FUNCTIONS

IGCSE Edexcel Syllabus: No Specification

·        IGCSE Cambridge: Syllabus Area 2

·        SPM: Form 4 Chapter 1

Some Words For IGCSE Edexcel Students

Your syllabus omits ‘functions’ as a dedicated segment in the syllabus while IGCSE Cambridge syllabus and SPM textbooks do not specify the 'concept of asymptotes'. As ‘functions’ is something that you come across as often as they do,  it is advisable that you too know:

·        What is a function?

o   Refers to 2 specific types of relations:

§  One-to-One relations

§  Many-to-One relations

What is a relation? It refers to the connection between one variable and another variable. The connection may be causal (due to cause and effect) - e.g. the extension (x) of a spring and the load (F) applied to it; or non-causal - e.g. area codes of  telephone lines and the areas.

How may a relation be represented?  A relation may be represented by:

• arrow diagram
• ordered pair - as in x and y coordinates like (x, y);
• graph
• mathematical formula or equation

The four (4) main types of relations (of which 2 of them are functions):

• 1-to-1 relation (a function): e.g. F = kx (Hooke's Law); growth against time;
• One-to-Many relation: y = x^(1/2)

• Many-to-One relation (a function): e.g. y = ax² + bx + c
• Many-to-Many relation: e.g. x² + y² = 1 (a unit circle, other circles, etc.)

·        Terminologies related to a function:

o   Domain, Co-domain and Range

o   Objects (inputs) and Images (Outputs)

o   Inverse Functions

o   Composite Functions

·        Notations related to functions:

o   f(x); f: x׀→

o   f^-1(x)

o   f(g(x))

o   f²(x) [= f(f(x))]

o   f’(x) [=dy/dx]

o   f’’(x) [=d²y/dx²]

·        Relationship between y = f(x) and y = ׀f(x)׀

·        What type of function has an inverse function?

·        How to find the inverse function of one-to-one function? Use sketch graphs to show that a function is the mirror image of its inverse function and vice versa in the line of reflection y = x.

·        Composite functions

(Bookmark this post for more ….relevant details as time goes) 

INDICES AND LOGARITHMS

SPM: Indices and Logarithms

Edexcel IGCSE (4PMO) Syllabus Area 1: Logarithmic Functions & Indices

Cambridge IGCSE (0606) Syllabus Area 7: Logarithmic & Exponential Functions

1.    Introduction: Some advice for SPM Form 4 and IGCSE Year 10 Students

Before you learn “logarithms”, it is advisable that you review “indices” that you learnt in your earlier years because “logarithm” is the inverse of “index” or "exponent".

2.    Indices (Exponents or Powers):

a.    If function f(x) (or number, y) = b^x

a.    then x is the index (or,  power or exponent) of the base number, b.

b.    The function f(x) = b^x is known as the exponential function.

c.    The number y = b^x is said to be in index or exponential format

b.    Indices: Integer, Fractional or Decimal Indices

a. Integer Index (Integer Power or Exponent) where x in b^x is an integer:

1.    +ve integer index - meaning:

·      2³ means 2 x 2 x 2 (repeated multiplication of three 2s)

·      b⁴ means b x b x b x b (repeated multiplication of four bs); or

2.    –ve integer index – meaning:

·      2^-3 means 1/ 2³ (i.e. one over or the reciprocal of two to the power of 3)

·      b^-x means 1/ b^x  (i.e. one over or the reciprocal of b to the power x); or

3.    0 as index – meaning:

·      2⁰ = 1; 3⁰ = 1; 4⁰ = 1;

·      b⁰ = 1, provided b ≠ 0

·      0⁰ is undefined; or

4.      1 as index – meaning:

·        2¹ = 2, 3¹ = 3, 4¹ = 4, …

·        b¹ = b

·        0¹ = 0

b.Fractional Index, n/d as in b^n/d (where n = numerator and d= denominator) – meaning:

1.    A) 4^1/2 = √4 (note: d = 2 is the root; n = 1 is the power, exponent or index);

B)  4^(-1/2) = 1 ÷ 4^1/2 = 1/√4 (note: -ve in the index means one over (i.e. 1/....))

