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Computation:
Indices
Clement
Radcliffe, Contributor
 |
| Action
from Fame-School Rules at Camperdown
High School in Janaury 2005. -
Toussaint Smith |
WE
WILL begin this week's lesson with a
review of the homework given last week.
1.
Calculate the simple interest if $4,000
is placed in a bank at eight per cent
per annum for 6 months.
SOLUTION
| Simple
Interest = |
Principal
x Time x Rate |
| |
100
|
Substituting
| ie.
Simple Interest = |
$4,000
x 1/2x 8
|
| |
100
|
N.B.:
Time must be in years.
=
$160.
2.
The simple interest on $12,000 invested
at eight per cent per annum is $6,720.
Calculate the number of years for
which the sum was invested.
SOLUTION
From
the above, the simple interest = $6,720,
Principal = $12,000 and Rate = 8%.
The
formula for number of years
| =
Simple Interest x 100 |
|
principal
x rate
|
|
= 6,720 x 100 |
=
7 years
|
|
12,000
x 8
|
|
3.
A woman bought a stove for $12,800.
After using it for two years she decided
to trade in the stove. The company
estimated a depreciation of 15 per
cent for the first year of its use
and a further 15 per cent on its reduced
value, for the second year.
(i)
Calculate the value of the stove after
the two years.
(ii)
Express the value of the stove after
two years as a percentage of the original
value.
SOLUTION
(i)
The cost of the stove is $12,800.
After one year, the depreciation is
15%. The stove has 85% of its value.
|
ie. 85% of $12,800 = |
$12,800
x 85
|
|
|
100
|
| =
$10,880 |
As
there is a reduced value of 15 per
cent after the second year, the reduced
value is
ie.
85% of $10,880, the value at the start
of the year.
|
ie. 85% of $10,880 = |
85
x 10,880
|
|
|
100
|
| $9,248 |
Answer
= $9,248
(ii)
The original value of the stove =
$12,800
The
value after two years = $9,248.
| ie.
percentage = |
$9,248
x 100 |
=
72.25%.
|
|
|
12,800
|
|
From
the above please note the following:
*
Care must be taken to ensure that
the value of the stove after two years
is used as the numerator in finding
the percentage.
*
The answer is not 70 per cent of the
original cost (as some students may
believe) since 15 per cent is deducted
twice.
The
lesson will continue with a review
of indices, an aspect of COMPUTATION.
INDICES
This
is the power of a number. For example,
16 may be expressed in the form
2
to the 4 power,
in this case 4 is the index or power
of 2.
Example:
Express 64 as a power of 2.
As
64 = 2 x 2 x 2 x 2 x 2 x 2
ie. 64 = 2 to the 6 power
Expressing
numbers in index form is fundamental
to solving certain problems.
Points
to note:
(a)
It is to your benefit to know the
value of some whole numbers raised
to powers, for example:
Powers
of 2 up to 2 to the 7 power, for example,
2 to the 1 power = 2, 2 to the 3 power
= 8, etc.
Powers
of 3 up to 3 to the 5 power, for example,
3 to the 2 power= 9, 3 to the 4 power
= 81, etc.
(b)
Denominator of a fractional power
represents root.
For
example, 81/3 is another way of writing
the cube root of 8, therefore 81/3
= 2.
N.B.
81/3 is the square of the cube root
of 8. It can be written also in the
form (81/3)2
(c)
Any number to the power zero is equal
to 1 for example 4 to the 0 power
= X to the 0 power = 1.
(d)
Negative power represents the reciprocal,
for example, 3 to the 2 power = 1/3²
= 1/9
Kindly
note that 3² is commonly misinterpreted
as -3² = -9. Avoid making this
error.
Repeating
2³ = (2-1)³ = (1/2)³
= 1/8
(e)
When we multiply numbers with the
same base, we add the indices.
e.g.,
2³ x 2 to the 4 power = 2³
+ 4 power = 2 to the 7 power
Could
you say why this is so?
It
can be shown to be true as follows.
2³
x 2 to the 4 power= 2 x 2 x 2 x 2
x 2 x 2 x 2 = 2 to the 7 power
(f)
To divide numbers with the same base
we subtract the indices.
e.g.,
8 to the 6 power ÷ 8 to the
4 power = 8 to the 6 power - 4 power
= 8²
NOTE:
This may be justified by expanding
and dividing.
Similarly
X²÷
X to the 5 power = X²-5
power = X-³
I
am sure that you have noted the importance
of directed numbers.
(g)
In attempting to simplify an expression
it is always necessary to express
each term in the form of its smallest
factor, for example:
Evaluate:
8²
x 4 to the 5 power
Given
that 8 = 2³ and 4 = 2²
ie.
8²
x 4 to the 5 power = (2³)2 x
(2²)
3 = 2 to the 6 power x 2 to the 10
power = 2 to the 16 power.
Let
us apply these to the following examples:
1.
2a²b
x -4ab³
| (A)
-2ab2 |
(B)
8a³b to the 4 power |
(C)
-8a³b to the 4 power |
(D)
-4a-3b4 |
SOLUTION
2
x -4= -8, a²
x a = a³, b x b³ = b to
the 4 power
Answer
is (C)
2.
Simply 811/2 x 27-1/3
As
81 = 34 and 27 = 33 then
81
1/2 x 27-1/3 = (34)1/2 x (3³)-1/3
As
4 x 1/2 = 2 and 3 x - 1/3 = -1
(3
to the 4 power)1/2 x (3³)-1/3
= 32 x 3-¹ =
3
3.
Solve the following equation for x.
9²x
= 1/27
(3²)²x
= 27-¹ = (3³)-¹
3
to the 4x power = 3-³
Since
4x and -3 are both powers of 3 then
4x
= -3
ie
x = -3/4
This
topic, as I said before, is a crucial
one and I want you to absorb the information
given in this lesson. Reinforce the
concepts by doing the following for
homework.
Simplify
the following:
1.
a) 4x³ x 5x²y
x 2xy³
b)
9x-4y²÷
3x³y to the -5 power
2.
Find the values of:
a)
491/2 b) 27-2/3 c) 811/4
3.
4³x-1 = 64 x 4 to the x power
*
Clement Radcliffe is principal
of Glenmuir High School in Clarendon.
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