Indices and Logrithma PDF December 1 2008-3-16 Pm 263kqwe
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INDICES AND LOGARITHMS
ADDITIONAL
MATHEMATICS
MODULE 8
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CHAPTER 5 : INDICES AND LOGARITHMS
CONTENTS PAGE
5.0 INDICES AND LOGARITHM
5.1 Concept Map 2
5.2 Indices and laws of Indices 3
Exercise 5.2
5.3 Simplify Algebraic Expressions 4 - 5
Exercise 5.3
5.4 Equation Involving Indices 5 - 7
Exercise 5.4
5.5 Past Year Questions 8
5.6 Assessment 9 - 10
Answers 11
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CHAPTER 5 : INDICES AND LOGARITHMS
5.1 CONCEPTUAL MAP
Indices
an index
base
Logarithms
If N = ax, then logaN = x
Fractional indices
a) a1/n
= ___________
b) am/n = ___________
= ___________
Laws of indices
a) a
m
x a
n
= _________
b) am an = _________
c) (am)n = _________
Integer indices
a)an = a x a x a x a x a
n factors
b) Negative index,a-n
= _________
c) Zero index, a0
= __________
Equations involving indices
a) If ax = ay, then _________
b) If mx
= nx, then ________
INDICES AND LOGARITHMS
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5.2 INDICES AND LAWS OF INDICES
Examples Hit the Buttons
Find the values of the following.
1. 63 = 216
2. 4-2 =2
4
1
=16
1
3. 32
8 = ( 3 8 )2 = 22
= 4
EXERCISE 5.2
a) Evaluate the following
1. 0.44
Solution
2. 32
8
27
Solution
3.
3
8
1
Solution
4.2
1
4
1
Solution
6 ^ 3 =
( 3 shift ^ 8
) =x2
1c
ba 4 =x
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5.3 SIMPLIFY ALGEBRAIC EXPRESSIONS
Examples Solution
Simplify
1. 2a2 x 3a3
= (2 x 3) x (a2 x a3)= 6 x a2 + 3
Sign : (+) x (+) = (+)
Number : 2 x 3 = 6Unknown/ : a2 x a3 = a5
base
2. 33n 2
3n 3
= 33n 2 - ( n 3)
= 33n 2 n + 3
= 3 2n + 1
am an = a m n
Use the laws of indices to simplify thedivision
EXERCISE 5.3
Simplify the following
1. (34)2n x 3n
Solution
2. 31 2n x 34 x 33n + 2
Solution
3. 121 48 nn
Solution
4. a2 x 2a3 (4a)2
Solution
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5.3n5
2n-32
p
pp
n
Solution
6.22
27
b6a
b18a
Solution
7. 3x 3x 1
Solution
8. 2n + 2 - 2n + 12(2n 1)
Solution
5.4 EQUATIONS INVOLVING INDICES
Examples Steps / Method
Solve each of the following equations
1. 33x = 8133x = 34
3 x = 4
x = 3
4
Express in the _________________ andcompare the indices
_____________________ the indices
2. 2x . 4x+1 =642x . 22 (x+1) = 26
x + 2x + 2 = 6
3x = 4
x =3
4
Use ______ as the same base throughout
_______________ the indices using the
laws of indices
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3. 0168 12
xx
022 1432
xx
143 222
xx
443
22
2 xx
3x2
= 4x + 4
3x2- 4x 4 = 0
(3x + 2)(x 2) = 0
x =3
2, x = 2
Use 2 as the same base throughout
______________________of a quadraticequation
_________________________________
EXERCISE 5.4
1. Solve 8x3
= 27
Solution
2. Solve x8 =
4
1
Solution
3. Solve 2x .3
x = 36
Solution
4. Solve 21
4x = 10
Solution
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5. Solve 14 2793 xx
Solution
6. Solve 53x 25x + 1 =
125
1
Solution
7. Solve 427
x =xx
93
13
Solution
8. Solve3
1
5
25125
x
x
Solution
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5.5 PAST YEAR QUESTIONS
1. Solve the equation 122 34 xx (SPM 05)
Solution
2. Solve the equation 684 432 xx (SPM 04)
Solution
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5.6 ASSESSMENT
Answer All the questions Time : 30 minutes
1. Solve the equation163x = 86x + 1
Solution
2. Solve the equation32x =
243
1
Solution
3.Solve the equation8x . 2x = 32
Solution
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4. Solve the equation 093 26
2
xx
Solution
5. Solve the equation 28 x =
xx
24
11
Solution
6.Solve the equation52x 52x 1 = 100
Solution
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ANSWERS
EXERCISE 1.1
1. 0.0256
2. 2.253. 512
4 2
EXERCISE 2.1
1. 39n
5. 12 n
2. 3n+7
6. 2a5
3.np 4
17.
