Transcript Section 4.4 - Canton Local

Chapter 4
Exponential and
Logarithmic
Functions
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All rights reserved
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SECTION 4.4
Rules of Logarithms
OBJECTIVES
1
2
3
Learn the rules of logarithms.
Change the base of a logarithm.
Apply logarithms in growth and decay.
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RULES OF LOGARITHMS
Let M, N, and a be positive real numbers
with a ≠ 1, and let r be any real number.
1. Product Rule
log a MN   log a M  log a N
The logarithm of the product of two (or
more) numbers is the sum of the logarithms
of the numbers.
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RULES OF LOGARITHMS
Let M, N, and a be positive real numbers
with a ≠ 1, and let r be any real number.
2. Quotient Rule
M
log a    log a M  log a N
N
The logarithm of the quotient of two (or
more) numbers is the difference of the
logarithms of the numbers.
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RULES OF LOGARITHMS
Let M, N, and a be positive real numbers
with a ≠ 1, and let r be any real number.
3. Power Rule
log a M  r log a M
r
The logarithm of a number to the power r is r
times the logarithm of the number.
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EXAMPLE 1
Using Rules of Logarithms to Evaluate
Expressions
Given that log 5 z = 3 and log 5 y = 2, evaluate
each expression.
7
b. log5 125y 
a. log5  yz 
c. log 5
d. log5  z1/30 y 5 
z
y
Solution
a. log 5  yz   log 5 y  log 5 z
 23
5
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EXAMPLE 1
Using Rules of Logarithms to Evaluate
Expressions
Solution continued
b. log 5 125 y 7   log 5 125  log 5 y 7
 log 5 53  7 log 5 y
 3  7  2   17
c. log 5

z
 log 5 
y

1/2
z
y 
1
  log 5 z  log 5 y 
2
1
1
 3  2 
2
2
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EXAMPLE 1
Using Rules of Logarithms to Evaluate
Expressions
Solution continued
d. log 5  z1/30 y 5   log 5 z1/30  log 5 y 5
1
 log 5 z  5 log 5 y
30
1
  3  5  2 
30
 0.1  10
 10.1
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EXAMPLE 2
Writing Expressions In Expanded Form
Write each expression in expanded form.
x  x  1
2
a. log 2
 2 x  1
3
b. ln x3 y 2 z 5
4
Solution
x  x  1
2
a. log 2
 2 x  1
3
4
 log 2 x  x  1  log 2  2 x  1
2
3
 log 2 x  log 2  x  1  log 2  2 x  1
2
3
4
4
 2log 2 x  3log 2  x  1  4log 2  2 x  1
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EXAMPLE 2
Writing Expressions In Expanded Form
Solution continued
b. ln x y z  ln  x y z
3
2 5
3

2 5 1/2
1
3 2 5
 ln  x y z 
2
1
  ln x 3  ln y 2  ln z 5 
2
1
  3ln x  2ln y  5ln z 
2
3
5
 ln x  ln y  ln z
2
2
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EXAMPLE 3
Writing Expressions in Condensed Form
Write each expression in condensed form.
a. log3x  log 4 y
1
b. 2ln x  ln  x 2  1
2
c. 2log 2 5  log 2 9  log 2 75
1
2

d. ln x  ln  x  1  ln  x  1 
3
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EXAMPLE 3
Writing Expressions in Condensed Form
Solution
 3x 
a. log 3 x  log 4 y  log 

