Test on exponential and logarithmic equations. Mathematics test on the topic "Logarithmic equations and inequalities

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Description of the presentation on individual slides:

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Scientific manual on algebra Topic: "Logarithmic and exponential equations and inequalities" Completed by: Manuilova L.N.-teacher of mathematics, MBOU secondary school No. 76, Izhevsk Udmurtia

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Contents: Chapter 1. 1.1. The concept of the logarithm 1.2. Properties of the logarithm 1.3. Logarithmic Equations BUT. Theoretical part B. Examples 1.4. Logarithmic inequalities A. Theoretical part B. Examples Chapter 2. 2.1. Power of a positive number 2.2. The exponential function 2.3. exponential equations A. Theoretical part B. Examples 2.4. Exponential inequalities A. Theoretical part B. Examples Chapter 3. 3.1. Test on the topic "Logarithmic equations and inequalities" I level of complexity II level of complexity III level of complexity 3.2. Test on the topic "Exponential equations and inequalities" I level of complexity II level of complexity III level of complexity

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1.1 The concept of the logarithm y x y \u003d b b M 1 0 n y \u003d ax (a\u003e 1) x y \u003d ax (0< a < 1) y= b M 1 0 b у Для любого положительного числа b существует, и притом только одно, число n, такое, что b = an . Это число называют логарифмом числа b по основанию a. n Логарифмом положительного числа b по основанию a (a >0, a ≠ 0) is a number n such that b = an The logarithm of a positive number b to base a (a > 0,a ≠ 1) is denoted as follows: n = loga b From the definition of the Logarithm, it obviously follows that for a > 0 , a ≠ 1, b > 0: a loga b = b

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Logarithmic function y y x x 1 2 2 1 1 1 -1 -1 2 2 -2 -2 3 0 0 y = log2 x y = log3 x y = log⅓x y = log½x The function y = loga x is called the logarithmic function. Properties of the function y = loga x, for a > 0: Continuous and increasing on the interval (0;+∞); If x→+∞, then y→+∞; if x→0, then y→ -∞. Since loga1=0, then it follows from property 1: if x > 1, then y > 0; if 0< х < 1 ,то у < 0. Свойства функции y = loga x, при 0 < a < 1: Непрерывна и убывает на промежутке (0;+∞); Если х→ +∞, то у→ -∞; если х→0, то у→+∞. Так как loga1=0, то из свойства 1 следует: если х >1, then< 0; если 0 < х < 1 ,то у >0.

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Let a, M and N be positive numbers, where a ≠ 1 and k is really a number. Then equalities are true: 1. loga (M N) = loga M + loga N - Logarithm of the product of positive numbers is equal to the sum logarithms of these numbers. 2. loga M = loga M - loga N - The logarithm of the partial positive numbers N is equal to the difference between the logarithms of the dividend and the divisor. 3. loga Mk = k loga M - The logarithm of the power of a positive number is equal to the product of the exponent and the logarithm of this number. 4. loga M = logb M → loga b = 1 - The formula for the transition of logarithms from one logb a logb a base to another. Special cases: 1. log10 b \u003d lg b - The logarithm of a positive number b to base 10 is called decimal logarithm numbers b. 2. loge b \u003d ln b - The logarithm of a positive number b to the base e is called natural logarithm numbers b 1.2 Properties of logarithms

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1. Let a be a given positive number not equal to 1, b be a given real number. Then the equation loga x = b is called the simplest logarithmic equation. For example, equations a) log3 x = 3 ; (1) b) log⅓ x = -2; (2) c) log25 x + 5 log4 x log3 x + 7 log22 x = 0 ; (3) are the simplest logarithmic equations. By the definition of the logarithm, if the number x0 satisfies the numerical equality loga x = b, then the number x0 is ab, and this number x0 = ab is unique. Thus, for any real number b, the equation loga x = b has a single root x0 = ab . 2. Equations that, after replacing the unknown, turn into the simplest logarithmic equations: a) log5 (4x - 3) = 2; (4) b) 2 + 1 = -1; (5) lg(3x + 1) + lg0.01 lg(3x + 1) 1.3 Equations (Theoretical part)

