TEACHER’S GUIDE
Subject:
MATHEMATICS
UNIT No. 5
TARGET GROUP: S. 2
TOPIC: INDICES AND LOGARITHMS
SUB-TOPIC: LAWS OF INDICES
When
a factor is multiplied by itself a number of times, we can use powers
to express this operation in shorter form. For instance, the
operation 2 x 2 x 2 may be expressed as 23,
where 2 is the number being multiplied and 3 is the number of times
we are multiplying it. Thus 2 is known as the base
and 3 as the power.
And we read 23
as ‘two raised to the power of 3.’
Another
name for power is index,
from which we get the plural indices.
As we have seen above, indices arise from multiplication of similar
factors a certain number of times.
Take
for instance, finding the area or volume of a figure of side 3cm.
From
the formulae Area = length x width and volume = length x width x
height
The
area is given as 3 x 3 = 32
cm2
and the volume is given as 3 x 3 x 3 = 33cm3
Note
that even the units are given the same index since they are also
multiplied together a given number of times.
Indices are used to
describe repeated multiplication. This is why they are very useful in
science when writing very large or very small numbers, for example,
the size of the HIV virus (about 400 nanometers or 4.0 x 10-7metres)
or the distance of the sun from the earth (about 149 million
kilometres or 1.49 x 108 km).
MAIN CONTENT AND CONCEPTS TO
EMPHASISE:
By
the end of this subtopic the learners should be able to:
state
and apply the laws of indices;
The
laws of indices are concerned with addition, subtraction,
multiplication and division of powers of similar numbers such as the
area and volume of the figure we saw above. We shall look at these
laws in regard to the four operations above.
Law
1: Adding indices (multiplying products of similar factors)
Have
the students find the product of 3 x 3 x 3 and 3 x 3. Then introduce
the indices for both groups and show how both approaches lead to the
same result.
Activity
1
Work
out the product of 3 x 3 x 3 and 3 x 3. We see clearly that 3 has
been multiplied 5 times, that is (3 x 3 x 3) x (3 x 3).
Since
in the first case 3 has been multiplied 3 times, we can write it as
33
and the multiplication in the second case can be written as 32
Use
the new expressions above to rewrite the original expression above,
i.e 33
x 32
What
can you say about the two expressions above?
From
the expressions 33
= 3 x 3 x 3 and 32
= 3 x 3 we see that
33
x 32
= (3 x 3 x 3) x (3 x 3)
3 x 3 x
3 x 3 x 3
But
as we saw above, 3 x 3 x 3 x 3 x 3 = 35
(since 3 is
multiplied 5 times)
33
x 32
= 35
If
you observe closely, you will notice that the powers on the left hand
side of the equal sign add up to the power on the right hand side of
the equal sign i.e. 3 + 2 = 5. This means that when similar factors
(or when similar numbers) are multiplied, their result is always that
number or factor but with its index as the sum of their indices.
Thus
if a
is a factor (or any natural number) then the product of any two of
its multiples am
and an
is given by am
x an
= am+n
where m
and n
are the indices
Law
2: Subtracting indices (dividing products of similar factors)
Have
the students find the quotient of 3 x 3 x 3 and 3 x 3. Then introduce
the indices for both groups and show how both approaches lead to the
same result.
Activity
2
Divide
3 x 3 x 3 by 3 x 3. We see clearly that the 3s in the denominator
cancel out two of the 3s in the numerator, leaving only 3.
Since
in the first case 3 has been multiplied 3 times, we can write it as
33
and the multiplication in the second case can be written as 32
Use
the new expressions above to rewrite the original expression above,
i.e 33
32
What
can you say about the two expressions above?
Again
from the expressions 33
= 3 x 3 x 3 and 32
= 3 x 3, we see that
33
32
= (3 x 3 x 3)
(3 x 3)
= 3
But
as we saw earlier, this means that 33
32
= 332
= 31
Here
we see that the powers on the left hand side of the equal sign are
subtracted to obtain that on the right hand side i.e. 3
2 = 1.
This
means that when products of similar factors are divided, the result
is always that factor having as its index the difference of their
indices.
Thus
if a
is a factor (or any natural number) then the ratio of any two of its
multiples am
and an
is given by am
an
=
am-n
where m
and n
are the indices.
Law
3: Multiplying indices (raising a product of similar factors to a
power)
Have
the students find the product of 32
and
32.
