees given at the beginning of the article:

Answer: x = 2.

Solution of exponential inequalities

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

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

Theorem 2. If a a \u003e 1, then the inequality a f(x) > a g(x) is equivalent to the inequality of the same meaning: f(x) > g(x). If 0< a < 1, то показательное неравенство a f(x) > a g(x) is equivalent to the inequality of the opposite meaning: f(x) < g(x).

Example 7.Solve the inequality:

Decision: we represent the original inequality in the form:

Divide both sides of this inequality by 3 2 x , moreover (due to the positivity of the function y= 3 2x) the inequality sign will not change:

Let's use the substitution:

Then the inequality takes the form:

So, the solution to the inequality is the interval:

passing to the reverse substitution, we get:

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

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

So, we finally get answer:

Example 8. Solve the inequality:

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

Let's introduce a new variable:

With this substitution, the inequality takes the form:

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

So, the following values \u200b\u200bof the variable satisfy the inequality t:

Then, passing to the reverse substitution, we get:

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

Finally we get answer:

Example 9. Solve the inequality:

Decision:

We divide both sides of the inequality by the expression:

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

t located in the interval:

Passing to the inverse substitution, we find that the original inequality splits into two cases:

The first inequality of solutions does not have due to the positiveness of the exponential function. We solve the second:

Example 10. Solve the inequality:

Decision:

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

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

At the same time, the function y = 3 x 2 -2x+2 on the right side of the equation. It reaches its smallest value at the same point as the parabola in the exponent, and this value is equal to 3 1 \u003d 3. So, the original inequality can be true only if the function on the left and the function on the right take the value at one point , equal to 3 (only this number is the intersection of the ranges of values \u200b\u200bof these functions). This condition is fulfilled at a single point x = 1.

Answer: x= 1.

To learn to solve exponential equations and inequalities, it is necessary to constantly train in solving them. In this difficult matter, you can be helped by various teaching aids, problem books in elementary mathematics, collections of competition problems, classes in mathematics at school, as well as individual lessons with a professional tutor. I sincerely wish you success in your preparation and excellent results on the exam.


Sergey Valerievich

P. S. Dear guests! Please do not write applications for solving your equations in the comments. Unfortunately, I have absolutely no time for this. Such messages will be deleted. Please read the article. Perhaps in it you will find answers to questions that did not allow you to solve your task on your own.

Solution of exponential equations. Examples.

Attention!
There are additional
materials in Special Section 555.
For those who are "not very ..."
And for those who are "very even ...")

What exponential equation? This is an equation in which the unknowns (x) and expressions with them are in indicators some degrees. And only there! It is important.

There you are examples of exponential equations:

3 x 2 x \u003d 8 x + 3

Note! In the bases of the degrees (below) - only numbers... IN indicators degrees (above) - a wide variety of expressions with x. If, suddenly, an x \u200b\u200bappears in the equation somewhere other than an indicator, for example:

this will already be a mixed-type equation. Such equations do not have clear rules for solving. We will not consider them yet. Here we will deal with solving exponential equations in its purest form.

In fact, even pure exponential equations are not always clearly solved. But there are certain types of exponential equations that can and should be solved. We will consider these types.

Solution of the simplest exponential equations.

To begin with, let's solve something very basic. For example:

Even without any theories, it is clear from a simple selection that x \u003d 2. No more, right !? No other x value rolls. Now let's take a look at the solution to this cunning exponential equation:

What have we done? In fact, we just threw out the same bases (threes). Thrown out completely. And, what pleases, hit the mark!

Indeed, if the exponential equation on the left and right contains the same numbers in any powers, these numbers can be removed and the exponents equated. Mathematics allows. It remains to solve a much simpler equation. Great, right?)

However, let us remember it ironically: you can remove the bases only when the base numbers on the left and right are in splendid isolation! Without any neighbors and coefficients. Let's say in the equations:

2 x + 2 x + 1 \u003d 2 3, or

deuces cannot be removed!

Well, the most important thing we have mastered. How to go from evil exponential expressions to simpler equations.

"These are the times!" - you say. "Who will give such a primitive on tests and exams !?"

I have to agree. Nobody will give. But now you know where to strive when solving confused examples. It is necessary to bring it to the form when the same base number is on the left - on the right. Then everything will be easier. Actually, this is the classics of mathematics. We take the original example and transform it to the desired one us mind. By the rules of mathematics, of course.

Let's look at examples that require some extra effort to make them simpler. Let's call them simple exponential equations.

Solving simple exponential equations. Examples.

When solving exponential equations, the main rules are - actions with degrees. Without knowledge of these actions, nothing will work.

Personal observation and ingenuity must be added to actions with degrees. Do we need the same base numbers? So we are looking for them in the example in explicit or encrypted form.

Let's see how this is done in practice?

