MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §9.3a Logarithms

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §9.2 → Inverse Functions  Any QUESTIONS About HomeWork §9.2 → HW MTH 55

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 3 Bruce Mayer, PE Chabot College Mathematics Logarithm → What is it?  Concept: If b > 0 and b ≠ 1, then y = log b x is equivalent to x = b y  Symbolically x = b y y = log b x The exponent is the logarithm. The base is the base of the logarithm.

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 4 Bruce Mayer, PE Chabot College Mathematics Logarithm Illustrated  Consider the exponential function f(x) = 3 x. Like all exponential functions, f is one-to-one. Can a formula for f −1 be found? Use the 4-Step Method f − 1 (x) ≡ the exponent to which we must raise 3 to get x. y = 3 x x = 3 y y ≡ the exponent to which we must raise 3 to get x. 4-Step

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 5 Bruce Mayer, PE Chabot College Mathematics Logarithm Illustrated  Now define a new symbol to replace the words “the exponent to which we must raise 3 to get x”: log 3 x, read “the logarithm, base 3, of x,” or “log, base 3, of x,” means “the exponent to which we raise 3 to get x.”  Thus if f(x) = 3 x, then f −1 (x) = log 3 x. Note that f −1 (9) = log 3 9 = 2, as 2 is the exponent to which we raise 3 to get 9

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 6 Bruce Mayer, PE Chabot College Mathematics Example  Evaluate Logarithms  Evaluate: a) log 3 81 b) log 3 1 c) log 3 (1/9)  Solution: a)Think of log 3 81 as the exponent to which we raise 3 to get 81. The exponent is 4. Thus, since 3 4 = 81, log 3 81 = 4. b)ask: “To what exponent do we raise 3 in order to get 1?” That exponent is 0. So, log 3 1 = 0 c)To what exponent do we raise 3 in order to get 1/9? Since 3 −2 = 1/9 we have log 3 (1/9) = −2

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 7 Bruce Mayer, PE Chabot College Mathematics The Meaning of log a x  For x > 0 and a a positive constant other than 1, log a x is the exponent to which a must be raised in order to get x. Thus, log a x = m means a m = x  or equivalently, log a x is that unique exponent for which

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example  Exponential to Log  Write each exponential equation in logarithmic form.  Soln

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example  Log to Exponential  Write each logarithmic equation in exponential form  Soln

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example  Evaluate Logarithms  Find the value of each of the following logarithms  Solution

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example  Evaluate Logarithms  Solution (cont.)

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example  Use Log Definition  Solve each equation for x, y or z  Solution

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 13 Bruce Mayer, PE Chabot College Mathematics Example  Use Log Definition  Solution (cont.)

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 14 Bruce Mayer, PE Chabot College Mathematics Inverse Property of Logarithms  Recall Def: For x > 0, a > 0, and a ≠ 1,  In other words, The logarithmic function is the inverse function of the exponential function; e.g.

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 15 Bruce Mayer, PE Chabot College Mathematics Show Log a a x = x

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example  Inverse Property  Evaluate:  Solution Remember that log 5 23 is the exponent to which 5 is raised to get 23. Raising 5 to that exponent, obtain

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 17 Bruce Mayer, PE Chabot College Mathematics Basic Properties of Logarithms  For any base a > 0, with a ≠ 1, Discern from the Log Definition 1.Log a a = 1 As 1 is the exponent to which a must be raised to obtain a (a 1 = a) 2.Log a 1 = 0 As 0 is the exponent to which a must be raised to obtain 1 (a 0 = 1)

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 18 Bruce Mayer, PE Chabot College Mathematics Graph Logarithmic Function  Sketch the graph of y = log 3 x  Soln: Make T-Table → xy = log 3 x(x, y) 3 –3 = 1/27–3(1/27, –3) 3 –2 = 1/9–2(1/9, –2) 3 –3 = 1/3–1(1/3, –1) 3 0 = 10(1, 0) 3 1 = 31(3, 1) 3 2 = 92(9, 2)

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 19 Bruce Mayer, PE Chabot College Mathematics Graph Logarithmic Function  Plot the ordered pairs and connect the dots with a smooth curve to obtain the graph of y = log 3 x

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example  Graph by Inverse  Graph y = f(x) = 3 x  Solution: Use Inverse Relation for Logs & Exponentials  Reflect the graph of y = 3 x in the line y = x to obtain the graph of y = f −1 (x) = log 3 x

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 21 Bruce Mayer, PE Chabot College Mathematics Domain of Logarithmic Fcns  Recall that the Domain of f(x) = a x is (−∞, ∞) Range of f(x) = a x is (0, ∞)  Since the Logarithmic function is the inverse of the Exponential function, Domain of f −1 (x) = log a x is (0, ∞) Range of f −1 (x) = log a x is (−∞, ∞)  Thus, the logarithms of 0 and negative numbers are NOT defined.

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example  Find Domain  Find the domain of each function.  Solution a. The Domain of a logarithmic function must be positive, that is,  Thus The domain of f is (−∞, 2).

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example  Find Domain  Find the domain of each function.  Solution b. The Domain of a logarithmic function must be positive, that is,  Need to Avoid Negative-Logs AND Division by Zero

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example  Find Domain  Soln b. (cont.)  Set numerator = 0 & denominator = 0  Construct a SIGN CHART x − 2 = 0 x + 1 = 0 x = 2 x = −1  The domain of f is (−∞, −1)U(2, ∞).

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 25 Bruce Mayer, PE Chabot College Mathematics Properties of Exponential and Logarithmic Functions Exponential Function f (x) = a x Logarithmic Function f (x) = log a x Domain (0, ∞) Range (–∞, ∞) Domain (–∞, ∞) Range (0, ∞) x-intercept is 1 No y-intercept y-intercept is 1 No x-intercept x-axis (y = 0) is the horizontal asymptote y-axis (x = 0) is the vertical asymptote

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 26 Bruce Mayer, PE Chabot College Mathematics Properties of Exponential and Logarithmic Functions Exponential Function f (x) = a x Logarithmic Function f (x) = log a x Is one-to-one, that is, log a u = log a v if and only if u = v Is one-to-one, that is, a u = a v if and only if u = v Increasing if a > 1 Decreasing if 0 < a < 1

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 27 Bruce Mayer, PE Chabot College Mathematics Graphs of Logarithmic Fcns f (x) = log a x (0 < a < 1)f (x) = log a x (a > 1)

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 28 Bruce Mayer, PE Chabot College Mathematics Graph Logs by Translation  Start with the graph of f(x) = log 3 x and use Translation Transformations to sketch the graph of each function  Also State the DOMAIN and RANGE and the VERTICAL ASYMPTOTE for the graph of each function

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 29 Bruce Mayer, PE Chabot College Mathematics Graph Logs by Translation  Solution Shift UP 2 Domain (0, ∞) Range (−∞, ∞) Vertical asymptote x = 0

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 30 Bruce Mayer, PE Chabot College Mathematics Graph Logs by Translation  Solution Shift RIGHT 1 Domain (1, ∞) Range (−∞, ∞) Vertical asymptote x = 1

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 31 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §9.3 Exercise Set 8, 18, 26, 38, 48  Logs & Exponentials Are Inverse Functions

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 32 Bruce Mayer, PE Chabot College Mathematics All Done for Today Inventor of Logarithms Born: 1550 in Merchiston Castle, Edinburgh, Scotland Died: 4 April 1617 in Edinburgh, Scotland John Napier

MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt 33 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –