Real Number and Numeration System

In this section, you will learn more about the subsets of the real number system and basic operations on them. Specifically, you will study surds, exponents (indices), logarithms and modular arithmetic. These are fundamental concepts in mathematics which have various applications in everyday life. Surds have applications in physics, number theory and finance. Indices and logarithms are applied in banking, population growth models and engineering. Modular arithmetic is used to design calendar systems and duty rosters. By the end of this section, you will be able to evaluate the relationships between the laws and properties of surds, indices and logarithms and apply them to solve problems in everyday life.

KEY IDEAS

• Applications of logarithms: Logarithms have several applications in real life.

• Concept of logarithms: Logarithm is the inverse of indices. That is, logarithm can be used to reverse indices.

• Indices in which the power is an integer: If the power of an indicial expression is an integer, three situations arise; the integer may be positive, zero or negative.

• Keywords: Pure surds, mixed surds, compound surds, radical, radicand, rationalisation, conjugate, conjugate pair, exponents, indices, logarithm, base, augment.

• Laws of logarithms: Like indices, there are laws, which guide us to perform operations on logarithms.

• Operations on surds: This involves the addition, subtraction, multiplication and division of surds.

• Order of operations involving indices: when several operations are involved in a computation, there are rules that will help you to do the computation.

• Perfect squares: A perfect square is a number whose square root is a natural number. In other words, when you square a natural number, you get a perfect square.

• Rationalisation: This is the process of making the denominator of a given surd rational.

• Rules of indices: There are rules that will guide you to perform operations on indices.

• Simplification of surds: This is the process of expressing a surd in its simplest form.

• Surd or radical: The square root of a number, which is not a perfect square, gives rise to a surd, also known as a radical. These are irrational numbers.

Do you remember the set of perfect squares you learnt in JHS? Write down at least six perfect squares and show them to a classmate. The set of perfect squares P = {1, 4, 9, 16, 25, 36, 49 …}. When we find the square root of a number, which is not a perfect square, we get a surd. Examples of surds include √2, √3, − 2 √5 and 2 − √2/3 .

Perform the activity below with in pairs or in a small group. This activity will help you to generate and simplify surds.

Activity 1.1: Generating Surds Using the Geodot or Graph Paper Materials needed: Geodot or graph paper, pencils and rulers.

Step 1: Take each small square of four dots on the geodot as a unit square.

If you are using a graph paper, take a one-centimetre square as a unit square.

Draw a unit square on your graph or geodot paper as shown below.

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Figure 1.1: Drawing unit square on geodot

Step 2: Draw a two-unit square on the graph or geodot paper. Count the number of unit squares in the square you have drawn. If you counted four-unit squares, you are correct.

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Figure 1.2: Drawiacng two-unit squares on geodot

Step 3: Draw a three-unit square on your graph or geodot paper and count the number of unit squares in it.

Step 4: If you draw a six-unit square, how many unit squares will you find in it? What are your observations? Share your observations with your group.

Step 5: Now, draw a rectangle on your geodot or graph paper and count the number of unit squares in the rectangle you have drawn. Is the number you obtained a perfect square? What conclusion can you draw from this activity?

Share your ideas with your group.

You can see that the first four steps of this activity assist you to generate the set of perfect squares. The fifth step gives you an example of a non-perfect square.

If you find the square root of non-perfect squares, you will get surds or radicals like √2, √3 and so on. Note that because 1, 4 and 9 are perfect squares, √1 = √1 × 1 = 1 , √4 = √2 × 2 = 2 and √9 = √3 × 3 = 3 .

Activity 1.2: Simplifying Surds Using the Geodot or Graph Paper Materials needed: Geodot or graph paper, pencils and rulers.

To simplify √12, using the geodot, draw a rectangle with 12 dots as shown in the diagram below. What is the largest possible square you can draw within the space of the 12 dots? How many unit squares are in the square you have drawn?

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Figure 1.3: Drawing 4-unit squares on geodot The square in the geodot has 4-unit squares within it. Thus, you can write √12 = √4 × 3 = √4 × √3 = 2 × √3 = 2 √3.

Similarly, to simplify √24 , express 24 as a product of two numbers, one of which must be a perfect square. That is, √24 = √4 × 6 = √4 × √6 = 2 × √6 = 2 √6.

Also, √32 = √16 × 2 = √16 × √2 = 4 × √2 = 4 √2.

Activity 1.3: Finding the Square Root of Perfect Squares Using Prime Factors The square root of a perfect square can be found by breaking the number down into its prime factors. If the number has an even number of prime factors, we can find the square root easily as shown in the examples below.

Example 1.1

√169 = √13 × 13 = 13.

Example 1.2

√36 = √__________ 2 × 2 × 3 × 3 = √2 × 2 × √3 × 3 = 2 × 3 = 6.

Example 1.3

√100 = √__________ 2 × 2 × 5 × 5 = √2 × 2 × √5 × 5 = 2 × 5 = 10.

There is yet another method for finding the square root of perfect squares.

You will be exposed to this method in the following activity.

Activity 1.4: Finding the Square Root of Perfect Squares Using Repeated Subtraction To use this method to find the square root of a perfect square, list the set of odd numbers and follow the steps below.

Step 1: Subtract the first odd number from the perfect square.

Step 2: Subtract the second odd number from the result of step 1.

Step 3: Subtract the third odd number from the result of step 2.

Step 4: Do this until you get zero.

Step 5: The number of times we subtract an odd number to get zero is the square root of the perfect square.

Example 1.4

Find the square root of 36, using the repeated subtraction method.

1. 36 − 1 = 35

2. 35 − 3 = 32

3. 32 − 5 = 27

4. 27 − 7 = 20

5. 20 − 9 = 11

6. 11 − 11 = 0 Since the subtraction was done six times to get zero, the square root of 36 is 6.

Example 1.5

Find the square root of 49, using the repeated subtraction method.

1. 49 − 1 = 48

2. 48 − 3 = 45

3. 45 − 5 = 40

4. 40 − 7 = 33

5. 33 − 9 = 24

6. 24 − 11 = 13

7. 13 − 13 = 0 Because it took seven steps to get zero, √49 = 7 .

