Recurso de Aprendizaje Green Bridge Para Indices, Logarithms And Surds

Resumen

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.

Objetivos

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

Nota de la lección

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.

Evaluación de la lección

Felicitaciones por completar la lección del Indices, Logarithms And Surds. Ahora que has explorado el conceptos e ideas clave, es hora de poner a prueba tus conocimientos. Esta sección ofrece una variedad de prácticas Preguntas diseñadas para reforzar su comprensión y ayudarle a evaluar su comprensión del material.

Te encontrarás con una variedad de tipos de preguntas, incluyendo preguntas de opción múltiple, preguntas de respuesta corta y preguntas de ensayo. Cada pregunta está cuidadosamente diseñada para evaluar diferentes aspectos de tu conocimiento y habilidades de pensamiento crítico.

Utiliza esta sección de evaluación como una oportunidad para reforzar tu comprensión del tema e identificar cualquier área en la que puedas necesitar un estudio adicional. No te desanimes por los desafíos que encuentres; en su lugar, míralos como oportunidades para el crecimiento y la mejora.

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. $49$

Libros Recomendados

	Mathematics for senior secondary schools 1 Subtítulo New Edition Editorial Longman Publishers Año 2020 ISBN 978-9788233023
	Advanced Mathematics Subtítulo Logarithms and Surds Editorial Pearson Education Año 2019 ISBN 978-9819764214

Preguntas Anteriores

¿Te preguntas cómo son las preguntas anteriores sobre este tema? Aquí tienes una serie de preguntas sobre Indices, Logarithms And Surds de años anteriores.

JAMB UTME - Mathematics - 2018 (Haz clic para realizar la evaluación completa)

Pregunta 1 Informe

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

\frac{1}{8}

\frac{1}{8}

\frac{2}{5}

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NECO SSCE - General Mathematics - 2005 (Haz clic para realizar la evaluación completa)

Pregunta 1 Informe

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

log 3

log 9

log 20

3 log 2

3 log 3

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WAEC SSCE - General Mathematics - 2020 (Haz clic para realizar la evaluación completa)

Pregunta 1 Informe

(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.

Respuesta

(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$\sqrt{2}$.

Detalles de la respuesta

(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$\sqrt{2}$.

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