Let us be given an example:

2 2x - 8x + 1 \u003d 0

The first keen glance is at grounds. They ... They are different! Two and eight. But it's too early to get discouraged. It's time to remember that

Two and eight are relatives in degree.) It is quite possible to write down:

8 x + 1 \u003d (2 3) x + 1

If you recall the formula from actions with powers:

(a n) m \u003d a nm,

it turns out great:

8 x + 1 \u003d (2 3) x + 1 \u003d 2 3 (x + 1)

The original example now looks like this:

2 2x - 2 3 (x + 1) \u003d 0

We transfer 2 3 (x + 1) to the right (no one canceled the elementary actions of mathematics!), we get:

2 2x \u003d 2 3 (x + 1)

That's practically all. We remove the bases:

We solve this monster and get

This is the correct answer.

In this example, knowing the powers of two helped us out. we identified in the eight is an encrypted two. This technique (encrypting common bases under different numbers) is a very popular technique in exponential equations! And in logarithms too. One must be able to recognize the powers of other numbers in numbers. This is extremely important for solving exponential equations.

The fact is that raising any number to any power is not a problem. Multiply, even on a piece of paper, and that's all. For example, everyone can raise 3 to the fifth power. 243 will work if you know the multiplication table.) But in exponential equations, it is much more often necessary not to raise to a power, but on the contrary ... what number to what degree is hidden behind the number 243, or, say, 343 ... No calculator will help you here.

You need to know the powers of some numbers by sight, yes ... Let's practice?

Determine what powers and what numbers are numbers:

2; 8; 16; 27; 32; 64; 81; 100; 125; 128; 216; 243; 256; 343; 512; 625; 729, 1024.

Answers (in disarray, naturally!):

5 4 ; 2 10 ; 7 3 ; 3 5 ; 2 7 ; 10 2 ; 2 6 ; 3 3 ; 2 3 ; 2 1 ; 3 6 ; 2 9 ; 2 8 ; 6 3 ; 5 3 ; 3 4 ; 2 5 ; 4 4 ; 4 2 ; 2 3 ; 9 3 ; 4 5 ; 8 2 ; 4 3 ; 8 3 .

If you look closely, you can see a strange fact. There are significantly more answers than tasks! Well, it happens ... For example, 2 6, 4 3, 8 2 are all 64.

Suppose that you have taken note of the information about familiarity with numbers.) Let me remind you that to solve exponential equations, we use whole stock of mathematical knowledge. Including those from the junior-middle classes. You didn't go to high school right away, did you?)

For example, when solving exponential equations, it often helps to put the common factor outside the parentheses (hello 7th grade!). Let's see an example:

3 2x + 4 -11 9 x \u003d 210

And again, at first glance - at the foundations! The bases of the degrees are different ... Three and nine. And we want them to be the same. Well, in this case, the desire is quite feasible!) Because:

9 x \u003d (3 2) x \u003d 3 2x

Following the same rules for dealing with degrees:

3 2x + 4 \u003d 3 2x 3 4

That's great, you can write:

3 2x 3 4 - 11 3 2x \u003d 210

We have led the example to the same grounds. So, what is next!? Threes must not be thrown away ... Dead end?

Not at all. Remember the most universal and powerful decision rule of all math tasks:

If you don't know what is needed, do what you can!

You look, everything will be formed).

What's in this exponential equation can to do? Yes, on the left side it is directly asking for parenthesis! The common factor of 3 2x clearly hints at this. Let's try, and then we will see:

3 2x (3 4 - 11) \u003d 210

3 4 - 11 = 81 - 11 = 70

The example keeps getting better and better!

We remember that to eliminate the grounds, we need a pure degree, without any coefficients. The number 70 gets in our way. So we divide both sides of the equation by 70, we get:

Oops! Everything worked out!

This is the final answer.

It happens, however, that taxiing on the same basis is obtained, but their elimination is not. This happens in exponential equations of another type. Let's master this type.

Change of variable in solving exponential equations. Examples.

Let's solve the equation:

4 x - 3 2 x +2 \u003d 0

First, as usual. Moving on to one base. To the deuce.

4 x \u003d (2 2) x \u003d 2 2x

We get the equation:

2 2x - 3 2 x +2 \u003d 0

And here we will freeze. The previous techniques will not work, no matter how cool. We'll have to get out of the arsenal of another powerful and versatile way. It is called variable replacement.

The essence of the method is surprisingly simple. Instead of one complex icon (in our case - 2 x) we write another, simpler (for example - t). Such a seemingly meaningless replacement leads to amazing results!) It just becomes clear and understandable!