In this lesson, you are going to perform the basic operations on surds. That is, you are going to learn how to add, subtract, multiply and divide surds.

Activity 1.5: Addition and subtraction of surds You can perform this activity alone or with a partner.

Materials needed: cardboard, board markers and a pair of scissors.

Step 1: Cut out about 20 pieces from the cardboard, making sure the cards you cut out have almost the same shape and size.

Step 2: Write √3 on ten of the cards and √6 on the remaining pieces.

Step 3: Take some of the cards with √6 on them. Ask your partner to take some of the cards with √6 on them. How many cards do you have? How will you represent that on paper? How many pieces does your partner have? How will you write what your partner has on paper?

Now, how many cards do you and your partner have altogether? How will you represent the operation you have just performed in mathematical symbols?

For example, it could be something like:

⎡ ⎢ ⎣ √6 √6 √6 ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ √6 √6 √6 √6 √6 √6 ⎤ ⎥ ⎦ = 3 √6 + 6 √6 = 9 √6

Step 4: Take some of the √3 cards. How many cards did you pick? Give some of your cards to your partner. How many do you have left? How will you represent the operation you have just performed in mathematical symbols?

Step 5: Take some of the cards with √3 on them. Let your friend take some of the cards with √6 on them. How many cards do you have in total? How will you represent the operation you have just performed on a piece of paper?

Show it to your friend.

In Activity 1.5, you have learnt that we can add or subtract like surds (surds with the same radicand) by counting forwards or backwards. However, unlike surds cannot be added or subtracted.

In general, if x and y are rational numbers and z is a non-perfect square, a x √z+ y √z = (x + y) √z b x √z − y √z = (x − y) √z c x √a ± y √b = x √a± y √b

Example 1.8

Simplify the following:

1. 3 √6+ 6 √6

2. 12 √3+ 6 √3

3. 3 √6− 6 √6

Solution

1. 3 √6+ 6 √6 = (3 + 6) √6 = 9 √6

2. 12 √3+ 6 √3 = (12 + 6) √3 = 18 √3

3. 3 √6− 6 √6 = (3 − 6) √6 = − 3 √6

Example 1.9

Simplify 3 √8 + 2 √18 − 6 √2

Solution

3 √8 + 2 √18 − 6 √2 = 3 √4 × 2 + 2 √9 × 2 − 6 √2 = 3 √4 × √2 + 2 √9 × √2 − 6 √2 = 3 × 2 × √2 + 2 × 3 × √2 − 6 √2 = 6 √2 + 6 √2 − 6 √2 = 6 √2

Example 1.10

Simplify 2 √24 + 3 √32 − 9 √2

Solution

2 √24 + 3 √32 − 9 √2 = 2 √4 × 6 + 3 √16 × 2 − 9 √2 = 2 √4 × √6 + 3 √16 × √2 − 9 √2 = 2 × 2 × √6 + 3 × 4 × √2 − 9 √2 = 4 √6 + 12 √2 − 9 √2 = 4 √6 + 3 √2

Activity 1.6: Multiplication of surds Whole class activity Materials needed: tables and chairs.

Step 1: Place three tables and three chairs in front of the class. Arrange the tables and chairs in such a way that if someone is facing the board, the chair will be on the left and the table will be on the right. Leave about a stride between the first table and chair, the second table and chair and the third table and chair.

Step 2: Two classmates to sit on the chairs and two others to squat under the first two tables.

Step 3: Ask the friends sitting on the chairs to walk together to sit on the third chair and let the friends squatting under the table walk together to squat under the third table.

This activity demonstrates that when multiplying two surds the numbers outside the square root sign multiply each other and the numbers under the square root sign multiply each other.

For example, a √b × c √d = ac √bd.

Note that two surds can be multiplied whether they are like or not.

Example 1.11

Express the following in the simplest form:

1. 2 √3 × 5 √2 = 2 × 5 √3 × 2 = 10 √6

2. 6 √2 × 2 √2 = 6 × 2 √2 × 2 = 12 √4 = 12 × 2 = 24 Division of Surds Division of surds is similar to multiplication of surds.

For example, a √b/c √d = a/c √_ b/d .

Example 1.12

Simplify the following, leaving your answer in surd form:

1. √18___ √6 = √__ 18/6 = √3

2. √12___ √3 = √__ 12/3 = √4 = 2

3. √__ 16___ √_ 9 = √__ 16/9 = √__ 16___ √_ 9 = 4/3 = 11/3 Rationalising Monomial Denominators Mathematicians do not like to see an irrational number as a denominator. For

example, mathematicians are uncomfortable with surds like 1___ √2 , 3___ √7 and 5+ √3/3 √5 .

Whenever this happens, they remove the irrational number from the denominator.

This is done in a process known as rationalisation. That is, the surd expression is manipulated until the irrational denominator becomes rational.

To rationalise a surd with an irrational monomial denominator, multiply both the numerator and denominator by the same irrational expression. When you simplify this expression, you will get an expression whose denominator is a rational number.

The number you must multiply numerator and denominator by is known as the conjugate of the irrational number you want to eliminate from the denominator.

The conjugate is the surd, which multiplies a given surd to make it a rational number. For example, 3 √2 × √2 = 3 × √2 × √2 = 3 × 2 = 6. Thus, the best conjugate of 3 √2 is √2 even though 3√2 is also a conjugate. This is because 3 √2 × 3 √2 = 9 × 2 = 18 .

Example 1.6

Rationalise 1___ √2 .

Solution

1_ √2 = 1___ √_ 2 × √_ 2___ √_ 2 = √_ 2/2

Example 1.7

Rationalise 3___ √7 .

Solution

3_ √7 = 3___ √_ 7 × √_ 7___ √_ 7 = 3√_ 7/7

If 2 is multiplied by itself six times, you can write 2 × 2 × 2 × 2 × 2 × 2. If 2 is multiplied by itself 100 times, imagine how tedious it would be to write it out.

Consider the time it will take and the space it will occupy. To simplify this, when 2 is multiplied by itself 100 times, mathematicians express this as 2¹⁰⁰, read as ‘two to the power 100’. This abbreviation makes writing it far less tedious and time consuming. In the expression 2¹⁰⁰, 2 is known as the base and 100 is the power or exponent.