Then raise the first expression to the power of the second expression
to show how both approaches lead to the same result.
Activity
3
Work
out the product of 3 x 3 and 3 x 3. We see clearly that 3 has been
multiplied 4 times, that is (3 x 3) x (3 x 3).
Since
in the first case 3 has been multiplied 2 times, we can write it as
32
and the multiplication in the second case can be written as 32
Use
the new expressions above to rewrite the original expression above,
i.e 32
x 32
What
can you say about the two expressions above?
We
see that 32
has been multiplied with itself once, that
is 32
x 32
By
expansion 32
x 32
= (3 x 3) x (3 x
3)
32
x 32
= 34
Now
what if we multiplied 32
twice, 32
x 32
x 32
By
expansion 32
x 32
x 32
= (3 x 3) x (3 x 3) x (3 x 3)
32
x 32
x
32
= 36
From
the examples above we see that
32
x 32
= (32)2
= 32
x
2
and 32
x 32
x 32
= (32)3
= 32
x 3
We
see that the powers on the left hand side are multiplied by the
number of times they appear to give the power on the right hand side.
Thus
if a multiple of a factor (or any natural number) am
is raised to a power n,
then the result is given by the factor a
raised to the power mn
(remember that in algebra we omit the multiplication sign)
(am)n
= am
x
n
or (am)n
= amn
Law
4: Dividing indices (dividing roots of a product of similar factors)
Have
the students find the quotient of 34
and
32.
Let them do it following the approach in activity 2 and then build on
their result as shown here below.
Activity
4
I
bet you now know that 3 x 3 = 32
and 32
x 32 =
34.
But how do we reverse this operation so that 34
becomes 32
again? Here is how we
go about it.
If
we wish to reverse the process of obtaining 34,
we simply raise it to a power that is the inverse of 4 i.e. ¼. So
(34)¼
= 34
x ¼
= 3
But
what if we do not wish to reverse it completely but only partly?
In
this case we raise the result 34
to a power which is the inverse of the power of 32.
So
34
is reversed by (34)½
=
32.
thus
when any number is multiplied with its inverse, it gives a result of
1. This inverse of a power is called the root of the number from
where we obtain the square root or cube root of a number.
Thus if a is a factor (or any natural number) raised to a
power m, then the nth root of am
is given by nam
= am
n,
where m and n are the indices.
Note
that m/n
is a fraction
and an index at the same time. So we call it a fractional
index.
Law
5: Negative indices (when the divisor is larger than the dividend)
Have
the students find the quotient of 32
and
33.
Let them do it following the approach in activity 2 and then build on
their result as shown here below.
Activity
5
We
saw earlier that 33
32
= 33
–
2
which gives a
positive power of 1. but
what if the divisor is larger than the dividend?
Consider
the expression 32
33.
The second law of indices gives 32
33
= 32
–
3
=
3-1
But
we know that 32
33 = (3 x 3)
xx
= 1/3
This
means that 3-1
= 1/3
Let
us look at another expression
Consider
33
35
Using the
second law of indices we obtain 33
35
= 335
= 32
But
we also know that
33
35
=
(3 x 3 x 3)/(3 x 3 x 3 x 3 x 3)
=
1/(3 x 3) = 1/32,
which means that 3-2
= 1/32
These
two examples show that negative indices represent proper fractions
with 1 as the numerator and the given number as the denominator.
But
1/32
is the reciprocal or the inverse of 3-2
since 32
x 3-2
= 1
So
if am
is a multiple of m
similar factors, then the negative power a-m
indicates a
reciprocal of am
such that a-m
=
1/am
or an inverse
of am
such that am
x a-m
= 1
Law
6: Zero Index (dividing two similar products or factors)
Have
the students divide two similar numbers such as 3
3 and
32
32.
Build upon their results by using the second law of indices to obtain
the zero index.
Activity
6
I
bet we all know that 3
3 = 1 and that 32
32
= 1 since we are dividing two similar members. But according to the
second law of indices 32
32
= 32
- 2
= 30
This
means that 30
= 1. Likewise, 53
53
= 1 or 53
53
= 53-3
= 50
and 50
= 1
This
means that any number with a power of zero equals 1.
So
if a
is a natural number, then a0
= 1.
Law
7: Indices of different bases with a common power
Have
the students multiply bases with common power such as (2 x 3)2.
Build upon their results as shown here below.
Activity
7
How
do we evaluate an expression like (2 x 3)2?
