Indices and Logarithms

Section 1.3 Indices and Logarithms

Indices, also known as exponents, represent the number of times a base is multiplied by itself. For example, in the expression \(a^n\text{,}\) \(a\) is the base and \(n\) is the exponent or index. This means that \(a\) is multiplied by itself \(n\) times.

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Logarithms, on the other hand, are the inverse of exponentiation. The logarithm of a number is the exponent to which a given base must be raised to obtain that number. For example, if \(b^x = y\text{,}\) then \(x = \log_b(y)\text{,}\) where \(b\) is the base and \(y\) is the number.

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Subsection 1.3.1 Expressing numbers in index form

Activity 1.3.1.

\(\textbf{Work in Groups}\)

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\(\textbf{What you need}\)

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\(\bullet\) Worksheet

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\(\bullet\) Pencil and erasers

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\(\bullet\) Optional: bottle tops for modeling

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\(\textbf{Instructions}\)

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Look at each number in the first column of the table Figure 1.3.1.
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Write the number as a product of the same factor repeated several times.
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Re-write your repeated multiplication as a power.
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\(\textbf{Note:} \) If the number cannot be written using one base only, express it as a product of powers of two (or more) different numbers.}
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Fill in your answers in the table below.

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Number	Product of repeated factors	Power form / Products of the power
\(8\)	\(2 \times 2 \times 2\)	\(2^3\)
\(27\)	\(\)	\(\)
\(36\)	\(\)	\(\)
\(64\)	\(\)	\(\)
\(72\)	\(2 \times 2 \times 2 \times 3 \times 3\)	\(2^3 \times 3^2\)
\(81\)	\(\)	\(\)
\(96\)	\(\)	\(\)
\(108\)	\(\)	\(\)
\(144\)	\(\)	\(\)

Figure 1.3.1.

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Share your work with your fellow learners.
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\(\textbf{Key Takeaway}\)

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Index form means writing a number using a base and an exponent (also called an index or power).

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\(\textbf{Base}\) is the number that is multiplied by itself, and \(\textbf{exponent}\) tells how many times the base is used in the multiplication.

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A number in index form is written as \(a^n\text{,}\) where \(a\) is the base and \(n\) is the exponent.

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\(\textbf{To express a number in index form, you can follow these steps:}\)

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Identify the base of the number you want to express in index form.
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Determine the exponent, which is the number of times the base is multiplied by itself to obtain the original number.
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Write the number in the form \(a^n\text{.}\)
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For example, the number \(8\) can be expressed in index form as \(2^3\text{.}\)
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Example 1.3.2.

Express the number \(32\) in index form.

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Solution.

First, we break down \(32\) into its prime factors by finding the prime numbers that multiply together to give \(32\text{.}\)

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To find the prime factors of \(32\text{,}\) we start with the smallest prime number, which is \(2\text{.}\) We check if \(32\) is divisible by \(2\text{.}\) Since \(32\div 2 = 16\text{,}\) it divides exactly, so \(2\) is our first prime factor.

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Next, we take \(16\) and divide it by \(2\) again: \(16 \div 2 = 8\text{.}\) It divides exactly, so \(2\) is our second prime factor.

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Now, we take \(8\) and divide it by \(2\) once more: \(8 \div 2 = 4\text{.}\) It divides exactly, so \(2\) is our third prime factor.

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Next, we take \(4\) and divide it by \(2\) again: \(4 \div 2 = 2\text{.}\) It divides exactly, so \(2\) is our fourth prime factor.

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Finally, we have \(2\text{,}\) which is already a prime number and cannot be divided further.

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\({\color{blue}2}\)	\(32\)
\({\color{blue}2}\)	16
\({\color{blue}2}\)	8
\({\color{blue}2}\)	4
\({\color{blue}2}\)	2
\({\color{blue}2}\)	1

Therefore, the prime factors of \(32\) are: \(2 \times 2 \times 2 \times 2 \times 2\)

\begin{equation*} =2^5\text{.} \end{equation*}

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Example 1.3.3.

Express each of the following numbers in index form:

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\(\displaystyle 2\times 2\times 2\times 2\times 2\)
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\(\displaystyle 64\)
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Solution.

