Chapter: Logarithms - Grade 12
In mathematics many ideas are related. We saw that addition and subtraction are related and that multiplication and division are related. Similarly, exponentials and logarithms are related.
Logarithms, commonly referred to as logs, are the inverse of exponentials. The logarithm of a number x in the base a is defined as the number n such that an=xan=x.
So, if an=xan=x, then:
log
a
(
x
)
=
n
log
a
(
x
)
=
n
(1)
When we say “inverse function” we mean that the answer becomes the
question and the question becomes the answer. For example, in the equation ab=xab=x the
“question” is “what is a raised to the power b.” The answer is “x.” The inverse function
would be logax=blogax=b or “by what power must we raise a to obtain x.” The answer is “b.”
The mathematical symbol for logarithm is loga(x)loga(x) and it is read “log to the base a of x”. For example, log10(100)log10(100) is “log to the base 10 of 100”.
Write the following out in words. The first one is done for you.
- log2(4)log2(4) is log to the base 2 of 4
-
log
10
(
14
)
log
10
(
14
)
-
log
16
(
4
)
log
16
(
4
)
-
log
x
(
8
)
log
x
(
8
)
-
log
y
(
x
)
log
y
(
x
)
The logarithm of a number is the index to which the base must be raised to give that number. From the first example of the activity log2(4)log2(4) (read log to the base 2 of 4) means the power of 2 that will give 4. Therefore,
log
2
(
4
)
=
2
log
2
(
4
)
=
2
(2)
The index-form is then 22=422=4 and the logarithmic-form is log24=2log24=2.
- Definition 1: Logarithms
If an=xan=x, then: loga(x)=nloga(x)=n, where a>0a>0; a≠1a≠1 and x>0x>0.
Find the value of:
-
log
7
343
log
7
343
R
e
a
s
o
n
i
n
g
:
7
3
=
343
t
h
e
r
e
f
o
r
e
,
log
7
343
=
3
R
e
a
s
o
n
i
n
g
:
7
3
=
343
t
h
e
r
e
f
o
r
e
,
log
7
343
=
3
(3)
-
log
2
8
log
2
8
-
log
4
1
64
log
4
1
64
-
log
10
1
000
log
10
1
000
Logarithms, like exponentials, also have a base and log2(2)log2(2) is not the same as log10(2)log10(2).
We generally use the “common” base, 10, or the natural base, e.
The number e is an irrational number between 2.712.71 and 2.722.72. It comes up surprisingly often in Mathematics, but for now suffice it to say that it is one of the two common bases.
The natural logarithm (symbol lnln) is widely used in the sciences.
The natural logarithm is to the base e which is approximately 2.71828183...2.71828183.... e is like π
and is another example of an irrational number.
While the notation log10(x)log10(x) and loge(x)loge(x) may be used, log10(x)log10(x) is often written log(x)log(x) in Science and loge(x)loge(x) is normally written as ln(x)ln(x) in both Science and Mathematics. So, if you see the loglog symbol without a base, it means log10log10.
It is often necessary or convenient to convert a log from one base to another. An engineer might need an approximate solution to a log in a base for which he does not have a table or calculator function, or it may be algebraically convenient to have two logs in the same base.
Logarithms can be changed from one base to another, by using the change of base formula:
log
a
x
=
log
b
x
log
b
a
log
a
x
=
log
b
x
log
b
a
(4)
where b is any base you find convenient. Normally a and b are known,
therefore logbalogba is normally a known, if irrational, number.
For example, change log212log212 in base 10 is:
log
2
12
=
log
10
12
log
10
2
log
2
12
=
log
10
12
log
10
2
(5)
Change the following to the indicated base:
- log2(4)log2(4) to base 8
- log10(14)log10(14) to base 2
- log16(4)log16(4) to base 10
- logx(8)logx(8) to base y
- logy(x)logy(x) to base x
Just as for the exponents, logarithms have some laws which make working with them easier. These laws are based on the exponential laws and are summarised first and then explained in detail.
log
a
(
1
)
=
0
log
a
(
a
)
=
1
log
a
(
x
·
y
)
=
log
a
(
x
)
+
log
a
(
y
)
log
a
x
y
=
log
a
(
x
)
-
log
a
(
y
)
log
a
(
x
b
)
=
b
log
a
(
x
)
log
a
x
b
=
log
a
(
x
)
b
log
a
(
1
)
=
0
log
a
(
a
)
=
1
log
a
(
x
·
y
)
=
log
a
(
x
)
+
log
a
(
y
)
log
a
x
y
=
log
a
(
x
)
-
log
a
(
y
)
log
a
(
x
b
)
=
b
log
a
(
x
)
log
a
x
b
=
log
a
(
x
)
b
(6)
Since
a
0
=
1
Then
,
log
a
(
1
)
=
log
a
(
a
0
)
=
0
by
definition
of
logarithm
in
Equation
Since
a
0
=
1
Then
,
log
a
(
1
)
=
log
a
(
a
0
)
=
0
by
definition
of
logarithm
in
Equation
(7)
For example,
log
2
1
=
0
log
2
1
=
0
(8)
and
log
2
51
=
0
log
2
51
=
0
(9)
Simplify the following:
-
log
2
(
1
)
+
5
log
2
(
1
)
+
5
-
log
10
(
1
)
×
100
log
10
(
1
)
×
100
-
3
×
log
16
(
1
)
3
×
log
16
(
1
)
-
log
x
(
1
)
+
2
x
y
log
x
(
1
)
+
2
x
y
-
log
y
(
1
)
x
log
y
(
1
)
x
Since
a
1
=
a
Then
,
log
a
(
a
)
=
log
a
(
a
1
)
=
1
by
definition
of
logarithm
in
Equation
Since
a
1
=
a
Then
,
log
a
(
a
)
=
log
a
(
a
1
)
=
1
by
definition
of
logarithm
in
Equation
(10)
For example,
log
2
2
=
1
log
2
2
=
1
(11)
and
log
25
25
=
1
log
25
25
=
1
(12)
Simplify the following:
-
log
2
(
2
)
+
5
log
2
(
2
)
+
5
-
log
10
(
10
)
×
100
log
10
(
10
)
×
100
-
3
×
log
16
(
16
)
3
×
log
16
(
16
)
-
log
x
(
x
)
+
2
x
y
log
x
(
x
)
+
2
x
y
-
log
y
(
y
)
x
log
y
(
y
)
x
Useful to know and remember
When the base is 10, we do not need to state it.
From the work done up to now, it is also useful to summarise the following facts:
-
log
1
=
0
log
1
=
0
-
log
10
=
1
log
10
=
1
-
log
100
=
2
log
100
=
2
-
log
1000
=
3
log
1000
=
3
The derivation of this law is a bit trickier than the first two. Firstly, we need to relate x and y to the base a. So, assume that x=amx=am and y=any=an. Then from Equation Equation 1, we have that:
log
a
(
x
)
=
m
and
log
a
(
y
)
=
n
log
a
(
x
)
=
m
and
log
a
(
y
)
=
n
(13)
This means that we can write:
log
a
(
x
·
y
)
=
log
a
(
a
m
·
a
n
)
=
log
a
(
a
m
+
n
)
Exponential
Law
Equation
=
log
a
(
a
log
a
(
x
)
+
log
a
(
y
)
)
From
Equation
and
Equation
=
log
a
(
x
)
+
log
a
(
y
)
From
Equation
log
a
(
x
·
y
)
=
log
a
(
a
m
·
a
n
)
=
log
a
(
a
m
+
n
)
Exponential
Law
Equation
=
log
a
(
a
log
a
(
x
)
+
log
a
(
y
)
)
From
Equation
and
Equation
=
log
a
(
x
)
+
log
a
(
y
)
From
Equation
(14)
For example, show that log(10·100)=log10+log100log(10·100)=log10+log100. Start with calculating the left hand side:
log
(
10
·
100
)
=
log
(
1000
)
=
log
(
10
3
)
=
3
log
(
10
·
100
)
=
log
(
1000
)
=
log
(
10
3
)
=
3
(15)
The right hand side:
log
10
+
log
100
=
1
+
2
=
3
log
10
+
log
100
=
1
+
2
=
3
(16)
Both sides are equal. Therefore, log(10·100)=log10+log100log(10·100)=log10+log100.
Write as seperate logs:
-
log
2
(
8
×
4
)
log
2
(
8
×
4
)
-
log
8
(
10
×
10
)
log
8
(
10
×
10
)
-
log
16
(
x
y
)
log
16
(
x
y
)
-
log
z
(
2
x
y
)
log
z
(
2
x
y
)
-
log
x
(
y
2
)
log
x
(
y
2
)
The derivation of this law is identical to the derivation of Logarithm Law 3 and is left as an exercise.
For example, show that log(10100)=log10-log100log(10100)=log10-log100. Start with calculating the left hand side:
log
(
10
100
)
=
log
(
1
10
)
=
log
(
10
-
1
)
=
-
1
log
(
10
100
)
=
log
(
1
10
)
=
log
(
10
-
1
)
=
-
1
(17)
The right hand side:
log
10
-
log
100
=
1
-
2
=
-
1
log
10
-
log
100
=
1
-
2
=
-
1
(18)
Both sides are equal. Therefore, log(10100)=log10-log100log(10100)=log10-log100.
Write as seperate logs:
-
log
2
(
8
5
)
log
2
(
8
5
)
-
log
8
(
100
3
)
log
8
(
100
3
)
-
log
16
(
x
y
)
log
16
(
x
y
)
-
log
z
(
2
y
)
log
z
(
2
y
)
-
log
x
(
y
2
)
log
x
(
y
2
)
Once again, we need to relate x to the base a. So, we let x=amx=am. Then,
log
a
(
x
b
)
=
log
a
(
(
a
m
)
b
)
=
log
a
(
a
m
·
b
)
(
Exponential
Law
in
Equation
)
But
,
m
=
log
a
(
x
)
(
Assumption
that
x
=
a
m
)
∴
log
a
(
x
b
)
=
log
a
(
a
b
·
log
a
(
x
)
)
=
b
·
log
a
(
x
)
(
Definition
of
logarithm
in
Equation
)
log
a
(
x
b
)
=
log
a
(
(
a
m
)
b
)
=
log
a
(
a
m
·
b
)
(
Exponential
Law
in
Equation
)
But
,
m
=
log
a
(
x
)
(
Assumption
that
x
=
a
m
)
∴
log
a
(
x
b
)
=
log
a
(
a
b
·
log
a
(
x
)
)
=
b
·
log
a
(
x
)
(
Definition
of
logarithm
in
Equation
)
(19)
For example, we can show that log2(53)=3log2(5)log2(53)=3log2(5).
log
2
(
5
3
)
=
log
(
5
·
5
·
5
)
=
log
2
5
+
log
2
5
+
log
2
5
(
∵
log
a
(
x
·
y
)
=
log
a
(
a
m
·
a
n
)
)
=
3
log
2
5
log
2
(
5
3
)
=
log
(
5
·
5
·
5
)
=
log
2
5
+
log
2
5
+
log
2
5
(
∵
log
a
(
x
·
y
)
=
log
a
(
a
m
·
a
n
)
)
=
3
log
2
5
(20)
Therefore, log2(53)=3log2(5)log2(53)=3log2(5).
Simplify the following:
-
log
2
(
8
4
)
log
2
(
8
4
)
-
log
8
(
10
10
)
log
8
(
10
10
)
-
log
16
(
x
y
)
log
16
(
x
y
)
-
log
z
(
y
x
)
log
z
(
y
x
)
-
log
x
(
y
2
x
)
log
x
(
y
2
x
)
The derivation of this law is identical to the derivation of Logarithm Law 5 and is left as an exercise.
For example, we can show that log2(53)=log253log2(53)=log253.
log
2
(
5
3
)
=
log
(
5
1
3
)
=
1
3
log
2
5
(
∵
log
a
(
x
b
)
=
b
log
a
(
x
)
)
=
log
2
5
3
log
2
(
5
3
)
=
log
(
5
1
3
)
=
1
3
log
2
5
(
∵
log
a
(
x
b
)
=
b
log
a
(
x
)
)
=
log
2
5
3
(21)
Therefore, log2(53)=log253log2(53)=log253.