Powers of integers The number to which we raise a base can be called an index (plural indices), a power or an exponent.

Positive exponents: A positive exponent, n, means that the base is multiplied by itself n times. For example, 2⁵= 2 × 2 × 2 × 2 × 2 = 32 and − 3⁴= − 3 × − 3 × − 3 × − 3 = 81 .

Zero exponents: Any number raised to the exponent of zero is always 1.

For example, 6⁰= 1 , − 4⁰= 1 and (3/4) 0 = 1.

Negative exponents: A negative exponent, n, means that the reciprocal of the base is raised to the absolute value of the exponent. For example, 3⁻³= 1/3³= 1/27, − 6⁻²= 1___ − 6²= 1______

–6 × − 6 = 1/36.

Scientific calculators can be used to verify answers, but if the base is negative, be sure to put it in brackets before raising to the exponent.

Order of Operations with Exponents

Even though BODMAS is usually interpreted as bracket, of, division, multiplication, addition and subtraction, of can also be thought of as order. Order represents roots and exponents. Thus, BODMAS implies that, moving from left to right, you have to expand the brackets first before coming to the order (roots and exponents), followed by division/ multiplication, and then addition/subtraction. That is, if mixture of these operations is found in the same expression.

For example, 3 − √16 ÷ 2 = 3 − 4 ÷ 2 = 3 − 2 = 1.

Or, 1 − 2 + 3 × 4 ÷ 5²= 1 − 2 + 3 × 4 ÷ 25 = 1 − 2 + 3 × 4/25 = 1 − 2 + 12/25 = − 1+ 12/25 = −13/25 Rules of Exponents There are important exponent rules to remember when simplifying expressions involving indices or exponents.

They are the product rule, the quotient rule and the power of a power rule.

The Product Rule

6³× 6⁵= (6 × 6 × 6) × (6 × 6 × 6 × 6 × 6) = 6⁸= 6³⁺⁵.

In general terms, mᵃ× mᵇ= mᵃ+b.

This is the product rule which tells us that when multiplying terms with the same base you add the exponents.

For example, 3²× 3⁶= 3⁶⁺²= 3⁸and 7⁻⁴× 7⁶= 7⁻⁴⁺⁶= 7²The Quotient Rule 5⁷÷ 5⁴= 5 × 5 × 5 × 5 × 5 × 5 × 5/5 × 5 × 5 × 5 = 5³= 5⁷⁻⁴In general terms, mᵃ÷ mᵇ= mᵃ−b.

This is the quotient rule which tells us that when dividing terms with the same base you subtract the exponents.

For example, 5³÷ 5⁶= 5³⁻⁶= 5⁻³= 1/5³= 1/125 and 9⁻⁴÷ 9⁶= 9⁻⁴⁺⁶= 9²= 81.

The Power of a Power Rule

(2³)⁴= 2³× 2³× 2³× 2³= 2³+ 3 + 3 + 3 = 2³× 4 = 2¹².

In general terms, (aᵐ)ⁿ= aᵐⁿ.

For example, (4³)²= 4³× 2 = 4⁶and (3⁻¹)⁴= 3⁻⁴= 1/3⁻⁴= 1/3⁴= 1/81.

Power of a Product

In general terms, (ab)ᵐ= aᵐ× bᵐFor example: (2 × 3)²= 2²× 3²= 4 × 9 = 36 Exponential Equations An exponential equation is an equation in which the variable appears in the exponent.

Form:

aˣ= b or aᶠ(x) = g(x) These equations require logarithms, common bases, or trial and error to solve.

Worked Examples

Example 1

Solve:

2ˣ= 8

Solution

Write 8 as a power of 2:

8 = 2³⇒ 2ˣ= 2³⇒ x = 3

Example 2

Solve:

5ˣ+ 1 = 125

Solution

125 = 5³⇒ 5ˣ⁺¹= 5³⇒ x + 1 = 3 ⇒ x = 2

Example 3

Solve:

3ˣ= 1_ 27

Solution

1_ 27 = 3⁻³⇒ 3ˣ= 3⁻³⇒ x = − 3 The following examples will help you to apply the rules of indices.

Example 1.13

Evaluate the following:

1. 81³__ 4

2. 8²__ 3

3. 64⁻¹___ 2 × 2 16²__ 3

4. ( 27/216) 1/3

Solution

1. 81³__ 4 = (3⁴)³__ 4 = 3⁴× 3/4 = 3³= 27.

2. 8²__ 3 = (2³)²__ 3 = 2³× 2/3 = 2²= 4

3. 64⁻¹___ 2 × 2 16²__ 3 = (2⁶)⁻¹___ 2 × (6³)²__ 3 = 2⁶×⁻¹__ ₂ × 6³× 2/3 = 2⁻³× 6²= 1/2³× 36 = 1/8 × 36 = 36/8 = 4.5

4. ( 27/216) 1/3 = (3³__ 6³) 1/3 = [(3/6) 3 ] 1/3 = [(1/2) 3 ] 1/3 = (1/2) 3 × 1/3 = (1/2) 1 = 1/2 Real-Life Applications of Indices (Exponents) Exponents are used in population growth, compound interest, computing, medicine, and radioactive decay.

Example 1: Compound Interest (Banking)

You invest GH¢1,000 at an annual interest rate of 10% compounded yearly. What will the value be after 3 years?

Formula:

A = P (1 + r)ᵗWhere:

• P = 1000

• r = 0.10

• t = 3

Solution

A = 1000 (1 + 0.10)³= 1000 (1.1)³= 1000(1.331) = GH¢1331

Example 2: Bacteria Growth (Biology)

A bacteria population doubles every hour. If you start with 200 bacteria, how many will there be after 4 hours?

Formula:

Population = 200 × 2⁴= 200 × 16 = 3200

Example 3: Computer Memory (ICT)

Each time you upgrade your memory card, its capacity doubles. If your original card is 8 GB, what will be the capacity after 3 upgrades?

Capacity = 8 × 2³= 8 × 8 = 64 GB

Example 4: Radioactive Decay (Physics)

A radioactive substance decays such that half remains every 5 hours. If you start with 80 grams, how much remains after 15 hours?

