Logarithms Text

Logarithms
mc-TY-logarithms-2009-1
Logarithms appear in all sorts of calculations in engineering and science, business and economics.
Before the days of calculators they were used to assist in the process of multiplication by replacing
the operation of multiplication by addition. Similarly, they enabled the operation of division to
be replaced by subtraction. They remain important in other ways, one of which is that they
provide the underlying theory of the logarithm function. This has applications in many ﬁelds, for
example, the decibel scale in acoustics.
In order to master the techniques explained here it is vital that you do plenty of practice exercises
so that they become second nature.
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.
Contents
1. Introduction 2
2. Why do we study logarithms ? 2
3. What is a logarithm ? if x = an
then loga x = n 3
4. Exercises 4
5. The ﬁrst law of logarithms loga xy = loga x + loga y 4
6. The second law of logarithms loga xm
= m loga x 5
7. The third law of logarithms loga
x
y
= loga x − loga y 5
8. The logarithm of 1 loga 1 = 0 6
9. Examples 6
10. Exercises 8
11. Standard bases 10 and e log and ln 8
12. Using logarithms to solve equations 9
13. Inverse operations 10
14. Exercises 11
www.mathcentre.ac.uk 1 c mathcentre 2009

1. Introduction
In this unit we are going to be looking at logarithms. However, before we can deal with logarithms
we need to revise indices. This is because logarithms and indices are closely related, and in order
to understand logarithms a good knowledge of indices is required.
We know that
16 = 24
Here, the number 4 is the power. Sometimes we call it an exponent. Sometimes we call it an
index. In the expression 24
, the number 2 is called the base.
Example
We know that 64 = 82
.
In this example 2 is the power, or exponent, or index. The number 8 is the base.
2. Why do we study logarithms ?
In order to motivate our study of logarithms, consider the following:
we know that 16 = 24
. We also know that 8 = 23
Suppose that we wanted to multiply 16 by 8.
One way is to carry out the multiplication directly using long-multiplication and obtain 128.
But this could be long and tedious if the numbers were larger than 8 and 16. Can we do this
calculation another way using the powers ? Note that
16 × 8 can be written 24
× 23
This equals
27
using the rules of indices which tell us to add the powers 4 and 3 to give the new power, 7. What
was a multiplication sum has been reduced to an addition sum.
Similarly if we wanted to divide 16 by 8:
16 ÷ 8 can be written 24
÷ 23
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.
If we had a look-up table containing powers of 2, it would be straightforward to look up 27
and
obtain 27
= 128 as the result of ﬁnding 16 × 8.
Notice that by using the powers, we have changed a multiplication problem into one involving
addition (the addition of the powers, 4 and 3). Historically, this observation led John Napier
(1550-1617) and Henry Briggs (1561-1630) to develop logarithms as a way of replacing multi-
plication with addition, and also division with subtraction.

3. What is a logarithm ?
Consider the expression 16 = 24
. Remember that 2 is the base, and 4 is the power. An alternative,
yet equivalent, way of writing this expression is log2 16 = 4. This is stated as ‘log to base 2 of 16
equals 4’. We see that the logarithm is the same as the power or index in the original expression.
It is the base in the original expression which becomes the base of the logarithm.
The two statements
16 = 24
log2 16 = 4
are equivalent statements. If we write either of them, we are automatically implying the other.
Example
If we write down that 64 = 82
then the equivalent statement using logarithms is log8 64 = 2.
Example
If we write down that log3 27 = 3 then the equivalent statement using powers is 33
= 27.
So the two sets of statements, one involving powers and one involving logarithms are equivalent.
In the general case we have:
Key Point
if x = an
then equivalently loga x = n
Let us develop this a little more.
Because 10 = 101
we can write the equivalent logarithmic form log10 10 = 1.
Similarly, the logarithmic form of the statement 21
= 2 is log2 2 = 1.
In general, for any base a, a = a1
and so loga a = 1.
Key Point
loga a = 1