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1.3 Examples log3 x = 3 Rewrite the equation as follows: log3 x = log3 27 Then it is obvious that this equation has a unique root x0 = 27. Answer: 27. b) log1/3 x = -2 This equation has a unique root x0 = ( ⅓)-2 \u003d 9 Answer: 9. c) log25 x + 5 log4 x log3 x + 7 log22 x = 0 (1) Reducing all logarithms to the same base, we rewrite the equation in the form: 0 (2) log25 x log5 4 log5 3 log25 2 Therefore, equation (1), and hence equation (2), are equivalent to the equation log25 x = 0, which has a unique root x0 = 1. Therefore, equation (1) has a unique root x0 = 1. Answer: 1. a, b – simplest equations; c - an equation that, after transformations, turns into the simplest log. the equation

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1.3 Examples a) log5 (4x – 3) = 2 (1) Entering a new known t = 4x – 3, we rewrite the equation in the form: log5 t = 2. This equation has a single root t1 = 52 =25. To find the root of equation (1), you need to solve the equation: 4x - 3 = 25. (2) It has a single root x1 =7. Therefore, equation (1) also has a single root x1=7. Answer: 7. b) 2 + 1 = -1 (1) lg(3x + 1) + lg0.01 lg(3x + 1) Introducing a new unknown t = lg (3x + 1) and considering that lg 0.01 = -2, we rewrite equation (1) in the form: 2 + 1 = -1 (2) t - 2 t roots of equation (1), it is necessary to combine the roots of the two equations lg(3x + 1) = -2 and lg(3x + 1) = 1. The first equation is equivalent to the equation 3x + 1 = 10-2 , which has a single root x1 = -0.33. The second equation is equivalent to 3x + 1 = 10, which also has a single root x2 = 3. Answer: -0.33; 3. a, b - equations that are reduced to the simplest by replacing the unknown

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1.4 Inequalities (Theoretical part) Let a be a given positive number not equal to 1, b be a given real number. Then the inequalities: logа x > b (1) logа x< b (2) являются простейшими логарифмическими неравенствами. Неравенства (1) и (2) можно переписать в виде: loga x >log x0 (3) log x< loga x0 (4) , где x0 = ab . Если a >1, then the function y = loga x increases over its entire domain of definition, i.e. on the interval (0;+∞). Therefore, for any number x > x0, numerical inequality loga x > loga x0 , and for any number x from the interval 0< x < x0 справедливо числовое неравенство logа x < logа x0 . Кроме того, равенство logа x = logа x0 справедливо лишь при х = х0 . Таким образом, при а >1 and any real number b, the set of all solutions of inequality (3) is the interval (x0 ;+ ∞), and the set of all solutions of inequality (4) is the interval (0; x0) . If 0< a < 1, то функция y = loga х убывает. Поэтому для любого числа x >x0 the numerical inequality loga x< loga x0 , а для любого числа х из промежутка 0 < x < x0 справедливо числовое неравенство loga x >log x0 . In addition, the equality loga x = loga x0 is valid only for x = x 0 . Thus, at 0< a < 1 и любом действительном числе b множество всех решений неравенства (3) есть интервал (0; х0) , а множество всех решений неравенства (4) есть интервал (х0 ;+∞).

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1.4 Inequalities (Theoretical part) coordinate plane xOy consider the graphs of the function y = loga x and y = b. The line y = b intersects the graph of the function y = loga x at a single point x0 = ab . If a > 1, then for each x > x0 the corresponding point of the graph of the function y = loga x is above the line y = b, i.e. for each x > x0 the corresponding ordinate y = ax is greater than the ordinate ax0 , and for each x from the interval 0< x < x0 соответствующая точка графика функции y = loga x находится ниже прямой y = b. Если же 0 < a <1, то, наоборот, для каждого x >x0 the corresponding point of the graph of the function y = loga x is below the line y = b, and for each x from the intervals 0< x < x0 соответствующая точка графика функции y = loga x находится выше прямой y = b. у у х х 1 1 1 1 х0 0 0 y = b y = loga x (a >1) y = b y = log x (0< a < 1) х0

