Green Bridge Learning Resource For Indices, Logarithms And Surds

Overview

Indices, logarithms, and surds are fundamental concepts in General Mathematics that play a crucial role in various calculations and problem-solving scenarios. Understanding these topics is essential for students to navigate through complex mathematical operations efficiently. This course material will delve deep into the intricacies of indices, logarithms, and surds, providing a comprehensive overview of their principles, applications, and interrelationships.

The primary objective of this course material is to equip students with the necessary skills to apply the laws of indices in calculations effectively. Indices, also known as exponents, govern the way numbers are raised to powers, leading to efficient computations across different numerical scenarios. By mastering the laws of indices, students will be able to simplify complex expressions, manipulate variables with ease, and solve equations involving powers and roots proficiently.

Furthermore, this course material aims to establish a clear relationship between indices and logarithms to enhance students' problem-solving abilities. Logarithms serve as powerful tools that help convert exponential equations into linear form, simplifying calculations and facilitating the solving of intricate mathematical problems. Understanding how logarithms and indices correlate enables students to tackle complex equations, evaluate functions, and analyze growth and decay processes effectively.

In addition to exploring indices and logarithms, this course material will focus on solving problems in different bases using logarithmic functions. Students will learn how to manipulate numbers across various number bases ranging from 2 to 10, understanding the significance of base transformations and their impact on mathematical operations. By mastering logarithmic computations in different bases, students will enhance their numerical fluency and problem-solving skills across diverse mathematical contexts.

Moreover, this course material will delve into the realm of surds, emphasizing the importance of simplifying and rationalizing these irrational numbers. Surds often appear in mathematical expressions involving roots and provide a unique challenge that requires careful manipulation to simplify and integrate seamlessly into calculations. By mastering basic operations on surds, students will develop the skills to simplify square roots, manipulate radical expressions, and solve equations involving irrational numbers efficiently.

Objectives

Solve Problems In Different Bases In Logarithms
Apply The Laws Of Indices In Calculation
Perform Basic Operations On Surds
Simplify And Rationalize Surds
Establish The Relationship Between Indices And Logarithms In Solving Problems

Lesson Note

In this lesson, we will delve into the concepts of Indices, Logarithms, and Surds, which are fundamental topics in General Mathematics. Understanding these topics helps build a strong foundation for more advanced mathematical problems and applications. Each section will cover definitions, laws, and problem-solving techniques to help you master these concepts.

Lesson Evaluation

Congratulations on completing the lesson on Indices, Logarithms And Surds. Now that youve explored the key concepts and ideas, its time to put your knowledge to the test. This section offers a variety of practice questions designed to reinforce your understanding and help you gauge your grasp of the material.

You will encounter a mix of question types, including multiple-choice questions, short answer questions, and essay questions. Each question is thoughtfully crafted to assess different aspects of your knowledge and critical thinking skills.

Use this evaluation section as an opportunity to reinforce your understanding of the topic and to identify any areas where you may need additional study. Don't be discouraged by any challenges you encounter; instead, view them as opportunities for growth and improvement.

Simplify the expression $4^{3} \times 4^{-2}$. A. $16^{5}$ B. $16$ C. $\frac{1}{16}$ D. $1$ Answer: C. $\frac{1}{16}$
Solve for $x$ in the equation $2^{x} = 8$. A. $2$ B. $3$ C. $4$ D. $8$ Answer: B. $3$
Evaluate $\log_{2} 32$. A. $4$ B. $5$ C. $3$ D. $6$ Answer: B. $5$
Simplify $\sqrt{50}$. A. $5\sqrt{2}$ B. $10\sqrt{5}$ C. $5\sqrt{10}$ D. $25$ Answer: A. $5\sqrt{2}$
If $\log_{3} y = \frac{1}{2}$, then $y$ is equal to: A. $3$ B. $\frac{3}{2}$ C. $9$ D. $\frac{9}{2}$ Answer: C. $9$
Simplify $\sqrt{75} + \sqrt{27}$. A. $12$ B. $11$ C. $10$ D. $9$ Answer: A. $12$
If $\log_{5} x = 2$, then $x$ is equal to: A. $25$ B. $10$ C. $125$ D. $5$ Answer: A. $25$
Evaluate $\frac{2^3 \times 3^2}{2^2 \times 3^3}$. A. $3$ B. $\frac{3}{2}$ C. $2$ D. $\frac{2}{3}$ Answer: D. $\frac{2}{3}$
Simplify $\sqrt{128} - \sqrt{32}$. A. $4\sqrt{2}$ B. $6\sqrt{2}$ C. $8\sqrt{2}$ D. $10\sqrt{2}$ Answer: B. $6\sqrt{2}$
If $\log_{7} z = 2$, then $z$ is equal to: A. $14$ B. $21$ C. $49$ D. $28$ Answer: C. $49$

