Revision of Logarithm of Numbers Greater than One and Logarithm of Numbers Less than one; Reciprocal and Accuracy of Results Using Straight Calculation.

Subject: 

MATHEMATICS

Term:

FIRST TERM

Week:

WEEK 1

Class:

JSS 3 / BASIC 9

Topic:

Revision of Logarithm of Numbers Greater than One and Logarithm of Numbers Less than one; Reciprocal and Accuracy of Results Using Straight Calculation.

Previous lesson: 

The pupils have previous knowledge of

 SS 1 THIRD TERM EXAMINATION MATHEMATICS

that was taught as a topic in the previous lesson

Behavioural objectives:

At the end of the lesson, the learners will be able to

• solve revision questions on Numbers Greater than One
• give examples of various Logarithm of Numbers Less than one
• explain the importance of Reciprocal
• point out the need to master the Accuracy of Results Using Straight Calculation.

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

Reference Materials:

• Scheme of Work
• Online Information
• Textbooks
• Workbooks

Content:

WEEK ONE

TOPIC: REVISION OF LOGARITHM OF NUMBERS GREATER THAN ONE AND LOGARITHM OF NUMBERS LESS THAN ONE.

CONTENT

• Standard forms
• Logarithm of numbers greater than one
• Multiplication and divisions of numbers greater than one using logarithm
• Using logarithm to solve problems with roots and powers (no > 1)
• Logarithm of numbers less than one.
• Multiplication and division of numbers less than one using logarithm
• Roots and powers of numbers less than one using logarithm

STANDARD FORMS

A way of expressing numbers in the form A x 10^x where 1< A < 10 and x is an integer, is said to be a standard form. Numbers are grouped into two. Large and small numbers. Numbers greater than or equal to 1 are called large numbers. In this case the x, which is the power of 10 is positive. On the other hand, numbers less than 1 are called small numbers. Here, the integer is negative.

Numbers such as 1000 can be converted to its power of ten in the form 10^x where x can be termed as the number of times the decimal point is shifted to the front of the first significant figure i.e. 10000 = 10⁴

Number Power of 10

1. 10²
2. 10¹
3. 10⁰
   1. 10^-3
   2. 10^-1

Note: One tenth; one hundredth, etc are expressed as negative powers of 10 because the decimal point is shifted to the right while that of whole numbers are shifted to the left to be after the first significant figure.

Examples

1. Express in standard form (i) 0.08356 (ii) 832.8 in standard form

Solution

i 0. 08356 = 8.356 x 10^-2

ii 832.8 = 8.328 x 10²

2. Express the following in standard form

(a) 39.32 = 3.932 x 10¹

(b) 4.83 = 4.83 x 10⁰

(c) 0.005321 = 5.321 x 10^-3

WORKING IN STANDARD FORM

Example

Evaluate the following leaving your answer in standard form

1. 4.72 x 10³ + 3.648 x 10³

(ii)6.142 x 10⁵ + 7.32 x 10⁴

(iii) 7.113 x 10^-5– 8.13 x 10^-6

solution

i. 4.72 x 10³ + 3.648 x 10³

= [ 4.72 + 3.648 ] x 10³

= 8.368 x 10 ³

ii. = 6.142 x 10⁵+ 7.32 x 10⁴

= 6.142 x 10⁵+ 0.732 x 10⁵

= [6.142 + 0.732 ] x 10⁵

= 6.874 x 10⁵

iii. = 7.113 x 10^-5 – 8.13 x 10^-6

= 7.113 x 10^-5 – 0.813 x 10^-5

= [ 7.113 – 0.813 ] x 10^-5

= 6.3 x 10^-5

Example: Simplify : √[P/Q], leaving your answer in standard form given that P = 3.6 x 10^-3 and

Q = 4 x 10^-8.

Solution

= √[P/Q]

3.6 x 10^-3

= 4 x 10^-8

= / 36 x 10^-4

√ 4 x 10^-8

= √ 9 x 10^{-4 –(-8)}

= 3 x (10⁴) ^½

= 3 x 10²

EVALUATION

1. Evaluate 2.5 x 10^-3 + 3.2 x 10^-2

2. Without using table, evaluate the following leaving your answer in standard form,

i. 4ab given that a= 3.5 x 10^-3 and b = 2.3 x 10⁶ ii. 0.08 x 0.000025

0.0005

LOGARITHM OF NUMBERS GREATER THAN ONE

Base ten logarithm of a number is the power to which 10 is raised to give that number e.g.

