8.1 Understanding Logarithms
Recall the definition of an inverse function.
y  f ( x) 
x  f ( y)
1
y  f ( x)
Write the inverse
y  2x 1
y  2x
Isolate y variable
x  2 y 1
x 1
y
2
x  2y
y  log 2 x
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The logarithm function is the inverse of an exponential
function.
x, y  y, x 
x -3 -2 -1 0
y = 2x
(x =
2y)
y = log2x
y
1
2
3
4
y
1
8
1
4
1
2
1
2
4
8
16
x
1
8
1
4
1
2
1
2
4
8
16
y -3 -2 -1 0
1
2
3
4
log x
log 2
Change of base formula:
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What is Log?
log c b
log a b 
log c a
2
y  log 2 x
y  2x
x
log2x
y
0
log20
?
1
log21
0
2
log22
4
log24
8
log28
1
2
3
log 2 1  0
log 2 20
log 2 2  1
log 2 21
log 2 4  2
log 2 22
log 2 8  3
log 2 23
log b b  m
m
Vertical Asymptote at x = 0
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Comparing Exponential and Logarithmic Function Graphs
y = 2x
y = log2x
The y-intercept is 1.
There is no x-intercept.
The domain is {x | x  R}.
The range is {y | y > 0}.
There is no y-intercept.
The x-intercept is 1.
The domain is {x | x > 0}.
The range is {y | y  R}.
There is a horizontal asymptote
at y = 0.
There is a vertical asymptote
at x = 0.
The graph of y = 2x has been reflected in the line of y = x,
to give the graph of y = log2x.
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Graphing the Logarithmic Function
y=x
y = 2x
y = log2x
log x
y
log 2
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Base is ½
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y  log 3 x
y  3x
x
log2x
y
0
log30
?
1
log31
0
3
log33
9
log39
1
2
 x | x > 0, x 
 y | y 
Vertical Asymptote at
x0
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Characteristics of the Graphs of Logarithmic
Functions of the Form f(x) = logbx
•
•
•
•
The x-intercept is 1. There is no y-intercept.
The y-axis is a vertical asymptote. (x = 0)
If 0 < b < 1, the function is decreasing. If b > 1, the function is increasing.
The graph is smooth and continuous. It has no sharp corners or edges.
6
5
6
5
4
4
3
3
2
-2
-1
-1
-2
2
3 4
5
6
f (x) = logb x
0<b<1
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2
-2
-1
-1
-2
2
3 4
5
6
f (x) = logb x
b>1
8
Converting Exponential Expressions to Logarithms
Multiplication is a shortcut for addition.
4444444  74
Exponents are a shortcut for multiplication.
4  4  4  4  4  4  4  47
Logarithms are a shortcut for exponents.
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52  25
log5 25  2
9
Notation
Index
Exponent
Logarithm
Base
102 = 100
Result
“10 raised to the exponent 2 gives 100”
“The exponent to which the base 10 must be raised to give 100 is 2”
“The logarithm to the base 10 of 100 is 2”
Log10100 = 2
“To what exponent must 10 be raised to get 100? The exponent is 2
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Conversions
logby = x
is the inverse of
y = bx
Logarithm
Base
2
10
= 100
Result
Base
Logarithm
Log10100 = 2
Result
by = x is written as y = logbx
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Conversions
by = x is written as y = logbx
2 to what exponent is
8?
23 = 8
Log28 = 3
3 to what exponent is
81?
34 = 81
Log381 = 4
Log525 =2
52 = 25
Log93 = 1/2
91/2 = 3
The logarithm answer tells us the exponent to which a
given base must be raised to achieve a certain value.
If there is no base b written in the log expression, the base
is assumed to be 10, called a common log.
log x  log10 x
log100  2
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103 = 1000
log101000 = 3
p = q2
logqp = 2
24 = 16
log216 = 4
xy = 2
logx2 = y
104 = 10,000
log1010000 = 4
pq = r
logpr = q
32 = 9
log39 = 2
logxy = z
xz = y
42 = 16
log416 = 2
loga5 = b
ab = 5
10-2 = 0.01
log100.01 = -2
logpq = r
pr = q
log464 = 3
43 = 64
c = logab
b = ac
log327 = 3
33 = 27
log366 = 1/2
361/2 = 6
log121= 0
120 = 1
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103 = 1000
log1000 = 3
p = q2
logqp = 2
24 = 16
log216 = 4
xy = 2
logx2 = y
104 = 10,000
log10000 = 4
pq = r
logpr = q
32 = 9
log39 = 2
logxy = z
xz = y
42 = 16
log416 = 2
loga5 = b
ab = 5
10-2 = 0.01
log0.01 = -2
logpq = r
pr = q
log464 = 3
43 = 64
c = logab
b = ac
log327 = 3
33 = 27
log366 = 1/2
361/2 = 6
log121= 0
120 = 1
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Convert to exponential form: more than one step
m log b n  q
2 log b 5  9
q
log b n 
m
9
log b 5 
2
q
m
b n
9
2
b 5
1
3log 2 r 
2
1
log 2 r      3
2
1
1