Solution

15 hours = 3 half-lives Remaining = 80 × (1/2) 3 = 80 × 1/8 = 10 grams

Example 5: Sound Intensity (Science/Decibels)

Every 10 dB increase in sound intensity is 10 times more powerful. If a sound is 30 dB more intense than a baseline, how many times more powerful is it?

Solution

10³⁰__ 10 = 10³= 1000 times

Just as subtraction can be used to reverse addition and division can be used to reverse multiplication, there is a concept that can be used to reverse indices. This concept is called logarithm. Thus, indices and logarithms are closely related. The concept of logarithms is an important tool in many fields. For example, banks use logarithms to calculate the compound interest of their customers; biologists use it to determine the rate at which the population of organisms like bacteria grow in a given medium; chemists use it to find the pH of substances and geologists use it to measure the magnitude of an earthquake.

Here you will learn the rules which govern the concept of logarithms.

Laws of Logarithms

When you see logₐy , a is the base and y is the augment. This is read as logarithm of y to the base a. In short, you can say, logy base a. Logarithms to base 10 are known as common logarithms. For common logarithms, sometimes the base is omitted. Thus, logy = log₁₀y. Common logarithms were very important in computations before the calculator was invented.

1. Since logarithms reverse indices, aˣ= y ⟺ logₐy = x.

2. logₐ(x × y) = logₐx+ logₐy. You can use this law, known as the addition

-product law, if, and only if, the augment is a product.

3. logₐ(x ÷ y) = logₐx − logₐy . You can apply this law, known as the subtraction

– quotient law, if, and only if, the augment is a quotient.

4. logₐyˣ= x logₐy. You can apply this law, known as the exponent law, if, and only if, the augment has an exponent or power.

5. logₐa = 1. If the augment is the same as the base the answer is always 1

i. as a¹= a .

6. logₐ1 = 0. No matter the base, the logarithm of one is always zero as a⁰= 1

7. logₐy = logₓy_____ logₓa . This law is used to change the base of a logarithm.

Applying the Laws of Logarithms

Apply the laws of logarithms to solve these problems.

Example 1.14

Given that log₂3 = 1.58 and log₂5 = 2.32, find the value of:

1. log₂15

2. log₂45

3. log₂(5/3).

Solution

1. log₂15 = log₂(3 × 5) = log₂3+ log₂5 = 1.58 + 2.32 = 3.90

2. log₂45 = log₂(9 × 5) = log₂9 + log₂5 = log₂3²+ log₂5 = 2 log₂3 + log₂5 = 2(1.58) + 2.32 = 3.16 + 2.32 = 5.48

3. log₂(5/3) = log₂5 − log₂3 = 2.32 − 1.58 = 0.74

Example 1.15

Solve for x : log₂(x − 3) = 4

Solution

log₂(x − 3) = 4 ⟹ × − 3 = 2⁴⟹ × − 3 = 16 ⟹ × = 16 + 3 = 19

Example 1.16

Simplify: log₂8 + log₂4

Solution

log₂8 + log₂4 = log₂2³+ log₂2²= 3 log₂2 + 2 log₂2 = 3(1) + 2(1) = 3 + 2 = 5 Alternatively, log₂8 + log₂4 = log₂(8 × 4) = log₂32 = log₂2⁵= 5 log₂2 = 5 × 1 = 5.

Example 1.17

Find the value of x given that log₃x + 4 logₓ3 = 5.

Solution

log₃x+ 4 logₓ3 = 5 ⟹ log₃x+ logₓ3⁴= 5 ⟹ log₃x + logₓ81 = 5 ⟹ log₃x + log₃81_____ log₃x = 5 ⟹ log₃x + log₃3⁴_____ log₃x = 5 ⟹ log₃x + 4 log₃3_____ log₃x = 5 ⟹ log₃x + 4____ log₃x = 5 At this point, let log₃x = y.

⟹ y+ 4/y = 5 ⟹ y(y)+ 4/y × y = 5y ⟹ y²+ 4 = 5y ⟹ y²− 5y+ 4 = 0 ⟹ y²− y − 4y + 4 = 0 ⟹ y(y − 1) − 4(y − 1) = 0 ⟹ (y − 1)(y − 4) = 0 ⟹ y − 1 = 0 or y − 4 = 0 ∴ y = 1 or y = 4 ⟹ log₃x = 1 or log₃x = 4 ⟹ × = 3¹or x = 3⁴⟹ × = 3 or x = 81 .

Time for a quick recap of some of the key concepts of indices and logarithms.

Indices or Exponents or Powers:

1. These are mathematical operations that involve raising a number to a specific power known as the index/exponent. This power or exponent or index represents the number of times a base is multiplied by itself.

2. For example, the expression aᵇ, where ‘a’ is the base and ‘b’ is the exponent, means we multiply a by itself b number of times.

Logarithms:

1. Logarithms are the inverse operation of indices.

2. The logarithm of a number with respect to a given base tells us what exponent we need to raise the base to in order to obtain that number. For example, if bˣ= a , then the logarithm is expressed as log_(b)a = x and is read as “the logarithm of a to the base b is x” meaning the logarithm of a to base b is the exponent (x) to which we raise the base (b) to get the number (a).

Let us consider some key relationships between indices and logarithms.

1. Logarithms are the inverse of indices.

2. The laws of logarithms are derived from the laws of indices.

3. The logarithmic function “cancels out” or reverses the effect of exponents in expressions or equations. It allows us to find the unknown exponent when we know the base and the result.

4. Logarithms are useful in solving exponential equations, analysing growth rates and working with data that follow exponential patterns.

5. Indices and logarithms are two sides of the same mathematical coin. They complement each other and play essential roles in various fields including science, engineering and finance.

Let us explore areas in which indices and logarithms are applied.

In Banking and Finance

Indices can be used:

1. to calculate interest rates on investments or loans.

For example, the formula for Compound Interest, A = P (1 + r/n)ⁿᵗ2. in inflation rates to determine the change in prices of items.

3. in analysing stock market prices.

Logarithms can be used:

1. to determine the time it takes an investment to reach a specific value/amount.

2. in assessing financial risk and uncertainty.