So let

Then 2 2x \u003d 2 x2 \u003d (2 x) 2 \u003d t 2

Replace all powers with x in our equation with t:

Well, it dawns?) Have you forgotten the quadratic equations yet? We solve through the discriminant, we get:

Here, the main thing is not to stop, as it happens ... This is not the answer yet, we need X, not t. We return to the Xs, i.e. we make a return replacement. First for t 1:

That is,

One root was found. We are looking for the second, from t 2:

Um ... Left 2 x, right 1 ... A problem? Not at all! It is enough to remember (from actions with powers, yes ...) that one is any number to the zero degree. Anyone. We will deliver what is needed. We need a deuce. Means:

Now that's it. We got 2 roots:

This is the answer.

When solving exponential equations sometimes we end up with some awkward expression. Type:

From seven, two through a prime degree does not work. They are not relatives ... How to be here? Someone may be confused ... But the person who read on this site the topic "What is a logarithm?" , only smiles sparingly and writes down with a firm hand the absolutely correct answer:

There can be no such answer in tasks "B" on the exam. There, a specific number is required. But in tasks "C" - easily.

This lesson provides examples of solving the most common exponential equations. Let's highlight the main thing.

Practical advice:

1. First of all, we look at grounds degrees. We consider whether it is possible to make them the same. We try to do this by actively using actions with degrees. Do not forget that numbers without x can also be converted to powers!

2. We try to reduce the exponential equation to the form when the left and right are the same numbers in any degree. We use actions with degrees and factorization.What can be counted in numbers - we count.

3. If the second tip didn't work, we try to apply variable replacement. The end result is an equation that can be easily solved. Most often it is square. Or fractional, which also boils down to square.

4. To successfully solve exponential equations, you need to know the powers of some numbers "by sight".

As usual, at the end of the lesson you are asked to decide a little.) On your own. From simple to complex.

Solve exponential equations:

More difficult:

2 x + 3 - 2 x + 2 - 2 x \u003d 48

9 x - 8 3 x \u003d 9

2 x - 2 0.5x + 1 - 8 \u003d 0

Find the product of roots:

2 3-x + 2 x \u003d 9

Happened?

Well, then the most complicated example (solved, however, in the mind ...):

7 0.13x + 13 0.7x + 1 + 2 0.5x + 1 \u003d -3

What's more interesting? Then here's a bad example. Quite drawn to increased difficulty. I will hint that in this example, ingenuity and the most universal rule for solving all math problems save.)

2 5x-1 3 3x-1 5 2x-1 \u003d 720 x

An example is simpler, for rest):

9 2 x - 4 3 x \u003d 0

And for dessert. Find the sum of the roots of the equation:

x 3 x - 9x + 7 3 x - 63 \u003d 0

Yes Yes! This is a mixed equation! Which we did not consider in this lesson. And that they should be considered, they must be solved!) This lesson is quite enough for solving the equation. Well, savvy is needed ... And may the seventh grade help you (this is a hint!).

Answers (in a mess, separated by semicolons):

1; 2; 3; 4; no solutions; 2; -2; -five; 4; 0.

Is everything all right? Fine.

There is a problem? No problem! In Special Section 555, all of these exponential equations are solved with detailed explanations. What, why, and why. And, of course, there is additional valuable information on working with all kinds of exponential equations. Not just these.)

One last funny question to consider. In this tutorial, we worked with exponential equations. Why didn't I say a word about ODZ here? In equations, this is a very important thing, by the way ...

If you like this site ...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Instant validation testing. Learning - with interest!)

you can get acquainted with functions and derivatives.

1º. Exponential equationsare called equations containing a variable in the exponent.

The solution of exponential equations is based on the property of degree: two degrees with the same base are equal if and only if their exponents are equal.

2º. Basic methods for solving exponential equations:

1) the simplest equation has a solution;

2) an equation of the form by logarithm to the base a reduce to the mind;

3) an equation of the form is equivalent to an equation;

4) an equation of the form is equivalent to an equation.

5) an equation of the form through a change is reduced to an equation, and then a set of the simplest exponential equations is solved;

6) an equation with mutually reciprocal values by replacement, reduce to an equation, and then solve a set of equations;

7) equations that are homogeneous with respect to a g (x) and b g (x) given that kind through the replacement, they are reduced to an equation, and then the set of equations is solved.

Classification of exponential equations.

1. Equations Solved by Going to One Base.

Example 18. Solve the equation .

Solution: Let's use the fact that all bases of degrees are powers of 5:.

2. Equations Solved by Going to One Exponent.

These equations are solved by transforming the original equation to the form , which is reduced to the simplest using the proportion property.

Example 19. Solve the equation:

3. Equations Solved by Taking the Common Factor outside the parentheses.

If in the equation each exponent differs from the other by a certain number, then the equations are solved by placing the exponent with the smallest exponent outside the brackets.

Example 20. Solve the equation.