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1.4 Examples Let's solve the inequality log1/3 x > -2. (1) Since -2 = log⅓ 9 , inequality (1) can be rewritten as log ⅓x > log ⅓ 9 (2) Since ⅓< 1, то функция y = log⅓ x убывающая. Поэтому множество всех решений неравенства (2), а значит и неравенства (1), есть интервал 0 < x <9. Ответ: (0;9). 2. Решим неравенство log4 x >½. (3) Since ½ = log4 2, inequality (3) can be rewritten as log4 x > log4 2 (4) Since 4 > 1, the function y = log4 x is increasing. Therefore, the set of all solutions of inequality (4), and hence of inequality (3), is the interval (2;+∞). Answer: (2;+∞). (see Fig. 1) x y 1 2 3 4 1 -1 0 Fig. 1 y = ½ y = log4 x

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1.4 Examples Let's solve the inequality log3 x – 3log9 x – log81 x > 1.5. (5) Since log9 x = (log3 x) / (log3 9) = (log3 x) / 2 = ½ (log3 x), log81 x = (log3 x) / (log3 81) = (log3 x) / 4 = ¼ (log3 x), then inequality (5) can be rewritten as: (1 – 1.5 – ¼) log3 x > 1.5 or as log3 x< log3 1/9. (6) Так как 3 >1, then the function y = log3 x is increasing. Therefore, the set of all solutions of inequality (6), and hence inequality (5), is the interval 0< x < 1/9 (рис.2) Ответ: (0 ; 1/9). y x 1 2 0 -1 y = log3 x y = -2 (рис.2) 1/9

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2.1 Power of a positive number Power c rational indicator Let a be a positive number and p/q be rational number(q ≥ 2). By definition, the number a to the power p/q is the arithmetic root of the power q of a to the power p, i.e. ap/q = q√ap . THEOREM. Let a be a positive number, p an integer, k and q integers, q ≥ 2, k ≥ 2. Then the equalities a) ap/q = (a1/p)p ; b) ap/q = a pk / qk ; c) ap = a pq / q ; Properties of a power with a rational exponent THEOREM 1. A positive number a in a power with any rational exponent r is positive: ar > 0 THEOREM 2. Let a be a positive number, and r1 , r2 and r be rational numbers. Then the following properties are true: 1. When multiplying powers with rational exponents of the same positive number, the exponents are added: ar1 ∙ ar2 = ar1 + r2 . 2. When dividing degrees with rational exponents of the same positive number, the exponents are subtracted: ar1: ar2 = ar1 - r2. 3. When raising a power with a rational exponent of a positive number in rational degree the exponents are multiplied: (a r1) r2 = a r1∙ r2 . THEOREM 3. Let a and b be positive numbers and r a rational number. Then the following properties of a degree with a rational exponent are true: The degree with a rational exponent of a product of positive numbers is equal to the product of the same powers of the factors: (ab)r = ar ∙ br . The degree with a rational exponent of the quotient of positive numbers is equal to the quotient of the same powers of the dividend and divisor: (a / b)r = ar / br. THEOREM 4. Let a > 1 and r be a rational number. Then ar > 1 for r > 0 0< ar < 1 при r < 0 ТЕОРЕМА 5. Пусть число a >1, and the rational numbers r1 and r2 satisfy the inequality r1< r2 . Тогда a r1 < a r2 . ТЕОРЕМА 6. Пусть число a принадлежит интервалу (0;1), а рациональные числа r1 и r2 удовлетворяют неравенству r1< r2 . Тогда a r1 < a r2 .