Simplify the following:
-
log
2
(
8
4
)
log
2
(
8
4
)
-
log
8
(
10
10
)
log
8
(
10
10
)
-
log
16
(
x
y
)
log
16
(
x
y
)
-
log
z
(
y
x
)
log
z
(
y
x
)
-
log
x
(
y
2
x
)
log
x
(
y
2
x
)
Simplify, without use of a calculator:
3
log
2
+
log
125
3
log
2
+
log
125
(22)
Step: Try to write any quantities as exponents
125 can be written as 5353.
Step: Simplify
3
log
2
+
log
125
=
3
log
2
+
log
5
3
=
3
log
2
+
3
log
5
∵
log
a
(
x
b
)
=
b
log
a
(
x
)
3
log
2
+
log
125
=
3
log
2
+
log
5
3
=
3
log
2
+
3
log
5
∵
log
a
(
x
b
)
=
b
log
a
(
x
)
(23)
Step: Final Answer
The final answer does not have to be that simple.
We cannot simplify any further. The final answer is:
3
log
2
+
3
log
5
3
log
2
+
3
log
5
(24)
Simplify, without use of a calculator:
8
2
3
+
log
2
32
8
2
3
+
log
2
32
(25)
Step: Try to write any quantities as exponents
8 can be written as 2323. 32 can be written as 2525.
Step: Re-write the question using the exponential forms of the numbers
8
2
3
+
log
2
32
=
(
2
3
)
2
3
+
log
2
2
5
8
2
3
+
log
2
32
=
(
2
3
)
2
3
+
log
2
2
5
(26)
Step: Determine which laws can be used.
We can use:
log
a
(
x
b
)
=
b
log
a
(
x
)
log
a
(
x
b
)
=
b
log
a
(
x
)
(27)
Step: Apply log laws to simplify
(
2
3
)
2
3
+
log
2
2
5
=
(
2
)
3
2
3
+
5
log
2
2
(
2
3
)
2
3
+
log
2
2
5
=
(
2
)
3
2
3
+
5
log
2
2
(28)
Step: Determine which laws can be used.
We can now use logaa=1logaa=1
Step: Apply log laws to simplify
(
2
)
3
2
3
+
5
log
2
2
=
(
2
)
2
+
5
(
1
)
=
4
+
5
=
9
(
2
)
3
2
3
+
5
log
2
2
=
(
2
)
2
+
5
(
1
)
=
4
+
5
=
9
(29)
Step: Final Answer
The final answer is:
8
2
3
+
log
2
32
=
9
8
2
3
+
log
2
32
=
9
(30)
Write 2log3+log2-log52log3+log2-log5 as the logarithm of a single number.
Step: Reverse law 5
2
log
3
+
log
2
-
log
5
=
log
3
2
+
log
2
-
log
5
2
log
3
+
log
2
-
log
5
=
log
3
2
+
log
2
-
log
5
Step: Apply laws 3 and 4
=
log
3
2
×
2
÷
5
=
log
3
2
×
2
÷
5
Step: Write the final answer
=log3,6=log3,6
In grade 10 you solved some exponential equations by trial and error, because you did not know the great power of logarithms yet. Now it is much easier to solve these equations by using logarithms.
For example to solve x in 25x=5025x=50 correct to two decimal places you simply apply the following reasoning. If the LHS = RHS then the logarithm of the LHS must be equal to the logarithm of the RHS. By applying Law 5, you will be able to use your calculator to solve for x.
Solve for x: 25x=5025x=50 correct to two decimal places.
Step: Taking the log of both sides
log
25
x
=
log
50
log
25
x
=
log
50
Step: Use Law 5
x
log
25
=
log
50
x
log
25
=
log
50
Step: Solve for x
x
=
log
50
÷
log
25
x
=
log
50
÷
log
25
x
=
1
,
21533
.
.
.
.
x
=
1
,
21533
.
.
.
.
Step: Round off to required decimal place
x=1,22x=1,22
In general, the exponential equation should be simplified as much as possible. Then the aim is to make the unknown quantity (i.e. x) the subject of the equation.
