Chapter 10 Exploring Exponential and Logarithmic Functions By Kathryn Valle

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 bx1 = bx2 if and only if x1 = x2.

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 = 24n – 1 53n + 2 > 625

27 = 24n – 1 53n + 2 > 54

7 = 4n – 1 3n + 2 > 4

8 = 4n 3n > 2

n = 2 n > ²/³

10-1 Practice

1. Simplify each expression:a. (23)6 c. p5 + p3

b. 7√4 + 7√3 d. (k√3)√3

2. Solve each equation or inequality.a. 121 = 111 + n c. 343 = 74n – 1

b. 33k = 729 d. 5n2 = 625

Answers: 1)a) 218 b) 7√4 + √3 c) p8 d) k3 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 = log6 x

³/2 = log2 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: log5 53 = 3

2log2 (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 x1 = logb x2 if and only if x1 = x2.

– Example: log8 (k2 + 6) = log8 5k

k2 + 6 = 5k

k2 – 5k + 6 = 0

(k – 6)(k + 1) = 0

k = 6 or k = -1

10-2 Practice

1. Evaluate each expression.a. log3 ½7 c. log5 625

b. log7 49 d. log4 64

2. Solve each equation.a. log3 x = 2 d. log12 (2p2) – log12

(10p – 8)

b. log5 (t + 4) = log5 9t e. log2 (log4 16) = x

c. logk 81 = 4 f. log9 (4r2) – log9(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 log2 5 ≈ 2.322, find log2 80:

log2 80 = log2 (24 · 5) = log2

24 + log2 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 log3 6 ≈ 1.6309, find log3 6/81:

log3 6/81 = log3 6/34 = log3 6 – log3 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 log4 16– 2·log4 8 = log4 x

½ log4 16– 2·log4 8 = log4 x

log4 161/2 – log4 82 = log4 x

log4 4 – log4 64 = log4 x

log4 4/64 = log4 x

x = 4/64

x = 1/16

10-3 Practice1. Given log4 5 ≈ 1.161 and log4 3 ≈ 0.792, evaluate

the following:

a. log4 15 b. log4 192

c. log4 5/3 d. log4 144/25

2. Solve each equation.a. 2 log3 x = ¼ log2 256

b. 3 log6 2 – ½ log6 25 = log6 x

c. ½ log4 144 – log4 x = log4 4

d. 1/3 log5 27 + 2 log5 x = 4 log5 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

10-4 Common Logarithms

• Logarithms in base 10 are called common logarithms. They are usually written without the subscript 10.– Example: log10 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 d. log 3.6

b. characteristic of log 3600 e. 10 3.5563

c. antilog 3.5563 f. mantissa of log 0.036

2. Find the antilogarithm of each.a. 2.498 c. -1.793

b. 0.164 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

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.a. ln 6.94 e. antiln -3.24

b. ln 0.632 f. antiln 0.493

c. ln 34.025 g. antiln -4.971

d. ln 0.017 h. 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

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: 5x = 18

log 5x = 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 aFor this formula a, b, and n are positive numbers where a ≠ 1 and b ≠ 1.- Example: log7 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. log7 19 c. log3 91

b. log12 34 d. log5 48

2. Use logarithms to solve each equation. Round to three decimal places.

a. 13k = 405 c. 5x-2 = 6x

b. 6.8b-3 = 17.1 d. 362p+1 = 14p-5

Answers: 1)a) 1.513 b) 1.419 c) 4.106 d) 2.405 2)a) k = 2.341

b) B = 4.481 c) x = -17.655 d) p = -3.705

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 = e10·k

ln 1.16452 = ln e10·k

0.152311 = 10·k

k = 0.015231

equation: y = 129,180·e0.015231·t

10-7 Practice Solution

b. Use your equation to predict the population of Bloomington-Normal in 2015.

y = 129,180·e0.015231·t

y = 129,180·e(0.015231)(25)

y = 129,180·e0.380775

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·e0.015231·t

223,525 = 129,180·e0.015231·t

1.73034 = e0.015231·t

ln 1.73034 = ln e0.015231·t

0.548318 = 0.015231·t

t = 36 years

1990 + 36 = 2026