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.

Answer is $428,000.

Alternative Solution

Since the value of the land is increased by 7%, the total value is 107%. Therefore, the value may also be found as $400,000 x ¹⁰⁷⁄₁₀₀ = $428,000.

Answer is $428,000.

2. In Jamaica, electricity charges are calculated based on the following table:

(i) Calculate the electricity charges for a customer who used 1200 kwH.

There is a government tax of 17.5% on the electricity charges.

(ii) Calculate the tax on the customer's electricity charges, giving your answer to the nearest cent.

(iii) Calculate the total amount paid by the customer.

Solution

(1) The electricity charge is computed as, Fixed Charge + Charge for KwH used.

As 1200 KwH was used, then the charge is:

= $ 3.50 + 1200 x 15 cents

= $ 3.50 + ^{( $1200 x 15 )}⁄₁₀₀ = $ 3.50 + $ 180.00

= $ 183.50

(11) Since there is a Government tax on the customer's electricity charges,

The charge is = $ 183.50,
17.5% tax = ^{( $183.50 x 17.5 )}⁄₁₀₀ = $32.11

Tax = $ 32.11

(111) The total amount paid by the customer is:

Electricity Charge + Tax
$ 183.50 + $ 32.11
Total amount is $215.61

Answer is $215.61

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, 8¹⁄₃ is another way of writing the cube root of 8, therefore 8¹⁄₃ = 2.

NB 8²⁄₃ 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 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.

e.g., 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.
e.g., 3⁶ / 3⁴ = 3^{6 - 4} = 3²

NOTE: This may be justified by expanding and dividing.
Similarly X² / X⁵ = X^{2 - 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.

A number of rules were stated above. Please review these and then attempt the following examples.

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: 81¹⁄₂ x 27^-1⁄₃

Solution

As 81 = 3⁴ and 27 = 3³, then

81¹⁄₂ 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: 4^{3x - 1} = 64 x 4^x

Solution

Expressing all terms as a power of 4:

64 = 4³

... 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.

1. Simplify the following:

(a) 5a³b x 4a²b x 7ab³

(b) 12x^-4y² / 3x³y^-5

2. Find the values of:

(a) 64^-5⁄₆

(b) 27^-2⁄₃

3. Solve the following equation for x.

4^2x = ¹⁄₃₂

Clement Radcliffe is an independent contributor. Send questions and comments to kerry-ann.hepburn@gleanerjm.com