For example, the equation
2
(
x
+
2
)
=
1
2
(
x
+
2
)
=
1
(31)
is solved by moving all terms with the unknown to one side of the equation and taking all constants to the other side of the equation
2
x
·
2
2
=
1
2
x
=
1
2
2
2
x
·
2
2
=
1
2
x
=
1
2
2
(32)
Then, take the logarithm of each side.
log
(
2
x
)
=
log
(
1
2
2
)
x
log
(
2
)
=
-
log
(
2
2
)
x
log
(
2
)
=
-
2
log
(
2
)
Divide
both
sides
by
log
(
2
)
∴
x
=
-
2
log
(
2
x
)
=
log
(
1
2
2
)
x
log
(
2
)
=
-
log
(
2
2
)
x
log
(
2
)
=
-
2
log
(
2
)
Divide
both
sides
by
log
(
2
)
∴
x
=
-
2
(33)
Substituting into the original equation, yields
2
-
2
+
2
=
2
0
=
1
2
-
2
+
2
=
2
0
=
1
(34)
Similarly, 9(1-2x)=349(1-2x)=34 is solved as follows:
9
(
1
-
2
x
)
=
3
4
3
2
(
1
-
2
x
)
=
3
4
3
2
3
-
4
x
=
3
4
3
-
4
x
=
3
4
·
3
-
2
3
-
4
x
=
3
2
take
the
logarithm
of
both
sides
log
(
3
-
4
x
)
=
log
(
3
2
)
-
4
x
log
(
3
)
=
2
log
(
3
)
divide
both
sides
by
log
(
3
)
-
4
x
=
2
∴
x
=
-
1
2
9
(
1
-
2
x
)
=
3
4
3
2
(
1
-
2
x
)
=
3
4
3
2
3
-
4
x
=
3
4
3
-
4
x
=
3
4
·
3
-
2
3
-
4
x
=
3
2
take
the
logarithm
of
both
sides
log
(
3
-
4
x
)
=
log
(
3
2
)
-
4
x
log
(
3
)
=
2
log
(
3
)
divide
both
sides
by
log
(
3
)
-
4
x
=
2
∴
x
=
-
1
2
(35)
Substituting into the original equation, yields
9
(
1
-
2
(
-
1
2
)
)
=
9
(
1
+
1
)
=
3
2
(
2
)
=
3
4
9
(
1
-
2
(
-
1
2
)
)
=
9
(
1
+
1
)
=
3
2
(
2
)
=
3
4
(36)
Solve for x in 7·5(3x+3)=357·5(3x+3)=35
Step: Identify the base with x as an exponent
There are two possible bases: 5 and 7. x is an exponent of 5.
Step: Eliminate the base with no x
In order to eliminate 7, divide both sides of the equation by 7 to give:
5
(
3
x
+
3
)
=
5
5
(
3
x
+
3
)
=
5
(37)
Step: Take the logarithm of both sides
log
(
5
(
3
x
+
3
)
)
=
log
(
5
)
log
(
5
(
3
x
+
3
)
)
=
log
(
5
)
(38)
Step: Apply the log laws to make x the subject of the equation.
(
3
x
+
3
)
log
(
5
)
=
l
o
g
(
5
)
divide
both
sides
of
the
equation
by
log
(
5
)
3
x
+
3
=
1
3
x
=
-
2
x
=
-
2
3
(
3
x
+
3
)
log
(
5
)
=
l
o
g
(
5
)
divide
both
sides
of
the
equation
by
log
(
5
)
3
x
+
3
=
1
3
x
=
-
2
x
=
-
2
3
(39)
Step: Substitute into the original equation to check answer.
7
·
5
(
-
3
2
3
+
3
)
=
7
·
5
(
-
2
+
3
)
=
7
·
5
1
=
35
7
·
5
(
-
3
2
3
+
3
)
=
7
·
5
(
-
2
+
3
)
=
7
·
5
1
=
35
(40)
Solve for x:
-
l
o
g
3
x
=
2
l
o
g
3
x
=
2
-
10
l
o
g
27
=
x
10
l
o
g
27
=
x
-
3
2
x
-
1
=
27
2
x
-
1
3
2
x
-
1
=
27
2
x
-
1
Logarithms are part of a number of formulae used in the Physical Sciences. There are formulae that deal with earthquakes, with sound, and pH-levels to mention a few. To work out time periods is growth or decay, logs are used to solve the particular equation.
A city grows 5% every 2 years. How long will it take for the city to triple its size?
