Chapter 10 Exploring Exponential and Logarithmic Functions By

Chapter 10 Exploring Exponential and Logarithmic Functions By Kathryn Valle

10 -1 Real Exponents and Exponential Functions • An exponential function is any equation in the form y = a·bx where a ≠ 0, b > 0, and b ≠ 1. b is referred to as the base. • Property of Equality for Exponential Functions: If in the equation y = a·bx, b is a positive number other than 1, then bx 1 = bx 2 if and only if x 1 = x 2.

10 -1 Real Exponents and Exponential Functions • Product of Powers Property: To simplify two like terms each with exponents and multiplied together, add the exponents. – Example: 34 · 35 = 39 5√ 2 · 5√ 7 = 5√ 2 + √ 7 • Power of a Power Property: To simplify a term with an exponent and raised to another power, multiply the exponents. – Example: (43)2 = 46 (8√ 5)4 = 84·√ 5

10 -1 Examples • Solve: 128 = 24 n – 1 27 = 24 n – 1 7 = 4 n – 1 8 = 4 n n=2 53 n + 2 > 625 53 n + 2 > 54 3 n + 2 > 4 3 n > 2 n > ²/³

10 -1 Practice 1. Simplify each expression: a. (23)6 b. 7√ 4 + 7√ 3 c. p 5 + p 3 d. (k√ 3)√ 3 2. Solve each equation or inequality. a. 121 = 111 + n b. 33 k = 729 c. 343 = 74 n – 1 d. 5 n 2 = 625 Answers: 1)a) 218 b) 7√ 4 + √ 3 c) p 8 d) k 3 2)a) n = 1 b) n = 2 c) n = 1 d) n = ± 2

10 -2 Logarithms and Logarithmic Functions • A logarithm is an equation in the from logbn = p where b ≠ 1, b > 0, n > 0, and bp = n. • Exponential Equation Logarithmic Equation n = bp p = logbn exponent or logarithm base number – Example: x = 63 can be re-written as 3 = log 6 x ³/2 = log 2 x can be re-written as x = 23/2

10 -2 Logarithms and Logarithmic Functions • A logarithmic function has the from y = logb, where b > 0 and b ≠ 1. • The exponential function y = bx and the logarithmic function y = logb are inverses of each other. This means that their composites are the identity function, or they form an equation with the form y = logb bx is equal to x. – Example: log 5 53 = 3 2 log 2 (x – 1) = x – 1

10 -2 Logarithms and Logarithmic Functions • Property of Equality for Logarithmic Functions: Given that b > 0 and b ≠ 1, then logb x 1 = logb x 2 if and only if x 1 = x 2. – Example: log 8 (k 2 + 6) = log 8 5 k k 2 + 6 = 5 k k 2 – 5 k + 6 = 0 (k – 6)(k + 1) = 0 k = 6 or k = -1

10 -2 Practice 1. Evaluate each expression. a. log 3 ½ 7 b. log 7 49 c. log 5 625 d. log 4 64 2. Solve each equation. a. log 3 x = 2 b. log 5 (t + 4) = log 5 9 t d. log 12 (2 p 2) – log 12 (10 p – 8) e. log 2 (log 4 16) = x c. logk 81 = 4 f. log 9 (4 r 2) – log 9(36) Answers: 1)a) -3 b) 2 c) 4 d) 3 2)a) 9 b) ½ c) 3 d) 1, 5 e) 1 f) -3, 3

10 -3 Properties of Logarithms • Product Property of Logarithms: logb mn = logb m + logb n as long as m, n, and b are positive and b ≠ 1. – Example: Given that log 2 5 ≈ 2. 322, find log 2 80: log 2 80 = log 2 (24 · 5) = log 2 24 + log 2 5 ≈ 4 + 2. 322 ≈ 6. 322 • Quotient Property of Logarithms: As long as m, n, and b are positive numbers and b ≠ 1, then logb m/n = logb m – logb n - Example: Given that log 3 6 ≈ 1. 6309, find log 3 6/81: log 3 6/81 = log 3 6/34 = log 3 6 – log 3 34 ≈ 1. 6309 – 4 ≈ -2. 3691

10 -3 Properties of Logarithms • Power Property of Logarithms: For any real number p and positive numbers m and b, where b ≠ 1, logb mp = p·logb m – Example: Solve ½ log 4 16– 2·log 4 8 = log 4 x log 4 161/2 – log 4 82 = log 4 x log 4 4 – log 4 64 = log 4 x log 4 4/64 = log 4 x x = 4/64 x = 1/16

10 -3 Practice 1. Given log 4 5 ≈ 1. 161 and log 4 3 ≈ 0. 792, evaluate the following: a. log 4 15 b. log 4 192 c. log 4 5/3 d. log 4 144/25 2. Solve each equation. 2 log 3 x = ¼ log 2 256 3 log 6 2 – ½ log 6 25 = log 6 x ½ log 4 144 – log 4 x = log 4 4 1/3 log 5 27 + 2 log 5 x = 4 log 5 3 Answers: 1)a) 1. 953 b) 3. 792 c) 0. 369 d) 1. 544 2)a) x = ± 2 b) x = 8/5 c) x = 36 d) x = 3√ 3 a. b. c. d.

