Transcript No Slide Title

Copyright © 2012 Pearson Education, Inc.
Slide 10- 1
9.4
Properties of
Logarithmic
Functions
■
■
■
■
Logarithms of Products
Logarithms of Powers
Logarithms of Quotients
Using the Properties Together
Copyright © 2012 Pearson Education, Inc.
Logarithms of Products
The first property we discuss is related to
the product rule for exponents:
m
n
a a  a
m n
.
Copyright © 2012 Pearson Education, Inc.
Slide 9- 3
The Product Rule for
Logarithms
For any positive numbers M, N and a (a  1),
loga (MN )  loga M  loga N.
(The logarithm of a product is the sum of the
logarithms of the factors.)
Copyright © 2012 Pearson Education, Inc.
Slide 9- 4
Example
Express as an equivalent expression
that is a sum of logarithms: log3(9 · 27).
Solution
log3(9 · 27) = log39 + log327.
As a check, note that
log3(9 · 27) = log3243 = 5
35 = 243
and that
log39 + log327 = 2 + 3 = 5.
Copyright © 2012 Pearson Education, Inc.
32 = 9 and 33 = 27
Slide 9- 5
Example
Express as an equivalent expression
that is a single logarithm: loga6 + loga7.
Solution
loga6 + loga7 = loga(6 · 7)
= loga(42).
Copyright © 2012 Pearson Education, Inc.
Using the product
rule for logarithms
Slide 9- 6
Logarithms of Powers
The second basic property is related to the
power rule for exponents:
 
a
n
m
 a mn .
Copyright © 2012 Pearson Education, Inc.
Slide 9- 7
The Power Rule for Logarithms
For any positive numbers M and a (a  1),
and any real number p,
log a M p  p log a M .
(The logarithm of a power of M is the
exponent times the logarithm of M.)
Copyright © 2012 Pearson Education, Inc.
Slide 9- 8
Example
Use the power rule to write an equivalent
expression that is a product:
a) loga6–3;
b) log 4 x .
Solution
6-3
a) loga
= –3loga6
Using the power rule
for logarithms
b) log 4 x = log4x1/2
= ½ log4x
Copyright © 2012 Pearson Education, Inc.
Using the power rule
for logarithms
Slide 9- 9
Logarithms of Quotients
The third property that we study is similar
to the quotient rule for exponents:
a
m
a
n
a
mn
.
Copyright © 2012 Pearson Education, Inc.
Slide 9- 10
The Quotient Rule for
Logarithms
For any positive numbers M, N and a (a  1),
M
log a
 log a M  log a N .
N
(The logarithm of a quotient is the logarithm
of the dividend minus the logarithm of the
divisor.)
Copyright © 2012 Pearson Education, Inc.
Slide 9- 11
Example
Express as an equivalent expression
that is a difference of logarithms: log3(9/y).
Solution
log3(9/y) = log39 – log3y.
Copyright © 2012 Pearson Education, Inc.
Using the quotient
rule for logarithms
Slide 9- 12
Example
Express as an equivalent expression
that is a single logarithm: loga6 – loga7.
Solution
loga6 – loga7 = loga(6/7)
Copyright © 2012 Pearson Education, Inc.
Using the quotient
rule for logarithms
“in reverse”
Slide 9- 13
Using the Properties Together
Copyright © 2012 Pearson Education, Inc.
Slide 9- 14
Example
Express as an equivalent expression
using individual logarithms of x, y, and z.
3
x
a) log 4
yz
b) logb 3
xy
z7
Solution
3
x
a) log 4
= log4x3 – log4 yz
yz
= 3log4x – log4 yz
= 3log4x – (log4 y + log4z)
= 3log4x –log4 y – log4z
Copyright © 2012 Pearson Education, Inc.
Slide 9- 15
Solution continued
1/ 3
 xy 
b) logb 3
 logb  
 z7 
z7
xy
1
xy
  logb
3
z7

1
7
 logb xy  logb z
3

1
  logb x  logb y  7logb z 
3
Copyright © 2012 Pearson Education, Inc.
Slide 9- 16
CAUTION! Because the product and
quotient rules replace one term with two, it is
often best to use the rules within parentheses,
as in the previous example.
Copyright © 2012 Pearson Education, Inc.
Slide 9- 17
Example
Express as an equivalent expression
1
that is a single logarithm. logb x  2logb y  logb z
3
Solution
1
logb x  2logb y  logb z
3
= logbx1/3 – logb y2 + logbz
 logb
 logb
x1/ 3
y2
 logb z
z3 x
y
2
Copyright © 2012 Pearson Education, Inc.
Slide 9- 18
The Logarithm of the Base to
an Exponent
For any base a,
k
log a a  k .
(The logarithm, base a, of a to an exponent is
the exponent.)
Copyright © 2012 Pearson Education, Inc.
Slide 9- 19
Example
Simplify: a) log668
b) log33–3.4
Solution
a) log668 =8
8 is the exponent to which you
raise 6 in order to get 68.
b) log33–3.4 = –3.4
Copyright © 2012 Pearson Education, Inc.
Slide 9- 20
For any positive numbers M, N, and a
(a  1) :
loga (MN )  loga M  loga N;
log a M
p
 p  log a M ;
M
loga
 loga M  loga N ;
N
k
log a a  k .
Copyright © 2012 Pearson Education, Inc.
Slide 9- 21