Solution: Take the degree with the smallest exponent outside the brackets on the left side of the equation:



Example 21. Solve the equation

Solution: We group separately on the left side of the equation the terms containing degrees with base 4, on the right side - with base 3, then put the degrees with the smallest exponent outside the brackets:

4. Equations Reducing to Quadratic (or Cubic) Equations.

The equations are reduced to a quadratic equation with respect to the new variable y:

a) the type of substitution, while;

b) kind of substitution, while.

Example 22. Solve the equation .

Solution: Let's change the variable and solve the quadratic equation:

.

Answer: 0; 1.

5. Equations homogeneous with respect to exponential functions.

An equation of the form is a homogeneous second-degree equation with respect to unknowns a x and b x ... Such equations are reduced by the preliminary division of both parts by and subsequent substitution to quadratic equations.

Example 23. Solve the equation.

Solution: Divide both sides of the equation by:

Putting, we get a quadratic equation with roots.

Now the problem is reduced to solving a set of equations ... From the first equation we find that. The second equation has no roots, since for any value x.

Answer: -1/2.

6. Equations rational with respect to exponential functions.

Example 24. Solve the equation.

Solution: Divide the numerator and denominator of the fraction by 3 x and instead of two we get one exponential function:

7. Equations of the form .

Such equations with a set of permissible values \u200b\u200b(ODV), determined by the condition, by taking the logarithm of both sides of the equation are reduced to an equivalent equation, which in turn are equivalent to a combination of two equations or.

Example 25. Solve the equation:.

.

Didactic material.

Solve the equations:

1. ; 2. ; 3. ;

4. ; 5. ; 6. ;

9. ; 10. ; 11. ;

14. ; 15. ;

16. ; 17. ;

18. ; 19. ;

20. ; 21. ;

22. ; 23. ;

24. ; 25. .

26. Find the product of the roots of the equation .

27. Find the sum of the roots of the equation .

Find the meaning of the expression:

28., where x 0 - root of the equation ;

29., where x 0 - whole root of the equation .

Solve the equation:

31. ; 32. .

Answers:ten; 2. -2/9; 3. 1/36; 4.0, 0.5; 50; 6.0; 7. -2; 8.2; 9. 1, 3; 10.8; 11.5; 12. 1; 13. ¼; 14. 2; 15. -2, -1; 16. -2, 1; 17.0; 18.1; 19.0; 20. -1.0; 21. -2, 2; 22. -2, 2; 23.4; 24. -1, 2; 25. -2, -1, 3; 26. -0.3; 27.3; 28.11; 29.54; 30. -1, 0, 2, 3; 31.; 32..

Topic number 8.

Exponential inequalities.

1º. An inequality containing a variable in the exponent is called exponential inequality.

2º. The solution to exponential inequalities of the form is based on the following statements:

if, then inequality is equivalent;

if, then the inequality is equivalent.

When solving exponential inequalities, the same techniques are used as when solving exponential equations.

Example 26. Solve inequality (method of transition to one base).

Solution: Since , then the specified inequality can be written as: ... Since, then this inequality is equivalent to the inequality .

Having solved the last inequality, we get.

Example 27. Solve the inequality: ( the method of taking the common factor outside the brackets).

Solution: Let's take out the brackets on the left side of the inequality, on the right side of the inequality and divide both sides of the inequality by (-2), changing the sign of the inequality to the opposite:

Since, then when passing to inequality of indicators, the sign of inequality again changes to the opposite. We receive. Thus, the set of all solutions to this inequality is an interval.

Example 28. Solve the inequality ( by introducing a new variable).

Solution: Let it be. Then this inequality takes the form: or whose solution is the interval.

From here. As the function increases, then.

Didactic material.

Indicate the set of solutions to the inequality:

1. ; 2. ; 3. ;

6. At what values x do the points of the function graph lie below the straight line?

7. At what values x do the points of the function graph lie not below the straight line?

Solve the inequality:

8. ; 9. ; 10. ;

13. Indicate the largest integer solution to inequality .

14. Find the product of the largest integer and the smallest integer solution to the inequality .

Solve the inequality:

15. ; 16. ; 17. ;

18. ; 19. ; 20. ;

21. ; 22. ; 23. ;

24. ; 25. ; 26. .

Find the scope of the function:

27. ; 28. .

29. Find the set of argument values \u200b\u200bfor which the values \u200b\u200bof each of the functions are greater than 3:

and .

Answers: 11.3; 12.3; 13. -3; 14. 1; 15. (0; 0.5); sixteen. ; 17. (-1; 0) U (3; 4); 18. [-2; 2]; 19. (0; + ∞); 20. (0; 1); 21. (3; + ∞); 22. (-∞; 0) U (0.5; + ∞); 23. (0; 1); 24. (-1; 1); 25. (0; 2]; 26. (3; 3.5) U (4; + ∞); 27. (-∞; 3) U (5); 28.)

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