Removing
the brackets by multiplying through with the common power gives:
(2
x 3)2
= 22
x 32
=
4 x 9
=
36
Alternatively,
we can consider the values in brackets first: that is, (2 x 3)2
Evaluating the
values inside the brackets gives
(2
x 3)2
= 62
6 x
6 = 36
So
the product of two bases a
and b
raised to power n
can be evaluated as:
(a1b1)n
= anbn
When
1 becomes another power m,
it leads to property or law 7 of indices:
(ambm)n
=
amnbmn
SUB-TOPIC:
STANDARD FORM (SCIENTIFIC NOTATION)
TIME REQUIRED:
80 minutes
BRIEF DESCRIPTION OF SUBTOPIC:
In
this method we write any number as a product of two factors, the
first factor being a number between 1 and 10 (with 1 inclusive) and
the second factor being a power of ten. Numbers in standard form are
written in the form p
x 10n,
where p
is the first factor and n
is an index. For large numbers n
is positive and for small numbers it is negative.
MAIN CONTENT AND CONCEPTS TO
EMPHASISE:
By
the end of this subtopic the learners should be able to:
- apply the laws of indices to standard form;
- write large and small numbers in standard form.
TEACHING/LEARNING MATERIALS,
ACTIVITIES AND GUIDANCE:
Follow
the procedure below when writing numbers in standard form:
Determining
the leading factor
The
first factor (sometimes called a leading factor) should be a whole
number or a mixed fraction greater than or equal to 1 but less than
10, i.e. 1
p
< 10, where p
is the leading factor.
Activity
1
Show
the leading factor for each of the following numbers:
(a) 0.0062
(b) 0.062
(c) 0.62
(d) 6.2
(e) 62,000
(f) 620,000
- 6,200,000
Solution
(a) 0.0062
becomes 6.2 x …when we move 3 paces to the right
(b) 0.062
becomes 6.2 x … when we move 2 paces to the right
(c) 0.62
becomes 6.2 x … when we move 1 pace to the right
(d) 6.2
becomes 6.2 x … when we move no pace on either side
(e) 62,000
becomes 6.2 x … when we move 4 paces to the left
(f) 620,000
becomes 6.2 x … when we move 5 paces to the left
(g) 6,200,000
becomes 6.2 x … when we move 6 paces to the left
Note
that for whole numbers (those without a decimal) we assume that they
have a decimal point to their right e.g. 65 may be written as 65.0.
Remember
that a decimal point is moved until we get a leading factor between 1
and 10 i.e. 1
p
10 (one is included but 10 is not).
All
digits in the leading factor, p,
are significant.
Determining
the second factor
When
a decimal point is moved past values, all values to the left of a
decimal point carry a positive power and those on the right carry a
negative power. For example, 0.002 becomes 2.0 x 10-3
while 2000 becomes 2.0 x 103
when a decimal point moves three paces to the right and left
respectively.
Activity
2
Show
the second factor for each of the following numbers:
(a) 0.0062
(b) 0.062
(c) 0.62
(d) 6.2
(e) 62,000
(f) 620,000
- 6,200,000
Solution
(a) 0.0062
gives 10-3
because we move 3 paces to the right
(b) 0.062
gives 10-2
because we move 2 paces to the right
(c) 0.62
gives 10-1
because we move 1 pace to the right
(d) 6.2
gives … x 100
because we move no pace on either side
(e) 62,000
gives … x 104
because we move 4 paces to the left
(f) 620,000
gives … x 105
because we move 5 paces to the left
(g) 6,200,000
gives … x 106
because we move 6 paces to the left
Note
that all the digits moved past are counted and their total number
gives the power of ten (the second factor). The sign of this power
depends on the direction as we have seen above; right is negative,
left is positive.
Writing
figures in standard form
After
obtaining the first and second factors, we join them together using a
multiplication sign. Why this sign? It is because the two factors are
the respective parts of the number.
