Transcript Elementary Algebra

Chapter 4
Section 4.2
Logarithmic Functions
and Models
Logarithmic Functions

What Is a Logarithm ?

Recall the exponential function ax with base a , a > 0 , a ≠ 1

We define a new function, the logarithm with base a , as
loga x = y
ay = x
where logax is the power of the base a that equals x

For example
log232 = 5
and log101000 = 3
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since 25 = 32
since
Section 4.7 v5.2.1
103 = 1000
2
Logarithmic Functions
loga x = y
ay = x

So we can think of the logarithm as an exponent

Since y IS the logarithm, we could also write
ay = alogax = x
and logaay = logax = y
So, acting on variable x or y or … with either function
… and then on that result with the other function…
yields the original variable value
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Section 4.7 v5.2.1
3
Logarithmic Functions

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 = a logax
Since we showed that
logaay = logax = y
we call this inverse the logarithm function with base a

Since logax is an inverse of ax then
loga(ax) = x and a logax = x
Question: What are the domain and range of ax ?
What are the domain and range of logax ?
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Section 4.7 v5.2.1
4
Logarithmic Functions

A Second Look
Graphs of f(x) = ax and f–1(x) = logax
y
f(x) = ax , for a > 1
are mirror images with respect
to the line y = x
●
(0, 1)
Note that y = f–1(x) if and
f–1(x) = logax
(1, a)
●
only if f(y) = ay = x
●(a, 1)
●(1, 0)
x
y = logax iffi ay = x
Remember:
logax is the power of a that yields x
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Section 4.7 v5.2.1
5
Logarithmic Functions

A Third Look

y
f(x) = ax, for a > 1
loga x = y means ay = x
What if x = a ? Then logaa = y
1
●
● (a, 1)
●(1, 0)
logaa = 1

●
(0, 1)
So for any base a > 0, a ≠ 1,
f–1(x) = logax
(1, a)
and a = a = a so that y = 1 WHY?
y
x
What if x =1 ?
Then loga1 = y so ay = 1
and thus y = 0 WHY?
So for any base a > 0 , a ≠ 1 ,
loga1 = 0
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Question: Can logax = ax ?
Section 4.7 v5.2.1
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Logarithmic Functions

A Fourth Look
 What if 0 < a < 1 ?
y
f(x) = ax, for a < 1
Then what is f–1(x) ?
Are the graphs still symmetric
with respect to line y = x ?
YES !
Is it still true that
(0, 1)●
(a, 1)
●
● ●(1, a)
●
(1, 0)
x
f(1) = a and f–1(x) = 1 ?

YES !
Is it possible that
loga x = ax ?
YES !
on the line y = x
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f–1(x) = logax
Question: For what x does logax = ax ?
Section 4.7 v5.2.1
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Logarithmic Functions

Example
 Let a = ½
y
f(x) = ax, for a = ½
Then f(x) = ax = (½)x
Then what is f–1(x) ?
f–1(x)
(0,1) ●
= logax = log½x
It is still true that
(½,1)
● (1,½)
●●
●
(1,0)
x
f(1) = (½)1 = ½
and f–1(½) = log½(½) = 1
Where do the graphs intersect?
y = (½)x = log½x = x
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f–1(x) = log1/2x
WHY ?
Section 4.7 v5.2.1
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Exponential/Logarithmic Comparisons

Compare ax with logax
Exponential
102 = 100
34 = 81
(½)–5 = 32
25 = 32
5–3 = 1/125
e3 ≈ 20.08553692

Common Logarithm



Base 10
Log10 x = Log x
Natural Logarithm


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Inverse Function
log10100 = 2
log381 = 4
log1/232 = –5
log232 = 5
log5(1/125) = –3
loge(20.08553692) ≈ 3
Base e
Loge x = ln x
Section 4.7 v5.2.1
9
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=3
Solution set : { 3 }
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Section 4.7 v5.2.1
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Solving Equations

Solve

1. log3 x = –3
With inverse function:
With definition:
3log 3 x = 3–3
1
27
x=
Solution set: {
4

2. log36  6 = x
4
=6
62x = 61/4
2x = 41
36x
x=
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3log 3 x = 3–3
1
8
1
27
3–3 = x
1
x
27 =
}
This is already solved for x , so simplify
By definition the logarithm (i.e. x) is the
4
power of the base that yields  6
Solution set: {
Section 4.7 v5.2.1
1
8
}
11
Solving Equations

