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) = bx
a. then x is the index (or, power or exponent) of the base number, b.
b.
The function f(x) = bx is known as the exponential function.
c.
The number y = bx 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 bx 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/ bx (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 bn/d (where n = numerator
and d= denominator) – meaning:
1.
A) 41/2 = √4 (note: d = 2 is the root; n = 1 is the power, exponent or index);
B) 4(-1/2)
= 1 ÷ 41/2 = 1/√4 (note: -ve in the index means one over (i.e. 1/....))
2.
A) b1/x = x√b (note again: the denominator x implies the x root);
B) b(-1/x) = 1/b1/x = 1/x√b
B) b(-1/x) = 1/b1/x = 1/x√b
3.
A) 34/3 = 3√34
B) 3(-4/3)
= 1/ 3√34
4.
A) bn/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):
·
40.5 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. bm x bn = b(m + n)
Proof: 23 x 24 = (2x2x2)x(2x2x2x2) = (2x2x2x2x2x2x2)
= 27 = 2(3 + 4)
Therefore: bm x bn = b(m + n)
b. bm ÷ bn = b(m - n)
Proof: 23 ÷ 24 = (2x2x2)/(2x2x2x2) = 1/2 = 2(-1) = 2(3 – 4)
Therefore: bm ÷ bn = b(m - n)
c. (bm)n = bmn
Proof: (23)4 = (2x2x2)(2x2x2)(2x2x2)(2x2x2)
= (2x2x2x2x2x2x2x2x2x2x2x2) = 212
= 2(3x4)
Therefore: (bm)n
= bmn
d. am x bm = (ab)m
Proof: 53 x 23 = (5x5x5)(2x2x2) = (5x2)(5x2)(5x2)
= (5x2)3
Therefore: am x bm = (ab)m
e. am ÷ bm = (a/b)m
Proof: 53
÷ 23 = (5x5x5)/(2x2x2) = (5/2)(5/2)(5/2) = (5/2)3
Therefore: am ÷ bm = (a/b)m
Practice Questions:
1. 271/3 x 31/2
2. 271/2 x 31/2
3. 323/5 x 161/4
4. 2-3 x 163/4
5. 4x = 8
6. 4(x +1) = 0.25
7. 2(2x + 3) + 2(x + 3) = 1 + 2x
8. Given that y = axb – 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:
3x x 92y = 27
2x x 4-y = 1/8
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3.
Logarithms:
a.
In the exponential
function f(x) = y = bx, 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 = bx.
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 = bx.
b.
Logarithm was
formulated / invented by John Napier (1550 – 1617) - a mathematician from
Scotland:
Thus, the inverse of f(x) i.e. f-1(x) = logb 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 = logb x can be seen here. As you can see from the graph:
- when 0 < x < 1, y or logb 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")
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 logb x = 0
- when x> 1, y or logb x > 1
·
The exponential function graph of y = bx and
its inverse, the logarithmic function graph of y = logb 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), 5x = 2, etc. which
you often encounter in questions involving Geometric Progressions (GP).
For example:
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
Tn = arn-1 =
(a/r)(rn)
Let
the last term be Tn, thus, Tn = 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
Practice Questions:
1. logb xy = logb x +
logb y
Proof: Let x = bm then, m = logb x; and,
y = bn ⇔
n = logb y
Therefore: logb (xy) = logb (bm x bn)
= logb b(m + n) = m + n
= logb
x + logb y
Practice Questions:
Q1. Evaluate:
a)
log4 2 + log4 8
b)
log6 8 + log6 27
c)
log8 16 + log8 4
2. logb (x/y) = logb x - logb
y
Proof: Let x = bm , then, m = logb x; and,
y = bn ⇔
n = logb y
Therefore: logb (x/y) = logb (bm ÷ bn)
= logb b(m - n) = m – n
= logb
x - logb y
Practice Questions:
Q2. Evaluate:
a)
log3 18 - log3 6
b)
log2 6 - log2 3
c)
log4 3 - log4 48
3. logb xn = nlogb x
Proof: Let x = bm , then, m = logb x;
Therefore: logb xn = logb (bm)n
= logb bmn = mn = n(m) = n(logb x)
Practice Questions:
Q3.
Evaluate:
a) log5 25
b) log3 81
c) log4 (1/64)
d) logb (1/x)
4.
logb
a = logc a / logc b (change of base of logarithm)
Proof: Let y = logb a then, a = by
Therefore: logc both sides: logc a
= logc by
logc a = ylogc
b
y = logc a / logc b
logb a = logc a / logc
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. logb a = 1/ loga b (change
of base – special case)
Proof: Let y = logb a; then, a = by
loga both sides: loga a
= loga by
loga a = y loga
b
y = loga a / loga b
Therefore loga a = loga a /
loga b
= l / loga 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:
·
logb
(1/x) = logb x-1 = -logb x
·
logb
b = 1; loga a = 1; logx x = 1
·
logb
bn = n (thus, log2 23 = 3)
·
logb
1 = 0 (thus, logb x = 0, then x = 1)
2. More Practice Questions:
Q6. Solve
the following equations:
a) 32x = 1000
b) 4x / 25-x = 24x /
8x-3
c) 3x+1 + 32-x = 28
d) 23x = 8 + 23x – 1
e) 3x + 2 – 3x = 8/9
f) 162x-3 = 84x
g) 9(3n – 1) = 27n
h) 3n – 3 x 27n =243
i) 2x + 4 – 2x + 3 = 1
j) 82x-3 = 1/√(4x + 2)
k) 27(32x + 4) = 1
Q7. Given
that logp x = 9 and logp y = 6, find:
a)
logp √x (1)
b)
logp 1/x (1)
c)
logp xy (2)
d)
logp x/y (2)
Q8. Solve
the equations:
a)
logp 243 = 5 (1)
b)
log4 (3q – 4) = 3 (2)
c)
f(x) = 2xlogx3 – 5logx9 – x + 5
i) Find
the value of a and the value of b such that
f(x) = (x – 5)(alogx3 – 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 logb32
= 5 in index form (1)
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More on Applications of Logarithms
· Logarithm is useful to solve such equations as: 5x = 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
Tn = arn-1 = (a/r)(rn)
Let the last term be Tn, thus, Tn = 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:
Hence, n = log (Lc/Lo) ÷ log (1/2)
n = log (Lc/Lo) ÷ log (1/2)
where, n = number of half-lives
Lo = Original amount of radioactive substance or original level of radioactivity
Lc = Current amount of radioactive substance or current level of radioactivity
(1/2)n x Lo = Lc
(1/2)n = Lc/Lo
Log both sides:
log (1/2)n = log (Lc/Lo)
· Logarithms are also used in the measurements of:
o Acidity or Alkalinity - in pH value:
pH = - log10 [CH+], ….Eqn (1)
(where [CH+] = Concentration of hydrogen ions in mol/dm3)
From Eqn (1):
10(pH) = [CH+]-1
[CH+] = 1 ÷ 10(pH)
Hence, when pH = 1, [CH+] = 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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