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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
10/26/2012 Section 4.7 v5.2.1 3
Logarithmic Functions
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
10/26/2012 Section 4.7 v5.2.1 4
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
Logarithmic Functions
10/26/2012 Section 4.7 v5.2.1 5
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
Logarithmic Functions
10/26/2012 Section 4.7 v5.2.1 6
Logarithmic Functions
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
Logarithmic Functions
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 ?
10/26/2012 Section 4.7 v5.2.1 8
Logarithmic Functions
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
10/26/2012 Section 4.7 v5.2.1 9
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 }
10/26/2012 Section 4.7 v5.2.1 11
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
10/26/2012 Section 4.7 v5.2.1 12
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 ?
10/26/2012 Section 4.7 v5.2.1 16
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
10/26/2012 Section 4.7 v5.2.1 18
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
Earthquake Comparison Ratios
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
Relative Strength Ratios
10/26/2012 Section 4.7 v5.2.1 21
Earthquake Comparison Ratios
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
Relative Strength Ratios
10/26/2012 Section 4.7 v5.2.1 22
Think about it !