In computer science and IT

1. Logarithmic procedures can be used to compress data.

2. Indices are used to calculate compression ratios.

3. Logarithms are used in calculations to secure online transactions. E.g., SSL

4. Google’s search procedures use logarithmic scaling to rank web pages.

Science and Engineering

1. In Physics, indices are used to calculate forces, energies, and velocities

a. (e.g., kinetic energy = 1 __ 2 m v²)

2. Sound intensity is measured in decibels(dB) using logarithmic scales.

3. In Chemistry, pH levels of acids and bases are measured using logarithms.

4. Logarithms are used in signal processing in the compression/decompression in audio and image processing.

5. In epidemic modelling, indices are used to calculate the spread and growth rates of diseases (e.g., exponential growth)

6. The decay of radioactive substances follows an exponential pattern.

7. Half-life (the time for half of a substance to decay) is determined using logarithms.

In our everyday life

1. Logarithmic scales are used to measure sound frequencies in music.

2. In photography, logarithmic exposure compensation adjusts image brightness.

This shows how useful and applicable indices and logarithms are. Explore the internet to look for other areas where they can be applied.

Applications of Common Logarithms

Before calculators were invented, common logarithms were used to make computations because it is easy to work in base 10.

Let us consider some questions and see how to solve them using the concept of common logarithms.

Example 1.18: Compound interest Mrs. Agbenyegah invested GH¢15 000.00 at 12% interest monthly, how much will she have after 3years? Use the formula: Amount( A) = P (1 + r/n)ⁿᵗSolution Amount( A) = P (1 + r __ n ) ⁿᵗwhere P = GH¢15 000, r = 0.12, n = 12, t = 3 A = 15 000 1 + 0.12/12 ¹²× 3 A = 15 000 (1.01)³⁶A = 15,000(1.430768784) A ≈ GH¢ 21461.53 Therefore, Mrs. Agbenyegah will have a total amount of GH¢ 21461.53 after 3years.

Example 1.19: Sound (decibels)

A sound wave has an intensity of 0.01 watts/m². What is its decibel level?

Use the formula: Decibel (dB) level = 10 × log₁₀( I__ I0) Where I is the intensity of sound and I0 is the reference intensity, typically 10⁻¹²W/m

Solution

Decibel Level (L) = 10 × log₁₀( I__ I0) L = 10log( 0.01/10⁻¹²) = 10(log₁₀0.01 − log₁₀10⁻¹²) = 10(log₁₀10⁻²−log₁₀10⁻¹²) = 10( − 2 log₁₀10 − ( − 12)log₁₀10) = 10( − 2 + 12) = 10(10) = 100 decibels(dB) Therefore, the decibel level of the sound wave is 100dB.

Example 1.20: Sound

A data compression algorithm reduces a file size from 100MB to 10MB.

What is the compression ratio?

Use the formula: Common Ratio = log( Original Size____________ Compressed Size)

Solution

Compression ratio = log(100/10 ) = log(100) − log(10) = log₁₀10²− log₁₀10 = 2 log₁₀10 − log₁₀10 = log₁₀10(2 − 1) ≈ 1 Alternately, Compression ratio = log(100/10 ) = log(100) − log(10) = log₁₀10²− log₁₀10 = 2 log₁₀10 − log₁₀10 since log₁₀10 = 1 = 2(1) − 1 ≈ 1 Or, another alternative, Compression ratio = log(100/10 ) = log(10) = log₁₀10 = 1 Therefore, the compression ratio is approximately 1

Example 1.21: Biology and Medicine

In a particular country, a population of bacteria grows from 100 to 1000 cells in 3 hours. What is the growth rate of the bacteria?

Use the formula: r = log(P(t)___ Po )_______ t , where:

r is the growth rate (as a decimal) P(t) is the population at time t Po is the initial population t is time (in hours)

Solution

Growth rate (r) = log(P(t)___ Po )_______ t r = log(1000/100 )________ 3 r = log10/3 = 1/3 ≈ 0.3333 Therefore, the growth rate (r) is approximately 0.3333 per hour.

In primary school, you were taught sorting and the divisibility rules of integers.

Hopefully you can recall that the divisibility rule of integers is a method used to determine whether one integer is divisible by another without performing the full division.

Some of these rules are;

1. Divisibility by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).

2. Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. For example (0, 3, 21, 63, 84,102, 258,1095, 8631, etc.)

3. Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5.

4. Divisibility by 10: A number is divisible by 10 if its last digit is 0.

5. Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4. E.g. (112, 224, 1092, 9908, etc.)

6. Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. For example (9, 27, 72, 81, 117, 333, 225, 3141, 4203, 7110, 8001, 9032031, etc.)

Divisibility rules works perfectly because one integer divides another without a remainder and they are a quick way to determine whether an integer is divisible by another. Do you know any other divisibility rules? Discuss them with a classmate.

But what happens when one integer is not divisible by another? This results in a remainder.

For example, dividing 8 by 3. By the divisibility rule, a number can only be divisible by 3 if the sum of its digits is 3. Therefore, we know that this is not divisible and there will be remainder.

8/3 = 22/3. This mean that 8 divides 3, 2 times with a remainder of 2.

This concept is closely related to modular arithmetic, which deals with the remainder when one integer is divided by another.

Introduction and Basic Operations with Modulo

Arithmetic With the help of clocks/watches (analogue and digital), perform this activity.

Activity 1.7: Clock arithmetic For this activity you need both an analogue clock/watch and a digital clock/ watch.

Figure 1.4: A clock

Step 1: Identify the various sections or digits on the surface of the clock/ watch.

Step 2: Set the time to 12:00 or 00:00 on the analogue clock and 00:00 on the digital clock/watch.

Step 3: Turn the clock forward by 13 hours.

Step 4: Identify the numbers obtained on both the analogue and digital clock/ watch and write them down.

Step 5: Repeat steps 2 to 4 using other number of hours.

Step 6: Discuss your findings/ observations with a classmate.

You will have observed that +13hrs is 1:00 on the analogue clock but 13:00 on the digital clock.

If you went forward by 21hrs from the starting point you have 9:00 on the analogue clock but 21:00 on the digital clock. This is because, the numbers wrap round a fixed number, 12:00 (the starting point), on the analogue clock.

