Review of indices
Clement Radcliffe, Contributor

We will begin this week with the solution to last week's practice exercise.

1. Mr Williams bought a plot of land for $400,000. The value of the land appreciated by 7% each year. Calculate the value of the land after one year.

Solution

The increase in value of the land after the first year is: $400,000 x ⁷/100 = $28,000

... The value after the first year is: $400,000 + $28,000 = $428,000.

Since the value is increased by 7%, the total value is 107%. Therefore, the value may also be found as $40,000 x 107/100 = $428,000.

2. In a certain country, electricity charges are made up of a fixed fuel charge of 45 cents per unit and an energy charge computed under THREE schemes as follows:

The meter reading of a certain business place is as follows:

Calculate:

(i) The number of units used

(ii) The energy charge

(iii) The fuel charge

(iv) The total amount paid by the business place.

Solution

(i) The number of units used

= present reading - previous reading

= 39,421 - 18,368 units

= 21,053 units

(ii) As the rate for energy used for business places is 30 cents per unit

... Total energy charge

= 21,053 x 30 cents = $6, 315.90

(iii) The fixed fuel charge is 45 cents per unit

... Total fuel charge

= 21,053 x 45 cents = 943,785 cents

= $9,437.85

(iv) The total electricity charge

= fuel charge + energy charge

= $6,315.90 + $9,437.85

= $15,789.75

We will continue this lesson 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 power of 4, that is 2⁴ . In this case 4 is the index or power of 2.

Example: Express 32 as a power of 2.

As 32 = 2 x 2 x 2 x 2 x 2

... 32 = 2⁵

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 2⁷, for example, 2¹ = 2, 2³ = 8, 2⁴ = 16, etc.

Powers of 3 up to 3⁵, for example, 3² = 9, 3⁴ = 81, etc.

(b) Denominator of a fractional power represents root.

For example, 8^1/3 is another way of writing the cube root of 8, therefore, 8^1/3 = 2.

Note briefly that 8^2/3 is the square of the cube root of 8. It can be written also in the form (8^1/3)²

(c) Any number to the power zero is equal to 1 for example 4⁰ = X⁰ = 1.

(d) Negative power represents the reciprocal, for example, 3^-2 = 1/3² = 1/9

3^-2 is commonly misinterpreted as -3² = -9. Avoid making this error.

Repeating 2^-3 = (2^-1) 3 = (1/2)³ = 1/8

(e) When we multiply numbers with the same base, we add the indices.

for example: 4³ x 4⁴ = 4³ + 4⁴ = 4⁷

Can you say why this is so?

It can be shown to be true as follows:

4³ x 4⁴ = 4 x 4 x 4 x 4 x 4 x 4 x 4 = 4⁷

(f) To divide numbers with the same base, we subtract the indices.

for example: 3⁶ ÷ 3⁴ = 3^6-4 = 3²

NOTE: This may be justified by expanding and dividing.

Similarly, X² ÷ X⁵ = X²-5 = 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 4⁵

Given that 8 = 2³ and 4 = 2²

... 8² x 4⁵ = (23)² x (2²)⁵ = 2⁶ x 2¹⁰ = 2¹⁶.

Let us apply these to the following examples:

1. 3a²b x - 4ab³

(A) -3ab²

(B) 12a³b⁴

(C) -12a³b⁴

(D) -a-3b⁴

Solution

3 x -4 = -12, a² x a = a³, b x b³ = b⁴

... The product is -12 x a³ x b⁴ = -12a³b⁴

Answer is (C)

2. Simplify: 81^1/2 x 27-^1/3

Solution

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

(3⁴)^1/2 x (3³)^-1/3 = 3² x 3^-1 = 3

3. Solve the equation: 43x-1 = 64 x 4x

Solution

Expressing all terms as a power of 4:

... 4^3x-1 = 4³ x 4^x = 4^3+x

Since 4^3x-1 = 4^3+x, equating the indices:

3x - 1 = 3 + x Transposing

... 2x = 4

... x = 2

This topic, as I said before, is a crucial one and I want you to absorb the information given in this lesson. You will reinforce the concepts when you do the following for homework.

Simplify the following:

1. (a) 5a³b x 4a²b x 7ab³

(b) 12x ^-4y² ÷ 3x³y^-5

2. Find the values of:

(a) 64^-5/6

(b) 27^-2/3

(c) 81^3/4

3. Solve the following equation for x.

4^2x = ¹/32

Clement Radcliffe is the principal of Glenmuir High School in May Pen.