In \(2\times 2\times 2\times 2\times 2\text{,}\) the factor \(2\) is multiplied by itself \(5\) times.
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This implies that the base is \(2\) and the exponent is \(5\text{.}\)
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Therefore, in index form it is written as:

\begin{equation*} 2^5\text{.} \end{equation*}

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\(64\) can be expressed as a product of repeated factors as:
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\begin{align*} 64=\amp 2 \times 2 \times 2 \times 2 \times 2 \times 2\\ =\amp 2^6 \end{align*}

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Therefore, in index form it is written as: \(2^6\text{.}\)
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Example 1.3.4.

Given bases \(2\text{,}\) \(3\) and \(5\text{,}\) express the following numbers in index form using the bases provided:

a) \(512\)

b) \(243\)

c) \(125\)

Solution.

a) To express \(512\) in index form using base \(2\text{,}\) we find the prime factorization:

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\begin{align*} 512 = \amp 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \\ =\amp 2^9 \end{align*}

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b) To express \(243\) in index form using base \(3\text{,}\) we find the prime factorization:

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\begin{align*} 243 =\amp 3 \times 3 \times 3 \times 3 \times 3\\ =\amp 3^5 \end{align*}

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c) To express \(125\) in index form using base \(5\text{,}\) we find the prime factorization:

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\begin{align*} 125=\amp 5 \times 5 \times 5\\ =\amp 5^3 \end{align*}

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Work in Groups

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Instructions

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Use numbers between \(1\) and \(9\) to form at least three pairs of numbers that have the same base but different powers.
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Multiply each pair that you have formed.
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For example: \(2^2 \times 2^3\)
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Simplify each expression using the multiplication law of indices.
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For example: \(2^2 \times 2^3 = 2^{2+3} = 2^5\)
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Discuss your work with your fellow learners.
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Key Takeaway

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The multiplication law of indices states that when you multiply two numbers with the same base, you add their exponents. Mathematically this can be expressed as: \(a^m \times a^n = a^{m+n}\text{,}\) where \(a\) is the base and \(m\text{,}\) \(n\) are the exponents.

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Example 1.3.5.

Simplify the following expressions:

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a) \(2^3 \times 2^4\)

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b) \(8^3 \times 8^6\)

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Solution.

a) Since the bases in \(2^3 \times 2^4\) are the same, we apply the multiplication law of indices:

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\begin{align*} a^m \times a^n= \amp a^{m+n}\\ 2^3 \times 2^4= \amp 2^{3+4}\\ =\amp 2^7 \end{align*}

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b) For \(8^3 \times 8^6\text{,}\) we have:

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\begin{align*} 8^3 \times 8^6= \amp 8^{3+6}\\ =\amp 8^9 \end{align*}

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Example 1.3.6.

Evaluate: \(4x^3 \times 7x^2\)

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Solution.

First, we multiply the coefficients (the numbers in front):

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\(4 \times 7 = 28\)

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Next, we apply the multiplication law of indices to the variable \(x\text{:}\)

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\(x^3 \times x^2 = x^{3+2} = x^5\)

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Therefore,

\begin{equation*} \end{equation*}

\begin{align*} 4x^3 \times 7x^2 = \amp (4 \times 7)x^{3+2}\\ = \amp 28x^5 \end{align*}

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Example 1.3.7.

Simplify the expression \(10^3 \times 10^{-2}\text{.}\)

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Solution.

To simplify this expression, we use the rule: \(a^m \times a^n = a^{m+n}\text{.}\)

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\begin{align*} 10^3 \times 10^{-2} = \amp 10^{3+-2}\\ = \amp 10^1\\ = \amp 10 \end{align*}

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Therefore, \(10^3 \times 10^{-2} = 10\text{.}\)

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Subsubsection 1.3.2.2 Division law of indices

Activity 1.3.3.

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Work in Groups

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Instructions

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Using the division law of indices, work with the following pairs of numbers:
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i) \(5^2 , 5^1\)
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ii) \(2^3, 2^5\)
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iii) \(4^4, 4^8\)
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iv) \(6^6, 6^3\)
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For each pair above, form a division expression.
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For example: \(5^2 \div 5^1\)
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Apply the division law of indices to simplify each expression you created.
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Create one division expression of your own using the law of indices. Simplify and calculate it.
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Share your work with your fellow learners.
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Key Takeaway

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The division law of indices states that when you divide two numbers with the same base, you subtract their exponents. Mathematically this can be expressed as: \(\frac{a^m}{a^n} = a^{m-n}\text{,}\) where \(a\) is the base and \(m\) and \(n\) are the exponents.

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Example 1.3.8.

