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Chabot Mathematics
§9.4a
Logarithm Rules
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Review § 9.3
MTH 55
 Any QUESTIONS About
• §9.3 → Common & Natural Logs
 Any QUESTIONS About HomeWork
• §9.3 → HW-45
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Bruce Mayer, PE
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Product Rule for Logarithms
 Let M, N, and a be positive real
numbers with a ≠ 1, and let r be any real
number. Then the PRODUCT Rule
log a MN   log a M  log a N
 That is, The logarithm of the product of
two (or more) numbers is the sum of the
logarithms of the numbers.
Chabot College Mathematics
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Bruce Mayer, PE
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Quotient Rule for Logarithms
 Let M, N, and a be positive real
numbers with a ≠ 1, and let r be any real
number. Then the QUOTIENT Rule
M
log a    log a M  log a N
N
 That is, The logarithm of the quotient of
two (or more) numbers is the difference
of the logarithms of the numbers
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Power Rule for Logarithms
 Let M, N, and a be positive real
numbers with a ≠ 1, and let r be any real
number. Then the POWER Rule
loga M  r log a M
r
 That is, The logarithm of a number to
the power r is r times the logarithm of
the number.
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Product Rule
 Express as an equivalent expression
that is a single logarithm: 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. 32 = 9 and 33 = 27
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Product Rule
 Express as an equivalent expression
that is a single logarithm: loga6 + loga7
 Solution
loga6 + loga7 = loga(6·7)
= loga(42).
Chabot College Mathematics
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Using the product
rule for logarithms
Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Quotient Rule
 Express as an equivalent expression
that is a single logarithm: log3(9/y)
 Solution
log3(9/y) = log39 – log3y.
Chabot College Mathematics
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Using the quotient
rule for logarithms
Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Quotient Rule
 Express as an equivalent expression
that is a single logarithm: loga6 − loga7
 Solution
loga6 – loga7 = loga(6/7)
Chabot College Mathematics
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Using the
quotient rule for
logarithms “in reverse”
Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Power Rule
 Use the power rule to write an
equivalent expression that is a product:
a) loga6−3
b) log 4 x .
 Solution
a)
loga6−3
= −3loga6
Using the power
rule for logarithms
b) log 4 x = log4x1/2
= ½ log4x
Chabot College Mathematics
10
Using the power
rule for logarithms
Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Use The Rules
 Given that log5z = 3 and log5y = 2,
evaluate each expression.
a. log 5 yz 
c. log 5
 Solution
z
y

b. log 5 125y 7
 301 5 
d. log 5  z y 


a. log 5 yz   log 5 y  log 5 z
 23
5
Chabot College Mathematics
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
Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Use The Rules

 Solution b. log 5 125y
7
 log 125  log
5
5
y
7
 log 5 5 3  7 log 5 y
 3  7 2   17
 Soln
c. log 5
1
2
 z
z
1
 log 5    log 5 z  log 5 y 
 y
y
2
1
1
 3  2  
2
2
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Use The Rules
 Soln
1
 301 5 
5
30
d. log 5  z y   log 5 z  log 5 y


1

log 5 z  5 log 5 y
30
1

3  5 2 
30
 0.1  10
 10.1
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Use The Rules
 Express as an equivalent expression
using individual logarithms of x, y, & z
x3
xy
a) log 4
b) logb 3
yz
z7
3
x
3 – log yz
a)
log
=
log
x
4
 Soln
4
4
yz
a)
= 3log4x – log4 yz
= 3log4x – (log4 y + log4z)
= 3log4x –log4 y – log4z
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[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Use The Rules
1/ 3
 Soln
xy
 xy 
3
b)
log
 logb  
b
b)
7
7
z 
z
1
xy
  logb
3
z7

1
 logb xy  logb z 7
3

1
  logb x  logb y  7logb z 
3
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Caveat on Log Rules
 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
1
xy
  logb
3
z7
1
  logb x  logb y  7logb z 
3
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Expand by Log Rules
 Write the expressions in expanded form
x x  1
3
2
a. log 2
2x  1
4
3 2 5
b. log c x y z
 Solution a)
3
2
x x  1
3
4
2
a. log 2
4  log 2 x x  1  log 2 2x  1
2x  1
 log 2 x  log 2 x  1  log 2 2x  1
2
3
4
 2 log 2 x  3log 2 x  1  4 log 2 2x  1
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Expand by Log Rules

1
3 2 5 2
b. log c x y z  log c x y z
3 2 5
 Solution
b)




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
1
log c x 3 y 2 z 5
2
1
3
2
5
log c x  log c y  log c z
2
1
3log c x  2 log c y  5 log c z 

2
3
5
log c x  log c y  log c z
2
2




Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Condense Logs
 Write the expressions in condensed form
a. log 3x  log 4y
1
b. 2 ln x  ln x 2  1
2


c. 2 log 2 5  log 2 9  log 2 75
1
2

d.  ln x  ln x  1  ln x  1 
3

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
Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Condense Logs
 Solution a)
 3x 
a. log 3x  log 4y  log  
 4y 
 Solution b)
1
b. 2 log x  ln x 2  1  ln x 2  ln x 2  1
2




 ln x
Chabot College Mathematics
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2
x 1
2
1
2


Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Condense Logs
 Solution c)
c. 2 log 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
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Condense Logs
 Solution d)
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 


x
x

1


 ln 3 2
x 1
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

Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Log of Base to Exponent
 For any
Base a
k
log a a  k .
 That is, the logarithm, base a, of a
to an exponent is the exponent
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Example  Log Base-to-Exp
 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.
 Solution b)
log33−3.4 = −3.4
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Summary of Log Rules
 For any positive numbers M, N,
and a with a ≠ 1
log a ( MN )  log a M  log a N ;
log a M
p
 p log a M ;
M
log a
 log a M  log a N ;
N
k
log a a  k .
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Typical Log-Confusion
 Beware that Logs do NOT behave
Algebraically. In General:
log a ( MN )  (log a M )(log a N ),
M log a M
log a

,
N log a N
log a ( M  N )  log a M  log a N ,
log a ( M  N )  log a M  log a N .
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
WhiteBoard Work
 Problems From §9.4 Exercise Set
•
24, 30, 36, 58, 60
 Condense
Logarithm
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Bruce Mayer, PE
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All Done for Today
Mathematical
Association
Log Poster
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-62_sec_9-4a_Log_Rules.ppt