Chapter 4 Section 4.2 Logarithmic Functions and Models.

Chapter 4

Section 4.2

Logarithmic Functions and Models

10/26/2012 Section 4.7 v5.2.1 2

What Is a Logarithm ? Recall the exponential function ax

We define a new function, the logarithm with base a , as

where logax is the power of the base a

For example

loga x = y

Logarithmic Functions

ay = xthat equals x

log232 = 5 since 25 = 32

log101000 = 3 since 103 = 1000and

with base a , a > 0 , a ≠ 1

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loga x = y

So we can think of the logarithm as an exponent

Since y IS the logarithm, we could also write

So, acting on variable x or y or … with either function

ay = x

= x logaay = logax = yandalogaxay =

… and then on that result with the other function… yields the original variable value

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Consider the exponential function f(x) = ax , 0 < a ≠ 1 We can show that f(x) is a 1–1 function Hence f(x) does have an inverse function f–1(x)

f–1(f(x)) = f–1(ax) = x Since we showed that

we call this inverse the logarithm function with base a

Since logax is an inverse of ax then

loga(ax) = x and logaxa = x

Question: What are the domain and range of ax ?

What are the domain and range of logax ?

= a logax

logaay = logax = y


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A Second Look

Graphs of f(x) = ax and f–1(x) = logax

are mirror images with respect

to the line y = x

Note that y = f–1(x) if and

only if f(y) = ay = x

y = logax iffi ay = x

x

y

●

●(0, 1)

(1, 0)

y =

x

f(x) = ax

f–1(x) = logax

Remember:logax is the power of a that yields x

●

●

(1, a)

(a, 1)

, for a > 1


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A Third Look loga x = y means ay = x

What if x = a ?

and ay = a

So for any base a > 0, a ≠ 1,

logaa = 1

What if x =1 ?

Then loga1 = y so ay = 1

and thus y = 0

So for any base a > 0 , a ≠ 1 ,

loga1 = 0

(1, 0)

(a, 1)

Then logaa = y

WHY?

WHY?

= a1

so that y = 1 f–1(x) = logax

x

y

●

●(0, 1)

y =

x

f(x) = ax

●

●

(1, a)

, for a > 1

Question: Can logax = ax ?

10/26/2012 Section 4.7 v5.2.1 7


A Fourth Look What if 0 < a < 1 ?

Then what is f–1(x) ?

Are the graphs still symmetricwith respect to line y = x ?

YES !

Is it still true that

f(1) = a and f–1(x) = 1 ?

YES ! Is it possible that

loga x = ax ?

YES !

on the line y = x

(1, 0)

(a, 1)

f–1(x) = logax

x

y

●

●(0, 1)

y =

x

f(x) = ax

●●

(1, a)

, for a < 1

●

Question: For what x does logax = ax ?

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Example Let a = ½

Then f(x) = ax

Then what is f–1(x) ?

f–1(x) = logax = log½x

It is still true that

f(1) = (½)1 = ½

and f–1(½) = log½(½) = 1

Where do the graphs intersect?

y = (½)x = log½x = x

(1,0)

(½,1)

f–1(x) = log1/2x

x

y

●

●(0,1)

y =

x

f(x) = ax

●●

(1,½)

, for a = ½

●

WHY ?

= (½)x

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Exponential/Logarithmic Comparisons

Compare ax with logax

Exponential Inverse Function 102 = 100

34 = 81 (½)–5 = 32

25 = 32 log232 = 5

5–3 = 1/125 log5(1/125) = –3 e3 ≈ 20.08553692

Common Logarithm Base 10 Log10 x = Log x

Natural Logarithm Base e Loge x = ln x

2log10100 =log381 = 4log1/232 = –5

loge(20.08553692) ≈ 3

10/26/2012 Section 4.7 v5.2.1 10

One-to-One Property Review

Since ax and loga x are 1-1

1. If x = y then ax = ay

2. If ax = ay then x = y

3. If x = y then loga x = loga y

4. If log xa = loga y then x = y

These facts can be used to solve equations

Example: Solve 10x – 1 = 100

10x – 1 = 102

x – 1 = 2

x = 3Solution set : { 3 }

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Solving Equations

Solve 1. log3 x = –3

With inverse function:

2.

With definition:

3log x3 = 3–3

3–3 = x

271

= xSolution set: { } 27

1

64

= xlog36 This is already solved for x , so simplify

By definition the logarithm (i.e. x) is the

power of the base that yields 64

64

36x =

61/462x =

41

=2x

271

=x

81

=x Solution set: { } 81

3log x3 = 3–3

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Solving Equations

Solve 3. logx 5 = 4

By inverse function:

5 = x4

x logx 5 = x4

= x

4

x44

5 =

=x , for x > 0–x , for x < 0

By definition:

( loga x = b iffi ab = x )

x4 = 5

x = 54

Solution set: 54{ }

Question: Since x2 = 5 yields x = 5 , then why

? 54doesn’t x4 = 5 yield x =

Note: If x =

5– then logx 5 is defined with a negative base !!