3
23x
4. 3a
8
18. 92n
EQUATION INVOLVING INDICES1. same base, compare
2. 2, Add
3. General form, solving by factorization
EXERCISE 3.1
1. x =2
35. x = 6
2. x =3
26. x = -1
3. x = 2 7. x = -2
4. x = 6.25 8. x = 1
PAST YEAR QUESTIONS
1. x = -1
2. x = 3
ASSESSMENT
1. x =2
54. x = 6, -2
2. x =2
1 5. x = 2
3. x =4
56. x =
2
3
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INDICES AND LOGARITHMS
ADDITIONAL
MATHEMATICS
MODULE 9
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__________________________________________
CHAPTER 5 : INDICES AND LOGARITHMS
CONTENTS PAGE
5.0 INDICES AND LOGARITHM
5.1 Conceptual Map 2
5.2 Logarithm and laws of logarithm 3 4
Exercise 5.2
5.3 Finding Logarithm of numbers 4 6
Exercise 5.3
5.4 Changing the base of logarithm 7 - 9
Exercise 5.4.1
Exercise 5.4.2
Exercise 5.4.3
5.5 Equation Involving logarithm 10 11
Exercise 5.5
5.6 Past Year Questions 11 - 12
5.7 Assessment 13
Answers 14
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CHAPTER 5 : INDICES AND LOGARITHMS
5.1 CONCEPTUAL MAP
Index
an index
base
Logarithms
If N = ax, then logaN = x
Laws of Logarithms
d) Logaxy = _________________
e) Logax/y = _________________
f) Logaxm = _________________
Change of base
of logarithms
a) Logab = __________
b) Logab = __________
Equations involving
Logarithms
If logaX = logaY, then
__________________
INDICES AND LOGARITHMS
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5.2 LOGARITHM AND LAWS OF LOGARITHM
Find the value of the following
Examples Solution
1. log4641
Let. log464
1= x
Then 4x = 4-3
x = -3
2. log5= 5 Let. log5 5 = x
Then 5x = 21
5
x =2
1
3. log2x = -5 Given log2x = - 5Then x = 2
-5
=52
1 =321
EXERCISE 5.2
Find the value of each of the following
1. log2128 2 log5 125
3. . log864
14. log10
3
5
2
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5. log10010 6. log 644
7. log9x = -1 8. logx 225
1
5.3 FINDING LOGARITHM OF NUMBERS
Examples Solution
Evaluate the following
1. log318 log36
= log36
18
= log33
= 1
2. log42 + log4 8= log4(2 x 8)
= log442
= 2 log44
= 2(1)= 2
3. Given thatloga2 = 0.123 andloga3 = 0.256, calculate the following
a) loga12 = loga(2 x 2 x 3)= loga2
2 + loga3
= 2loga 2 + loga3
= 2(0.123) + 0.256= 0.502
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b) loga 6 = loga 21
6
=2
1loga(2 x 3)
=2
1[ loga2 + loga3]
=2
1[0.123 + 0.256]
= 0.1895
4. Given thatlogx2 = aandlogx3 = b, writelogx 29
4
xin term of a and b.
Logx 29
4
x= logx4 logx 29x
= logx22
- [ logx 9 + logxx2
]= 2 logx2 - logx3
2 - 2 logxx= 2a - 2logx 3 - 2(1)
= 2a - 2b - 2
EXERCISE 5.3
Simplify each of the following logarithms expression
1. log42 + log432 2. log248 + log23 log29
3. 2 log510 + 3 log52 log532 4. 4logx2 3logx 4 + logx6
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5. Expresslogac
b2in terms oflogab
andlogac
Solution
6. Given that log26 = 2.59 andlog25 = 2.32 find the value of
log1030
Solution
7. Given log32 = 0.631 and log35 = 1.465, find the values ofa) log320
b) log35
4
c) log37.5
Solution
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5.4 CHANGING THE BASE OF LOGARITHM
Examples
Find the value of the logarithm
1. log35 =
3log
5log
10
10
=4771.0
699.0
= 1.465
EXERCISE 5.4.1
Find the value of the logarithm by changing the base of the logarithm
1. Log4 3 2. Log310
3. Log20.1 4. Log50.048
5.Log4
5
1 6.Log2
4
3
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Examples
Find the value of the logarithm
1. log45 =4log
5log
2
2
= 2log2
322.2
2
=2
322.2
= 1.161
EXERCISE 5.4.2
Given that log25 = 2.322 and log23 = 1.585, find the value of the logarithm
1. Log3 5 2. Log815
3. Log9125 4. Log825
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Examples
Find the value of the logarithm
1. log10p2 =
16log
log
4
2
4 p
=4log2
log2
4
4 p
=2
2m
=m
EXERCISE 5.4.3
Given that log4p = m. