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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

- Simplify each expression:
- (23)6 c. p5 + p3
- 7√4 + 7√3 d. (k√3)√3
- Solve each equation or inequality.
- 121 = 111 + n c. 343 = 74n – 1
- 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

- Evaluate each expression.
- log3½7c. log5 625
- log7 49 d. log4 64
- Solve each equation.
- log3 x = 2 d. log12 (2p2) – log12(10p – 8)
- log5 (t + 4) = log5 9t e. log2 (log4 16) = x
- 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)

= log224 + 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 log36/81:

log36/81 = log36/34= 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

log44/64 = log4 x

x = 4/64

x = 1/16

10-3 Practice

- Given log4 5 ≈ 1.161 and log4 3 ≈ 0.792, evaluate the following:
- log4 15 b. log4 192
- log4 5/3 d. log4144/25
- Solve each equation.
- 2 log3 x = ¼ log2 256
- 3 log6 2– ½ log6 25 = log6 x
- ½ log4 144– log4 x = log4 4
- 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

- If log 3600 = 3.5563, find each number.
- mantissa of log 3600 d. log 3.6
- characteristic of log 3600 e. 10 3.5563
- antilog 3.5563 f. mantissa of log 0.036
- Find the antilogarithm of each.
- 2.498 c. -1.793
- 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

- Find each value rounded to four decimal places.
- ln 6.94 e. antiln -3.24
- ln 0.632 f. antiln 0.493
- ln 34.025 g. antiln -4.971
- 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 a

For this formula a, b, and n are positive numbers where a ≠ 1 and b ≠ 1.

- Example: log7 196

log 196change of base formula

log 7 a = 7, n = 196, b = 10

≈ 2.7124

10-6 Practice

- Find the value of the logarithm to 3 decimal places.
- log7 19 c. log3 91
- log12 34 d. log5 48
- Use logarithms to solve each equation. Round to three decimal places.
- 13k = 405 c. 5x-2 = 6x
- 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

- 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.
- Use this information to write a growth equation for Bloomington-Normal, where t is the number of years after 1990.
- Use your equation to predict the population of Bloomington-Normal in 2015.
- Use your equation to find the amount year when the population of Bloomington-Normal reaches 223,525.

10-7 Practice Solution

- 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

- 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

- 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

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