Figure 1.5: A clock In modular arithmetic, numbers “wrap around” upon reaching a given fixed quantity (this given quantity is known as the modulus) to leave a remainder. Other examples are the days of the week, calendar months, etc.

Modular arithmetic, often referred to as remainder arithmetic, is a mathematical system where an integer is expressed by its remainder after being divided by another integer.

The modulo of any integer is determined by dividing that integer by a specified modulus and identifying the remainder as the result. The term “mod” is used as a notation for modulo. That is, “a mod b” represents the remainder when a is divided by b.

For example, 31(mod 7) signifies the remainder obtained when 31 is divided by 7.

31 (mod 7) = 31 − 7 = 24 − 7 = 17 − 7 = 10 − 7 = 3 9 (mod 4) signifies the remainder obtained when 9 divided by 4. Thus, 9 (mod 4) = 9 − 4 = 5 − 4 = 1. Therefore, 9(mod 4) = 1 and can also be written as:

9 = 1(mod 4) The remainder can be obtained either by:

1. repeatedly subtracting the modulo number from the given number until subtracting any further will result in negative numbers.

For example, 7(mod 4) = 7 − 4 = 3 . If we try and subtract 4 again, we will have a negative number ∴ 7(mod 4) = 3 or 7 = 3(mod 4)

2. dividing the given number by the modulo number and finding the remainder.

For example, if d/c = Ab/c, where A is the integer part, c is the divisor and b is the remainder, so the value of b is the answer.

For example, 7(mod 4) = 7/4 = 13/4. The 3 is the remainder.

⟹ 7(mod 4) = 3 Examples 1.22

1. 7 (mod 4) = 3 (remainder after dividing 7 by 4)

2. 2 (mod 10) =2 (remainder after dividing 2 by 10)

3. 9 (mod 3) = 0 (remainder after dividing 9 by 3)

4. 81 (mod 5) =1(remainder after dividing 81 by 5) Hints in finding the modulo:

1. If the dividend is less than the divisor (modulo number), then the dividend remains the answer. For example:

a. 3 (mod 7) = 3

b. 1 (mod 5) = 1

2. Modulo 5 of all numbers ending with any of these numbers; 0, 1, 2, 3, 4 is the last or ending digit. For example:

a. 1 (mod 5) = 1

b. 12 (mod 5) = 2

c. 170 (mod 5) = 0

d. 1094 (mod 5) = 4 etc

3. Modulo 5 of all numbers ending with any of these numbers; 5, 6, 7, 8, 9 is found by subtracting 5 from the last digit. For example:

a. 19(mod 5) = 9 – 5 = 4. Therefore, 19(mod 5) = 4

b. 36 (mod 5) = 6 − 5 = 1 Integers for a given Modulo Let’s take for example, the division of any integer by 3:

1_ 3 has a aremainder of 1 2_ 3 has a aremainder of 2 3_ 3 has a aremainder of 0 4_ 3 has a aremainder of 1 It is seen that the possible remainders when any integer divided by 3 are 0, 1 and

2. The divisor (3), is called the modulus and the arithmetic is said to be in modulo

3. Modulo 3 can therefore take one of the values {0, 1, 2}. Likewise, modulo 5 can take one of the values {0, 1, 2, 3, 4}, Modulo 6 can take one of the values {0, 1, 2, 3, 4, 5} Modulo 9 can take one of the values {0, 1, 2, …, 8} In general, modulo n can take one of the values {0, … , n − 1} You can try this with several other integers and discuss your observations with your classmates and teacher.

Modulo of Negative Numbers

The modulo of any negative number is determined by adding the modulo to the number successively until a positive number, which is the answer, is reached.

This implies that we repeatedly add the modulus to the negative number until it gives an answer which is positive.

For example, − 5 (mod 4) = − 5 + 4 = − 1 + 4 = 3 Since 3 is the first positive number, we end the process that is the answer.

Therefore, − 5 (mod 4) = 3

Example 1.23

Simplify − 8 (mod 3)

Solution

− 8(mod 3) = − 8 + 3 = − 5 = − 5 + 3 = − 2 = − 2 + 3 = 1 ∴ − 8(mod 3) = 1

Example 1.24

What is the value of − 33 in modulo 11?

Solution

− 33 (mod 11) = − 33 + 11 = − 22 = − 22 + 11 = − 11 = − 11 + 11 = 0 ∴ − 33 (mod 11) = 0 Equivalent or Congruent Modulo In modular arithmetic, two integers a and b are said to be equal modulo n, written as a ≡ b(mod n) if they leave the same remainder under the same mod n.

In other words, a and b are equivalent if their difference is a multiple of n.

For example:

1. 30 (mod 7) = 2 and 23(mod 7) = 2. Therefore, 30 and 23 are said to be equivalent and written as: 30 (mod 7) ≡ 23(mod 7) = 2 or 30 ≡ 23(mod 7)

2. If 16 (mod 5) = 1 and 91(mod 5) = 1 Then 16 ≡ 91(mod 5) Basic Operations with Modulo The basic operations are addition (+), subtraction (−), multiplication (×) and division (÷). The act of performing these operations is known as simplification of modular arithmetic.

This simplification of modular arithmetic is done by;

1. performing the operation first. (+ , − , × , ÷)

2. converting the number obtained to the given modulo.

Addition The sum of two or more numbers in a given modulo is found by adding the numbers first before converting to the given modulo.

Example 1.25

1. 4 + 5 ( mod 5) = 9 (mod 5) = 4

2. 8 + 12 (mod 12) = 20 (mod12) = 8

3. − 3 + 31 + 4 (mod 9) = 32 (mod 9) = 5

4. (5 mod 3)+ (2 mod 3) = (5 + 2)( mod 3) = 7 (mod 3) = 1 Subtraction The difference of two or more numbers in a given modulo is found by first finding the difference between the numbers before converting to the given modulo.

Example 1.26

1. 99 (mod 5) − 34 (mod 5) = (99 − 34)(mod5) = 65 (mod 5) = 0

2. (7 mod 4)− (2 mod 4) = (7 − 2)( mod 4) = 5 (mod 4) = 1

3. 12 − 14 (mod 6) = − 2 (mod 6) = 4 Multiplication The product of two or more numbers in a given modulo is obtained by multiplying the numbers first before converting to the given modulo.