Activity
3
Write
the following numbers in standard form:
(a) 0.0062
(b) 0.062
(c) 0.62
(d) 6.2
(e) 62,000
(f) 620,000
- 6,200,000
Solution
(a) 0.0062
becomes 6.2 x 10-3
i.e. 3 paces to the right
(b) 0.062
becomes 6.2 x 10-2
i.e. 2 paces to the right
(c) 0.62
becomes 6.2 x 10-1
i.e. 1 pace to the right
(d) 6.2
becomes 6.2 x 100
i.e. no pace on either side
(e) 62,000
becomes 6.2 x 104
i.e. 4 paces to the left
(f) 620,000
becomes 6.2 x 105
i.e. 5 paces to the left
(g) 6,200,000
becomes 6.2 x 106
i.e. 6 paces to the left
SUB-TOPIC: LOGARITHMS
TIME REQUIRED:
80 minutes
BRIEF DESCRIPTION OF SUBTOPIC:
The
term logarithm was first used by a Scottish mathematician John Napier
(1550 - 1617) in reference to the power or index of a real number. It
is believed to have come from two Greek words: logo
meaning ratio
and arithmos,
which means
number.
The name therefore, suggests that he must have used ratios to obtain
the powers or indices of numbers.
Given
that 102
= 100, the logarithm or ratio of the numbers 102
and 100 is obtained as follows:
Step
1: introduce logarithms on both sides i.e log 102
= log 100
Step
2: separate the power from the base i.e 2 log 10 = log 100
The
translation reads: “twice the power of 10 equals once the power of
100.” This literal translation is obviously not clear. A better
translation reads: “The power of 10 is twice that of 100.”
Step
3: divide through by log10 to obtain the ratio i.e
2
log 10/log 10 = log 100/log 10
This
gives 2 = log 100/log 10
This
could be read as: “The ratio of the powers of 100 and 10 is 2 : 1.”
Step
4: simplify the ratio, i.e 2 = log10
100
Here
10 is the base (or the factor when the number is factorised to obtain
power) and 100 is the multiple (or product).
In
general, given that ab
= c,
where a
is the base, b
is the power and c
is the product, we may say that:
(i) log ab
= log c (ii) b
log a
= log c (iii) b
= log c/log a
and (iv) b
= loga
c
Note
that a
is the base in both the exponential equation and the log equation.
Note also that b
and c
switch places when we change from index form to log form.
MAIN CONTENT AND CONCEPTS TO
EMPHASISE:
By
the end of this subtopic the learners should be able to:
relate
indices to logarithms;
relate
the laws of indices to those of logarithms.
TEACHING/LEARNING MATERIALS,
ACTIVITIES AND GUIDANCE:
The
laws of logarithms do not differ from those of indices, given the
fact that logarithms are actually ratios of indices of real numbers.
We shall look at these laws in relation with addition, subtraction,
multiplication and division of indices.
Law
1: Adding logarithms (The logarithms of a multiple of factors)
Here
we want to show that the power of a multiple of similar factors is
equal to the sum of logarithms of the same factors.
Activity
1
Consider
the expression 3 x 3 = 32
We
see that 32
is a multiple (actually a
square) of 3
Step
1: Take logs in
base 3 on both sides; i.e. log3
(3 x 3) = log3
32
Step
2: Separate the power from the factor. We saw earlier that log3
32
= 2 log3
3
log3
(3 x 3) = 2 log33
log3
(3 x 3) = 2
(since log 3/log 3 = 1)
What
we have on the left is a logarithm (to base 3) of two factors.
We
now know that log3
(3 x 3) = log3
32
=
2 log 3/log 3
=
2
But
there is another interesting finding; you will realise that log33
+ log33
is also equal
to 2
i.e.
log33
+ log33
= log 3/log
3 + log 3/log 3
=
1 + 1
=
2
This
suggests that log3
(3 x 3) = log3 3
+ log3
3
So
the logarithm of a product of factors is equal to the sum of the
logarithms of each of these factors.
In
general, given that p
and q
are factors of a multiple pq,
then,
logb
pq
= logb
p
+ logb
q
Law
2: Subtracting logarithms (The logarithm of a ratio of two numbers)
Here
we want to show that the power of a quotient of similar factors is
equal to the difference of logarithms of the same factors.
Activity
2
Now
consider the expression 33
32.
This is a ratio of two numbers i.e. 33/32.
