Logarithmic and Exponential Functions

Chapter 12

§12.2

The Logarithmic Function

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #3

Logarithms

y = logbx x = by

The logarithm, base b, of a positive number x is the power (exponent) to which the base b must be raised to produce x. That is, y = logbx is the same as x = by, where b > 0 and b 0.

Exponent = y

Base = b

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #4

Exponential Equations in Log Form

Exponential Form Logarithmic Form

50 = 1 log51= 0

23 = 8 log28= 3

41 1=2 16 1 2

1log = 416

2 16 =36

6

1log = 236

Example:Write the following exponential equations in logarithmic form.

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #5

Log Equations in Exponential Form

Exponential FormLogarithmic Form

101 = 10log1010= 1

25 = 32log232= 5

21 1=3 9

1 31log = 29

14 =64

34

1log = 364

Example:Write the following logarithmic equations in exponential form.

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #6

Solving Logarithmic Equations

Example: Solve the following logarithmic equation.

Convert into an equivalent exponential equation.

a) logx 81 = 4 b) log10 x = 2

a) logx 81 = 4

x4 = 81

Solve the exponential equation.x4 = 34

x = 3

Convert into an equivalent exponential equation.

b) log10 x = 2

10– 2 = x

Simplify.1

100x

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #7

Graphing a Logarithmic Function

Example: Graph y = log2 x.

y

x2–2

2

2

24

12

01

1

2

yx1412

The exponential form of the function is x = 2y.

y = log2 x

y = 2x

y = log2 x is the inverse function of y = 2x.

§12.3

Properties of Logarithms

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #9

The Logarithm of a Product

Property 1: The Logarithm of a Product

For any positive real numbers M and N and any positive base b 1,

logbMN = logbM + logbN.

Example: Write the following logarithm as a sum of logarithms

a) log5(4 · 7) b) log10(100 · 1000)

a) log5(4 · 7)

b) log10(100 · 1000)

= log54 + log57

= log10100 + log101000 = 2 + 3 = 5

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #10

The Logarithm of a Product

Example: Write the following logarithms as single logarithms

a) log24 + log2x b) log 50 + log x + log 2

a) log24 + log2x

b) log 50 + log x + log 2

= log24x

= log 50(x)(2) = log 100x

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #11

The Logarithm of a Quotient

Property 2: The Logarithm of a Quotient

For any positive real numbers M and N and any positive base b 1,

Example: Write the following as a difference of logarithms.

log log log . b b bM M NN

245a) log11 2b) log

3

zz

245a) log11 2 2=log 45 log 11

2b) log3

zz =log ( 2) log 3 z zz

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #12

The Logarithm of a Quotient

Example: Write the following as a single logarithm.

5 5a) log 12 log 2 2b) log logc cw y

5 5a) log 12 log 2 512=log2

2b) log logc cw y2

=log

cwy

5=log 6

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #13

The Logarithm of a Number Raised to a Power

Property 3: The Logarithm of a Product

For any positive real numbers M, and any real number p, and any positive base b 1,

logbMp = plogbM.

Example: Write the following as a single logarithm.

a) 3 log8 5 log8 z b) 2 log 100 + 3 log 10

a) 3 log8 5 log8 z

b) 2 log 100 + 3 log 10

= log8 53 log8 z

= log 1002 + log 103

= log8 53 log8 z 8125=log

z

log[ 10000 (1000)]

log10000000 7

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #14

Solving Logarithm Equations

The following properties are true for all positive values of b 1, and all positive values of x and y.

Property 4 logb b = 1

Property 5 logb 1 = 0

Property 6 If logb x = logb y, then x = y.

Example: a) Evaluate log9 9.

b) Evaluate log8 1.

c) Find y if log4 y = log4 19.

a) log9 9 = 1 b) log8 1 = 0 c) If log4 y = log4 19, then y = 19.

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #15

Solving Logarithm Equations

Example: Solve for x.

log5 1 = log5 x – log5 8

log5 1 = log5 x – log5 8

log5 1 + log5 8 = log5 x Isolate the variable.

Property 50 + log5 8 = log5 x

Property 68 = x

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #16

Solving Logarithm Equations

Example: Solve for x.

log3(4x + 6) – log3(x – 1) = 2

Property 1

Convert to exponential form.

Simplify.

log3(4x + 6) – log3 (x – 1) = 2

34 6log 2

1

xx

2 4 631

xx

4 691

xx

Solve for x.9( 1) 4 6 x x

9 9 4 6 x x5 15x

3x

§12.5

Exponential and Logarithmic Equations

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #18

Solving Logarithmic Equations

When solving logarithmic equations, we generally try to get all of the logarithms on one side of the equation and the numerical values on the other side. Then the properties of logarithms are used to obtain a single logarithmic expression on one side.

Step 1: If an equation contains some logarithms and some terms without logarithms, try to get one logarithm alone on one side and one numerical value on the other.

Step 2: Convert to an exponential equation using the definition of a logarithm.

Step 3: Solve the equation.

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #19

Solving Logarithmic Equations

Example:Solve log24 + log2(x – 1) = 5.

Property 1

log24 + log2(x – 1) = 5

log2[4(x – 1)] = 5

Simplify.log2(4x – 4) = 5

Write in exponential form.4x – 4 = 25

Solve for x.4x – 4 = 32

4x = 36

x = 9Check:

log24 + log2(9 – 1) = 5

2 + 3 = 5

log24 + log28 = 5

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #20

Solving Logarithmic Equations

Example:Solve log(x + 3) + log x = log 4.

Property 1

log(x + 3) + log x = log 4

log(x + 3)x = log 4

Property 6(x + 3)x = 4

Simplify.x2 + 3x = 4

Solve for x.x2 + 3x – 4 = 0

(x + 4)(x – 1) = 0

x = – 4 or x = 1Check:log (4 + 3) + log(4) = log 4

log (1 + 3) + log 1 = log 4

log 4 + 0 = log 4

Stop! It is not possible to take the logarithm of a negative number.

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #21

Solving Exponential EquationsProperty 7

If x and y > 0 and x = y, then logbx = logby, where b > 0 and b 1.

Notice that this is the reverse of Property 6.Property 7 is also referred to as “taking the logarithm of each side of the equation.”

Example:Solve 5x = 23.

log 5x = log 235x = 23

Property 3

Property 7

x log 5 = log 23

Divide both sides by log 5.log23 1.948log5

x

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #22

Solving Exponential Equations

Example:Solve 153x – 2 = 230.

153x – 2 = 230

log 153x – 2 = log 230 Property 7

(3x – 2)log 15 = log 230 Property 3

Divide each side by log 15.log2303 2log15

x

3x – 2 = 2.008116 Simplify.

3x = 4.008116 Solve for x.

x 1.336

Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #23

Example:In 2000, a forest had 300 male deer. The growth in the number of deer is estimated by the function g(t) = 300e0.07t where t is the number of years after 2000. How many deer will be in the forest in a) 2010? b) 2020?

In the year 2000, t = 0. (Notice that f(0) =300e0.07(0) = 300e0 = 300, the original number of deer.)

In the year 2010, t = 10. g(10) = 300e0.07(10) = 300e0.7 300(2.01375) 604 deer 2010.

Solving Applications

In the year 2020, t = 20. g(20) = 300e0.07(20) = 300e1.4 300(4.0552) 1217 deer 2010.