Simplify the following expressions:

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\(\displaystyle \frac{5^6}{5^2} \)
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\(\displaystyle \frac{10^5}{10^3}\)
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Solution.

To solve for \(\frac{5^6}{5^2}\text{,}\) we apply the division law of indices \(\frac{a^m}{a^n} = a^{m-n}\)
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Therefore,

\begin{equation*} \end{equation*}

\begin{align*} \frac{5^6}{5^2} =\amp 5^{6-2}\\ = \amp 5^4 \end{align*}

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To solve \(\frac{10^5}{10^3}\text{,}\) we have:
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\begin{align*} \frac{10^5}{10^3}= \amp 10^{5-3}\\ = \amp 10^2 \end{align*}

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Example 1.3.9.

Evaluate: \(\frac{3x^5}{9x^2}\)

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Solution.

First, we simplify the coefficients:

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\(\frac{3}{9} = \frac{1}{3}\)

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Next, we apply the division law of indices to the variable \(x\text{:}\)

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\(\frac{x^5}{x^2} = x^{5-2} = x^3\)

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Therefore,

\begin{equation*} \end{equation*}

\begin{align*} \frac{3x^5}{9x^2} = \amp \frac{1}{3}^{5-2}\\ =\amp \frac{1}{3}x^3 \end{align*}

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Subsubsection 1.3.2.3 Product of two powers

Activity 1.3.4.

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Work in Groups

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Instructions

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Consider the following expressions.
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1. \(\displaystyle (2^3)^2\)
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2. \(\displaystyle (3^2)^3\)
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3. \(\displaystyle (5^1)^4\)
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4. \(\displaystyle (4^2)^1\)
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5. \(\displaystyle (10^2)^3\)
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6. \(\displaystyle (7^3)^2\)
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Simplify the above expressions.
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Share your work with your fellow learners.
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Key Takeaway

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Product of two powers happen when a term inside a bracket is raised to a power, and the whole bracket is also raised to another power.

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Rule:

\begin{equation*} \end{equation*}

\((a^m)^n = a^{m \times n} = a^{mn}\)

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For example: \((a^2)^3 = a^{2 \times 3} = a^6\)

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Example 1.3.10.

Simplify the expression \((5^3)^3\text{.}\)

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Solution.

To simplify \((5^3)^3\) multiply the powers as shown below

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\begin{align*} (5^3)^3 = \amp 5^{3 \times}3\\ = \amp 5^9 \end{align*}

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Hence, \((5^3)^3 = 5^9 \)

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Example 1.3.11.

Solve the expression \((4y^2)^3\)

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Solution.

Recall that \((a^m)^n = a^{m \times n} = a^{mn}\text{.}\)

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\begin{align*} (4y^2)^3 = \amp 4^3 \times (y^2)^3 \\ = \amp 4^3 \times y^{2 \times 3} \\ = \amp 4^3 \times y^6\\ = \amp 64y^6 \end{align*}

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Subsection 1.3.3 Zero indices

Activity 1.3.5.

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Work in Groups

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Instructions

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Choose any \(3\) natural numbers from the list below:
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\(2, 5, 7, 10, 25, 100\)
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For each chosen number:
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1. Write two equal powers of the number.e.g \(2^3 \div 2^3\)
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2. Simplify the expression formed.
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Look at the pattern below and complete the missing values:
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\(2^4 = 16\)

\begin{equation*} \end{equation*}

\(2^3 = 8\)

\begin{equation*} \end{equation*}

\(2^2 = 4\)

\begin{equation*} \end{equation*}

\(2^1 = 2\)

\begin{equation*} \end{equation*}

\(2^0 = ?\)

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What happens when the exponent reaches zero?
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Discuss your work with your fellow learners.
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Key Takeaway

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For any natural number \(a\text{,}\) if it is raised to the power of zero, the result is always \(1\text{.}\)

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This means that \(a^0 = 1\) for any \(a \neq 0\text{.}\)

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For example,

\begin{equation*} \end{equation*}

\begin{align*} \frac{4^2}{4^2} = \amp 4^2 \div 4^2\\ = \amp 4^{2-2}\\ = \amp 4^0 \end{align*}

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Example 1.3.12.

Simplify \(7^2 \times 7^0\)

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Solution.

By definition, \(7^0 = 1\text{.}\)

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Therefore,

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\begin{align*} 7^2 \times 7^0 = \amp 7^2 \times 1\\ =\amp 7^2 \\ = \amp 7 \times 7 \\ = \amp 49 \end{align*}

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Subsection 1.3.4 Negative indices

Activity 1.3.6.