Step
1: apply the law
of indices; i.e. 33
32
= 33
- 2 =
3
33
32
= 3
Step
2: take
logarithms in base 3 on both sides:
log3
(33
32)
= log3
3
log3
(33/32)
= log3
3
You
realise that the value on the right is 1, implying that: log3
(33/32)
=
1
An
interesting finding here is that
log3
(33/32)
=
log3
33
– log3
32
=
3 log3
3
– 2 log3
3
=
3 –
2
=
1
This
means that the logarithm of a ratio of two numbers is the same as the
difference between their separate logarithms. This is in line with
the second law of indices. In general, given that p
and q are
real numbers in the ratio p/q,
then:
logb
p/q
= logb
p –
logb
q
Law
3: Multiplying a logarithm with a real number (The logarithm of a
number carrying an index)
Here
we want to show that when a logarithm is multiplied with a real
number, this number becomes the power of the factor in the logarithm
(that is, this number tells us how many similar factors are in the
multiple).
Activity
3
We
know that 10 x 10 = 100 and that
102
= 100
Step
1: Introduce
logs on both sides; i.e. log (10 x 10) =
log 100
or
log 102
= log 100.
Step
2: Separate the
powers; i.e. 2 log 10 = log 100.
This
we read as: ‘the power of 10 is twice that of 100’ or ‘the
ratio of powers of 10 and 100 is 2 :1’
Here
we see clearly that log 100 = log 102
and that 2 log 10 =
log 102.
In
general if p
is a multiple of n
factors, then
n
logb
p
= logb
pn
Activity
Law
4: Dividing logarithms (The ratio of 2 logarithms)
Here
we want to show that when logarithms are divided, the number in the
dividend becomes the base of the resultant logarithm.
Activity
4
Given
that log 102
= log 100,
Step
1: apply the
third law of logarithms to the left hand side; i.e. 2 log 10 = log
100
Step
2: Divide
through by log 10; i.e. 2 log 10/log 10 = log 100/log 10
Step
3: Simplify;
i.e. 2 = log10
100
where
the base of the logarithm is actually the denominator in the ratio.
So to find the power or logarithm of a number all we need is its
base.
In
general, if p and
b
are real numbers then: log
p/
log b
= logb
p
Example
1
Use
the laws above to express the following problems as a single
logarithm.
(a) log3
7 + log3
5 (b) log2
7 log2
5 (c) log4
256
(d) log5
12 + log6
18 (e)
log 27/log 3
Solution
(a) From
the first law
log3
7 + log3
5 =
log3
(7 x 5)
= log3
35
(b) From
the second law
log2
7
log2
5
= log2
7/5
(c) 4
is a 4th
root of 256
So
log4
256 = log4
44
= 4 log4
4
(d) log512
+ log618.
This cannot be expressed as a single logarithm because the base is
different in the two logarithms.
(e) From
the fourth law
log 27/log
3 = log3 27
= log3
33
= 3 log33
= 3
Example
2
Given
that x
= log8
16, what
problem was converted into logarithms?
Solution
(a) From
x
= log8
16 we obtain:
x
= log 16/log 8
Multiplying
through by the denominator gives
x
log 8
= (log 16/log 8) x log 8
x
log 8 = log 16
But
we know that x
log 8 = log 8x
(from the third law of logarithms)
So
we get log 8x
= log 16
And
when we remove the logs we get 8x
= 16 as the
converted problem.
SUB-TOPIC:
LOGARITHMS TO BASE TEN
BRIEF DESCRIPTION OF SUBTOPIC:
We
have seen that logarithms can be obtained in any base; after all, a
logarithm is a ratio of powers (or indices) of real numbers, be it
positive or negative. But just as we saw with counting systems, base
ten has been identified as the most ideal base to use when working
with logarithms. It is for this reason that we convert all logarithms
that cannot be evaluated in other bases into base ten.
MAIN CONTENT AND CONCEPTS TO
EMPHASISE:
By
the end of this subtopic the learners should be able to:
- find the logarithm of numbers in base ten;
- convert logarithms given in non-decimal bases to decimal base.
TEACHING/LEARNING MATERIALS,
ACTIVITIES AND GUIDANCE:
Consider
a problem such as x
= log816.
It is obvious that we cannot evaluate this problem in base 8 because
8 is not a root of 16. This means that x
(the power of 8) is not a whole number but a ratio of two numbers or
a decimal number.
This
kind of index has the property of first increasing the number (base)
before decreasing it by finding its root.
Such
a problem is easier to evaluate in base ten whereby we express it as
a ratio of two logarithms such that the base becomes a denominator.
Activity
1: Finding logarithms in base ten
Step
1: use the
fourth law of logarithms to write the R.H.S as a fraction; i.e
x
= log8
16
=
log 16/log 8
Step
2: Use the log
tables to find the values of the numerator and denominator. These
tables show the logarithms of the factors in the standard form
p x 10n
(where p
is the leading factor and n
is the power of 10).
