college algebra fifth edition james stewart lothar redlin saleem watson
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College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson. Exponential and Logarithmic Functions. 5. Laws of Logarithms. 5.3. Laws of Logarithms. In this section, we study properties of logarithms. - PowerPoint PPT PresentationTRANSCRIPT
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College AlgebraFifth EditionJames Stewart Lothar Redlin Saleem Watson
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Exponential and Logarithmic Functions5
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Laws of Logarithms5.3
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Laws of Logarithms
In this section, we study properties
of logarithms.
• These properties give logarithmic functions a wide range of applications—as we will see in Section 5.5.
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Laws of Logarithms
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Laws of Logarithms
Since logarithms are exponents,
the Laws of Exponents give rise to
the Laws of Logarithms.
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Laws of Logarithms
Let a be a positive number, with a ≠ 1.
Let A, B, and C be any real numbers
with A > 0 and B > 0.
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Law 1
• The logarithm of a product of numbers is the sum of the logarithms of the numbers.
log ( ) log loga a aAB A B
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Law 2
• The logarithm of a quotient of numbers is the difference of the logarithms of the numbers.
log log loga a a
AA B
B
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Law 3
• The logarithm of a power of a number is the exponent times the logarithm of the number.
log logCa aA C A
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Laws of Logarithms—Proof
We make use of the property
loga ax = x
from Section 5.2.
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Let loga A = u and loga B = v.
• When written in exponential form, these equations become:
• Thus,
Law 1—Proof
andu va A a B
alog ( ) log ( ) log ( )
log log
u v u va a
a a
AB a a a
u v
A B
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Using Law 1, we have:
• Thus,
Law 2—Proof
log log log loga a a a
A AA B B
B B
log log loga a a
AA B
B
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Let loga A = u.
Then, au = A.
• So,
Law 3—Proof
log log log
log
CC u uCa a a
a
A a a
uC C A
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Evaluate each expression.
E.g. 1—Using the Laws to Evaluate Expressions
4 4
2 2
13
(a) log 2 log 32
(b) log 80 log 5
(c) log8
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E.g. 1—Evaluating Expressions
3
4 4
4
4
Law 1
Because 64 = 4
log 2 log 32
log (2 32)
log 64
3
Example (a)
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E.g. 1—Evaluating Expressions
4
2 2
2
2
Law 2
Because16 = 2
log 80 log 5
80log
5
log 16
4
Example (b)
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E.g. 1—Evaluating Expressions
13
1/ 3
12
Law 3
Property of negative exponents
Calculator
log8
log8
log
0.301
Example (c)
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Expanding and Combining Logarithmic Expressions
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The laws of logarithms allow us to write
the logarithm of a product or a quotient as
the sum or difference of logarithms.
• This process—called expanding a logarithmic expression—is illustrated in the next example.
Expanding Logarithmic Expressions
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Use the Laws of Logarithms to expand
each expression.
E.g. 2—Expanding Logarithmic Expressions
2
3 65
3
(a) log (6 )
(b) log
(c) ln
x
x y
ab
c
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E.g. 2—Expanding Log Expressions
2 2 2
3 6 3 65 5 5
5 5
Law 1
Law 1
Law 3
log (6 ) log 6 log
log log log
3 log 6 log
x x
x y x y
x y
Examples (a) & (b)
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E.g. 2—Expanding Log Expressions
3
3
1/ 3
13
Law 2
Law 1
Law 3
ln ln( ) ln
ln ln ln
ln ln ln
abab c
c
a b c
a b c
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The laws of logarithms also allow us to
reverse the process of expanding done
in Example 2.
• That is, we can write sums and differences of logarithms as a single logarithm.
• This process—called combining logarithmic expressions—is illustrated in the next example.
Combining Logarithmic Expressions
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Combine 3 log x + ½ log(x + 1) into
a single logarithm.
E.g. 3—Combining Logarithmic Expressions
12
3 1/ 2
1/ 23
(Law 3)
(Law 1)
3log log( 1)
log log( 1)
log 1
x x
x x
x x
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E.g. 4—Combining Logarithmic Expressions
Combine 3 ln s + ½ ln t – 4 ln(t2 + 1)
into a single logarithm.