628000 = 6.28 x10⁵

628000 = 10^0.7980 x 10⁵

= 10^{0.7980+ 5}

= 10^5.7980

Log 628000 = 5.7980

Integer Fraction (mantissa)

If a number is in its standard form, its power is its integer i.e. the integer of its logarithm e.g. log 7853 has integer 3 because 7853 = 7.853 x 10³

Examples:

Use tables (log) to find the complete logarithm of the following numbers.

(a) 80030 (b) 8 (c) 135.80

Solution

(a) 80030 = 4.9033

(b) 8 = 0.9031

(c) 13580 = 2.1329

Evaluation

Use table to find the complete logarithm of the following:

(a) 183 (b) 89500 (c) 10.1300 (d) 7

Multiplication and Division of numbers greater than one using logarithm

To multiply and divide numbers using logarithms, first express the number as logarithm and then apply the addition and subtraction laws of indices to the logarithms. Add the logarithm when multiplying and subtract when dividing.

Examples

Evaluate using logarithm.

1. 4627 x 29.3

2. 8198 ÷ 3.905

3. 48.63 x 8.53

15.39

Solutions

1. 4627 x 29.3

No Log

4627 3.6653

X 29.3 + 1.4669

antilog → 135600 5.1322

∴ 4627 x 29.3 = 135600

To find the Antilog of the log 5.1322 use the antilogarithm table:

Check 13 under 2 diff 2 (add the value of the difference) the number is 0.1356. To place the decimal point at the appropriate place, add one to the integer of the log i.e. 5 + 1 = 6 then shift the decimal point of the antilog figure to the right (positive) in 6 places.

= 135600

2. 819.8 x 3.905

No Log

819.8 2.9137

3.905 0.5916

antilog → 209.9 2.3221

∴ 819.8 ÷ 3.905 = 209. 9

3. 48.63 X 8.8.53

15.39

No Log

48.63 1.6869

8.53 + 0.9309

2.6178

÷ 15.39 – 1.1872

antilog → 26.95 1.4306

∴48.63 ÷ 8.53 = 26.96

15.39

Evaluation: Use logarithm to calculate. 3612 x 750.9

113.2 x 9.98

USING LOGARITHM TO SOLVE PROBLEMS WITH POWERS AND ROOT (NO. GREATER THAN ONE)

Examples:

Evaluate:

(a) 3.53³ (b) 4 40000 (c) 94100 x 38.2 to 2 s.f

5.683³ x 8.14

Solution

1. 3.53³

No. Log_____

3.53³ 0.5478 x 3

44.00 1.6434

∴ 3.53³ = 44.00

(b) 4 4000

No. Log_____

4 4000 3.6021 ÷ 4

7.952 0.9005

∴4 4000 = 7.952

(c) 94100 x 38.2

5.683³x 8.14

Find the single logarithm representing the numerator and the single logarithm representing the denominator, subtract the logarithm then find the antilog.

No Log

94100 4.9736 ÷ 2 = 2.4868

38.2 1.5821

Numerator 4.0689→ 4.0689

5.68³ 0.7543 x 3 = 2.2629

8.14 0.9106

Denominator 3.1735→ 3.1735

7.859 0.8954

∴94100 x 38.2 = 7.859 ~ 7.9 (2.sf)

5.68³ x 8.14

LOGARITHM OF NUMBERS LESS THAN ONE

To find the logarithm of number less than one, use negative power of 10 e. g.

0.037 = 3.7 x 10^-2

= 10 ^{0.5682 x} 10^-2

= 10 ^{0.5682 + (-2)}

= 10^{-2 5682}

Log 0.037 = 2 . 5682

2 . 5682

Integer decimal fraction (mantissa)

Example: Find the complete log of the following.

