Week 1 – SS2 First Term Mathematics Notes

SUBJECT: MATHEMATICS CLASS: SS2

SCHEME OF WORK

WEEK	TOPIC
1	Revision of Logarithm of Numbers Greater than One and Logarithm of Numbers Less than one; Reciprocal and Accuracy of Results Using Straight Calculation.
2	Approximations; Calculations Using Standard Form; Significant Figures; and Percentage Error.
3	Sequence and Series: Concept of Sequence and Series; Terms of Arithmetic Progressions and Sum ; Solving problem on A.P
4	Geometric Progressions: The nth Term and Sum of the First n-terms. Problem Solving on G.P and Geometric Mean.
5	Construction of Quadratic Equation from Sum and Product of Roots. Word Problem Leading to Quadratic Equation.
6	Review of the Half Term Work and Periodic Test.
7	Simultaneous Equations: Solving Simultaneous Equations Using Elimination and Substitution Method; Word Problem Leading to Simultaneous Equations.
8	Simultaneous Equations: Solving Equations Involving One Linear and One Quadratic; Using Graphical Method to Solve Quadratic Equations.
9	Straight Line Graphs: Gradient of a Straight Line; Gradient of a Curve; Drawing of Tangents to a Curve.
10	Revision.

REFERENCE BOOKS

New General Mathematics SSS2 by M.F. Macrae etal.
Essential Mathematics SSS2 by A.J.S. Oluwasanmi.

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 = 10⁴Number Power of 10

10²

10¹

10⁰
10^-3
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^-2ii 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

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 10³ii. = 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}
= 10^5.7980Log 628000 = 5.7980

IntegerFraction (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

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

.7675 + 2.4536
6.8053 – 4.1124
2.4423 x 3
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

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