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Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

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Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2
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Page 1: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Warm UpSolve.

1. log16

x =

2. logx1.331 = 3

3. log10,000 = x

64

1.1

4

32

Page 2: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve exponential and logarithmic equations and equalities.

Solve problems involving exponential and logarithmic equations.

Objectives

Page 3: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

exponential equationlogarithmic equation

Vocabulary

Page 4: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

An exponential equation is an equation containing one or more expressions that have a variable as an exponent. To solve exponential equations:

• Try writing them so that the bases are all the same.

• Take the logarithm of both sides.

Page 5: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

When you use a rounded number in a check, the result will not be exact, but it should be reasonable.

Helpful Hint

Page 6: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve and check.

98 – x = 27x – 3

(32)8 – x = (33)x – 3 Rewrite each side with the same base; 9 and 27 are powers of 3.

316 – 2x = 33x – 9 To raise a power to a power, multiply exponents.

Example 1A: Solving Exponential Equations

16 – 2x = 3x – 9 Bases are the same, so the exponents must be equal.

x = 5 Solve for x.

Page 7: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Example 1A Continued

Check 98 – x = 27x – 3

98 – 5 275 – 3

93 272

729 729

The solution is x = 5.

Page 8: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve and check.4x – 1 = 5

log 4x – 1 = log 5 5 is not a power of 4, so take the log of both sides.

(x – 1)log 4 = log 5 Apply the Power Property of Logarithms.

Example 1B: Solving Exponential Equations

Divide both sides by log 4.

Check Use a calculator.

The solution is x ≈ 2.161.

x = 1 + ≈ 2.161log5log4

x –1 = log5log4

Page 9: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve and check.

32x = 27

(3)2x = (3)3 Rewrite each side with the same base; 3 and 27 are powers of 3.

32x = 33 To raise a power to a power, multiply exponents.

Check It Out! Example 1a

2x = 3 Bases are the same, so the exponents must be equal.

x = 1.5 Solve for x.

Page 10: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Check 32x = 27

32(1.5) 2733 27

27 27

The solution is x = 1.5.

Check It Out! Example 1a Continued

Page 11: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve and check.

7–x = 21

log 7–x = log 21 21 is not a power of 7, so take the log of both sides.

(–x)log 7 = log 21 Apply the Power Property of Logarithms.

Check It Out! Example 1b

Divide both sides by log 7.

x = – ≈ –1.565log21log7

–x = log21log7

Page 12: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Check Use a calculator.

The solution is x ≈ –1.565.

Check It Out! Example 1b Continued

Page 13: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve and check.

23x = 15

log23x = log15 15 is not a power of 2, so take the log of both sides.

(3x)log 2 = log15 Apply the Power Property of Logarithms.

Check It Out! Example 1c

Divide both sides by log 2, then divide both sides by 3.

x ≈ 1.302

3x = log15 log2

Page 14: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Check Use a calculator.

The solution is x ≈ 1.302.

Check It Out! Example 1c Continued

Page 15: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Suppose a bacteria culture doubles in size every hour. How many hours will it take for the number of bacteria to exceed 1,000,000?

Example 2: Biology Application

Solve 2n > 106

At hour 0, there is one bacterium, or 20 bacteria. At hour one, there are two bacteria, or 21 bacteria, and so on. So, at hour n there will be 2n bacteria.

Write 1,000,000 in scientific annotation.

Take the log of both sides.log 2n > log 106

Page 16: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Example 2 Continued

Use the Power of Logarithms.

log 106 is 6.nlog 2 > 6

nlog 2 > log 106

6log 2n > Divide both sides by log 2.

60.301n > Evaluate by using a calculator.

n > ≈ 19.94 Round up to the next whole number.

It will take about 20 hours for the number of bacteria to exceed 1,000,000.

Page 17: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Example 2 Continued

Check In 20 hours, there will be 220 bacteria.

220 = 1,048,576 bacteria.

Page 18: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

You receive one penny on the first day, and then triple that (3 cents) on the second day, and so on for a month. On what day would you receive a least a million dollars.

Solve 3n – 1 > 1 x 108

$1,000,000 is 100,000,000 cents. On day 1, you would receive 1 cent or 30 cents. On day 2, you would receive 3 cents or 31 cents, and so on. So, on day n you would receive 3n–1 cents.

Write 100,000,000 in scientific annotation.

Take the log of both sides.log 3n – 1 > log 108

Check It Out! Example 2

Page 19: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Use the Power of Logarithms.

log 108 is 8.(n – 1)log 3 > 8

(n – 1) log 3 > log 108

8log 3n – 1 > Divide both sides by log 3.

Evaluate by using a calculator.

n > ≈ 17.8 Round up to the next whole number.

Beginning on day 18, you would receive more than a million dollars.

Check It Out! Example 2 Continued

8log3n > + 1

Page 20: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Check On day 18, you would receive 318 – 1 or over a million dollars.

317 = 129,140,163 cents or 1.29 million dollars.

Check It Out! Example 2

Page 21: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

A logarithmic equation is an equation with a logarithmic expression that contains a variable. You can solve logarithmic equations by using the properties of logarithms.