2.    A) b^1/x = ^x√b (note again: the denominator x implies the x root);
     B)  b^(-1/x) = 1/b^1/x = 1/^x√b

3.    A) 3^4/3 = ³√3⁴

B)  3^(-4/3) = 1/ ³√3⁴

4.    A) b^n/d = ^d√bⁿ; (denominator d denotes 'd root' and numerator n denotes 'to the power')

B)  b^(-n/d) = 1/ ^d√bⁿ

c. Decimal Index (Decimal Power or Exponent):

·        4^0.5 means 4 to the power or exponent of 0.5 (which you can use calculator to find easily).

c.    Law of Indices: You need to know these before you learn logarithms.

a.    b^m  x bⁿ = b^{(m + n)}

Proof:        2³ x 2⁴ = (2x2x2)x(2x2x2x2) = (2x2x2x2x2x2x2) = 2⁷ = 2^{(3 + 4)}

Therefore: b^m  x bⁿ = b^{(m + n)}

b.    b^m  ÷ bⁿ = b^{(m - n)}

Proof:        2³ ÷ 2⁴ = (2x2x2)/(2x2x2x2) =  1/2 = 2^(-1) = 2^{(3 – 4)}

Therefore: b^m  ÷ bⁿ = b^{(m - n)}

c.    (b^m)ⁿ = b^mn

Proof:        (2³)⁴ = (2x2x2)(2x2x2)(2x2x2)(2x2x2) = (2x2x2x2x2x2x2x2x2x2x2x2) = 2¹²

                         = 2^(3x4)

Therefore: (b^m)ⁿ = b^mn

d.    a^m  x b^m = (ab)^m

Proof:        5³ x 2³ = (5x5x5)(2x2x2) = (5x2)(5x2)(5x2) = (5x2)³

    

Therefore: a^m  x b^m = (ab)^m

e.    a^m  ÷ b^m = (a/b)^m

Proof:        5³ ÷ 2³ = (5x5x5)/(2x2x2)  = (5/2)(5/2)(5/2) = (5/2)³

Therefore: a^m  ÷ b^m = (a/b)^m

Practice Questions:

1.       27^1/3 x 3^1/2

2.       27^1/2 x 3^1/2

3.       32^3/5 x 16^1/4

4.       2^-3 x 16^3/4

5.       4^x = 8

6.       4^{(x +1)} = 0.25

7.       2^{(2x + 3)} + 2^{(x + 3)} = 1 + 2^x

8.       Given that y = ax^b – 5 and that y = 7 when x = 2 and y = 22 when x = 3, find the value of a and the value of b.

9.       Solve the simultaneous equations:

3^x x 9^2y = 27

2^x x 4^-y = 1/8

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3.    Logarithms:

a.    In the exponential function f(x) = y = b^x, where b is a real number > 0 and b ≠ 1:

a. For whatever values of x (+ve or –ve), f(x) or y > 0

b.When x = 0, f(x) or y = 1;

c. When x < 0 (i.e. x is a -ve real number), then: 0 < f(x) or y < 1

d. As x increases above 1, y increases sharply or exponentially

e. As x approaches -ve infinity, y approaches 0 but never becomes zero - the x-axis is the horizontal asymptote to the exponential function f(x) = y = b^x.

f.  The above relationship can be seen in the exponential graph of f(x) or y = b^x  here:

b.    Logarithm was formulated / invented by John Napier (1550 – 1617) - a mathematician from Scotland:

·      Logarithm is the inverse of exponential function: If f(x) or y = b^x,  then expressing x in terms of y to get the inverse of f(x), we obtain x = log_b y
   
    Thus, the inverse of f(x) i.e. f^-1(x) = log_b x (the image or output y of the original exponential function now becomes the object or input x of the inverse function). The graph of logarithmic function y = log_b x can be seen here. As you can see from the graph:

•          when 0 < x < 1, y or log_b x = -ve number. 

                   Eg. log 0.00001, log 0.25, log (1/5), log 0.99999 are all -ve numbers although you                       don't see the -ve signs in them! 
                   
                   Thus, in this inequality: n log 0.25 > log 0.0005. 
                            To find the greatest integer value of n,
                            we divide both sides of the inequality by log 0.25; and
                            since log 0.25 is a negative number (-0.602059991),
                            we must therefore reverse the inequality sign to become:
                            n < log 0.0005/log 0.25
                            n < 5.482892142
                           Therefore, the greatest integer value of n = 5 
                           (This skill is tested in Q9(f) of Edexcel PMO 2013 Jan Paper 2 on "Series")      

•         when x = 1,  y or log_b x = 0
•         when x> 1,   y or log_b x > 1

·      The exponential function graph of y = b^x and its inverse, the logarithmic function graph of y = log_b x are mirror images of one another in the mirror line y = x as can be seen here.