212
3 1/ 2 2 4
3 1/ 2 2 4
3
42
(Law 3)
(Law 1)
(Law 2)
3ln ln 4ln( 1)
ln ln ln( 1)
ln( ) ln( 1)
ln1
s t t
s t t
s t t
s t
t
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Warning
Although the Laws of Logarithms tell us how
to compute the logarithm of a product or
a quotient, there is no corresponding rule for
the logarithm of a sum or a difference.
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Warning
For instance,
• In fact, we know that the right side is equal to loga(xy).
log log loga a ax y x y
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Warning
Also, don’t improperly simplify quotients
or powers of logarithms.
• For instance,
3
2 2
log6 6log
log2 2
log 3logx x
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Applying Laws of Logarithms
Logarithmic functions are used to
model a variety of situations involving
human behavior.
• One such behavior is how quickly we forget things we have learned.
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Applying Laws of Logarithms
For example, if you learn algebra at a certain
performance level (say 90% on a test) and
then don’t use algebra for a while, how much
will you retain after a week, a month, or a
year?
• Hermann Ebbinghaus (1850–1909) studied this phenomenon and formulated the law described in the next example.
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E.g. 5—The Law of Forgetting
Ebbinghaus’ Law of Forgetting states that,
if a task is learned at a performance level P0,
then, after a time interval t, the performance
level P satisfies
log P = log P0 – c log(t + 1)
where:• c is a constant that depends on the type of task.
• t is measured in months.
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E.g. 5—The Law of Forgetting
(a) Solve for P.
(b) If your score on a history test is 90,
what score would you expect to get
on a similar test after two months?
After a year?
• Assume c = 0.2.
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E.g. 5—The Law of Forgetting
We first combine the right-hand side.
Example (a)
0
0
0
0
(Law 3)
(Law 2)
(Because log is one-to-one)
log log log( 1)
log log log( 1)
log log( 1)
( 1)
c
c
c
P P c t
P P t
PP
t
PP
t
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E.g. 5—The Law of Forgetting
Here, P0 = 90, c = 0.2, and t is measured in
months.
• Your expected scores after two months and one year are 72 and 54, respectively.
Example (b)
0.2
0.2
90In two months: 2 and 72
2 1
90In one year: 12 and 54
12 1
t P
t P
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The Law of Forgetting
Forgetting what we’ve learned
depends logarithmically on how long
ago we learned it.
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Change of Base
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Change of Base
For some purposes, we find it useful to
change from logarithms in one base to
logarithms in another base.
• Suppose we are given loga x and want to find logb x.
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Change of Base
Let
y = logb x
• We write this in exponential form and take the logarithm, with base a, of each side—as follows.
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Change of Base
• This proves the following formula.
a
Exponential form
Taking log of each side
Law 3
log log
log log
log
log
y
ya a
a a
a
a
b x
b x
y b x
xy
b
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Change of Base Formula
loglog
loga
ba
xx
b
• In particular, if we put x = a, then loga a = 1, and this formula becomes:
1log
logba
ab
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Change of Base
We can now evaluate a logarithm to any
base by:
1. Using the Change of Base Formula to express the logarithm in terms of common logarithms or natural logarithms.
2. Using a calculator.
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E.g. 6—Evaluating Logarithms
Use the Change of Base Formula and
common or natural logarithms to evaluate
each logarithm, correct to five decimal places.
(a) log8 5
(b) log9 20
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E.g. 6—Evaluating Logarithms
We use the Change of Base Formula with
b = 8 and a = 10:
108
10
log 5log 5
log 8
0.77398
Example (a)
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E.g. 6—Evaluating Logarithms
We use the Change of Base Formula with
b = 9 and a = e:
9
ln20log 20
ln91.36342
Example (b)
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E.g. 7—Graphing a Logarithmic Function
Use a graphing calculator to graph
f(x) = log6 x
• Calculators don’t have a key for log6.
• So, we use the Change of Base Formula to write:
6
ln( ) log
ln6
xf x x
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E.g. 7—Graphing a Logarithmic Function
Since calculators do have an LN key,
we can enter this new form of the function
and graph it.