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Work in Groups

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What you need

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\(\bullet\) Pen and small pieces of paper

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Instructions

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Write the following negative index expressions on separate pieces of paper.
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\(2^{-3}\text{,}\) \(5^{-2}\text{,}\) \(3^{-4}\text{,}\) \(10^{-1}\)
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Place all the papers on the table and mix them up.
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On other pieces of paper, write their positive index equivalents:
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\(\frac{1}{25}\text{,}\) \(\frac{1}{8}\text{,}\) \(\frac{1}{10}\text{,}\) \(\frac{1}{81}\)
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Take turns to:
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1. Pick one paper with a negative index expression.
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2. Simplify it to find its value.
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3. Match it with the correct positive index equivalent.
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Record your results in the table below:
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Negative Index Expression Positive Index Equivalent

\(2^{-3}\) \(\frac{1}{8}\)

\(5^{-2}\)

\(3^{-4}\)

\(10^{-1}\)

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Discuss your work with your fellow learners.
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Definition 1.3.13.

Negative indices are used to represent the reciprocal of a number raised to a positive exponent.

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When a base is raised to a negative exponent, it indicates that the base should be taken as the reciprocal and then raised to the absolute value of the exponent.

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Key Takeaway

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If we have a base \(a\) and a negative exponent \(-n\text{,}\) it can be expressed as:

\begin{equation*} a^{-n} = \frac{1}{a^n}\text{.} \end{equation*}

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Example 1.3.14.

Simplify the expression \(3^{-2} \times 3^4\text{.}\)

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Solution.

By definition, \(3^{-2} = \frac{1}{3^2}\text{.}\)

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Therefore,

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\begin{align*} 3^{-2} \times 3^4 = \amp \frac{1}{3^2} \times 3^4\\ = \amp \frac{3^4}{3^2}\\ = \amp 3^4 \div 3^2 \\ = \amp 3^{4-2} \\ = \amp 3^2 \end{align*}

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Subsection 1.3.5 Fractional indices

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Activity 1.3.7.

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Work in Groups

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Investigate what happens when a number is raised to a fractional exponent.

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Instructions

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Consider the number \(25^\frac{1}{2}\text{.}\)
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1. Express it as a product of its prime factors in index form .
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2. Apply the law of indices to simplify the above expression.
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Work out the following:
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1. \(\displaystyle 8^\frac{2}{3}\)
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2. \(\displaystyle 81^\frac{3}{4}\)
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3. \(\displaystyle 27^\frac{1}{3}\)
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Discuss your work with your fellow learners.
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Definition 1.3.15.

Fractional indices, also known as rational indices, are used to represent roots of numbers in index form.

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A fractional index indicates that the base is raised to a power and then the root is taken.

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Key Takeaway

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Given a base \(a\) and a fractional index \(\frac{m}{n}\text{,}\) it can be expressed as:

\begin{equation*} a^{\frac{m}{n}} = \sqrt[n]{a^m} \end{equation*}

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This means that we first raise \(a\) to the power of \(m\) and then take the \(n\)-th root of the result.

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Example 1.3.16.

Evaluate:

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\(\displaystyle 16^\frac{3}{4}\)
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\(\displaystyle 4^{-\frac{1}{2}}\)
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Solution.

First, express \(16^{\frac{3}{4}}\) as a product of its prime factors:
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Therefore,
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\begin{align*} 16^{\frac{3}{4}} = \amp (4^2)^\frac{3}{4}\\ = \amp 4^{2 \times \frac{3}{4}}\\ =\amp 4^ \frac{3}{2}\\ = \amp (\sqrt{4})^3\\ = \amp 2^3 \end{align*}

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Expressing \(4^{-\frac{1}{2}}\) as a product of its prime factors we have:
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\begin{align*} 4^{-\frac{1}{2}} = \amp (2^2)^{-\frac{1}{2}}\\ = \amp 2^{2 \times (-\frac{1}{2})}\\ = \amp 2^{-1}\\ = \amp \frac{1}{2} \end{align*}

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Subsection 1.3.6 Powers of 10 and common logarithms

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Activity 1.3.8.

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Work in Groups

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Instructions

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Imagine you are climbing a mountain and each step you take is a jump by a power of \(10\text{.}\) The first step is \(10^0\text{,}\) the second step is \(10^1\text{,}\) and so on. Each step represents a significant increase in height, just like how powers of \(10\) represent large numbers.

