Mathematics for Business and Economics - I

CHP4 - Logarithmic Functions (Lecture 14)

The inverse function of the exponential function with base b is called the logarithmic function with base b. • For x > 0, and b > 0, b 1

y = logbx if and only if by = x • The equation y = logbx and by = x are different ways

of expressing the same thing. The first equation is the logarithmic form; the second is the exponential form.

Logarithmic: y = logbx

Exponent

Base

Exponential: by = x

Exponent

Base

Example: Write each equation in its equivalent exponential form.

a. 2 = log5 x b. 3 = logb 64 c. log3 7 = y

Solution With the fact that y = logb x means by = x,

c. log3 7 = y or y = log3 7 means 3y = 7.

a. 2 = log5 x means 52 = x.

Logarithms are exponents.

b. 3 = logb 64 means b3 = 64.

Logarithms are exponents.

Note :

a. log5 x = 2 means 52 = x So, x = 25

b. logb64 = 3 means b3 = 64 So, b = 4 since 43 = 64

c. log216 = x means So, x = 4 since 24 = 16

d. log255 = x means So, x = ½ since the square root of 25 = 5!

2x = 16

25x = 5

a. 122 = x means log12x = 2

b. b3 = 9 means logb9 = 3

c. c4 = 16 means

d. 72 = x means logc16 = 4

log7x = 2

Basic Logarithmic Properties Involving One:

because b1 = b. because b0 = 1

Inverse Properties of Logarithms:

because bx = bx

because b raised to the log of some number x (with the same base) equals that number

because

log 1b b

log 1 0b

log xb b x

logb xb x

xx

aa log1

log

1log loga ax x

General

Properties

Common

Logarithms

(base 10)

Natural

Logarithms

(base e)

10log 10 1

10log 1 0

10log 10x x

10log10

xx

log10 1

log1 0

log10x x

log10 x x

ln 1e

ln1 0

ln xe x

ln xe x

The Product Rule:

(The logarithm of a product is the sum of the

logarithms)

Example: log4(7 • 9) = log47 + log49

Example: log (10x) = log10 + log x

Example: log8(13 • 9) =

Example: ln(1000x) =

log813 +log89

ln1000 + lnx

log log logb b bMN M N

ln( ) ln lnMN M N

The Quotient Rule

(The logarithm of a quotient is the difference of the logs)

Example:

Example:

log log logb b b

MM N

N

log log log 22

xx

14ln

x

ln14 ln x

ln ln lnM

M NN

log log logb b bM N M N

ln( ) ln lnM N M N

log log logb b bM N M N

ln( ) ln lnM N M N

The Power Rule:

(The log of a number with an exponent is the product of the exponent and the log of that number)

Example: Simplifying (using Properties) ◦ log x2 = 2 log x

◦ ln 74 = 4 ln 7

◦ log359 =

-

9log35 1

21

ln ln2

x x

log logpb bM p M

ln lnpM p M

Ex1: log94 + log96 = log9(4 • 6) = log924

Ex2: log 146 = 6log 14

Ex3:

Ex4: log1636 - log1612 =

Ex5: log316 + log24 =

Ex6: log 45 - 2 log 3 =

3ln 3 ln 2 ln

2

log163

Impossible!

log 5

Example: Expand

Solution:

Use exponential notation

Use the product rule

Use the power rule

1

2 2

1

2 2

log

log log

12log log

2

b

b b

b b

x y

x y

x y

2logb x y

use exponent for square root

3

2

3ln

3

x

x

13 2

2

3ln

3

x

x

exponent rule

3

2

1 3ln

2 3

x

x

3 21ln3 ln( 3)

2x x

quotient rule

3 21ln3 ln ln( 3)

2x x

product rule

21ln 3 3ln ln( 3)

2x x

exponent rule

21 3 1ln3 ln ln( 3)

2 2 2x x

multiply

Example: Expand: Solution:

1

3

6 4

1

436 6

1

436 6 6

6 6 6

6 6

log36

log log 36

log log 36 log

1log log 36 4log

3

1log 2 4log

3

x

y

x y

x y

x y

x y

Example: Expand Solution:

3

6 4log

36

x

y

Example: Expand the following log

Example: express as a single logarithm.

Solution: Product Rule

Power Rule Quotient Rule

3

3

log 3log

log log

log

b b

b b

b

MN P

MN P

MN

P

log log 3logb b bM N P

xx ln2ln1ln21

x

x

xxxx

12ln

1ln12lnln2ln1ln

21

Example: express as a single logarithm.

Solution:

1

2

1

2

1log

2

log

log log

b

b

b b

MN

P

MN

P

MN MNor

P P

Example: express as a single logarithm. Solution:

1

log log log2

b b bM N P

Express as the logarithm of a single quantity:

yx loglog23log

yx loglog3log 2

yx log3log 2

y

x23log

42log PnPn loglog 2log4

3010.04

2040.1

Example: Evaluate 4.185.3log 85.3log4.1

5855.4.1

8196.0

3 916.0log

31

916.0

31

916.0log

916.0log3

1

0381.03

1

0127.0

Example: Evaluate

30log

21

30

21

30log

30log2

1

4771.12

1

7386.0

82.6

80451.6log

82.6log804log51.6log

82.6log804log2

151.6log

8338.04526.18136.1

4324.2

Example log58 =

This is also how you graph in another base. Enter y1=log(8)/log(5). Remember, you don’t have to enter the base when you’re in base 10!

log

log.

8

512900

b

logMlog M

logb

The log of zero is undefined for all bases !

log 0a undefined

ln0 undefined

1. Solve for x:

2. Product rule

3. Special product

4. Definition of log

5. x can be +10 only 6. Why?