10 -4 Common Logarithms • Logarithms in base 10 are called common logarithms. They are usually written without the subscript 10. – Example: log 10 x = log x • The decimal part of a log is the mantissa and the integer part of the log is called the characteristic. – Example: log (3. 4 x 103) = log 3. 4 + log 103 = 0. 5315 + 3 mantissa characteristic

10 -4 Common Logarithms • In a log we are given a number and asked to find the logarithm, for example log 4. 3. When we are given the logarithm and asked to find the log, we are finding the antilogarithm. – Example: log x = 2. 2643 x = 10 2. 2643 x = 183. 78 – Example: log x = 0. 7924 x = 10 0. 7924 x = 6. 2

10 -4 Practice 1. If log 3600 = 3. 5563, find each number. a. mantissa of log 3600 b. characteristic of log 3600 c. antilog 3. 5563 d. log 3. 6 e. 10 3. 5563 f. mantissa of log 0. 036 2. Find the antilogarithm of each. c. -1. 793 d. 0. 704 – 2 Answers: 1)a) 0. 5563 b) 3 c) 3600 d) 0. 5563 e) 3600 f) 0. 5563 2)a) 314. 775 b) 1. 459 c) 0. 016 d) 0. 051 a. 2. 498 b. 0. 164

10 -5 Natural Logarithms • e is the base for the natural logarithms, which are abbreviated ln. Natural logarithms carry the same properties as logarithms. • e is an irrational number with an approximate value of 2. 718. Also, ln e = 1.

10 -5 Practice 1. Find each value rounded to four decimal places. ln 6. 94 ln 0. 632 ln 34. 025 ln 0. 017 e. f. g. h. antiln -3. 24 antiln 0. 493 antiln -4. 971 antiln 0. 835 Answers: 1)a) 1. 9373 b) -0. 4589 c) 3. 5271 d) -4. 0745 e) 0. 0392 f) 1. 6372 g) 0. 0126 h) 2. 3048 a. b. c. d.

10 -6 Solving Exponential Equations • Exponential equations are equations where the variable appears as an exponent. These equations are solved using the property of equality for logarithmic functions. – Example: 5 x = 18 log 5 x = log 18 x · log 5 = log 18 x = log 18 log 5 x = 1. 796

10 -6 Solving Exponential Equations • When working in bases other than base 10, you must use the Change of Base Formula which says loga n = logb n logb a For this formula a, b, and n are positive numbers where a ≠ 1 and b ≠ 1. - Example: log 7 196 log 196 change of base formula log 7 a = 7, n = 196, b = 10 ≈ 2. 7124

10 -6 Practice 1. Find the value of the logarithm to 3 decimal places. a. log 7 19 b. log 12 34 c. log 3 91 d. log 5 48 2. Use logarithms to solve each equation. Round to three decimal places. a. 13 k = 405 b. 6. 8 b-3 = 17. 1 c. 5 x-2 = 6 x d. 362 p+1 = 14 p-5 b) B = 4. 481 c) x = -17. 655 d) p = -3. 705 Answers: 1)a) 1. 513 b) 1. 419 c) 4. 106 d) 2. 405 2)a) k = 2. 341

10 -7 Growth and Decay • The general formula for growth and decay is y = nekt, where y is the final amount, n is the initial amount, k is a constant, and t is the time. • To solve problems using this formula, you will apply the properties of logarithms.

10 -7 Practice 1. Population Growth: The town of Bloomington. Normal, Illinois, grew from a population of 129, 180 in 1990, to a population of 150, 433 in 2000. a. Use this information to write a growth equation for Bloomington-Normal, where t is the number of years after 1990. b. Use your equation to predict the population of Bloomington-Normal in 2015. c. Use your equation to find the amount year when the population of Bloomington-Normal reaches 223, 525.

10 -7 Practice Solution a. Use this information to write a growth equation for Bloomington-Normal, where t is the number of years after 1990. y = nekt 150, 433 = (129, 180)·ek(10) 1. 16452 = e 10·k ln 1. 16452 = ln e 10·k 0. 152311 = 10·k k = 0. 015231 equation: y = 129, 180·e 0. 015231·t

10 -7 Practice Solution b. Use your equation to predict the population of Bloomington-Normal in 2015. y = 129, 180·e 0. 015231·t y = 129, 180·e(0. 015231)(25) y = 129, 180·e 0. 380775 y = 189, 044

10 -7 Practice Solution c. Use your equation to find the amount year when the population of Bloomington-Normal reaches 223, 525. y = 129, 180·e 0. 015231·t 223, 525 = 129, 180·e 0. 015231·t 1. 73034 = e 0. 015231·t ln 1. 73034 = ln e 0. 015231·t 0. 548318 = 0. 015231·t t = 36 years 1990 + 36 = 2026