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2.2 The exponential function Consider a function y = a (1) , where a > 0 and a ≠ 0, on the set of rational numbers. For every rational number r, the number ar is defined. By this, the function (1) is still defined on the set of rational numbers. The graph of this function in the x0y coordinate system is a set of points (x; ax) , where x is any rational number. For a > 1, this graph is schematically shown in Figure (1), and for 0< a < 1 – на рисунке (2). у у x x 1 2 1 2 3 -2 -1 -2 -1 0 1 1 2 2 Рис. 1 Рис. 2 Её называют exponential function with base a.

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2.3 Exponential equations (Theoretical part) 1. Let a be a given positive number not equal to 1, b be a given real number. Then the equation ax = b (1) is called the simplest exponential equation. For example, the equations 2x = 8, (1/3)x = 9, 25x = -25 are the simplest exponential equations. The root (or solution) of an equation with unknown x is the number x0, when substituting it into the equation instead of x, the correct numerical equality is obtained. To solve an equation means to find all its roots or show that there are none. Since ax0 > 0 for any real number x0 for which the numerical equality ax0 = b would be true satisfies singular x0 = log b. Thus, equation (1) : For b ≤ 0, it has no roots; For b > 0, it has a single root x0 = loga b. 2. Equations that, after replacing the unknown, turn into the simplest exponential equations.

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2.3 Examples Let's solve the equation (1/2) x = 2 (2) Since 2 > 1, this equation has a single root x0 = log½ 2 = -1. Answer: -1. We solve the equation 3x \u003d 5 (3) Since 5\u003e 0, then this equation has a single root x0 \u003d log3 5. Answer: log3 5. We solve the equation 25x \u003d -25 Since -25< 0, то это уравнение не имеет корней. Ответ: нет корней. Для отыскания корня уравнения ax = b (1) при b >0 this equation is often written as ax = aα, where α = loga b . Then it is obvious that the only root of this equation, and hence of equation (1), is the number α. Since equation (2) can be written as (1/2) x = (1/2) -1, then its only root is x0 = -1. Since equation (3) can be written as 3x = 3log 35 , its only root is x0 = log3 5.

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2.3 Examples Now let's consider the equations, which, after simple transformations, turn into the simplest exponential equations. We solve the equation 5x+2 - 2 5x - 3 5x+1 = 200 (4) Since 5x+2 = 25 5x, 5x+1 = 5 5x, equation (4) can be rewritten as 5x ( 25 - 2 - 15) = 200 or in the form 5x = 52 (5) Obviously, equation (5), and hence equation (4), have a single root x0 = 2. Answer: 2. Solve the equation 4 3x - 9 2x = 0 (6) Since 2x ≠ 0 for any real number, dividing equation (6) by 2x, we obtain the equation 4 (3/2)x - 9 = 0, (7) equivalent to equation (6 ). Equation (7) can be rewritten as (3/2)x = (3/2)2. (8) Since equation (8) has a unique root x0 = 2, then the equivalent equation (6) has a unique root x0 = 2. Answer: 2.

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2.3 Examples Let's solve the equation 9 2x2-4x + 2 - 2 · 34x2 – 8x + 3 -1 = 0 . (9) Rewriting equation (9) as 34x2 – 8x + 3 = 1 , we introduce a new unknown t = 4x2 – 8x + 3. Then equation (9) can be rewritten as 3t = 1. (10) Since equation (10 ) has a single root t1 = 0, then in order to find the roots of equation (9), it is necessary to solve the equation 4x2 - 8x + 3 = 0. This equation has two roots x1 = 1/2, x2 = 3/2, so the equation (9) has the same roots. Answer: 1/2; 3/2. Now consider the solution of equations that, after introducing a new unknown t, turn into quadratic or rational equations with an unknown t. We solve the equation 4x - 3 2x + 2 = 0. (11) Since 4x = (2x)2 , equation (11) can be rewritten in the form (2x)2 - 3 2x + 2 = 0. Introducing a new unknown t = 2x, we get the quadratic equation t2 - 3t + 2 = 0, which has two roots t1 = 1, t2 = 2. Therefore, to find all the roots of equation (11), we must combine all the roots of the two equations 2x = 1 and 2x = 2 After solving these simplest exponential equations, we obtain that all roots of equation (11) are x1 = 0; x2 = 1. Answer: 0; one .