 4y 
1/2
1
2
2
2
b. 2 ln x  ln  x  1  ln x  ln  x  1
2

 ln x 2 x 2  1
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
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EXAMPLE 3
Writing Expressions in Condensed Form
Solution continued
c. 2log 2 5  log 2 9  log 2 75
 log 2 5  log 2 9  log 2 75
2
 log 2  25  9   log 2 75
25  9
 log 2
75
 log 2 3
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EXAMPLE 3
Writing Expressions in Condensed Form
Solution continued
1
d. ln x  ln  x  1  ln  x 2  1 
3
1
2

 ln x  x  1  ln  x  1 
3
1  x  x  1 
 ln  2

3  x 1 
 ln 3
x  x  1
x2  1
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CHANGE-OF-BASE FORMULA
Let a, b, and x be positive real numbers
with a ≠ 1 and b ≠ 1. Then logb x can be
converted to a different base as follows:
log a x
log b x 
log a b
(base a )

log x
log b

ln x
ln b
(base 10) (base e)
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EXAMPLE 4
Using a Change of Base to Compute
Logarithms
Compute log513 by changing to a. common
logarithms and b. natural logarithms.
Solution
log13
a. log 5 13 
log 5
 1.59369
ln 13
b. log 5 13 
ln 5
 1.59369
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EXAMPLE 5
Matching Data to an Exponential Curve
Find the exponential function of the form
f (x) = aebx that passes through the points
(0, 2) and (3, 8).
Solution
Substitute (0, 2) into f (x) = aebx.
2  f  0   ae
b 0 
 ae0  a  1  a
So a = 2. Now substitute (3, 8) into the
equation.
b 3
3b
8  f  3  2e  2e
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EXAMPLE 5
Matching Data to an Exponential Curve
Solution continued
Now solve for b.
8  2e 3 b
4  e 3b
ln 4  3b
1
b  ln 4
3
Thus f  x   2e
1

ln
4

x
3


is the desired function.
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STANDARD GROWTH FORMULA
Exponential growth (or decay) occurs when a
quantity grows (or decreases) at a rate
proportional to its size. The standard growth
formula is
kt
A  t   A0e
A(t) = amount at time t
A0 = A(0), the initial amount
k = relative rate of growth (k > 0) or
decay (k < 0)
t = time
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HALF-LIFE FORMULA
The half-life of any quantity whose value
decreases with time is the time required for the
quantity to decay to half its initial value.
The half-life of a substance undergoing
exponential decay at a rate k (k < 0) is given by
the formula
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EXAMPLE 6
Finding the Half-Life of a Substance
In an experiment, 18 grams of the radioactive
element sodium-24 decayed to 6 grams in 24
hours. Find its half-life to the nearest hour.
Solution
So
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EXAMPLE 6
Finding the Half-Life of a Substance
Solution continued
Use the formula
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RADIOCARBON DATING
Carbon-14 (14C) is a form of carbon that decays
radioactively with a half-life (time required for
half of any given mass to decay) of 5700 years.
After the organism dies, the age of its remains
can be calculated by determining how much
carbon-14 has decayed.
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EXAMPLE 7
King Tut’s Treasure
In 1960, a group of specialists investigated
whether a piece of art containing organic
material found in Tutankhamun’s tomb had
been made during his reign or whether it
belonged to an earlier period.
We know that King Tut died in 1346 B.C. and
ruled Egypt for 10 years. What percent of the
amount of carbon-14 originally contained in the
object should be present in 1960 if the object
was made during Tutankhamun’s reign?
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EXAMPLE 7
King Tut’s Treasure
Solution
The half-life of carbon-14 is approximately
5700 years, so we rewrite A(t) = A0ekt as
Now find
k.
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EXAMPLE 7
King Tut’s Treasure
Solution continued
Substituting this value into the equation yields,
A  t   A0e0.0001216t .
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EXAMPLE 7
King Tut’s Treasure
Solution continued
The time t that elapsed between King Tut’s
death and 1960 is t = 1960 + 1346 = 3306.
The percent of the original amount of carbon14 remaining in the object (after 3306 years)
is 66.897%.
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EXAMPLE 6
King Tut’s Treasure
Solution continued
King Tut ruled Egypt for 10 years; the time t1
that elapsed from the beginning of his reign
to 1960 is t1 = 3306 + 10 = 3316.
Thus, if the piece of art was made during King
Tut’s reign, the amount of carbon-14 remaining in
1960 should be between 66.816% and 66.897%.
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