It
is for this reason that the values in the table lie between 1 and 10
i.e. 1
p
10.
So
the numbers in the tables range from 1.00 to 9.99, although the
decimal points is not shown in the table.
Table1:
Logarithms of
numbers, p
log10
p
The
logarithms of p
(log p)
in the table is found by first locating the given values in the table
(i.e. find the values in the first column and first row that
correspond with the given value). Then the row and column in which
they are found are made to intersect at right angles. Where they
intersect is the required logarithm.
Finding
logarithms of positive numbers less than 10
So
from our expression x
= log16/ log 8 we can find log 8 as follows:
Step
3: Look for 80
in the first column; this is the value 8.0
Step
4: Look for 0 in
the first row; this is the value 0.00. This means that we now have
the value 8.00
Step
5: Make the row
of 80 and column of 0 to intersect; where they intersect is the
logarithm of 8.00 i.e. log 8 = 0.9031 since 8 is written as 8.0 x 100
in standard form and log 100
is zero (which means that there is nothing to add to it).
Finding
logarithms of positive numbers greater than 10
We
have seen that logarithms of the leading factor in the standard form
give us a decimal fraction. This is because when expressed in
standard form, all numbers below 10 carry zero as the power of 10
e.g. 8.0 above becomes 8.0 x 100.
It
is this power of 10 that becomes the whole number in the logarithm.
Consider
a number such as 16. In standard form it becomes 1.6 x 101.
Taking logarithms gives log10
(1.6 x 101).
Now
from the first law of logarithms
log10
(1.6 x 101)
= log10
1.6 + log10
101
Since
we cannot evaluate log10
1.6 easily, we use log tables as follows:
Step
1: Look for 16
in the first column; this is the value 1.6.
Step
2: Look for 0 in
the first row; this is the value 0.00.
Step
3: Make the row
from 16 and column from 0 to intersect; where they meet is the
logarithm for 1.60. This is given as 0.2041.
So
log 1.6
+ log10
101
or log10
1.6 + 1 log10
10
=
0.2041 + 1
=
1.2041
So
x
= log 16/log 8
=
1.2041 – 0.9031
=
0.3010
Do
you realise that 1.2041 is the power of 10 when 16 is expressed as a
product of two factors (1.6 and 101)?
Finding
logarithms of positive numbers less than 1
When
positive numbers less than 1 are expressed in standard form, the
power of 10 (the characteristic) is given as negative. This is
because the decimal point is moved past the given digits towards the
right when trying to obtain a leading factor in the range of 1 to 10.
Note that the logarithm (or power) of the leading factor (the
mantissa) is positive since the number is positive.
Consider
a number 0.01. In standard form the number becomes 1.0 x 10-2.
The logarithm of the leading factor log10
1.0 = 0, implying that
log10
(1.0 x 10-2)
= log10
1.0 + log10
10-2
=
log10
(100
x 10-2)
=
102
+ 0.00
=
2 + 0.00
=
2.000
notice
how we write a negative characteristic; we put a bar on top to
signify that the power of ten is negative or that the actual number
that was expressed as a logarithm is less than 1 (we read a negative
logarithm as a bar of the number e.g. bar 2 point zero).
Activity:
find log 0.02
Step
1: Convert each
value to standard form
Log
0.02 = log (2.0 x 10-2)
Step
2: Find the
logarithm of the leading factor (the mantissa) from the logarithm
tables. The logarithm of the second factor is its power (this is the
characteristic).
Log
2.0 = 0.3010 and log 10-2
= -2
Step
3: Write the
logarithm by adding the mantissa to the characteristic. Do
not subtract!
Simply place a bar on top of the characteristic to show that it is
negative.
2.3010
SUB-TOPIC:
MULTIPLYING AND DIVIDING NUMBERS USING LOGARITHMS
TIME REQUIRED:
40 minutes
BRIEF DESCRIPTION OF SUBTOPIC:
We
have seen that the laws of logarithms make use of multiplication and
division. Therefore we can use logarithms to multiply and divide
numbers.
MAIN CONTENT AND CONCEPTS TO
EMPHASISE:
By
the end of this subtopic the learners should be able to:
- multiply and divide using logarithms
TEACHING/LEARNING MATERIALS, ACTIVITIES
AND GUIDANCE:
Multiplying numbers using logarithms
According
to the first law of logarithms: log (2 x 3) = log 2 + log 3
So
we can easily multiply numbers according to the first law of
logarithms using the steps outlined here.
