WEEK SCHEME OF WORK

  1. Theory of Logarithms: Laws of Logarithms and application of Logarithmic equations and indices
  2. Surds: Rational and Irrational numbers; basic operations with surds and conjugate of binomial surds
  3. Application of surds to trigonometrical ratios. Draw the graphs of sine and cosine for angles 00< x < 3600
  4. Matrices and Determinant: Types, order, Notation, basic operations, transpose, determinants of 2 x 2 and 3 x 3 matrices, Inverse of 2 x 2 matrix and application to simultaneous equation
  5. Linear and Quadratic Equations: Application, one linear-one quadratic, word problems leading to one linear-one quadratic
  6. Surface areas and volume of spheres and hemispheres (solid and hollow sphere and hemisphere)
  7. Longitude and Latitude: Identification of longitude and latitude, North and south, meridian, equator. Calculation of length of parallel of latitude.
  8. Longitude and Latitude: Calculation of distance between two points on the latitude, longitude, time or speed of aircraft
  9. Arithmetic Finance: Simple Interest, Compound Interest, Annuities, Depreciation and Amortization
  10. Revision of the term’s work

DATE: 04/09/2023 WEEK: ONE SUBJECT: MATEMATICS CLASS: SSS THREE LESSON TITLE: Basic Concept of Laws of Indices Application of Laws of Indices PERIOD: 1 and 2 DURATION: 80 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; (i) Interpret the practical problems of Indices. (ii) Solve problems on Indices. LEARNING RESOURCES: Audio Visual Resources. - Object of different length or height - Model Mathematical Set Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by Association of Mathematics teachers

CONTENT:

Lessons objectives: At the end of the lesson students should be able to; Interpret the practical problems of Indices Basic Concept of Laws of Indices A number of the form am where a is a real number, a is multiplied by itself m times The number a is called the base and the super script m is called the index (plural indices) or exponent. 1. A m x A n = Am + n ——————–Multiplication law Example: X3 xX2 =( X x Xx X) x (X x X) = X 5 Or X3 x X2 = X3 + 2 = X5 2. Am ÷ An=Am-n ———————Division law Example: X6 ÷ X4 = X6-4 = X2 3. (a m )n = amn —————-Power law Example: (x3)2 = X3 x X3 = X3+3 =X6 Or X3X2 = X6 4. am ÷ am = am-m = a0 =1 am ÷am = am/am = ao = 1 a0………………………………….Zero Index :. Any number raised to power zero is 1 Example: 3o = 1, co = 1, yo = 1 5. (ab)m =ambm ————-Product power law e.g. (2xy)2= 4x2y2 6. Negative index
a –m = 1/am ————- Negative Index Example: 2 -1 = ½, and 3 -2 = 1/3 2 = 1/9 7. a1/n =n√a ————–Root power law Example : 9 ½ =√9=3

27 1/3 =3√27 = 3 ie (3)3 = 3 8. a m/n = (a 1/n) m = (n√a)m ———–Fraction Index or a m/n = (am) 1/n = (n√a)m Example : 272/3 = 3√27=32=9.

EVALUATION 1. 275/3 2. 10000000000 Application of Laws of Indices Solve the following (i) 32 3/5 (ii) 343 2/3 (iii) 64 2/3 (iv) 0.001 (v) 14 0 Solution: i) 32 3/5 = (32 1/5) 3 = (5√32) 3 = 2 3 = 8 ii) 343 2/3 = (343 1/3 )2 = (3√343)2 = (7 3)1/3)2 = 72 = 49 iii) 64 2/3 = (64 1/3)2 = (4 3)1/3)2 = 4 2 iv) (0.001)3 = (1/100)3 = (1/10)3)3 = (10 -3)3 = 10 -9 = 1/10 9 v) 14 0 = 1

EVALUATION 1) Simplify the following a) 216 4/3 b) 25 1.5 c) (0.00001)2 d) 32 2/5 e) 81 ¾

Exponential Equation of Linear Form Solve the following exponential equations a) (1/2) x = 8 b) (0.25) x+1 = 16 c) 3x = 1/81 d) 10 x = 1/0.001 e) 4/2x = 64 x Solution a) (1/2) X = 8 (2 -1) x = 2 3 2 –x = 2 3 -x = 3 X = - 3 b) (0.25) x+1 = 16 (25/100) x+1 = 16 (1/4) x+1 = 4 2 (4 -1) x+1 = 4 2 - x – 1 = 2 - x = 2 + 1 - x = 3 X = - 3 (c) 3x = 1/81 3x = 1/34 3x = 3 -4 X = -4 (d) 10x = 1/0.001 10 x = 1000 10 x = 10 3 10x = 10 3 X = 3 EVALUATION Solve the following exponential equations a) 2 x = 0.125 b) 25 (5x) = 625 c) 10 x = 1/100000

