# 6 Rational Inequalities, (In)equalities with Absolute value; Exponents and Logarithms

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1 AAU - Business Mathematics I Lecture #6, March 16, Rational Inequalities, (In)equalities with Absolute value; Exponents and Logarithms 6.1 Rational Inequalities: x + 1 x 3 > 1, x + 1 x 2 3x + 5 < 0, x 2 x 1 2x 2 + 4x 3 > 5,... Problem: Solve 2x x+2 > 1 Note that we can not just multiply the inequality by (x + 2) and solve the resulting linear inequality 2x > x + 2, because we do not know whether x + 2 is positive or negative and hence we do not know whether we should change the sign of inequality or not. So we distinguish two cases: x + 2 > 0 x > x > x + 2 x > 2 x + 2 < 0 x < x < x + 2 x < 2 Alternative solution: 2x x + 2 > 1 2x x > 0 x 2 x + 2 > 0 Here, we need the ratio of two numbers be positive. This is the case if both numerator and denominator are positive or if both numerator and denominator are negative: x 2 > 0 and x + 2 > 0 x > 2 and x > 2 x > 2 OR x 2 < 0 and x + 2 < 0 x < 2 and x < 2 x < 2 Hence the solution to the problem is x (, 2) (2, ). Problem: Solve and x2 3x 10 1 x 2 x 2 3x 10 2 x 2 3x x 0 33

2 x 2 x 12 0 This fraction is greater or equal to 0 if both numerator and denominator are positive or if both are negative. In this problem, numerator can be equal to 0 as well. Denominator can never be equal to 0! (x + 3)(x 4) 0 (x + 3)(x 4) 0 and > 0 x (, 3] [4, ) and x < 1 x (, 3] OR (x + 3)(x 4) 0 and < 0 x [ 3, 4] and x > 1 x (1, 4] Therefore, the solution to this problem is x (, 3] (1, 4]. 6.2 Equations and Inequalities with Absolute Value - if a is some number, than absolute value of a, a, is the distance of a from 0. x = { x if x 0 x if x < 0 Examples: 4 = 4, 5 = 5, 1 2 = 2 1,... - if a and b are some numbers, than absolute value of a b, a b, is the distance of a from b, or the distance between a and b. It holds, that a b = b a. Examples: 9 4 = 5 = 5, 4 9 = 5 = 5, 5 = 5 0 = 5,... Problem: Solve x 1 = 2 We are looking for such number(s) x that the distance of x from 1 is equal to 2. It s clear that there are 2 such numbers: -1 and 3. Formally: x a = b x a = b or x a = b, i.e. x a = ±b. Then it follows that x = a ± b. Problem: Solve x + 4 = 1 Note that x + 4 = 1 can be written as x ( 4) = 1. We are looking for such number(s) x that the distance of x from -4 is equal to 1. It s clear that there are 2 such numbers: -5 and -3. Note: In the problem x + 4 = 1 we are looking for number(s) such that their distance from -4 is equal to 1. Not distance from 4 is equal to 1!!! 34

3 Problem: Solve 3x 7 = 2 3x 7 = ±2 3x = 7 ± 2 x = 7 ± 2 3 x = 3, 5 3 Problem: Solve 2x + 5 = 3 2x + 5 = 3 2x + 5 = ±3 2x = 5 ± 3 x = 5 ± 3 2 x = 4, 1 If we have variable x on both sides of the equation, solution is not that easy any more: Problem: Solve x + 4 = 3x 8 We distinguish two cases: x x 4... x + 4 = 3x 8 x = 6 x x < 4... x 4 = 3x 8 x = 1 In the first case, the initial condition is x and x = 6 satisfies this condition. Hence, x = 6 is a solution to our equation. In the second case, the initial condition is x But corresponding solution x = 1 does not satisfy this condition. Hence, x = 1 is not a solution to our equation. Problem: Solve x + 4 = 2x 6 Absolute value keeps the expression the same if it is positive, in changes the sign of the expression if it is negative. Generally, any equation with any number of absolute values can be solved by getting rid of absolute values. To do so, we need to divide the problem into subcases for which we can eliminate the absolute value: x + 4 is positive for x 4 and negative for x 4 2x 6 is positive for x 3 and negative for x 3 Put the two together: if x (, 4], both expressions are negative 35