Activity:
find the product of 2 and 3 using logarithms
Step
1: Express the
product as a sum of logarithms i.e. log (2 x 3) = log 2 + log 3
Step
2: Find the
logarithms of the factors in the sum using the log tables
i.e.
log 2 = 0.3010 and log 3 = 0.4771
Step
3: Add the
logarithms obtained i.e. 0.3010 + 0.4771 = 0.7781
or
100.3010
x 100.4771
= 100.3010
+ 0.4771
= 100.7781
Step
4: Look for the
logarithm obtained as a sum in step 3 from the log tables. Then read
off the first values, first along the row and then along the column
intersecting at this point e.g. along the row, the value is 60 which
is 6.0 (we take 0.7781 ~ 0.7782 because it is nearer to our value)
and along the column, the value is 0 which is 0.00.
Step
5: Add up the
values obtained. The sum is the required number
so
100.7781
= 6.00. A quick check shows
that 2 x 3 = 6
Example
Evaluate
245 x 35
Solution
Using
the standard form: 245
becomes 2.45 x 102
35
becomes 3.5 x 101
Applying
logarithms gives
log
(245 x 35) = log 245 + log 35
=
log (2.45 x 102)
+ log (3.5 x 101)
=
log 2.45 + log 102
+ log 3.5 + log 101
=
0.3892 + 2 + 0.5441 + 1
= 3.9333
Now
expressing the sum of logarithms as powers of 10 gives
103.9333
= 103
x 100.9333
Using
log tables 100.9333
lies between 100.9330
and 100.9335.
So
we find their average i.e. (0.9330 + 0.9335)/2
But
100.9330
= 8.57 while 100.9335
= 8.58.
So
their average is 8.57 + 8.58 = 8.575
Thus
we have 8.575 x 103
or 8575. A quick check shows that 245 x 35 = 8575
Note
that we use a lot of approximation and estimation while dealing with
logarithms, especially while trying to find the leading factor in the
standard form.
Dividing
numbers using logarithms
We
can also divide numbers using logarithms and the laws of indices.
Consider the expression 3.5
1.45.
Activity:
evaluate 3.5
1.45 using logarithms
Step
1: Apply the
second law of logarithms
log
(3.5
1.45) = log 3.5 – log 1.45
Step
2: Find the
logarithms from the log tables
log
3.5 = 0.5441 and log 1.45 = 0.1614
Thus
log 3.5 – log 1.45 = 0.5441 – 0.1614
Step
3: express the
logs as powers of 10
100.5441
100.1614
=
100.5441 - 0.1614
=
100.3827
And
the equivalent value of 100.3827
in the log tables is
about 2.41 (since 0.3827 is nearer 0.3820 than 0.3838).
Example
Evaluate:
380
61
Solution
These
problems are quite hard to evaluate using the method of long division
but become a lot easier when we use logarithms
(i) 380
61 becomes
log
(380
61) when we introduce logs
but
log 380 = log 3.8 + log 102
and log 61 = log 6.1 + log 101
Now
from the log tables log 3.8 = 0.5798 and log 6.1 = 0.7853
So
log 380 = log 3.8 + log 102
= 0.5978 + 2 or 2.5798
and
log 61 = log 6.1 + log 101
= 0.7853 + 1
= 1.7853
Expressing
these logs as powers of base 10,
log
(380
61) = 102.5798
101.7853
And
the law of indices gives:
102.5798
101.7853
=
102.5798
1.7853
= 100.7945
And
the equivalent value in the log tables for 100.7945
is 6.23.
SUB-TOPIC:
ANTILOGARITHMS
TIME REQUIRED:
40 minutes
BRIEF
DESCRIPTION OF SUBTOPIC:
In
Mathematics the original value of a logarithm is known as the
antilogarithm
of that log. It is the value we obtain by
subjecting the
base to the logarithm i.e. by finding the value of the base raised to
the given logarithm (or power).
We
should look at the antilogarithm (or antilog in short) as a process
of reversing or removing logarithms from the problems so that they
appear as if we had not applied logarithms to them.
