Chabot Mathematics
§9.3a
Logarithms
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Review § 9.2
MTH 55
 Any QUESTIONS About
• §9.2 → Inverse Functions
 Any QUESTIONS About HomeWork
• §9.2 → HW-43
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Logarithm → What is it?
 Concept: If b > 0 and b ≠ 1, then
y = logbx is equivalent to x = by
 Symbolically
The exponent is the logarithm.
x = by
y = logbx
The base is the base of the logarithm.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Logarithm Illustrated
 Consider the exponential function
f(x) = 3x. Like all exponential functions,
f is one-to-one. Can a formula for f−1
be found? Use the 4-Step Method
y = 3x
4-Step
x = 3y
y ≡ the exponent to which we
must raise 3 to get x.
f −1(x) ≡ the exponent to which we must
raise 3 to get x.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Logarithm Illustrated
 Now define a new symbol to replace the
words “the exponent to which we must
raise 3 to get x”:
log3x, read “the logarithm, base 3, of x,”
or “log, base 3, of x,” means “the
exponent to which we raise 3 to get x.”
 Thus if f(x) = 3x, then f−1(x) = log3x.
Note that f−1(9) = log39 = 2, as 2 is the
exponent to which we raise 3 to get 9
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Evaluate Logarithms

Evaluate:
a) log381

b) log31
c) log3(1/9)
Solution:
a) Think of log381 as the exponent to which we
raise 3 to get 81. The exponent is 4. Thus,
since 34 = 81, log381 = 4.
b) ask: “To what exponent do we raise 3 in order
to get 1?” That exponent is 0. So, log31 = 0
c) To what exponent do we raise 3 in order to get
1/9? Since 3−2 = 1/9 we have log3(1/9) = −2
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
The Meaning of logax
 For x > 0 and a a positive constant
other than 1, logax is the exponent to
which a must be raised in order to get x.
Thus,
logax = m means am = x
 or equivalently, logax is that unique
exponent for which
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a
log a x
 x.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Exponential to Log
 Write each exponential equation in
logarithmic form.
4
2
1
1

3
c. a  7
a. 4  64
b.   
 2
16
3
 Soln
a. 4  64  log 64  3
4
4
1
1
 1
b.   
 log1 2
4
 2
16
16
c. a 2  7  log a 7  2
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Log to Exponential
 Write each logarithmic equation in
exponential form
a. log 3 243  5 b. log 2 5  x c. log a N  x
 Soln
a. log 3 243  5  243  35
b. log 2 5  x  5  2
x
c. log a N  x  N  a
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x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Evaluate Logarithms
 Find the value of each of the following
logarithms
c. log1 3 9
a. log 5 25
b. log 2 16
1
d. log 7 7
e. log 6 1
f. log 4
2
 Solution
a. log 5 25  y  25  5 or 5  5
y
y2
b. log 2 16  y  16  2 or 2  2
y
y4
y
y
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2
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Evaluate Logarithms
 Solution (cont.)
y
 1
2
y
c. log1 3 9  y  9    or 3  3
 3
d. log 7 7  y  7  7 y or 71  7 y
e. log 6 1  y  1  6 or 6  6
y
0
y
y 1
y0
1
1
y
1
2y
f. log 4  y   4 or 2  2
2
2
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y  2
1
y
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Use Log Definition
 Solve each equation for x, y or z
1
a. log 5 x  3
b. log 3
y
27
2
d. log 2 x  6x  10  1
c. log z 1000  3

 Solution
a. log 5 x  3
3
x5
1
1
x 3 
5
125
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
 1 
The solution set is 
.
125 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Use Log Definition
 Solution (cont.)
1
b. log 3
y
27
1
 3y
27
33  3y
3  y
c. log z 1000  3
1000  z
3
10  z
10  z
The solution set is 10.
3
3
The solution set is 3.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Inverse Property of Logarithms
 Recall Def: For x > 0, a > 0, and a ≠ 1,
y  log a x if and only if
xa .
y
 In other words, The logarithmic function
is the inverse function of the
exponential function; e.g.
log a a  x
x
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a
loga x
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Show
Logaax = x
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Inverse Property
 Evaluate: log5 23
5
.
 Solution
Remember that log523 is the exponent
to which 5 is raised to get 23. Raising 5
to that exponent, obtain
log5 23
5
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 23.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Basic Properties of Logarithms
 For any base a > 0, with a ≠ 1,
Discern from the Log Definition
1. Logaa = 1
• As 1 is the exponent to which a
must be raised to obtain a (a1 = a)
2. Loga1 = 0
• As 0 is the exponent to which a
must be raised to obtain 1 (a0 = 1)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Graph Logarithmic Function
 Sketch the graph of y = log3x
x
y = log3x
(x, y)
–3
(1/27, –3)
–2
(1/9, –2)
–1
(1/3, –1)
30 = 1
0
(1, 0)
31 = 3
1
(3, 1)
32 = 9
2
(9, 2)
 Soln:
3–3 = 1/27
Make
3–2 = 1/9
T-Table
–3 = 1/3
3
→
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Graph Logarithmic Function
 Plot the
ordered pairs
and connect
the dots with a
smooth curve
to obtain the
graph of
y = log3x
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Graph by Inverse
 Graph
y = f(x) = 3x
 Solution:
Use Inverse Relation
for Logs &
Exponentials
 Reflect the graph of
y = 3x in the line y = x
to obtain the graph of
y = f−1(x) = log3x
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Domain of Logarithmic Fcns
 Recall that the
• Domain of f(x) = ax is (−∞, ∞)
• Range of f(x) = ax is (0, ∞)
 Since the Logarithmic function is the
inverse of the Exponential function,
• Domain of f−1(x) = logax is (0, ∞)
• Range of f−1(x) = logax is (−∞, ∞)
 Thus, the logarithms of 0 and negative
numbers are NOT defined.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Find Domain
 Find the domain of each function.
a. f x   log 3 2  x 
 x  2
b. f x   log 3 
 x  1 
 Solution a.
The Domain of a
logarithmic function
must be positive, that is,
2x0
2x
 Thus The domain of f is (−∞, 2).
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Find Domain
 Find the domain of each function.
 x  2
b. f x   log 3 
 x  1 
 Solution b.
The Domain of a
logarithmic function
must be positive, that is,
x2
0
x 1
 Need to Avoid Negative-Logs AND
Division by Zero
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Example  Find Domain
 x  2
 Soln b. (cont.) b. f x   log 3 
 x  1 
 Set numerator = 0 & denominator = 0
x−2=0
x+1=0
x=2
x = −1
 Construct a SIGN CHART
 The domain of f is (−∞, −1)U(2, ∞).
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Properties of Exponential and
Logarithmic Functions
Exponential Function
f (x) = ax
Logarithmic Function
f (x) = loga x
Domain (–∞, ∞)
Range (0, ∞)
Domain (0, ∞)
Range (–∞, ∞)
y-intercept is 1
No x-intercept
x-intercept is 1
No y-intercept
x-axis (y = 0) is the
horizontal asymptote
y-axis (x = 0) is the
vertical asymptote
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Properties of Exponential and
Logarithmic Functions
Exponential Function
f (x) = ax
Logarithmic Function
f (x) = loga x
Is one-to-one , that is,
au = av
if and only if u = v
Is one-to-one, that is,
logau = logav
if and only if u = v
Increasing if a > 1
Decreasing if 0 < a < 1
Increasing if a > 1
Decreasing if 0 < a < 1
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Graphs of Logarithmic Fcns
f (x) = loga x (a > 1)
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f (x) = loga x (0 < a < 1)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Graph Logs by Translation
 Start with the graph of f(x) = log3x and
use Translation Transformations to
sketch the graph of each function
a. f x   log 3 x  2
b. f x   log 3 x  1
 Also State the DOMAIN and RANGE
and the VERTICAL ASYMPTOTE for
the graph of each function
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Graph Logs by Translation
 Solution
a. f x   log 3 x  2
• Shift UP 2
• Domain (0, ∞)
• Range (−∞, ∞)
• Vertical
asymptote
x=0
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Graph Logs by Translation
 Solution
b. f x   log 3 x  1
• Shift RIGHT 1
• Domain (1, ∞)
• Range (−∞, ∞)
• Vertical
asymptote
x=1
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
WhiteBoard Work
 Problems From §9.3 Exercise Set
• 8, 18, 26, 38, 48
 Logs &
Exponentials
Are Inverse
Functions
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
All Done for Today
Inventor
of
Logarithms
Born: 1550 in Merchiston
Castle, Edinburgh, Scotland
Died: 4 April 1617 in
Edinburgh, Scotland
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John Napier
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-60_Fa08_sec_9-3a_Intro-to-Logs.ppt