4 if x [ 4, 3], x + 4 is positive and 2x 6 is negative if x [3, ), both expressions are positive Now we solve following three problems: x (, 4] x 4 = 2x + 6 x = 10 x [ 4, 3] x + 4 = 2x + 6 x = 2 3 x [3, ) x + 4 = 2x 6 x = 10 In the firs equality, x = 10 does not satisfy the initial condition x (, 4) and therefore this is not a solution. So we have two solutions of our problem x = 2/3 and 10. INEQUALITIES WITH ABSOLUTE VALUE: Again, solving inequalities with absolute value is almost the same as solving equations with absolute value. Only multiplying or dividing by a negative number changes the sign of inequality. Problem: Solve x 5 < 1 We are looking for all x such that the difference of x from 5 is less than 1. It s clear that this is true for all x (4, 6). x 5 < 1 1 < x 5 < 1 4 < x < 6 Indeed, the inequality holds for all x (4, 6). Problem: Solve 3x 2 7 3x x x 9 5/3 x 3 The inequality holds for all x [ 5/3, 3]. 36

5 6.3 Exponents and logarithms Note that there is a difference between x 2 and 2 x. It makes a big difference whether a variable appears as a base with a constant exponent or as an exponent with a constant base. Exponential function: f(x) = b x, b > 0, b 0 f(x) defines an exponential function for each different constant b, called the base. The independent variable x may assume any real value. We require the base to be positive (b > 0) because if x is for example 1/2, we have f(x) = b x = b 1/2 = b and we only can have a non negative number under the square root. Exponential function properties: a x a y = a x+y (a x ) y = a xy (ab) x = a x b x ( a ) x a x a x = b b x a = y ax y a x = a y if and only if x = y for x 0, a x = b x if and only if a = b 0 x = 0, 1 x = 1, x 0 = 1 for all x Example: Simplify: ( ) (a) 3 4 ( ) = = 4(2 1) 3 (3 2) = 4 3 = 12 ( ) 2 2a 5 (b) 3b 2 3 ( ) 2 2a 5 3b 2 = 4a b 2 2 = a2 b 2 Example: Suppose \$4000 is invested at 10% annual rate compounded annually. How much money will be in the account in 1 year, 2 years, and in 10years? in one year: = 4000 ( ) in two years: 4000 ( ) ( ) = 4000 ( ) 2 in ten years: 4000 ( ) = Generally: If P is the amount of money invested (principal) at an annual rate r (expressed in decimal form), then the amount A in the account at the end of t years is given by: A = P (1 + r) t 37

6 Example: How much do you have to invest if you want to have \$ 5000 in 3 years at 5 % compounded annually? A = P (1 + r) t 5000 = P ( ) 3 P = = 4320 Example: Suppose you deposit \$1000 in a savings and loan that pays 8% compounded semiannually. How much will the savings and loan owe you at the end of 2 years? 10 years? in half a year: = 1000 ( ) 2 in a year: 1000 ( ) ( ) = 1000 ( )2 2 in two years: 1000 ( ) 4 in ten years: 1000 ( ) 20 Generally: If a principal P is invested at an annual rate r (expressed in decimal form) compounded n times a year, then the amount A in the account at the end of t years is given by: A = P ( 1 + r ) nt n Exponential function with base e: f(x) = e x, where e = Now, let s get back to the formula A = P ( 1 + r n) nt. What happens if n increases to infinity? In other words, what it an annual rate r is compounded continuously? A = P ( 1 + r ) ( nt = P ) (n/r)rt [( = P ) m ] rt = P e rt n n/r m For a very large values of m, m, (1 + 1/m) m e Continuous compound interest formula: If a principal P is invested at an annual rate r (expressed as a decimal) compounded continuously, then the amount A in the account at the end of t years is given by: A = P e rt This formula is widely used in business, banking and economics. 38

7 Definition of logarithmic function: For b > 0 and b 1, logarithmic form exponential form y = log b x is equivalent to x = b y For example, y = log 10 x is equivalent to x = 10 y y = log e x is equivalent to x = e y Example: Change each logarithmic form to an equivalent exponential form: log 2 8 = 3 is equivalent to 8 = 2 3 log 25 5 = 1/2 is equivalent to 5 = 25 1/2 log 2 1/4 = 2 is equivalent to 1/4 = 2 2 log 3 27 = 3 is equivalent to 27 = 3 3 log 36 6 = 1/2 is equivalent to 6 = 36 1/2 log 3 1/9 = 2 is equivalent to 1/9 = 3 2 Example: Change each exponential form to an equivalent logarithmic form: 49 = 7 2 is equivalent to log 7 49 = 2 3 = 9 is equivalent to log 9 3 = 1/2 1/5 = 5 1 is equivalent to log 5 1/5 = 1 16 = 4 2 is equivalent to log 4 16 = 2 3 = 27 1/3 is equivalent to log 27 3 = 1/3 4 = 16 1/2 is equivalent to log 16 4 = 1/2 39

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