10/26/2012 Section 4.7 v5.2.1 13

Solving Equations

Solve 4. log3 (x2 + 5) = 2

By inverse function:

9 – 5 = x2

5. 12 = 4x

log4 12 = log4 (4x)

= 3 log3 (x2 + 5)32

By definition:

( 2 is the power of 3 that yields x2 + 5 )

32 = 9 = x2 + 5x2 = 4

{ 2 }Solution set:

= x2 + 5

4=x

=x 2 x = 2

= x

Question: Now, how do we find log4 12 ?

10/26/2012 Section 4.7 v5.2.1 14

Solving Equations

Remember: ax and logax are inverses

1. To remove a variable from an exponent,

find the logarithm of the exponential form

2. To remove a variable from a logarithm,

exponentiate

10/26/2012 Section 4.7 v5.2.1 15

The Richter Scale

For fixed intensity x0 (as measured with a seismometer) the ratio of the seismic intensity x of an earthquake,relative to x0 , is measured by

This is the famous Richter Scale for measuring the relative “strength” of earthquakes

For x > x0 note that

R = log10 ( )xx0

xx0

> 1 so R = log10 ( )xx0

> 0

Question: How fast does R grow ?

If R increases by 1 what is the change in x ?

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Richter Scale Comparisons

In 1992 the Landers earthquake produced a Richter scale value of 7.3 compared with the 1994 Northridge earthquake which hit 6.7 on the Richter scale

How much more powerful was the Landers earthquake, expressed as a ratio ?

Let RL = Landers intensity and RN = Northridge intensity

So RL = log10

From the definition of logarithm

Thus L = x0107.3 and N = x0106.7

Now all we need is the ratio of L to N ...

Lx0

= 6.7RN = log10 Nx0

and= 7.3

Lx0

107.3 = andNx0

106.7 =

10/26/2012 Section 4.7 v5.2.1 17

The Richter Scale (continued)

The ratio is

Thus L = 3.981N , i.e. 3.981 times as strong as Northridge

How much stronger was Landers than the smallest recorded earthquake with Rs = 4.8 ?

Landers was 316 times stronger than the smallest quake

L

N=

x0 107.3

x0 106.7 = 107.3 – 6.7 = 100.6 ≈ 3.981

L

S=

x0 107.3

x0 104.8 = 107.3 – 4.8 = 102.5 ≈ 316

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Earthquake Comparison Ratios

2010 Haiti

1994 Northridge

2010 Haiti

1992 Landers

2010 Chile

2004 Indonesia

1960 Chile

3.98

7.94

3.98

199.53

2.00

2.00

31.62

3.16

63.10

100.0

501.2

Richter Scale

6.1

6.7

7.0

7.3

8.8

9.3

9.5

1.58

?

Relative Strength Ratios

10/26/2012 Section 4.7 v5.2.1 19

Earthquake Comparison Graph

4,0003,0002,000

R

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

1.0

x1,000

●●●

●●●

20x

10

R

8.0

6.0

7.0

●●

●

●

Haiti7.0

Intensityx 106

Landers7.3

Chile8.8

Indonesia9.3

Chile9.5

Haiti7.0

Haiti6.1

Northridge6.7

Landers7.3

Intensityx 106

10/26/2012 Section 4.7 v5.2.1 20


2011 Hawaii

2011 Hawaii

2010 Haiti

2010 Chile

2011 Japan

2004 Indonesia

1960 Chile

79.43

316.2

398.131,622.8

3.981

63.10

1.585

1.995

100.0

1,258.9

Richter Scale

4.5

6.4

7.0

8.8

9.0

9.3

9.5

1.585

?

3.162


10/26/2012 Section 4.7 v5.2.1 21


2011 Hawaii

1989 California

2010 Haiti

2010 Chile

2011 Japan

1964 Alaska

1960 Chile

3.162

3.981

125.9398.1

1.259

63.10

1.585

1.585

100.0398.1

Richter Scale

6.4

6.9

7.0

8.8

9.0

9.2

9.5

1.995

?

3.162


10/26/2012 Section 4.7 v5.2.1 22

Think about it !

Chapter 4 Section 4.2 Logarithmic Functions and Models.

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Transcript of Chapter 4 Section 4.2 Logarithmic Functions and Models.