Recommended Books

	Mathematics for senior secondary schools 1 Subtitle New Edition Publisher Longman Publishers Year 2020 ISBN 978-9788233023
	Advanced Mathematics Subtitle Logarithms and Surds Publisher Pearson Education Year 2019 ISBN 978-9819764214

Past Questions

Wondering what past questions for this topic looks like? Here are a number of questions about Indices, Logarithms And Surds from previous years

JAMB UTME - Mathematics - 2018 (Click to take full assessment)

Question 1 Report

Evaluate 5 $^{- 3}$ log $^{2}$ $\times$ 2 $^{2 \log 3}$

\frac{1}{8}

\frac{1}{8}

\frac{2}{5}

View Answer

NECO SSCE - General Mathematics - 2005 (Click to take full assessment)

Question 1 Report

Evaluate log 18 + log⁶ - log16$^{\frac{1}{2}}$

log 3

log 9

log 20

3 log 2

3 log 3

View Answer

WAEC SSCE - General Mathematics - 2020 (Click to take full assessment)

Question 1 Report

(a) In the diagram, AB is a tangent to the circle with centre O, and COB is a straight line. If CD//AB and < ABE = 40°, find: < ODE.

(b) ABCD is a parallelogram in which |$\overline{CD}$| = 7 cm, I$\overline{AD}$I = 5 cm and < ADC= 125°.

(i) Illustrate the information in a diagram.

(ii) Find, correct to one decimal place, the area of the parallelogram.

Answer

(a)

<ABE = <OBC (tangent and radius form a right angle)
<ABE = 40° (given)
Therefore, <OBC = 40°.
Since CD//AB, then <CDE = <ABE = 40° (alternate angles)
<ODE = <OBC + <CDE = 40° + 40° = 80°.

(b)

(i)

  A ----------- B
   \           / 
    \         /  
     \       /   
      \     /    
       \   /     
        \ /      
        D-------C
             (125°)

(ii)

The area of the parallelogram is given by A = base x height. Since AD || BC, then the height of the parallelogram is given by the perpendicular distance between AD and BC, which is 7 cm. To find the length of the base, we use the cosine rule:

|$\overline{AD}$|² = |$\overline{AB}$|² + |$\overline{BD}$|² - 2|$\overline{AB}$||$\overline{BD}$| cos(ADC)
5² = x² + (x+7)² - 2x(x+7)cos(125°)
25 = 2x² + 14x + 49 + 2x(x+7)(-0.5736)
25 = 2x² + 14x + 49 - 1.1472x² - 10.0317x - 20.7032
0 = 0.8528x² + 4.0317x + 4.7032

Using the quadratic formula: x = (-b ± ?(b² - 4ac))/2a, we have:
x = (-4.0317 ± ?(4.0317² - 4(0.8528)(4.7032)))/(2(0.8528))
x = (-4.0317 ± 4.9988)/1.7056
x = 0.6559 or x = -3.6582

Since x represents a length, we discard the negative solution. Therefore, x = 0.6559 cm.

So, the area of the parallelogram is A = base x height = 0.6559 cm x 7 cm = 4.5913 cm² (to one decimal place).

(c)

x = $\frac{1}{2}$(1 - $\sqrt{2}$)

2x² - 2x = 2$\left(\frac{1}{2}$(1 - $\sqrt{2}$)²\) - 2(1 - $\sqrt{2}$)
= (1 - 2$\sqrt{2}$ + 2) - 2 + 2$\sqrt{2}$
= 1 + 2$\sqrt{2}$.

Therefore, 2x² - 2x = 1 + 2$\sqrt{2}$.

Answer Details

(a)

(b)

(i)

  A ----------- B
   \           / 
    \         /  
     \       /   
      \     /    
       \   /     
        \ /      
        D-------C
             (125°)

(ii)

Using the quadratic formula: x = (-b ± ?(b² - 4ac))/2a, we have:
x = (-4.0317 ± ?(4.0317² - 4(0.8528)(4.7032)))/(2(0.8528))
x = (-4.0317 ± 4.9988)/1.7056
x = 0.6559 or x = -3.6582

Since x represents a length, we discard the negative solution. Therefore, x = 0.6559 cm.

So, the area of the parallelogram is A = base x height = 0.6559 cm x 7 cm = 4.5913 cm² (to one decimal place).

(c)

x = $\frac{1}{2}$(1 - $\sqrt{2}$)

2x² - 2x = 2$\left(\frac{1}{2}$(1 - $\sqrt{2}$)²\) - 2(1 - $\sqrt{2}$)
= (1 - 2$\sqrt{2}$ + 2) - 2 + 2$\sqrt{2}$
= 1 + 2$\sqrt{2}$.

Therefore, 2x² - 2x = 1 + 2$\sqrt{2}$.

View Answer