Solve

3. logx 5 = 4
By definition:
( loga x = b iffi ab = x )
x4 = 5
4
x= 5
By inverse function:
x logx 5 = x4
5 = x4
4
4
 5 =  x4
x , for x > 0
=  x  = –x , for x < 0
Solution set:
{ 4 5 }
Question: Since x2 = 5 yields x =   5 , then why
doesn’t x4 = 5 yield x =   5 ?
4
Note: If x = –  5 then logx 5 is defined with a negative base !!
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Section 4.7 v5.2.1
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Solving Equations

Solve

4. log3 (x2 + 5) = 2
By inverse function:
2
32 = 3 log3 (x + 5) = x2 + 5
9 – 5 = x2
x = 4
x = 2
Solution set: {  2 }
By definition:
( 2 is the power of 3 that yields x2 + 5 )
32 = 9 = x 2 + 5
x2 = 4
x=2
 5. 12 = 4x
log4 12 = log4 (4x) = x
Question: Now, how do we find log4 12 ?
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Section 4.7 v5.2.1
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Solving Equations

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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
Section 4.7 v5.2.1
14
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
R = log10 (


x
x0
)
This is the famous Richter Scale for measuring the
relative “strength” of earthquakes
x
For x > x0 note that x > 1 so R = log10
0
(
x
x0
) >0
Question: How fast does R grow ?
If R increases by 1 what is the change in x ?
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Section 4.7 v5.2.1
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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
L
N
So RL = log10 x = 7.3 and RN = log10
= 6.7
x0
0

From the definition of logarithm
Thus
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107.3
L
N
6.7
and 10 = x
=
x0
0
L = x0107.3 and N = x0106.7
Now all we need is the ratio of L to N ...
Section 4.7 v5.2.1
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The Richter Scale (continued)

The ratio is
x0 107.3
L
7.3 – 6.7
0.6 ≈ 3.981
10
10
=
=
=
N
x0 106.7
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 ?
x0 107.3
L
7.3 – 4.8
2.5 ≈ 316
10
10
=
=
=
S
x0 104.8
Landers was 316 times stronger than the smallest quake
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Section 4.7 v5.2.1
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Earthquake Comparison Ratios

2010 Haiti
Richter
Scale Relative Strength Ratios
6.1
3.98

1994 Northridge 6.7
7.94
2.00

2010 Haiti
7.0
3.98
501.2
2.00

1992 Landers
7.3
63.10
?
31.62

2010 Chile
8.8
199.53
100.0
3.16

2004 Indonesia 9.3
1.58

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1960 Chile
9.5
Section 4.7 v5.2.1
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Earthquake Comparison Graph
R
10.0
9.0
●
● Chile
R
8.8
8.0
●
●
7.0 ●
●
Indonesia
9.3
8.0
Landers
7.3
6.0
5.0
Chile
9.5
●
Haiti
7.0
● Haiti
7.0
●
4.0
3.0
6.0
Northridge
6.7
Landers
7.3
7.0
● Haiti
6.1
2.0
Intensity
x 106
1.0
20
10
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1,000
2,000
Section
4.7 v5.2.1
3,000
x
Intensity
x 106
4,000
x
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Earthquake Comparison Ratios

2011 Hawaii
Richter
Scale Relative Strength Ratios
4.5
79.43

2011 Hawaii
6.4
316.2
3.981

2010 Haiti
7.0
31,622.8
398.1
63.10

2010 Chile
8.8
100.0
1.585

2011 Japan
?
1,258.9
9.0
1.995

2004 Indonesia 9.3
3.162
1.585

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1960 Chile
9.5
Section 4.7 v5.2.1
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Earthquake Comparison Ratios

2011 Hawaii
Richter
Scale Relative Strength Ratios
6.4
3.162

1989 California
6.9
3.981
1.259

2010 Haiti
7.0
398.1
125.9
63.10

2010 Chile
8.8
1.585

2011 Japan
?
100.0
398.1
9.0
1.585

1964 Alaska
9.2
3.162
1.995

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1960 Chile
9.5
Section 4.7 v5.2.1
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Think about it !
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