·      Logarithm is useful to solve such equations as: -2ⁿ = -2048 (pl see below), 5^x = 2, etc. which you often encounter in questions involving Geometric Progressions (GP).

For example:

    The geometric progression 6, -12, 24, …, 6144 consists of n terms. Find the value of n.

                        In the GP, a = 6 and r = -12/6 = -2

                                    T_n = ar^n-1 = (a/r)(rⁿ)

                        Let the last term be T_n, thus, T_n = 6144 = (a/r)(rⁿ)

                                    6144 = (6/-2)(-2ⁿ) = -3(-2ⁿ)

                                    -2ⁿ = 6144/-3

                                    -2ⁿ = -2048

                        To log both sides, remove the -ve signs. Thus,

                                    n log 2 = log 2048

                                    n = log 2048 / log 2

                                    n = 11

·      Logarithms and Indices are also useful to simplify multiplications or divisions of very large or very small numbers that we often encounter in science and other fields of study.

·      Logarithms can also be used to solve “half-life” problems in atomic physics, particularly when the number of half-lives in reality is rarely the whole number which you encounter in your Form 5 science (details at the end of this post).

·       Logarithms are also used in the measurements of acidity and alkalinity in pH value; loudness of sound in decibels (dB) and magnitude of earthquake in Richter Scale (pl see end of this post)

c.    Laws of Logarithm 

1.    log_b xy = log_b x + log_b y

     Proof:             Let x = b^m     then, m = log_b x; and,

      y = bⁿ  ⇔ n = log_b y

     Therefore:      log_b (xy) = log_b (b^m x bⁿ) = log_b b^{(m + n)} = m + n

                                         = log_b x + log_b y   

          Practice Questions:

         

Q1. Evaluate:

a)      log₄ 2 + log₄ 8

b)      log₆ 8 + log₆ 27

c)      log₈ 16 + log₈ 4

2.    log_b (x/y) = log_b x - log_b y

     Proof:          Let x = b^m  , then, m = log_b x; and,

                              y = bⁿ  ⇔ n = log_b y

     Therefore:      log_b (x/y) = log_b (b^m ÷ bⁿ) = log_b b^{(m - n)} = m – n

                                      = log_b x - log_b y

          Practice Questions:

Q2. Evaluate:

a)      log₃ 18 - log₃ 6

b)      log₂ 6 - log₂ 3

c)      log₄ 3 - log₄ 48

3.    log_b xⁿ = nlog_b x

     Proof: Let x = b^m  , then, m = log_b x;

     Therefore:      log_b xⁿ = log_b (b^m)ⁿ = log_b b^mn = mn = n(m) = n(log_b x)

          Practice Questions:

Q3. Evaluate:

        a)      log₅ 25

        b)      log₃ 81

        c)      log₄ (1/64)

        d)   log_b (1/x)

                       

4.    log_b a = log_c a / log_c b (change of base of logarithm)

    Proof:       Let y = log_b a  then, a = b^y

    Therefore:    log_c both sides: log_c a = log_c b^y

                                                           log_c a = ylog_c b

                                                      y = log_c a / log_c b

                                              log_b a = log_c a / log_c b        

                 Practice Questions:

             Q4. Evaluate:

              a)      log₂ 9

              b)      log₃ 2

              c)      log₆ 12

              d)      log₈ 3

              e)      log₂₇ 9

              f)       log₂₅ 125

              g)      log₄ 32

5.    log_b a = 1/ log_a b (change of base – special case)

     Proof: Let y = log_b a;  then, a = b^y

     log_a both sides:  log_a a = log_a b^y

                                         log_a a = y log_a b

                               y = log_a a / log_a b

    Therefore     log_a a = log_a a / log_a b

                                  = l / log_a b

Practice Questions:

Q5. Evaluate:

a)      log₈ 2

b)      log₂₇ 3

c)      log₆₄ 4

d)      log₂₅ 5

e)      log₃₆ 6

f)       Show that log₁₆ 2 = log₈₁ 3.