Page 22: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Review the properties of logarithms from Lesson 7-4.

Remember!

Page 23: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve.

Example 3A: Solving Logarithmic Equations

Use 6 as the base for both sides.

log6(2x – 1) = –1

6 log6

(2x –1) = 6–1

2x – 1 = 1 6

7 12

x =

Use inverse properties to remove 6 to the log base 6.

Simplify.

Page 24: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve.

Example 3B: Solving Logarithmic Equations

Write as a quotient.

log4100 – log4(x + 1) = 1

x = 24

Use 4 as the base for both sides.

Use inverse properties on the left side.

100 x + 1log

4( ) = 1

4log4 = 41

100x + 1( )

= 4 100 x + 1

Page 25: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve.

Example 3C: Solving Logarithmic Equations

Power Property of Logarithms.

log5x 4 = 8

x = 25

Definition of a logarithm.

4log5x = 8

log5x = 2

x = 52

Divide both sides by 4 to isolate log5x.

Page 26: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve.

Example 3D: Solving Logarithmic Equations

Product Property of Logarithms.

log12

x + log12

(x + 1) = 1

Exponential form.

Use the inverse properties.

log12

x(x + 1) = 1

log12

x(x +1) 12 = 121

x(x + 1) = 12

Page 27: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Example 3 Continued

Multiply and collect terms.

Factor.

Solve.

x2 + x – 12 = 0

log12

x + log12

(x +1) = 1

(x – 3)(x + 4) = 0

x – 3 = 0 or x + 4 = 0 Set each of the factors equal to zero.

x = 3 or x = –4

log12

x + log12

(x +1) = 1

log12

3 + log12

(3 + 1) 1log

123 + log

124 1

log12

12 1

The solution is x = 3.1 1

log12

( –4) + log12

(–4 +1) 1

log12

( –4) is undefined.

x

Check Check both solutions in the original equation.

Page 28: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve.

3 = log 8 + 3log x

Check It Out! Example 3a

3 = log 8 + 3log x

3 = log 8 + log x3

3 = log (8x3)

103 = 10log (8x3)

1000 = 8x3

125 = x3

5 = x

Use 10 as the base for both sides.Use inverse properties on the right side.

Product Property of Logarithms.

Power Property of Logarithms.

Page 29: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Solve.

2log x – log 4 = 0

Check It Out! Example 3b

Write as a quotient.

x = 2

Use 10 as the base for both sides.

Use inverse properties on the left side.

2log( ) = 0 x 4

2(10log ) = 100

x 4

2( ) = 1 x 4

Page 30: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Watch out for calculated solutions that are not solutions of the original equation.

Caution

Page 31: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Use a table and graph to solve 2x + 1 > 8192x.

Example 4A: Using Tables and Graphs to Solve Exponential and Logarithmic Equations and Inequalities

Use a graphing calculator. Enter 2^(x + 1) as Y1 and 8192x as Y2.

In the table, find the x-values where Y1 is greater than Y2.

In the graph, find the x-value at the point of intersection.

The solution set is {x | x > 16}.

Page 32: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

log(x + 70) = 2log( )

In the table, find the x-values where Y1 equals Y2.

In the graph, find the x-value at the point of intersection.

x 3

Use a graphing calculator. Enter log(x + 70) as Y1 and 2log( ) as Y2. x

3

The solution is x = 30.

Example 4B

Page 33: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

In the table, find the x-values where Y1 is equal to Y2.

In the graph, find the x-value at the point of intersection.

Check It Out! Example 4aUse a table and graph to solve 2x = 4x – 1.

Use a graphing calculator. Enter 2x as Y1 and 4(x – 1) as Y2.

The solution is x = 2.

Page 34: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

In the table, find the x-values where Y1 is greater than Y2.

In the graph, find the x-value at the point of intersection.

Check It Out! Example 4bUse a table and graph to solve 2x > 4x – 1.

Use a graphing calculator. Enter 2x as Y1 and 4(x – 1) as Y2.

The solution is x < 2.

Page 35: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

In the table, find the x-values where Y1 is equal to Y2.

In the graph, find the x-value at the point of intersection.

Check It Out! Example 4cUse a table and graph to solve log x2 = 6.

Use a graphing calculator. Enter log(x2) as Y1 and 6 as Y2.

The solution is x = 1000.

Page 36: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Lesson Quiz: Part I

Solve.

1. 43x–1 = 8x+1

2. 32x–1 = 20

3. log7(5x + 3) = 3

4. log(3x + 1) – log 4 = 2

5. log4(x – 1) + log

4(3x – 1) = 2

x ≈ 1.86

x = 68

x = 133

x = 3

x = 5 3

Page 37: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Lesson Quiz: Part II

6. A single cell divides every 5 minutes. How long will it take for one cell to become more than 10,000 cells?

7. Use a table and graph to solve the equation 23x = 33x–1.

70 min

x ≈ 0.903

Page 38: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

Homework!

Chapter 7 Section 5Page 526 # 2-20, 21-36

Page 39: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.
Page 40: Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2.

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