Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve exponential and logarithmic equations and equalities. Solve problems involving

Holt Algebra 2

7-5 Exponential and Logarithmic Equations and Inequalities

Solve exponential and logarithmic equations and equalities.

Solve problems involving exponential and logarithmic equations.

Objectives

Holt Algebra 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.

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

Helpful Hint

Holt Algebra 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.

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

x = 5 Solve for x.

Holt Algebra 2


Check 98 – x = 27x – 3

98 – 5 275 – 3

93 272

729 729

The solution is x = 5.

Holt Algebra 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.

Divide both sides by log 4.

Check Use a calculator.

The solution is x ≈ 2.161.

x = 1 + ≈ 2.161log5log4

x –1 = log5log4

Holt Algebra 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.

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

x = 1.5 Solve for x.

Check

32x = 27

32(1.5) 2733 27

27 27

Holt Algebra 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

Check

Holt Algebra 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.

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

x ≈ 1.302

3x = log15 log2

Check

Holt Algebra 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.

Review the properties of logarithms from Lesson 7-4.

Remember!

Holt Algebra 2


Solve.

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.

Holt Algebra 2


Solve.

Write as a quotient.

log4100 – log4(x + 1) = 1

x = 24


Use inverse properties on the left side.

100 x + 1log

4( ) = 1

4log4 = 41

100x + 1( )

= 4 100 x + 1

Holt Algebra 2


Solve.

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.

Holt Algebra 2


Solve.

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

Holt Algebra 2


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.

Holt Algebra 2


Solve.

3 = log 8 + 3log x

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.

Holt Algebra 2


Solve.

2log x – log 4 = 0

Write as a quotient.

x = 2


Use inverse properties on the left side.

2log( ) = 0 x 4

2(10log ) = 100

x 4

2( ) = 1 x 4

Holt Algebra 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

Documents

Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve exponential and logarithmic equations and equalities. Solve problems involving