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Students should also be familiar with the following:

·        log_b (1/x) = log_b x^-1 = -log_b x

·        log_b b = 1; log_a a = 1; log_x x = 1

·        log_b bⁿ = n (thus, log₂ 2³ = 3)

·        log_b 1 = 0 (thus, log_b x = 0, then x = 1)

2.    More Practice Questions:

Q6.    Solve the following equations:

a)         3^2x = 1000

                                                                                

b)        4^x / 2^5-x = 2^4x / 8^x-3

                                                               

c)        3^x+1 + 3^2-x = 28

                                                                        

d)    2^3x = 8 + 2^{3x – 1}

e)    3^{x + 2} – 3^x = 8/9

f)     16^2x-3 = 8^4x

g)    9(3^{n – 1}) = 27ⁿ

h)    3^{n – 3} x 27ⁿ =243

i)     2^{x + 4} – 2^{x + 3} = 1

j)     8^2x-3 = 1/√(4^{x + 2})

k)    27(3^{2x + 4}) = 1

Q7.    Given that log_p x = 9 and log_p y = 6, find:

a)         log_p √x                                                                            (1)                   

b)        log_p 1/x                                                                                   (1)

c)         log_p xy                                                                                    (2)

d)        log_p x/y                                                                                   (2)

Q8.    Solve the equations:

a)         log_p 243 = 5                                                                         (1)

b)        log₄ (3q – 4) = 3                                                                   (2)

c)         f(x) = 2xlog_x3 – 5log_x9 – x + 5                                                                 

i)      Find the value of a and the value of b such that

                 f(x) = (x – 5)(alog_x3 – b)                              (3)

ii)  Hence, solve the equation f(x) = 0                                    (2)

Q9.    a) Given that:  f(x) = 5x ln x + 3 ln x – 10x – 6

i) Factorise f(x)                                                            (3)

ii) Hence, find x which satisfies f(x) = 0                                (3)

b) i) Write 125 = 5³ as a logarithmic equation                             (1)

    ii) Express log_b32 = 5 in index form                                       (1)

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More on Applications of Logarithms

·      Logarithm is useful to solve such equations as: 5^x = 2, -2ⁿ = -2048 (pl see below), etc. which you often encounter in solving questions involving Geometric Progressions (GP).

For example:

    The geometric progression 6, -12, 24, …, 6144 consists of n terms. Find the value of n.

                        In the GP, a = 6 and r = -12/6 = -2

                                    T_n = ar^n-1 = (a/r)(rⁿ)

                        Let the last term be T_n, thus, T_n = 6144

                                    6144 = (6/-2)(-2ⁿ) = -3(-2ⁿ)

                                    -2ⁿ = 6144/-3

                                    -2ⁿ = -2048

                        To log both sides, remove the -ve signs. Thus,

                                    n log 2 = log 2048

                                    n = log 2048 / log 2

                                    n = 11

·      Logarithms and Indices are also useful to simplify multiplications or divisions of very large or very small numbers that we often encounter in science and other fields of study.

·      Logarithms can also be used to solve “half-life” problems in atomic physics, particularly when the number of half-lives is rarely the whole number which you encounter in your Form 5 science:

                 n  = log (L_c/L_o) ÷ log (1/2)

      where,  n = number of half-lives

            L_o = Original amount of radioactive substance or original level of radioactivity

            L_c = Current amount of radioactive substance or current level of radioactivity

(1/2)ⁿ x L_o = L_c

                (1/2)ⁿ = L_c/L_o

Log both sides:

                 log (1/2)ⁿ = log (L_c/L_o)

            Hence, n  = log (L_c/L_o) ÷ log (1/2)  

·       Logarithms are also used in the measurements of:

o   Acidity or Alkalinity - in pH value:

pH = - log₁₀ [C_H+], ….Eqn (1)

(where [C_H+] = Concentration of hydrogen ions in mol/dm³)

From Eqn (1):

10^(pH) = [C_H+]^-1

[C_H+] = 1 ÷ 10^(pH)

Hence, when pH = 1, [C_H+] = 1 ÷ 10⁽¹⁾ = 0.1 mol/dm³

o   Loudness of Sound – in decibels, dB:

Loudness in dB = 10 log₁₀ (p x 10¹²),

(where p is the sound pressure)

o   The Magnitude of an Earthquake – in “Richter Scale”:

M (in “Richter Scale”) = log₁₀ A + B,

     (where A = amplitude (in mm) measured by seismograph and,

                 B = distance correction factor)

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Coming up......some notes on 'surds' for IGCSE students....hereunder: 

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