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Chapter 5: Exponential and Logarithmic Functions 5.1 5.2 5.3 5.4 5.5 Inverse Functions Exponential Functions Logarithms and Their Properties Logarithmic Functions Exponential and Logarithmic Equations and Inequalities 5.6 Further Applications and Modeling with Exponential and Logarithmic Functions Copyright © 2007 Pearson Education, Inc. Slide 5-2 5.3 Logarithms and Their Properties Logarithm For all positive numbers a, where a 1, a x y is equivalent to y log a x . A logarithm is an exponent, and loga x is the exponent to which a must be raised in order to obtain x. The number a is called the base of the logarithm, and x is called the argument of the expression loga x. The value of x will always be positive. Copyright © 2007 Pearson Education, Inc. Slide 5-3 5.3 Examples of Logarithms Exponential Form Logarithmic Form 2 8 log 2 8 3 3 16 1 4 2 log 16 4 1 2 5 5 log 5 5 1 1 log 1 0 1 3 0 4 Example Solution 3 4 Solve log 4 x 32 . log 4 x x4 x8 Copyright © 2007 Pearson Education, Inc. 3 2 3 2 Slide 5-4 5.3 Solving Logarithmic Equations Solve a) x log 8 4 Example Solution a) x log 8 4 x 16 8 4 3 x 2 2 3x 2 2 2 3x 2 x log x 16 4 4 x 2 b) b) log x 16 4 . x 4 16 x 2 Since the base must be positive, x = 2. 2 3 Copyright © 2007 Pearson Education, Inc. Slide 5-5 5.3 The Common Logarithm – Base 10 For all positive numbers x, log x log 10 x . Example Evaluate a) log 12 Solution Use a calculator. a) b) c) b) log .1 c) log 53 . log 12 1.07918124 6 log .1 1 3 log . 2218487496 5 Copyright © 2007 Pearson Education, Inc. Slide 5-6 5.3 Application of the Common Logarithm Example In chemistry, the pH of a solution is defined as pH log[ H 3 O ] where [H3O+] is the hydronium ion concentration in moles per liter. The pH value is a measure of acidity or alkalinity of a solution. Pure water has a pH of 7.0, substances with a pH greater than 7.0 are alkaline, and those less than 7.0 are acidic. a) b) Find the pH of a solution with [H3O+] = 2.5×10-4. Find the hydronium ion concentration of a solution with pH = 7.1. Solution a) pH = –log [H3O+] = –log [2.5×10-4] 3.6 b) 7.1 = –log [H3O+] –7.1 = log [H3O+] [H3O+] = 10-7.1 7.9 ×10-8 Copyright © 2007 Pearson Education, Inc. Slide 5-7 5.3 The Natural Logarithm – Base e For all positive numbers x, ln x log e x . • On the calculator, the natural logarithm key is usually found in conjunction with the e x key. Copyright © 2007 Pearson Education, Inc. Slide 5-8 5.3 The Graph of ln x and Some Calculator Examples Example Evaluate (a) ln 12, 10 (b) ln e . Solution (a) (b) ln 12 2 . 48490665 ln e 10 Copyright © 2007 Pearson Education, Inc. 10 Slide 5-9 5.3 Using Natural Logarithms to Solve a Continuous Compounding Problem Example Suppose that $1000 is invested at 3% annual interest, compounded continuously. How long will it take for the amount to grow to $1500? Analytic Solution A Pe rt 1500 1000 e 1 .5 e . 03 t . 03 t ln 1 . 5 . 03 t t Copyright © 2007 Pearson Education, Inc. ln 1 . 5 . 03 13 . 5 years Slide 5-10 5.3 Using Natural Logarithms to Solve a Continuous Compounding Problem Graphing Calculator Solution Let Y1 = 1000e.03t and Y2 = 1500. The table shows that when time (X) is 13.5 years, the amount (Y1) is 1499.3 1500. Copyright © 2007 Pearson Education, Inc. Slide 5-11 5.3 Properties of Logarithms For a > 0, a 1, and any real number k, 1. loga 1 = 0, 2. loga ak = k, log kk a 3. aa log = kk,. k > 0. a Property 1 is true because a0 = 1 for any value of a. Property 2 is true since in exponential form: a k a k . Property 3 is true since logak is the exponent to which a must be raised in order to obtain k. Copyright © 2007 Pearson Education, Inc. Slide 5-12 5.3 Additional Properties of Logarithms For x > 0, y > 0, a > 0, a 1, and any real number r, log a xy log a x log Product Rule Quotient Rule Power Rule log log x a y log a x log x r log r a a a a y. y. x. Examples Assume all variables are positive. Rewrite each expression using the properties of logarithms. 1. log 8 x log 8 log x 2. log 9 15 7 log 9 15 log 9 7 3. log 5 8 log 5 8 Copyright © 2007 Pearson Education, Inc. 1 2 1 2 log 5 8 Slide 5-13 5.3 Example Using Logarithm Properties Example Assume all variables are positive. Use the properties of logarithms to rewrite the expression 3 log n b x y z 5 m . 3 Solution log b n x y z m 5 x y log b m z 3 5 3 5 x y 1 log b m n z 1 n 1 log b x log b y log b z n 3 5 m 1 3 log b x 5 log b y m log b z n Copyright © 2007 Pearson Education, Inc. 3 5 m log b x log b y log b z n n n Slide 5-14 5.3 Example Using Logarithm Properties Example Use the properties of logarithms to write 2 3 1 log m log 2 n log m n as a single logarithm b b b 2 2 with coefficient 1. Solution 1 2 log b m 32 log b 2 n log b m n 2 log b m 1 log 2 1 log b m 2 2 n b 3 2 2 1 2 n 2 3 m 2 log b 2 3n m 3 Copyright © 2007 Pearson Education, Inc. 2 2 m n 3 log b 2 2 n log b m n 3 1 2 log b 8n 3 m Slide 5-15 5.3 The Change-of-Base Rule Change-of-Base Rule For any positive real numbers x, a, and b, where a 1 and b 1, log a x log b x . log b a Proof Let y log a x . a x y log b a log b x y y log b a log b x y Copyright © 2007 Pearson Education, Inc. log b x log b a log a x log b x log b a Slide 5-16 5.3 Using the Change-of-Base Rule Example Evaluate each expression and round to four decimal places. (a) log 5 17 (b) log 2 . 1 Solution Note in the figures below that using either natural or common logarithms produce the same results. (a) Copyright © 2007 Pearson Education, Inc. (b) Slide 5-17 5.3 Modeling the Diversity of Species Example One measure of the diversity of species in an ecological community is the index of diversity, where H P1 log 2 P1 P2 log 2 P2 Pn log 2 Pn and P1, P2, . . . , Pn are the proportions of a sample belonging to each of n species found in the sample. Find the index of diversity in a community where there are two species, with 90 of one species and 10 of the other. Copyright © 2007 Pearson Education, Inc. Slide 5-18 5.3 Modeling the Diversity of Species Solution Since there are a total of 100 members in the community, P1 = 90/100 = .9, and P2 = 10/100 = .1. log 2 . 9 ln . 9 . 152 and log 2 . 1 ln . 1 3 . 32 ln 2 ln 2 H . 9 log 2 . 9 . 1 log 2 . 1 . 9 ( . 152 ) . 1( 3 . 32 ) . 469 Interpretation of this index varies. If two species are equally distributed, the measure of diversity is 1. If there is little diversity, H is close to 0. In this case H .5, so there is neither great nor little diversity. Copyright © 2007 Pearson Education, Inc. Slide 5-19