Exponential Equation of Quadratic Form Some exponential equation can be reduced to quadratic form as can be seen below. Example : Solve the following equations. a) 2 2x – 6 (2 x) + 8 = 0 b) 5 2x + 4 X 5 x+1 – 125 = 0 c) 3 2x – 9 = 0 Solution a) 2 2x – 6 (2 x) + 8 = 0 (2 x)2 – 6 (2 x) + 8 = 0 Let 2 x = y. Then y2 – 6y + 8 = 0 Then factorize Y 2 – 4 y – 2y + 8 = 0 Y (y - 4) -2 (y -4) = 0 (y -2) (y - 4) = 0 Y – 2 = 0 or y – 4 = 0 Y = 2 or y= 4 Since 2 x = y, and y = 2 2 x = 2 2 x = 2 1 x = 1 Since 2 x = y and y = 4 2 x = 4 2 x = 2 2 x = 2 X = 1 and 2 b) 5 2x + 4 X 5 x+1 – 125 = 0 (5 x) 2 + 4 X (5 x X 5 1) – 125 = 0 Let 5 x = p P 2 + 4 X (p X 5) – 125 = 0 P2 + 4 (5p) – 125 = 0 P2 + 20p – 125 = 0 Then Factorise p2 + 25p – 5p – 125 = 0 P (p + 25) -5 (p + 25) = 0 (p - 5) (p + 25) = 0 P – 5 = 0 p + 25 = 0 P = 5 or p = - 25 Since 5x = p, p = 5 5 x = 5 1 X = 1 5x = -25 (Not simplified)
1) 3 2x – 9 = 0 (3 x) 2 - 9 = 0 Let 3 x = a a 2 – 9 = 0 a 2 = 9 a = ±√9 a = ± 3 a = 3 or – 3 Since 3 x = a, when a = 3 3 x = 3 1 X = 1 Since 3x = a, when a = -3 3 x = - 3 (Not a solution)
CLASS EVALUATION Solve the following exponential equations. a) 2 2x+ 1 – 5 (2 x) + 2 = 0 b) 3 2x – 4 (3 x+1) + 27 = 0

ASSIGNMENT 1) Evaluate 3 x = 1/8 1 (a) 4 (b) -4 (c) -2 (d) 2 2) Simplify 2r5 X 9r3 (a) P2 (b) 2p4 (c) P3 d)18r8 3) Solve 3-y = 243 (a) -5 (b) 5 (c) 3 (d) -3 4) Solve 25-5n = 625 (a) 1/5 (b) 2/5 (c) 1 1/5 (d) – 2/5 5) Simplify (0.0001)2= (a) 10-5 (b) 10 -3 (c) 10-8 (d) 10-10 Theory 1. 163/2 x 82/32. 3X2 x 4X3 321/56X7 DATE: 04/09/2023 WEEK: ONE SUBJECT: MATEMATICS CLASS: SSS THREE LESSON TITLE: APPLICATION OF LAWS OF LOGARITHM PERIOD: 3 and 4 DURATION: 80 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; (i) Application of Laws of Logarithm LEARNING RESOURCES: 1. Audio Visual Resources. - Object of different length or height - Model Mathematical Set 2. Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by CONTENT: Lessons objectives: At the end of the lesson students should be able to;  Solve problems on the Application of Laws of Logarithm

