8.3 Laws of Logarithms
For x > 0 and b  1,
Evaluation Logarithms
logb 1 = 0
logb b = 1
logb bm = m
b logb x = x
logb 0
Logax =
is not defined
Common Log
log 1 = 0
log 10 = 1
log 10s = s
10 log x = x
log 4 4 = 1
log 8 1 = 0
3 log 3 6 = 6
log 5 5 3 = 3
logb (-x) is not defined
2 log 2 7 = 7
Logbx
Logba
Examples
Log2x =
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Log x
Log 2
1
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Addition Law of Logarithms
loga(xy) = logax + logay
Convert x = ap and
y = aq
xy = ap+q
Convert p + q = loga(xy)
Let p = logax and
q = logay
p + q = logax + logay
loga(xy) = logax + logay
log 7x  log 7  log x
log 200  log 2  log100
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Express log 3 42 as the sum of two logarithms.
log 3 2  log 3 21
log 3 3  log 3 14
log 3 6  log 3 7
Express log 5 20 as the sum of two logarithms.
log 5 2  log 5 10
log 5 4  log 5 5
log 2  8  4   log 2  8   log 2  4 
log 2  8  4   log 2  8   log 2  4 
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Subtraction Law of Logarithms
loga(x/y) = logax - logay
Convert x = ap and
Let p = logax and
y = aq
q = logay
x/y = ap-q
Convert p - q = loga(x/y)
p - q = logax - logay
loga(x/y) = logax - logay
 x
log 3    log 3 x  log 3 2
 2
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Express
log 3 4 as the difference of two logarithms.
log 3 8  log 3 2
Express
log 3 12  log 3 3
log 5 2 as the difference of two logarithms.
log 5 10  log 5 5
log 5 20  log 5 10
8
log 2    log 2  8   log 2  4 
2
log 2  8  4   log 2  8   log 2  4 
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Comparing Exponent Laws to Laws of Logarithms
am.an = am+n
loga(xy) = logax + logay
am/an = am-n
loga(x/y) = logax - logay
Express
 27  3 
log 3 
as a sum and difference of logarithms
 9 
 27  3 
log 3 
 9  = log327 + log33 - log39
=3+1-2
=2
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Applying Laws of Logarithms
Express as a single log:
A 

log3A - log3B - log3C  log 3  
BC
AC 
B 
log3A - log3B + log3C  log 3 

Evaluate:
log210 + log212.8
= log2(10 x 12.8)
= log2(128)
= log2(27)
=7
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Simplify: log550 - log510
 50 
log 5  
 10 
log55
=1
9
Simplifying Logarithms
Simplify: log45a + log48a3 - log410a4
 5a  8a 3 
log 4 

4
10
a


 40a 4 
log 4 
4 
10
a


log44
=1
Given log79 = a, determine an expression in terms of a for log763.
log763 = log7(9 x 7)
= log79 + log77
=a+1
If log a 6  x and log a 4  y , express each of the following in
terms of x and y
log a 384
log a 9
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Power Law:
logb(xn) = nlogbx
logbmn = n logbm
n
d
n
log b m  log b m
d
 1
log b    log b x 1
 x
 log b x
logb m 3  logb (m  m  m)
 logb m  logb m  logb m
 3log b m
logb m  3logb m
log24
logbm a
3 log25
3
1/3
 logb m    logb m  logb m 
2
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Applying the Power Laws
Given that log3a = 6 and log3b = 5 determine the value of
log3(9ab2), where a, b > 0
2
log 3  9ab 2   log 3 9  log 3 a  log 3 b
 log 3  32   log 3 a  2 log 3 b
 2  6  2  5
 18
Write as a single logarithm, where x,1 y, z > 0
log z
2 log x 
 3log y  log x 2  log z 2  log y 3
2
 x2 y3 
2 3

x
y 
 log  1 
or log 

 2 
 z 
 z 
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Given log62 = a and log65 = b rewrite log 6 4 20 in terms of a and b.
1
log 6 20  log 6 2  2  5
4
1
1
log 6 20 4  4 log 6 2  log 6 2  log 6 5
4
1
 a  a  b 
4
2a  b

4
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The expression
log x 2
3
 3  3 
log x
9
log x
is equivalent to
log x 2
 log x 2
3
9
 log x 2
Evaluate:
3
a) 3 log2 16
 3log 2 2
4
 3
3
4
3
2
1

b) log 2 2 8  log 2 
8 
 log 2 2  log 22
 log 2
 log 2
= 4

2  log 2 
2  2 
8
3 2
8
6
2
8
6
 log 2 2 2
=2
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Assignment:
State whether the following are True or False for logarithms to
every base.
a) log 2 + log 3 = log 5
False
b) log 4 + log 3 = log 12
True
c) log 10 + log 10 = log 100 True
d) log 2 x log 3 = log 6
False
e) log 32 + log 3-2 = 0
True
5 log5
f) log 
3 log3
g) log 8  log4
log 2
False
False, think of change of base.
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Assignment
State whether the following are True or False for logarithms to
every base.
a) log 5-2 = -2log 5
b) log 4 
2
log 8
3
1
11
c) log11  log
3
3
1
d) log 5  log10
2
1
e) log  log5   log 25
5
f)
log10
2
 2log10
True
Page 400
1a,c, 2, 3, 5, 6, 8, 9,
10, 11, 12,
True
False
False
True
False
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