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2.4 Exponential inequalities (Theoretical part) Let a be a given positive number not equal to 1, b be a given real number. Then the inequalities ax > b (1) and ax< b (2) называют простейшими показательными неравенствами. Например, неравенства: 2x < 3 , (1/3)x >4√3, 25x< -25 являются простейшими показательными неравенствами. Решением неравенства с неизвестным х называют число х0 , при подстановке которого в неравенство вместо х получается верное числовое неравенство. Решить неравенство - значит найти все его решения или показать, что их нет. Поскольку a x0 >0 for any real number x0, then for b ≤ 0, the inequality a x0 > b is true for Any real number x0, but there is not a single real number x0 for which the numerical inequality a x0 would be true< b . Таким образом, если b ≤ 0 , то множество всех решений неравенства (1) есть интервал (-∞;+∞), а неравенство (2) решений не имеет. Если же b >0, then inequality (1) and (2) can be rewritten as ax > ax0 (1) and ax< ax0 , (2) , где х0 = loga b. Рассмотрим решение неравенств (3) и (4) сначала при а >1. Since the function y = ax is increasing for such a, then for any number x > > ax0, and for any number x > x0, the numerical inequality ax< ax0 . Кроме того, равенство ax = ax0 справедливо лишь при х = x0 .

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2.4 Exponential inequalities (Theoretical part) Thus, for b > 0 and a > 1, the set of all solutions of inequality (3) is the interval (x0 ;+∞), and the set of all solutions of inequality (4) is the interval (-∞; x0) , where x0 = log b. Let now 0< a < 1. Так как для такого а функция y = aх является убывающей, то для любого числа х >x0 the numerical inequality ax< ax0 . Кроме того, равенство ax = ax0 справедливо лишь при х = x0 . Таким образом, при b >0 and 0< a < 1 множество всех решений неравенства (3) есть интервал (-∞; x0), а множество всех решений неравенства (4) есть интервал (x0 ;+∞), где x0 = loga b. Приведенное выше решение простейших показательных неравенств можно дополнить графической иллюстрацией. Рассмотрим графики функций y = aх и y = b. Ясно, что при b ≤ 0 прямая y = b не пересекает график функции y = aх, так как расположена под кривой y = aх (а, б). Поэтому для любых х выполняется неравенство ax >b and there are no x for which the inequality ax< b . При b >0, the line y = b intersects the graph of the function y = ax at a single point x0 = loga b. 1 y y x x y = 0 y = 0 y = ax (a > 1) 0 1 y = b (b< 0) y = b (b < 0) 1 0 1 y = ax (0 < a < 1) a) б)

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2.4 Examples Solve the inequality 2x< 8 . (1) Так как 8 >0, then inequality (1) can be rewritten as 2x< 23. (2) Так как 2 >1, then the function y = 2x is increasing. Therefore, solutions to inequality (2), and hence inequality (1), are all x< 3. Ответ: (-∞; 3). Решим неравенство (1/3)х < 5 . (3) Так как 5 >0, then this inequality (3) can be rewritten as (1/3) x< (1/3) log⅓ 5 . (4) Так как 0 < 1/3 < 1, то функция y = (1/3)x убывающая. Поэтому решениями неравенства (4), а значит и неравенства (3), являются все х >log⅓ 5 . Answer: (log⅓ 5; +∞). Let us consider an inequality which, after the replacement of the unknown, becomes the simplest one exponential inequality. Solve the inequality 5 3x2 - 2x – 6< 1/5 . (5) Введя новое неизвестное t = 3x2 - 2x – 6, перепишем неравенство (5) в виде 5t < 5-1 . Так как 5 >1, then all solutions of this inequality are all t< -1. следовательно, все решения неравенства (5) есть решения неравенства 3x2 - 2x – 6 < -1. (6) Решив square inequality(6), find all its solutions: -1< x < 5/3 . Они являются решениями неравенства (5). Ответ: (-1 ; 5/3).