log 2 r  
6
2
r
6
 log 5 m  n
log 5 m  n
5 n  m
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Conversions
http://www.pinterest.com/pin/527062581117
57940/
pH of a solution

[ H ]  10
 pH


log
[
H
]
pH 
10
For any sound intensity, I, the decibel level, dB, is defined as follows:
 I  where I is the intensity of a barely
o
LdB  10 log  
 I o  audible sound
 I 
LdB
 log  
10
 Io 
I = Io(10)m
I
 10m
Io
10
dB
10
I

Io
I  I o10
dB
10
where m is the measure on the Richter scale (magnitude)
 I 
m  log10  
 Io 
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Evaluating logarithms
Definition:
to what exponent must the base be raised to get the number?
log 5 25
5 must be raised to the power of 2 to get 25
 log 5 52
2
logb b m
m
b
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logb m
m
17
What is the value of each logarithm? Benchmarks
log 3 27
3
log 3 33
log100
5
log 2 32
log 5 125
3
log 2 25
2
0
log b 1
log b b m
m
b0  1
log b 0
bx  0
not defined
log 2  4 
2 x  4
2  2
x
2
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not defined
log 2  x 
x>0
18
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Page 380
1, 2, 3, 4, 7, 8, 10, 17, 20
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Which came first exponents or logarithms?
Jobst Burgi
1552 – 1632
Switzerland
John Napier
1550 – 1617
Scotland
William
Oughtred
Henry Briggs
English
John
Wallis
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Johann
Bernoulli
21
Timeline of Logarithms
by Anthony Fogleman
Logarithms were invented to make calculations easier.
1550 John Napier was born in Edinburgh Scotland.
1552 Jobst Bürgi was born in Switzerland.
1588 Bürgi began working on his logarithms independent of Napier
~1594 John Napier started work on his tables and spent the next twenty years
completing. The tables were for trigonometric applications and gave the
logarithms for the sine of angles 30o to 90o. Although Napier did not actually use
in his logarithms it could be said his base was roughly 1/e.
1614 Napier published “Mirifici logarithmorum canonis descriptio” in which he
discusses his logarithms.
10 March 1615 Henry Briggs wrote a letter questioning Napier’s use of his base
(1/e) and why he did not use base 10 or why log 1 = 0. Napier replied that he too
had the idea but could not create the tables due to an illness.
Summer 1615 Henry Briggs visited John Napier and they spent a month working
on the tables for the logarithms to base 10.
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4 April 1617 John Napier passed away.
1617 Briggs published his “Logarithmorum Chilias Prima” which contained his tables for
logarithms to base 10.
1619 “Mirifici logarithmorum canonis constructio” is published in which the method
Napier used for constructing his logarithms is discussed.
1620 Bürgis’ work published in his “Arithmetische und Geometrische ProgressTabulen.” Bürgi’s work went unnoticed due to the beginning of the Thirty Years’ War.
1622 William Oughtred invented the slide rule, which offered an even quicker way of
calculating logarithms.
1632 Jobst Bürgi passed away.
1675 Newton discovers the fact that the d/dx ln x = 1/x.
1685 John Wallis realized that logarithms could be defined as exponents.
1694 Johann Bernoulli also realized that logarithms could be defined as exponents.
1694 to present Logarithms had reached their full potential and most of what was done
after 1694 was calculating logarithms to different bases.
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