We
have seen that logarithms are in essence ratios of powers or indices
of numbers (the base) the equivalent of which have until now been
referred to as the `original value. For instance, the original value
of log10100
= 2 is 100 since 102
= 100 (where 2 is the power and 10 is the base). Thus if log10
100 = 2, then 100 = 102
if we divide through by log10. Here 1/log 10 is the antilog.
MAIN CONTENT AND CONCEPTS TO
EMPHASISE:
By
the end of this subtopic the learners should be able to:
- find the antilogarithms of given logarithms.
TEACHING/LEARNING
MATERIALS, ACTIVITIES AND GUIDANCE:
From
the above example we have seen that log tables can be used to find
the antilog. This is because we know that the logs are in the middle
and the original numbers (the antilogs) are in the first column and
row.
But
we can also use the antilogarithm tables. These have been prepared in
such a way that the logarithms are the ones in the first column and
row while the antilogarithms are the values in the middle.
Note
that although the middle values (the antilogs) have been written
without the decimal point, they are actually numbers between 1 and
10. So the antilog is the actual value p
of the leading factor in the standard form p
x 10n
(and p
lies between 1 and 10 i.e. 1
p <
10).
In
the antilog tables the first two digits of the logarithm are in the
first column while the third digit is in the first row.
Table3:
Antilogarithms, p
10p
Activity:
To find the antilog of 0.124 and 4.124 from the antilog tables
Step
1 read the first
two values of the given logarithm from the first column of the
antilog tables. Here the first two digits are .12.
Step
2. read the
third value of the logarithm from the first row on top of the table.
This is the value 0.004 (represented by 4 in the table).
Step
3. extend the
row of the first two values to meet the column of the third value.
Where they meet is the required antilogarithm.
Step
4. express the
value obtained in such a way that it lies between 1 and 10. This is
because it is the value of p,
the leading factor in the standard form p
x 10n.
So 1330 is actually 1.330.
Step
5. if the
logarithm given is greater than 1 (if the characteristic is not zero
such as 4.124), we first express it as a sum of two indices or two
logarithms.
For
example, 104.124
= 104
+ 0.124 =
104
x 100.124
(from the
first law of indices). This is done because only the values of the
leading factor are obtained from the tables; the rest can be obtained
by simple multiplication e.g. from tables 100.124
= 1.33. so
logarithm of 4.124 is the value
1.33
x 104
= 1.33 x 10,000
= 13,300
LESSON PLAN
Date:
…………… Class: S. 2 Period: 1 Number of students: …...
TOPIC: INDICES
AND LOGARITHMS
SUBTOPIC: LAWS
OF INDICES
Lesson
objectives
By
the end of this subtopic the learners should be able to:
state
and apply the laws of indices.
.
Teaching
methods:
(i)
Guided discovery; (ii) demonstrations (iii) question and answer
Teaching
aids: chalkboard, log tables and manila paper.
References:
MK Secondary Mathematics (MKSM) Bk 2, SSM Bk 2, SMU Bk 2
|
STEPS
|
TIME
|
CONTENT
|
TEACHER’S ACTIVITY
|
STUDENTS’ ACTIVITY
|
|
I
|
5
minutes
|
Review
basic elements of algebra. Emphasise the four operations, namely:
addition, subtraction, multiplication and division.
|
Asks
students to do some basic algebra with similar values e.g 3 x 3 x
3
|
Do
as intructed
|
|
II
|
10
minutes
|
Introduction
of the topic of indices and related terms; base and index/indices
or power
|
Explains
the meaning of the different terms
|
Listen
attentively and then take notes
|
|
III
|
15
minutes
|
Introduction
of the first two laws of indices using activities
|
leads
students through the activities and states laws thereafter
|
Do
as instructed and take notes as given
|
|
IV
|
15
minutes
|
Introduction
of the next two laws of indices using activities
|
leads
students through the activities and states laws thereafter
|
Do
as instructed in their exercise books.
|
|
V
|
15
minutes
|
Introduction
of the next two laws of indices using
activities
|
leads
students through the activities and states laws thereafter
obtained above. Moves around to correct students
|
Do
as instructed and take notes as given
|
|
VI
|
15
minutes
|
Introduction
of the seventh law of indices and doing exercise
|
states
the seventh law and gives exercise thereafter. Moves around to
correct students
|
Do
as instructed
|
|
VI
|
5
minutes
|
Conclusion
and exercise
|
Recaps
the topic and allows students to ask any pending questions.
|
Attempt
exercise
|
No comments:
Post a Comment