Logarithmic equations, their types and methods of solution Concentration of attention: Concentration of attention is equal to N . N = (number of correct answers) x 0.125 x 100%. write down special case formulas for converting to the logarithm of another base Write down the formula for converting to the logarithm of another base What is the logarithm of the power of a number and base? What is the logarithm of the base power? What is the logarithm of the power of a number? What is the logarithm of the quotient? What is the logarithm of the product? Formulate the definition of the logarithm

Consider mutual arrangement graph of the function y = log a x (a > 0, a ≠ 1) and a straight line y = b . y = log a x (a>1) y x 0 y = log a x (0

Logarithmic equations, their types and methods for solving CONCLUSION: The graph of the function y = log a x (a > 0, a ≠ 1) and the line y = b intersect at a single point, i.e. the equation log a x = b, a > 0, a ≠ 1 , x > 0 has a unique solution x 0 = a b .

DEFINITION: The equation log a x = b , a > 0 , a ≠ 1 , x > 0 is called the simplest logarithmic equation. Logarithmic equations, their types and methods of solution Example:

Types and methods for solving logarithmic equations. DEFINITION: Equations are called logarithmic if they contain the unknown under the sign of the logarithm or in the base of the logarithm (or both at the same time). Logarithmic equations, their types and methods of solution

Types and methods for solving logarithmic equations. APPENDIX: When solving logarithmic equations, it is necessary to take into account: the area allowed values logarithm: only positive values can be under the sign of the logarithm; at the base of the logarithms - only positive values other than one; properties of logarithms; potentiation action. Logarithmic equations, their types and methods of solution

Logarithmic equations, their types and methods for solving Types and methods for solving logarithmic equations. 1) The simplest logarithmic equations. Example #1 Answer: Solution:

Logarithmic equations, their types and methods for solving Types and methods for solving logarithmic equations. 2) Logarithmic equations, reduced to the simplest logarithmic equations. Example #1 Answer: Solution:

Logarithmic equations, their types and methods for solving Types and methods for solving logarithmic equations. 3) Logarithmic equations reducing to quadratic equations. Example #1 Answer: Solution:

Logarithmic equations, their types and methods for solving Types and methods for solving logarithmic equations. 3) Logarithmic equations reducing to quadratic equations. Example No. 2 Answer: Solution: In the found range of acceptable values for the variable x, we transform the equation using the properties of logarithms. Taking into account the range of permissible values, we obtain: 10; one hundred

Logarithmic equations, their types and methods for solving Types and methods for solving logarithmic equations. 4) Logarithmic equations reducing to rational equations. Example No. 2 Answer: Solution: In the found range of acceptable values for the variable x, we transform given equation and we get: Let's return to the variable x:

Logarithmic equations, their types and methods for solving Types and methods for solving logarithmic equations. 5) Logarithmic equations with a variable in the base and under the sign of the logarithm. Example No. 1 Answer: Solution: In the found range of acceptable values for the variable x, we transform the equation and get: Taking into account the range of acceptable values for the variable x, we get:

Logarithmic equations, their types and methods for solving Types and methods for solving logarithmic equations. 5) Logarithmic equations with a variable in the base and under the sign of the logarithm. Example No. 2 Answer: Solution: In the found range of acceptable values for the variable x, the equation is equivalent to the set: Taking into account the range of acceptable values for the variable x, we get: 5;6.

Logarithmic equations, their types and methods of solution

When solving logarithmic equations and inequalities, they use the properties of logarithms, as well as the properties of the logarithmic function

y=log a x, a > 0, a 1:

1) Domain of definition: x > 0;

2) Range: y R ;

3) log a x 1 = log a x 2 x 1 = x 2;

4) For a>1, the function y=log a x increases, for 0< a < 1 функция y=log a x убывает при всех x >0, i.e.

a >1 and log a x 1 >log a x 2 x 1 >x 2 ,
0 log a x 2 x 1< x 2 ;

When passing from logarithmic equations (inequalities) to equations (inequalities) that do not contain the sign of the logarithm, one should take into account the range of permissible values (ODV) of the original equation (inequality).

