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Logarithms
mc-TY-logarithms-2009-1
Logarithms appear in all sorts of calculations in engineering and science, business and economics
...
Similarly, they enabled the operation of division to
be replaced by subtraction
...
This has applications in many fields, for
example, the decibel scale in acoustics
...
After reading this text and / or viewing the video tutorial on this topic you should be able to:
• explain what is meant by a logarithm
• state and use the laws of logarithms
• solve simple equations requiring the use of logarithms
...
Introduction
2
2
...
What is a logarithm ?
4
...
The first law of logarithms
loga xy = loga x + loga y
4
6
...
The third law of logarithms
loga
8
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Examples
6
10
...
Standard bases 10 and e
if x = an then loga x = n
3
4
x
y
= loga x − loga y
5
6
log and ln
8
12
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Inverse operations
10
14
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However, before we can deal with logarithms
we need to revise indices
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We know that
16 = 24
Here, the number 4 is the power
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Sometimes we call it an
index
...
Example
We know that 64 = 82
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The number 8 is the base
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Why do we study logarithms ?
In order to motivate our study of logarithms, consider the following:
we know that 16 = 24
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One way is to carry out the multiplication directly using long-multiplication and obtain 128
...
Can we do this
calculation another way using the powers ? Note that
24 × 23
can be written
16 × 8
This equals
27
using the rules of indices which tell us to add the powers 4 and 3 to give the new power, 7
...
Similarly if we wanted to divide 16 by 8:
16 ÷ 8
24 ÷ 23
can be written
This equals
21
or simply
2
using the rules of indices which tell us to subtract the powers 4 and 3 to give the new power, 1
...
Notice that by using the powers, we have changed a multiplication problem into one involving
addition (the addition of the powers, 4 and 3)
...
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What is a logarithm ?
Consider the expression 16 = 24
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An alternative,
yet equivalent, way of writing this expression is log2 16 = 4
...
We see that the logarithm is the same as the power or index in the original expression
...
The two statements
16 = 24
log2 16 = 4
are equivalent statements
...
Example
If we write down that 64 = 82 then the equivalent statement using logarithms is log8 64 = 2
...
So the two sets of statements, one involving powers and one involving logarithms are equivalent
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Because 10 = 101 we can write the equivalent logarithmic form log10 10 = 1
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In general, for any base a, a = a1 and so loga a = 1
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We can see from the Examples above that indices and logarithms are very closely related
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These are
developed in the following sections
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Exercises
1
...
001 g) 3−2 = 91
√
49 = 7
k) 272/3 = 9
e)
106 = 1000000
f)
i)
5−1 =
1
5
j)
c)
210 = 1024 d)
53 = 125
h)
60 = 1
l)
32−2/5 =
1
4
2
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The first law of logarithms
Suppose
x = an
and
y = am
and
loga y = m
then the equivalent logarithmic forms are
loga x = n
(1)
Using the first rule of indices
xy = an × am = an+m
Now the logarithmic form of the statement xy = an+m is loga xy = n + m
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This is the first law
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Suppose we raise both sides of x = an to the power
m:
xm = (an )m
Using the rules of indices we can write this as
xm = anm
Thinking of the quantity xm as a single term, the logarithmic form is
loga xm = nm = m loga x
This is the second law
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Key Point
loga xm = m loga x
7
...
x
= an ÷ am
y
= an−m
using the rules of indices
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The logarithm of 1
Recall that any number raised to the power zero is 1: a0 = 1
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9
...
This is the same as being asked ‘what is 512 expressed as a power of 2 ?’
Now 512 is in fact 29 and so log2 512 = 9
...
64
1
This is the same as being asked ‘what is
expressed as a power of 8 ?’
64
1
can be written 64−1
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So log8
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64
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Example
Suppose we wish to find log5 25
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Example
Suppose we wish to find log25 5
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So 25 2 = 5 and so log25 5 = 12
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Example
Consider log2 8
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What about log8 2 ? Now we are asking ‘what is 2 expressed as a power of 8 ?’ Now 23 = 8
√
1
and so 2 = 3 8 or 81/3
...
3
We see again
1
log8 2 =
log2 8
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Exercises
3 Each of the following expressions can be simplified to log N
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We have not explicitly written down the base
...
a) log 3 + log 5
b) log 16 − log 2
c) 3 log 4
d) 2 log 3 − 3 log 2
e)
log 236 + log 1
f)
g) 5 log 2 + 2 log 5
h)
i)
j) log 12 − 2 log 2 + log 3
log 128 − 7 log 2
k) 5 log 2 + 4 log 3 − 3 log 4
l)
log 236 − log 1
log 2 + log 3 + log 4
log 10 + 2 log 3 − log 2
11
...
These are
base 10 and base e
Logarithms to base 10, log10 , are often written simply as log without explicitly writing a base
down
...
Your calculator will
be pre-programmed to evaluate logarithms to base 10
...
The second common base is e
...
718
...
Base e is used because this constant occurs frequently in the mathematical
modelling of many physical, biological and economic applications
...
If you see an expression like ln x you can assume the base is
e
...
Your calculator will be
pre-programmed to evaluate logarithms to base e
...
Key Point
Common bases:
log means log10
ln means loge
where e is the exponential constant
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Suppose we wish to solve the equation 3x = 5
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Whilst logarithms to any base can be used, it is common practice to use base 10, as these
are readily available on your calculator
...
The unknown is no longer in the power
...
Example
Solve 3x = 5x−2
...
Take logs of both sides
...
x log 3 = (x − 2) log 5
Notice now that the x we are trying to find is no longer in a power
...
2 log 5 = x log 5 − x log 3
Factorise the right-hand side by extracting the common factor of x
...
And finally
x=
2 log 5
log 53
If we wanted, this value can be found from a calculator
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Suppose we pick a power, 8 say
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Suppose we now take logarithms to base 2 of 28
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So, raising the base 2 to a power, and then finding the logarithm to base 2 of the result are
inverse operations
...
Suppose we pick a number, 8 say
...
Suppose we now raise the base 2 to this power: 2log2 8
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Using the laws of logarithms this equals 23 log2 2
which equals 23 or 8, since log2 2 = 1
...
So raising a base to a power, and finding the logarithm to that base are inverse operations
...
Example
Suppose we are working in base e
...
If we follow this by
taking logarithms to base e we obtain
ln ex
Using the laws of logarithms this equals
x ln e
but ln e = 1 and so we are left with simply x again
...
Example
Suppose we are working in base 10
...
If we follow this
by taking logarithms to base 10 we obtain
log 10x
Using the laws of logarithms this equals
x log 10
but log 10 = 1 and so we are left with simply x again
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14
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1
1 x
1
e) 4x = 12 f) 3x = 2 g) 7x = 1
h)
= 100
2
1 x
i) π x = 10 j) ex = π k)
= 2 l) 10x = e2x−1
3
Answers to Exercises on Logarithms
1
...
001 = −3
i) log5 15 = −1
l) log32 14 = − 52
j) log49 7 =
2
...
a) 15
b)
8
e)
236
f)
236 g) 800
i)
24
j)
9
o)
1
l)
2
3
1
4
3
2
p) −3
c) 64
k)
log2 1024 = 10
2592
64
=
81
2
d)
9
8
h)
1
l)
45
4
...
699 b)
2
...
792
f)
0
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011
j)
1
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−0
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631
h)
−2
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644
l) −3