4 4

4

2

4

3 2

2

2

log ( 6) log ( 6) 3

log ( 6)( 6) 3

log 36 3

4 36

64 36

100

10

10

x x

x x

x

x

x

x

x

x

Solve for x. Obtain the exact solution of this equation in terms of e

(2.71828…)

Quotient property of logs Definition of (natural log) Multiply both sides by x

Collect x terms on left side Factor out common factor

Solve for x

ln (x + 1) – ln x = 1

1ln 1

x

x

1 1xe

xe

ex = x + 1

ex - x = 1

x(e - 1) = 1

1

1x

e

2x = 7 problem

ln2x = ln7 take ln both sides

xln2 = ln7 power rule

x = divide to solve for x

x = 2.807

ln7

ln2

Example2: Solving

• ex = 72 problem

• lnex = ln 72 take ln both sides

• x lne = ln 72 power rule

• x = 4.277 solution: because

ln e = ?

2ex + 8 = 20 problem

2ex = 12 subtract 8

ex = 6 divide by 2

ln ex = ln 6 take ln both sides

x lne = 1.792 power rule

x = 1.792 (remember: lne = 1)

Example 4 - One exponential expression.

1. Isolate the exponential expression.

3. Use the log rule that lets you

rewrite the exponent as a multiplier.

32x1 5 11

32x1 162. Take the log (log or ln) of both

sides of the equation. ln 32x1 ln 16

(2x 1)ln 3 ln16

4. Isolate the variable. 2x 1

ln16

ln 3

2x ln16

ln 31

x ln16

2 ln 3

1

2x 0.762

Example5 - Two exponential expressions.

1. The exponential expressions are

already isolated.

3. Use the log rule that lets you

rewrite the exponent as a multiplier

on each side..

2. Take the log (log or ln) of both

sides of the equation.

3x1 4 x2

ln 3x1 ln 4 x2 (x 1)ln 3 (x 2)ln 4

4. To isolate the variable, we need to

combine the ‘x’ terms, then factor out

the ‘x’ and divide.

x ln 3 ln 3 x ln 4 2 ln 4

x ln 3 x ln 4 ln 3 2 ln 4

x(ln 3 ln 4) (ln 3 2 ln 4)

x (ln 3 2 ln 4)

ln 3 ln 4

x 13.457

Example 6: Let 25(x + 2) = 5(3x – 4) Solve for x ◦ 52(x + 2) = 5(3x – 4)

◦ 2x + 4 = 3x – 4 ◦ x = 8

Example 7: Solve: 5 + (3)4(x – 1) = 12 ◦ 4(x – 1) = 7/3 ◦ (x – 1)ln4 = ln(7/3) ◦ x = ln(7/3)/ln(4) + 1 ≈ 1.6112

Example8

Solve for x:

133 x

Example9

Solve for t: 32.0te

Example10 Solve for x: 1411 x

In a Logarithmic Equation, the variable can be inside

the log function or inside the base of the log. There

may be one log term or more than one.

If the exponent is a variable, then take the natural log of both sides of the equation and use the appropriate property. (If the bases are the same on both sides, you can cancel the logs on both sides.)

Then solve for the variable.

Example 1 - Variable inside the log function.

Solve the given log equation

log4 2x 1 3 5

log4 2x 1 2

42 2x 1

16 2x 1

2x 17

x 8.5

1. Isolate the log expression.

2. Rewrite the log equation as a

power equation and solve for ‘x’.

Example 2 - Variable inside the log function, two log expressions.

Solve the given log equation

ln x ln(2x 1) 1

lnx

2x 1

1

e1 x

2x 1

e(2x 1) x

2ex e x

2ex x e

x(2e 1) e

x e

2e 1

1. To isolate the log expression, we 1st

must use the log property to combine a

difference of logs.

2. Rewrite the log equation as a power

equation (here, the base is ‘e’).

3. To solve for ‘x’ we must distribute

the ‘e’ and then collect the ‘x’ terms

together and factor out the ‘x’ and

divide.

x 0.613

Example 3 - Variable inside the base of the log.

Solve the given log equation

logx 3 2

x2 3

x2 1

2

3 1

2

x 1

3

x 3

3

1. Rewrite the log equation as a

power equation.

2. Solve the power equation.

log36 = log33 + log3x problem

log36 = log33x condense

6 = 3x drop logs

2 = x solution

Example5: Solve for x

• log 16 = x log 2 problem

• log 16 = log 2x condense

• 16 = 2x drop logs

• x = 4 solution

log4x = log44 problem

= log44 condense

= 4 drop logs

cube each side

X = 64 solution

3133 4x

1

3

1

34log x

1

3x

Example6: Solve for x

Example7

Solve for the equation:

154log8 x

Example8

Solve for the equation:

3ln24ln332ln x

7xlog25 = 3xlog25 + ½ log225

log257x = log25

3x + log225 ½

log257x = log25

3x + log251

7x = 3x + 1

4x = 1

1

4x

Solve: log77 + log72 = log7x + log7(5x – 3)

log714 = log7 x(5x – 3)

14 = 5x2 -3x

0 = 5x2 – 3x – 14

0 = (5x + 7)(x – 2)

7,2

5x

Do both answers work? NO!!

Example 11: Solve log4(x+3) = 2

42 = x+3

16 = x+3

13 = x

• Example 12: Solve 3ln(2x) = 12

• ln(2x) = 4

• Realize that our base is e, so

• e4 = 2x

• x ≈ 27.299

• You always need to check your answers because sometimes they don’t work!