(a) 0.004863 (b) 0. 853 (c) 0.293

Solution

Log 0.004863 = 3.6369

Log 0.0853 = 2.9309

Log 0.293 = 1.4669

Evaluation

1. Find the logarithm of the following:

(a) 0.064 (b) 0.002 (c) 0.802

2. Evaluate using logarithm.

95.3 x 318.4

1.29⁵ x 2.03

USING LOGARITHM TO EVALUATE PROBLEMS OF MULTIPLICATION, DIVISION, POWERS AND ROOTS WITH NUMBERS LESS THAN ONE

OPERATION WITH BAR NOTATION

Note the following when carrying out operations on logarithm of numbers which are negative.

i.The mantissa (fractional part) is positive, so it has to be added in the usual manner.

ii. The characteristic (integral part) is either positive or negative and should therefore be added or operated as directed numbers.

iii. For operations like multiplication and division, separate the integer from the characteristic before performing the operation.

Examples:

Simplify the following, leaving the answers in bar notation, where necessary

• 1. .7675 + 2.4536
  2. 6.8053 – 4.1124
  3. 2.4423 x 3
  4. 2.2337 ÷ 7

Solution

i. .7675 + 2.4536 ii. 6.8053 – 4.1124

.7675 6.8053

+ 2. 4536 – 4. 1124

6. 22112.6929

iii. 2.4423 x 3 iv. 2. 2337 ÷ 7

= 3( 2 + 0.4423) = 7 + 5.2337÷ 7

= 6 + 1.3269 = 1+ 0.7477

= 5.3269 = 1 + 0.7477

= 1.7477

Examples: Evaluate the following using the logarithm tables;

1. 0.6735 x 0.928

2. 0.005692 ÷ 0.0943

3. 0.6104³

4. 4 0.00083

5. 3 0. 06642

Solution

1. 0.6735 x 0.928

No. Log.__

0.6735 1.8283

0.928 1.9675

0.6248 1.7958

∴ 0.6735 x 0.928 = 0.6248

2. 0.005692 ÷ 0.0943

No Log

0.005692 3.7553

÷ 0.0943 2.9745

0.06037 2.7808

3. 0.6104³

No Log_____

0.6104³ 1.7856 x 3

0.2274 1.3568

∴ 0.6104³ = 0.2274

∴ 0.005692 ÷ 0.943 = 0.6037

4. 4 0.00083

No. Log._____

4 0.00083 4.9191 ÷ 4

0.1697 1.2298

∴ 4 0.06642 =0.1697

5. 3 0.06642

No. Log.____________

3 0.06642 2.8223 ÷ 3

3 ) 2 + 0.8223

3 ) 3 + 1.8223

1 + 0.6074

0.405 1.6074

30.6642 = 0.405

Note: 3 cannot divide 2 therefore subtract 1 from the negative integer and

add 1 to the positive decimal fraction so as to have 3 which is divisible

by 3 without remainder.

Evaluation: Use the logarithms table to evaluate

5 (0.1684)³

Presentation

The topic is presented step by step

Step 1:

The class teacher revises the previous topics

Step 2.

He introduces the new topic

Step 3:

The class teacher allows the pupils to give their own examples and he corrects them when the needs arise

EVALUATION

GENERAL EVALUATION / REVISION QUESTION

Use tables to evaluate the following, giving your answers correct to 3 s.f.

1. (0.897)³ 2.(0.896 × 0.791)³ 3. (800.9 × 87. 25)²

4. 8750000 × 8900 5. 80.4² × 78000

300.5 100.5 × 35.7

WEEKEND ASSIGNMENT

Use table to find the log of the following:

1. 900 (a) 3.9542 (b) 1.9542 (c) 2.9542 (d) 0.9542

2. 12.34 (a) 3.0899 (b) 1.089 (c) 2.0913 (d) 1.0913

3. 0.000197 (a) 4.2945 (b) 4.2945 (c) 3.2945 (d) 3.2945

4. 0.8 (a) 1.9031 (b) 1.9031 (c) 0.9031 (d) 2.9031

5. Use antilog table to write down the number whose logarithms is 3.8226.

(a) 0.6646 (b) 0.06646 (c) 0.006646 (d) 66.46

THEORY

Evaluate using logarithm.

1. 23.97 x 0.7124

3.877 x 52.18

2. 3 76.58

0.009523

Reading Assignment

Essential Mathematics for SSS2, pages 1-10, Exercise 1.8

Conclusion

The class teacher wraps up or concludes the lesson by giving out short notes to summarize the topic that he or she has just taught.

The class teacher also goes round to make sure that the notes are well copied or well written by the pupils.

He or she does the necessary corrections when and where the needs arise.