Tasks and tests on the topic "Logarithmic equations"

Logarithmic Equations
Lessons: 4 Assignments: 25 Tests: 1
Systems of exponential and logarithmic equations - Demonstrative and logarithmic functions Grade 11
Lessons: 1 Assignments: 15 Tests: 1
§5.1. Solving logarithmic equations
Lessons: 1 Assignments: 38
§7 Exponential and logarithmic equations and inequalities - Section 5. Exponential and logarithmic functions Grade 10
Lessons: 1 Assignments: 17
Equivalence of Equations - Equations and inequalities Grade 11
Lessons: 2 Assignments: 9 Tests: 1

When solving logarithmic equations, in many cases it is necessary to use the properties of the logarithm of the product, quotient, degree. In cases where there are logarithms with different bases in one logarithmic equation, the application specified properties possible only after the transition to logarithms with equal bases.

In addition, the solution of the logarithmic equation should begin with finding the area of \u200b\u200bpermissible values (O.D.Z.) given equation, because in the process of solving, the appearance of extraneous roots is possible. When completing the solution, do not forget to check the roots found for belonging to O.D.Z.

It is possible to solve logarithmic equations without using O.D.Z. In this case, verification is a mandatory element of the solution.

Examples.

Solve Equations:

a) log 3 (5x - 1) = 2.

Solution:

ODZ: 5x - 1 > 0; x > 1/5.
log 3 (5x - 1) = 2,
log 3 (5x - 1) = log 3 3 2,
5x - 1 \u003d 9,
x = 2.

the main objective when working with the offered tickets:

to teach students to see the commonality in solving the corresponding equations and inequalities and the difference when recording answers;
saving time;
ability to navigate the content of this material.

If the first goal does not raise questions, then the time savings are not immediately felt. Although it was the lack of time that affected the ticket structure. They are based on the same principle. Equations and inequalities are arranged so that it is easier to establish a correspondence between them.

And despite the teacher's recommendation: to solve the equation and immediately after it draw up the solution of the corresponding inequality, half of the students preferred to first solve all the equations from the first column, and then start solving the inequalities. When writing the answer, pay attention to the fact that, due to the lack of roots, the equation does not follow that the inequality will not have solutions.

When passing the second test, such problems did not arise, since many of them developed the ability to “see” and developed certain skills.

In each ticket, the material is selected so that, in addition to equations (inequalities) solved by definition and properties, equations (inequalities) are given that are solved by factorization; change of variables. And, of course, the decision is repeated quadratic equations and inequalities, second degree.

There are 26 tasks in the tickets. Therefore, the students were offered such norms: “5” - 26 ass. , “4” - 19–25 ass. , “3” - 14–18 ass. , “2” – less than 14 ass.

A student applying for a grade of “5” must have time to solve all the equations and inequalities for the lesson. The first fourteen tasks are the mandatory minimum. Of course, the report can also be retaken. But it is desirable that they fit within the allotted time.

When preparing for the exam, when the skills for solving equations (inequalities) are already formed, tasks can be replaced. For example, these:

indicate the sum (product) of the roots of the equation;
indicate the smallest (largest) root of the equation;
find the smallest (greatest) integer solution of the inequality;
find the sum (product) of entire solutions of the inequality.

Of course, each teacher can add to this list himself. Depending on the class, it becomes necessary to pay more attention to some tasks, less to others.

Tickets can be used both for tests and for independent work. Each ticket consists of two blocks: a basic level of(Level 1) and advanced (Level 2). The block consists of two parts: equations and inequalities, which are divided into two columns to make it easier for the student to match between them.

Below are six ticket options for each topic. Answers have been given to them.

Attachment 1. Logarithmic equations and inequalities.

Appendix 2. exponential equations and inequalities.

Appendix 3 Answers to tickets on algebra and the beginnings of analysis.