Example 1.27

1. (3 mod 5) × (2 mod 5) = (3 × 2)(mod 5) = 6(mod 5) = 1

2. 12 × 12 (mod 4) = 144 (mod 4) = 0

3. 6 × 4 (mod 7) = 24 (mod 7) = 3 Let us investigate the following properties of operation involving modulo arithmetic.

Properties of Modular Arithmetic

Commutative Property

Commutativity in operations involving modulo (e.g., addition, multiplication under mod) Addition and multiplication modulo a number are commutative:

• (a + b )mod n = (b + a )mod n

• (a × b )mod n = (b × a)mod n So commutativity applies to modular addition and multiplication.

Note!

But the statement:

“a mod b ≠ b mod a” is not about commutativity. That’s just comparing two different operations.

Let’s look at an example:

• 7mod 4 = 3

• 4mod 7 = 4 Clearly, 7mod 4 ≠ 4mod 7 So, modulo itself (i.e., a mod b) is not commutative.

Associative Property

This property is applicable to both addition and multiplication in modulo arithmetic.

That is:

(a + b) + c(mod n) = a + (b + c)(mod n) (a × b) × c(mod n) = a × (b × c)(mod n) Let us verify:

For example: (2 + 3) + 4(mod 5) = 2 + (3 + 4)(mod 5) Solving the Left-Hand Side (LHS) and Right- Hand Side (RHS) concurrently, 5 + 4(mod 5) = 2 + 7(mod 5) 9 (mod 5) = 9 (mod 5) 4 = 4 Since (LHS) = (RHS) = 4, the additive property holds.

Similarly (2 × 3) × 4 (mod 5) = 2 × (3 × 4)(mod 5) 6 × 4 (mod 5) = 2 × 12 (mod 5) 24 (mod 5) = 24 (mod 5) 4 = 4 Since (LHS) = (RHS) = 4, the multiplicative property also holds.

Therefore, associative property holds for modular arithmetic.

Identity This refers to the existence of an additive and multiplicative identity in modular arithmetic such that, when combined with any number under a specific operation, leaves that number unchanged.

1. The additive identity in modular arithmetic is 0 That is, a + 0 (mod n) = 0 + a(mod n) = a (mod n) For example: 3 + 0(mod 2) = 0 + 3(mod 2) = 3(mod 2) = 1

2. The multiplicative identity in modular arithmetic is 1 That is, a . 1(mod n) = 1 . a(mod n) = a (mod n) For example: 5 × 1(mod 3) = 1 × 5(mod 3) = 5(mod 3) = 2 Inverse This refers to the existence of additive and multiplicative inverses in modular arithmetic such that, when combined with any number under a specific operation, yields the identity element.

1. The additive inverse of modular arithmetic is the value that, when added to a number, gives a result of 0 (mod n). That is, a + b = 0 (mod n) , This means that b is the value that, when added to a, results in the identity element 0.

Example 1.28

a) 3 + 1 (mod 4) = 0 (mod 4) = 0, 1 is the additive inverse of 3 in mod 4

b) 5 + 2 (mod 7) = 7(mod 7) = 0, 2 is the additive inverse of 5 in mod 7

2. The multiplicative inverse is the value that, when multiplied by a number, gives a result of 1 (mod n). That is a . b = 1 (mod n) This means that b is the value that, when multiplied by a, gives a result that is congruent to 1 modulo n

Example 1.29

a) 3 × 5 = 1 (mod 7), therefore, 5 is the multiplicative inverse of 3 in mod 7

b) 4 × 3 = 1 (mod 11), so 3 is the multiplicative inverse of 4 in mod 11 Explore the internet for more examples, solve them and discuss them with your classmates. How can we find the modulo number if a number and its remainder are given or known but the modulo number is unknown? For example, 24 (mod x ) = 3 Let us go through the activity below to determine the missing value.

Activity 1.8: Finding the modulo number in a given equation Determine the value of × in 24 (mod x) = 3

Step 1: Find the difference between the numbers Thus, 24 − 3 = 21

Step 2: find all the factors of the value obtained Factors of 21 = {1, 3, 7, 21}

Step 3: Substitute each of the factors in the given equation.

when x = 1, 24 (mod 1) = 0, x ≠ 1 when x= 3, 24 (mod 3) = 0, x≠ 3 when x = 7, 24 (mod 7) = 3, x = 7 when x = 21, 24 (mod 21) = 3, x = 21

Step 4: identify the ones that make the statement true (ie, satisfy the statement), as the value of the variable.

x = 7 or 21 because 7 and 21 both satisfy the equation.

You can search for other methods on the internet.

Now, let’s try another example.

Example 1.30

Find p , if 52 = 3 (mod p) where 0 ≤ p ≤ 10

Solution

52 = 3(mod p) 52 − 3 = 49 Factors of 49 = {1, 7, 49} when p = 1, 52 (mod 1) = 0, p ≠ 1 when p = 7, 52 (mod 7) = 3, p = 7 when p = 49, 52(mod 49) = 3, p = 49 7 and 49 both satisfy the equation. However, the question states that 0 ≤ p ≤ 10, Therefore, p = 7

Example 1.31

You have 36 cookies and want to share them equally among your friends. If you have a remainder of 4 cookies after sharing, how many friends do you have?

Solution

Let n represent number of friends ⟹ 36 = 4 (mod n) 36 − 4 = 32 Factors of 32 = {1, 2, 4, 8, 16, 32} When n = 1, 36 (mod 1) = 0, n ≠ 1 When n = 2, 36 (mod 2) = 0, n ≠ 2 When n = 4, 36 (mod 4) = 0, n ≠ 4 When n = 8, 36 (mod 8) = 4, n = 8 When n = 16, 36 (mod 16) = 4, n = 16 When n = 32, 36 (mod 32) = 4, n = 32 n = 8 or 16 or 32 Therefore, the number of friends is either 8 or 16 or 32 Finding the Number (unknown value) Under a Given Modulo How do we determine the number which is being operated on by a given modulo?

For example, 7x = 1 (mod 2). Work through this activity to help you find the value of x.