Step: Use the formula
A=P(1+i)nA=P(1+i)n
Assume P=xP=x, then A=3xA=3x.
For this example n represents a period of 2 years, therefore the n is halved for this purpose.
Step: Substitute information given into formula
3
=
(
1
,
05
)
n
2
log
3
=
n
2
×
log
1
.
05
(
u
s
i
n
g
l
a
w
5
)
n
=
2
log
3
÷
log
1
,
05
n
=
45
,
034
3
=
(
1
,
05
)
n
2
log
3
=
n
2
×
log
1
.
05
(
u
s
i
n
g
l
a
w
5
)
n
=
2
log
3
÷
log
1
,
05
n
=
45
,
034
(41)
Step: Final answer
So it will take approximately 45 years for the population to triple in size.
- The population of a certain bacteria is expected to grow exponentially at a rate of 15 % every hour. If the initial population is 5 000, how long will it take for the population to reach 100 000 ?
- Plus Bank is offering a savings account with an interest rate if 10 % per annum compounded monthly. You can afford to save R 300 per month. How long will it take you to save up R 20 000 ? (Answer to the nearest rand)
I have R12 000 to invest. I need the money to grow to at least R30 000. If it is invested at a compound interest rate of 13% per annum, for how long (in full years) does my investment need to grow ?
Step: The formula to use
A
=
P
(
1
+
i
)
n
A
=
P
(
1
+
i
)
n
Step: Substitute and solve for n
30
000
<
12
000
(
1
+
0
,
13
)
n
1
,
13
n
>
5
2
n
log
1
,
13
>
log
2
,
5
n
>
log
2
,
5
÷
log
1
,
13
n
>
7
,
4972
.
.
.
.
30
000
<
12
000
(
1
+
0
,
13
)
n
1
,
13
n
>
5
2
n
log
1
,
13
>
log
2
,
5
n
>
log
2
,
5
÷
log
1
,
13
n
>
7
,
4972
.
.
.
.
(42)
Step: Determine the final answer
In this case we round up, because 7 years will not yet deliver the required R 30 000.
The investment need to stay in the bank for at least 8 years.
- Show that
logaxy=loga(x)-loga(y)logaxy=loga(x)-loga(y)(43)
- Show that
logaxb=loga(x)blogaxb=loga(x)b(44)
- Without using a calculator show that:
log7516-2log59+log32243=log2log7516-2log59+log32243=log2(45)
- Given that 5n=x5n=x and n=log2yn=log2y
- Write y in terms of n
- Express log84ylog84y in terms of n
- Express 50n+150n+1 in terms of x and y
- Simplify, without the use of a calculator:
- 823+log232823+log232
- log39-log55log39-log55
- 54-1-9-112+log392,1254-1-9-112+log392,12
- Simplify to a single number, without use of a calculator:
- log5125+log32-log8log8log5125+log32-log8log8
- log3-log0,3log3-log0,3
- Given: log36=alog36=a and log65=blog65=b
- Express log32log32 in terms of a.
- Hence, or otherwise, find log310log310 in terms of a and b.
- Given: pqk=qp-1pqk=qp-1
Prove: k=1-2logqpk=1-2logqp
- Evaluate without using a calculator: (log749)5+log51125-13log91(log749)5+log51125-13log91
- If log5=0,7log5=0,7, determine, without using a calculator:
- Given: M=log2(x+3)+log2(x-3)M=log2(x+3)+log2(x-3)
- Determine the values of x for which M is defined.
- Solve for x if M=4M=4.
- Solve: x3logx=10x2x3logx=10x2 (Answer(s) may be left in surd form, if necessary.)
- Find the value of (log273)3(log273)3 without the use of a calculator.
- Simplify By using a calculator: log48+2log327log48+2log327
- Write log4500log4500 in terms of a and b if 2=10a2=10a and 9=10b9=10b.
- Calculate: 52006-52004+2452004+152006-52004+2452004+1
- Solve the following equation for x without the use of a calculator and using the fact that 10≈3,16:10≈3,16:
2log(x+1)=6log(x+1)-12log(x+1)=6log(x+1)-1(46)
- Solve the following equation for x: 66x=6666x=66 (Give answer correct to 2 decimal places.)