We can see from the Examples above that indices and logarithms are very closely related. In
the same way that we have rules or laws of indices, we have laws of logarithms. These are
developed in the following sections.
4. Exercises
1. Write the following using logarithms instead of powers
a) 82
= 64 b) 35
= 243 c) 210
= 1024 d) 53
= 125
e) 106
= 1000000 f) 10−3
= 0.001 g) 3−2
= 1
9
h) 60
= 1
i) 5−1
= 1
5
j)
√
49 = 7 k) 272/3
= 9 l) 32−2/5
= 1
4
2. Determine the value of the following logarithms
a) log3 9 b) log2 32 c) log5 125 d) log10 10000
e) log4 64 f) log25 5 g) log8 2 h) log81 3
i) log3
1
27
j) log7 1 k) log8
1
8
l) log4 8
m) loga a5
n) logc
√
c o) logs s p) loge
1
e3
5. The first law of logarithms
Suppose
x = an
and y = am
then the equivalent logarithmic forms are
loga x = n and loga y = m (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. But n = loga x and
m = loga y from (1) and so putting these results together we have
loga xy = loga x + loga y
So, if we want to multiply two numbers together and find the logarithm of the result, we can do
this by adding together the logarithms of the two numbers. This is the first law.
Key Point
loga xy = loga x + loga y

6. The second law of logarithms
Suppose x = an
, or equivalently loga x = n. 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. It states that when ﬁnding the logarithm of a power of a number, this
can be evaluated by multiplying the logarithm of the number by that power.
Key Point
loga xm
= m loga x
7. The third law of logarithms
As before, suppose
x = an
and y = am
with equivalent logarithmic forms
loga x = n and loga y = m (2)
Consider x ÷ y.
x
y
= an
÷ am
= an−m
using the rules of indices.
In logarithmic form
loga
x
y
= n − m
which from (2) can be written
loga
x
y
= loga x − loga y
This is the third law.

Key Point
loga
x
y
= loga x − loga y
8. The logarithm of 1
Recall that any number raised to the power zero is 1: a0
= 1. The logarithmic form of this is
loga 1 = 0
Key Point
loga 1 = 0
The logarithm of 1 in any base is 0.
9. Examples
Example
Suppose we wish to ﬁnd log2 512.
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.
Example
Suppose we wish to ﬁnd log8
1
64
.
This is the same as being asked ‘what is
1
64
expressed as a power of 8 ?’
Now
1
64
can be written 64−1
. Noting also that 82
= 64 it follows that
1
64
= 64−1
= (82
)−1
= 8−2
using the rules of indices. So log8
1
64
= −2.

Example
Now 52
= 25 and so log5 25 = 2.
Example
We know that 5 is a square root of 25, that is 5 =
√
25. So 25
1
2 = 5 and so log25 5 = 1
2
.
Notice from the last two examples that by interchanging the base and the number
log25 5 =
1
log5 25
This is true more generally:
Key Point
logb a =
1
loga b
To illustrate this again, consider the following example.
Example
Consider log2 8. We are asking ‘what is 8 expressed as a power of 2 ?’ We know that 8 = 23
and so log2 8 = 3.
What about log8 2 ? Now we are asking ‘what is 2 expressed as a power of 8 ?’ Now 23
= 8
and so 2 = 3
√
8 or 81/3
. So log8 2 =
1
3
.
We see again
log8 2 =
1
log2 8

10. Exercises
3 Each of the following expressions can be simpliﬁed to log N.
Determine the value of N in each case. We have not explicitly written down the base. You
can assume the base is 10, but the results are identical whichever base is used.
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) log 236 − log 1
g) 5 log 2 + 2 log 5 h) log 128 − 7 log 2 i) log 2 + log 3 + log 4
j) log 12 − 2 log 2 + log 3 k) 5 log 2 + 4 log 3 − 3 log 4 l) log 10 + 2 log 3 − log 2
11. Standard bases
There are two bases which are used much more commonly than any others and deserve special
mention. These are
base 10 and base e
Logarithms to base 10, log10, are often written simply as log without explicitly writing a base
down. So if you see an expression like log x you can assume the base is 10. Your calculator will
be pre-programmed to evaluate logarithms to base 10. Look for the button marked log.
The second common base is e. The symbol e is called the exponential constant and has a
value approximately equal to 2.718. This is a number like π in the sense that it has an inﬁnite
decimal expansion. Base e is used because this constant occurs frequently in the mathematical
modelling of many physical, biological and economic applications. Logarithms to base e, loge,
are often written simply as ln. If you see an expression like ln x you can assume the base is
e. Such logarithms are also called Naperian or natural logarithms. Your calculator will be
pre-programmed to evaluate logarithms to base e. Look for the button marked ln.
Key Point
Common bases:
log means log10 ln means loge
where e is the exponential constant.
Useful results:
log 10 = 1, ln e = 1