Activity 1.9: Finding the unknown value in a given equation Find the possible value(s) of x such that 7x = 1 (mod 2)

Step 1: List all the possible remainders of the given modulo. They are {0,1}

Step 2: Substitute each of the remainder values into the given equation.

when x = 0, 7(0) (mod 2) = 0 (mod 2) = 0, x ≠ 0 when x = 1, 7(1) (mod 2) = 7(mod 2) = 1, x = 1

Step 3: Identify the value(s) of the variable (x) that makes the statement true.

Thus x = 1 makes the statement true.

Example 1.32

Find the value(s) of x in the given equation 5x = 2(mod 4).

Solution

5x = 2 (mod 4) or 5x (mod 4) = 2 In mod(4), x can take on values from 0 to n − 1 , that is {0,1,2,3} Substituting:

when x = 0, 5(0)mod 4 = 0 (mod 4) = 0 , x ≠ 0 when x = 1, 5(1)mod 4 = 5 (mod 4) = 1 , x ≠ 1 when x = 2, 5(2)mod 4 = 10 (mod 4) = 2 , x = 2 when x = 3, 5(3)mod 4 = 15 (mod 4) = 3 , x ≠ 3 ∴ x = 2

Example 1.33

Given that 2y + 1 = 3 (mod 7), what is the value of y?

Solution

2y + 1 = 3 (mod 7) In mod(7), y can take on values of {0,1, 2, 3, 4, 5, 6 } By substitution, let’s investigate:

When y = 0, 2(0) + 1 (mod 7) = 1 (mod 7) = 1, y ≠ 0 When y = 1, 2(1) + 1 (mod 7) = 3 (mod 7) = 3, y = 1 When y = 2, 2(2) + 1 (mod 7) = 5 (mod 7) = 5, y ≠ 2 When y = 3, 2(3) + 1 (mod 7) = 7 (mod 7) = 0, y ≠ 3 When y = 4, 2(4) + 1 (mod 7) = 9 (mod 7) = 2, y ≠ 4 When y = 5, 2(5) + 1 (mod 7) = 11 (mod 7) = 4 , y ≠ 5 When y = 6, 2(6) + 1 (mod 7) = 13 (mod 7) = 6 , y ≠ 6 ∴ y = 1 Search for other methods that can be used to solve the above examples and look for more examples from the extended reading materials provided below or from the internet.

Modular arithmetic is applicable in several areas and even in our everyday lives.

Let us consider some of these areas;

Applications of Modular Arithmetic

1. In our daily lives, modular arithmetic can be used to determine;

i. Time: Modular arithmetic can be used to determine what time it is after a number of hours. The 12-hour clock system is essentially modular arithmetic using modulo 12 (for example, 10 hours after 3pm is 1 am).

ii. Market days in our communities: Market days in our communities rotate in a cyclic manner. For instance, in Ho, market days are every five days counting from the recent market day. Since there are only 7 days in a week, the market days wrap around it.

The figure below shows how cyclic the days of the week are.

Figure 1.6: Days of the week clock For example:

a) If the recent market day is Saturday, then the next market day will be Wednesday.

b) If today is Thursday (represented by the number 4), what day will it be in 11 days?

Solution

Thursday (4) + 11 days = 15 ⇒ 15(mod 7) = 1 , From the cycle, 1 represents Monday. Therefore, 11days after Thursday is Monday.

c) The calendar is also designed using the concept of modular arithmetic (i.e. days of the week, months and year)

2. Finding the remainder: for example, 27 mod 4 = 3 (i.e. the remainder obtained when 27 is divided by 4)

3. In Computer Science: It is widely applied in various computer science areas like pseudo random number generation, finite field arithmetic, and generating permutations.

4. Error Detection and Correction: It’s used in checksums (adding digits and using the remainder for error detection) and cyclic redundancy checks (CRC) for data transmission reliability.

There are other areas you can look out for. Do search for them are discuss them with your classmates.

1. Which of the following is not a perfect square?

A. 225 B. 625 C. 652 D. 841

2. Simplify the following surds, if x is natural number:

a. √24

b. √45

c. √2 x²d. √4x.

3. The dimensions of a room are such that its length is √108 m longer than the width. If the width of the room is 3m, find the length of the room as a surd in its simplest form.

4. Working alone or with a friend, rationalise the following:

a. 6___ √3

b. 2 − √3/2 √6

c. 1___ √3 − 1___ √_ 2

5. Simplify:

a. √24 + √54

b. √200 − 3 √32+ √72

c. √18 − 3 √50+ √27

6. Expand and simplify ( √5 + √3)(2 √5 − √3), leaving your answer as a surd in its simplest form.

7. Simplify and leave your answer as a surd in its simplest form.

a √48 − √6 b √2 − √3 × √5 ÷ √6.

8. Given that √5 = 2.236 , evaluate, correct to one decimal place, 2 √5 (4− 3/√_ 5).

9. The length of the hypotenuse, AC, of a right-angled triangle ABC is 12 √5 cm. The length of the side AB is 8 √3cm. Find the length of the side BC, leaving your answer in surd form.

10. Simplify the following:

a. 125⁻¹___ 3

b. 8¹__ 3 × 4¹__ 2/32¹__ 5 × 16¹__ 2

11. Find the value of x if 2ˣ⁺²× 8ˣ= 1.

12. Simplify: 3⁵× 3 2 × 3⁻⁹__________ 3⁻⁷× 3 2

13. Find the value of y :

a. log_(y)8 = 1/3

b. log₂₇3²ʸ= 2

14. Without the use of four-figure tables or calculators, evaluate:

a. log₆36+ log₆6 √6 − log₆216

b. log₃₆6_____ log₄₉7

c. log₃27− log₃3 √3____________ log₃ 1___ √3 − log₃√3

15. Find the remainder when 100 is divided by 7

16. What is the remainder when 123456789 is divided by 9?

17. Find the modular multiplicative inverse of 5 (mod 6).

18. Show that 17 and 29 are congruent under modulo 4.

19. If today is Tuesday (represented by the number 3), what day will it be in 15 days?

20. Akosua can only access her results online 14 hours after it has been uploaded on their school’s website. If the results were uploaded at exactly 13:00, at what time will she be able to access it?

21. You have 19 cookies and want to share them equally among your friends.

If you have a remainder of 4 cookies after sharing, how many friends do you have?