Logarithm of numbers (Index & Logarithmic Form) The logarithm to base a of a number P, is the index x to which a must be raised to be equal to P. Thus if P = ax, then x is the logarithm to the base aof P. We write this as x = log a P. The relationship logaP = x and ax=P are equivalent to each other. ax=P is called the index form and logaP = x is called the logarithm form Conversion From Index to Logarithmic Form Write each of the following index form in their logarithmic form a) 26 = 64 b) 251/2 = 5 c) 44= 1/256 Solution a) 26 = 64 Log2 64 = 6 b) 251/2 = 5 Log255=1/2 c) 4-4= 1/256 Log41/256 = -4 Conversion From Logarithmic to Index form. a) Log2128 = 7 b) log10 (0.01) = -2 c) Log1.5 2.25 = 2 Solution a) Log2128 = 7 27 = 128 b) Log10(0.01) = -2 10-2= 0.01 c) Log1.5 2.25 = 2 1.52 = 2.25 Laws of Logarithm a) let P = bx, then logbP = x Q = by, then logbQ = y PQ = bx X by = bx+y (laws of indices) LogbPQ = x + y :. LogbPQ = logbP + LogbQ b) P÷Q = bx÷by = bx+y LogbP/Q = x –y :. LogbP/Q = logbP – logbQ c) Pn= (bx)n = bxn Logbpn = nbx :. LogPn = logbP d) b = b1 :. Logbb = 1 e) 1 = b0 Logb1 = 0 EXAMPLE - Solve each of the following:. a) Log327 + 2log39 – log354 b) Log313.5 – log310.5 c) Log28 + log23 d) given that log102 = 0.3010 log103 = 0.4771 and log105 = 0.699 find the log1064 + log1027 Solution a) Log327 + 2 log39 – log354 = log3 27 + log3 92 –log354 = log3 (27 X 92/54) = log3 (271 X 81/54) = log3 (81/2)
= log3 34/log32 = 4log3 3 – log3 2 4 X (1) – log3 2 = 4 – log3 2 = 4 - log3 2 b) log3 13.5 - log3 10.5 = log3 (13.5)- Log310.5 = log3 (135/105) = log3 (27/21) = log3 27 - log3 21 = log3 3 3 - log3 (3 X 7) = 3log3 3 - log3 3 -log37 = 2 - Log3 7 c) Log28 + Log33 = log223+ log33 = 2log22 + log33 2+1=3 d) log10 64 + log10 27 log10 26 + log1033 6 log10 2 + 3 log10 3 6 (0.3010) + 3(0.4771) 1.806 + 1.4314 = 3.2373. EVALUATION 1. Change the following index form into logarithmic form. (a) 63= 216 (b) 33 = 1/27 (c) 92 = 81 2. Change the following logarithm form into index form. (a) Log88 = 1 (b) log ½¼ = 2 3. Simplify the following a) Log512.5 + log52 b) ½ log48 + log432 – log42 c) Log381 4. Given that log 2 = 0.3010, log3 0.477 Log105 = 0.699, find the log10 6.25 + log10 Change of Base Let logbP = x and this means P = bx LogcP = logcbx = x logcb If x logcb = logcP X = logcP Logc b :. LogcP = logcP Logcb Example : Shows that logab X logba = 1 Logab = logcb Logca Logba = logca Logcb :. Logab X logba = logcb X logca Logca + logcb= 1 EVALUATION Solve the following logarithm equation. Log3 (x2 + 7x + 21) = 2 Log10 (x2 – 3x + 12) = 1 Reading Assignment : Further Maths Project Book 1(New third edition).Chapter 2 pg. 8- 10
ASSIGNMENT 1) If log81/64 = x, find the value of x (a) 2 (b) 1 (c) -3 (d) -4. 2) Solve 9(1 - x) = (1/27) x+1 (a) -5 (b) -1 (c) 1 (d) ½ 3) Simplify log7 49 (a) 1/7 (b) 2 (c) 7 (d) log 1/7 4) Solve the equation log216 = x (a) 8 (b) 4 (c) 2 (d) 2 5) Convert 52 = 25 into logarithm form (a) log525 = 2 (b) log 255 = 2 (c) log225 = 5 (d) None of the above Theory (1) Find the value of x for which log10 (4x2 + 1) -2 log10 x – log10 2 = 1 is valid. (2) Solve the logarithmic equation: Log4 (x2 + 6x + 11) = ½

DATE: 04/09/2023 WEEK: ONE SUBJECT: MATEMATICS CLASS: SSS THREE LESSON TITLE: APPLICATION OF LOGARITHM PERIOD: 5 DURATION: 40 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; (i) Interpret the practical problems of LOGARITHM LEARNING RESOURCES: 3. Audio Visual Resources. - Object of different length or height - Model Mathematical Set

  1. Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS
  1. New General Mathematics 2 published by
  2. MAN Mathematics 2 published by CONTENT: Lessons objectives: At the end of the lesson students should be able to;  Solve problems on Logarithm

Standard Forms Numbers such as 1000 can be converted to its power of ten in the form 10n where n can be term 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 10 100 102 10 101 1 100 0.01 10-3 0.10 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. A number in the form A x 10n, where A is a number between 1 and 10 i.e. 1<A<10 and n is an integer is said to be in standard form e.g. 3.835 x 103 and 8.2 x 10-5 are numbers in standard form. Examples : Express the following in standard form 1. 7853 2. 382 3. 0.387 4. 0.00104 Solutions 1. 7853 = 7.853 x 103 2. 382 = 3.82 x 102 3. 0.387 = 3.87 x 10-1 4. 0.00104 = 1.04 x 10-3 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 = 100.7980 x 105 = 100.7980+ 5 = 105.7980 Log 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

  1. 80030 = 4.9033
  2. 8 = 0.9031
  3. 13580 = 2.1329 Multiplication and Division of number 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

  4. 4627 x 29.3

        No    Log
    4627     3.6653
    X 29.3    + 1.4669

    Antilog → 1356005.1322 .
     4627 x 29.3 = 135600 Evaluation:

  5. Use table to find the complete logarithm of the following:

  1. 183 (b) 89500 (c) 10.1300 (d) 7 2 Use logarithm to calculate. 3612 x 750.9 113.2 x 9.9