12. Using logarithms to solve equations
We can use logarithms to solve equations where the unknown is in the power.
Suppose we wish to solve the equation 3x
= 5. We can solve this by taking logarithms of both
sides. Whilst logarithms to any base can be used, it is common practice to use base 10, as these
are readily available on your calculator. So,
log 3x
= log 5
Now using the laws of logarithms, the left hand side can be re-written to give
x log 3 = log 5
This is more straightforward. The unknown is no longer in the power. Straightaway
x =
log 5
log 3
If we wanted, this value can be found from a calculator.
Example
Solve 3x
= 5x−2
. Again, notice that the unknown appears in the power. Take logs of both sides.
log 3x
= log 5x−2
Now use the laws of logarithms.
x log 3 = (x − 2) log 5
Notice now that the x we are trying to ﬁnd is no longer in a power. Multiplying out the brackets
x log 3 = x log 5 − 2 log 5
Rearrange this equation to get the two terms involving x on one side and the remaining term on
the other side.
2 log 5 = x log 5 − x log 3
Factorise the right-hand side by extracting the common factor of x.
2 log 5 = x(log 5 − log 3)
= x log
5
3
using the laws of logarithms.
And ﬁnally
x =
2 log 5
log 5
3
If we wanted, this value can be found from a calculator.

13. Inverse operations
Suppose we pick a base, 2 say.
Suppose we pick a power, 8 say.
We will now raise the base 2 to the power 8, to give 28
.
Suppose we now take logarithms to base 2 of 28
.
We then have
log2 28
Using the laws of logarithms we can write this as
8 log2 2
Recall that loga a = 1, so log2 2 = 1, and so we have simply 8 again, the number we started with.
So, raising the base 2 to a power, and then finding the logarithm to base 2 of the result are
inverse operations.
Let’s look at this another way.
Suppose we pick a number, 8 say.
Suppose we find its logarithm to base 2, to evaluate log2 8.
Suppose we now raise the base 2 to this power: 2log2 8
.
Because 8 = 23
we can write this as 2log2 23
. Using the laws of logarithms this equals 23 log2 2
which equals 23
or 8, since log2 2 = 1. We see that raising the base 2 to the logarithm of a
number to base 2 results in the original number.
So raising a base to a power, and finding the logarithm to that base are inverse operations. Doing
one operation, and then following it by the other, we end up where we started.
Example
Suppose we are working in base e. We can pick a number x and evaluate ex
. 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. So, raising the base e to a power, and then
finding logarithms to base e are inverse operations.
Example
Suppose we are working in base 10. We can pick a number x and evaluate 10x
. 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. So, raising the base 10 to a power, and
then finding logarithms to base 10 are inverse operations.

Key Point
ln ex
= x, eln x
= x
Similarly,
log 10x
= x, 10log x
= x
These results will be useful in doing calculus, especially in solving diﬀerential equations.
14. Exercises
4 Use logarithms to solve the following equations
a) 10x
= 5 b) ex
= 8 c) 10x
= 1
2
d) ex
= 0.1
e) 4x
= 12 f) 3x
= 2 g) 7x
= 1 h) 1
2
x
= 1
100
i) πx
= 10 j) ex
= π k) 1
3
x
= 2 l) 10x
= e2x−1
Answers to Exercises on Logarithms
1. a) log8 64 = 2 b) log3 243 = 5 c) log2 1024 = 10
d) log5 125 = 3 e) log10 1000000 = 6 f) log10 0.001 = −3
g) log3
1
9
= −2 h) log6 1 = 0 i) log5
1
5
= −1
j) log49 7 = 1
2
k) log27 9 = 2
3
l) log32
1
4
= −2
5
2. a) 2 b) 5 c) 3 d) 4
e) 3 f) 1
2
g) 1
3
h) 1
4
i) −3 j) 0 k) −1 l) 3
2
m) 5 n) 1
2
o) 1 p) −3
3. a) 15 b) 8 c) 64 d) 9
8
e) 236 f) 236 g) 800 h) 1
i) 24 j) 9 k) 2592
64
= 81
2
l) 45
4. All answers are correct to 3 decimal places
a) 0.699 b) 2.079 c) −0.301 d) −2.303
e) 1.792 f) 0.631 g) 0 h) 6.644
i) 2.011 j) 1.145 k) −0.631 l) −3.305

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