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1. In groups of \(5\text{,}\) study and fill the table below.

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Power of \(10\)	Numerical Value	Common Logarithm	Logarithm Value
\(10^0\)	1	\(\log_{10}(1)\)	0
\(10^1\)		\(\log_{10}(10)\)
\(10^2\)		\(\)	\(2\)
\(\)	\(1000\)	\(\log_{10}(1000)\)
\(10^4\)	\(10000\)	\(\)	\(4\)
\(10^5\)		\(\)	\(\)

2. Now, consider what happens if you take steps in the opposite direction—down the mountain—using negative powers of \(10\) such as \(10^{-1}\text{,}\) \(10^{-2}\text{,}\) and \(10^{-3}\text{.}\)

What are the numerical values of these negative powers of \(10\text{?}\)

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What are their common logarithms?

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Fill in a similar table for these negative powers of \(10\text{.}\)

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3. What patterns or rules did you notice about powers of \(10\) and their common logarithms? Write your observations and explanations.

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4. Discuss your work with your classmates.

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Definition 1.3.17.

Powers of 10 are numbers that represent a base of \(10\) raised to an exponent.

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Common logarithm of a number \(x\) is defined as the exponent to which the base \(10\) must be raised to obtain \(x\text{.}\) In mathematical terms, if \(y = \log_{10}(x)\text{,}\) then \(10^y = x\text{.}\)

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Key Takeaway

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When a number is rewritten as a power of \(10\text{,}\) its common logarithm is just the exponent.

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Number	Re-written as a power	Number’s logarithm
\(1\)	\(10^0\)	\(\log_{10}(1) = 0\)
\(10\)	\(10^1\)	\(\log_{10}(10) = 1\)
\(100\)	\(10^2\)	\(\log_{10}(100) = 2\)
\(1000\)	\(10^3\)	\(\log_{10}(1000) = 3\)
\(10000\)	\(10^4\)	\(\log_{10}(10000) = 4\)
\(0.1\)	\(10^{-1}\)	\(\log_{10}(0.1) = -1\)
\(0.01\)	\(10^{-2}\)	\(\log_{10}(0.01) = -2\)
\(0.001\)	\(10^{-3}\)	\(\log_{10}(0.001) = -3\)
\(0.0001\)	\(10^{-4}\)	\(\log_{10}(0.0001) = -4\)
\(0.00001\)	\(10^{-5}\)	\(\log_{10}(0.00001) = -5\)

Example 1.3.18.

Express the following in logarithmic notation.

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\(\displaystyle 0.00000001 = 10^{-8}\)
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\(\displaystyle 1000000000 = 10^9\)
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Solution.

If \(0.00000001 = 10^{-8}\text{.}\)
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Then, the logarithmic notation of \(0.00000001= 10^{-8}\) is:
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\(\log_{10}(0.00000001) = -8\text{.}\)
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Since \(1000000000 = 10^9\text{.}\)
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Its logarithm is:
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\(\log_{10}(1000000000) = 9\text{.}\)
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Example 1.3.19.

Determine numbers whose logarithms are equal to the following values:

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\(\displaystyle 5\)
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\(\displaystyle -3\)
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\(\displaystyle 8\)
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Solution.

To find the numbers whose logarithms are equal to the given values, we use the definition of logarithms:

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If \(y = \log_{10}(x)\text{,}\) then \(x = 10^y\text{.}\)

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For \(5\text{,}\) we have:
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\(x = 10^5 = 100000\text{.}\)
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Therefore, the number whose logarithm is \(5\) is \(100000\text{.}\)
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For \(-3\text{,}\) we have:
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\(x = 10^{-3} = \frac{1}{1000} = 0.001\text{.}\)
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Hence, the number whose logarithm is \(-3\) is \(0.001\text{.}\)
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For \(8\text{,}\) we have:
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\(x = 10^8 = 100000000\text{.}\)
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The number is \(100000000\text{.}\)
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Subsection 1.3.7 Using IT to learn more on indices and logarithms

In this section, we will explore how Information Technology (IT) can be utilized to enhance our understanding of indices and logarithms. IT tools such as calculators, computer software, and online resources can provide valuable assistance in learning and applying these mathematical concepts.

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Example of sources that can be used include:

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1. \(\textbf{Video tutorials explaining logarithm concepts and applications}\)

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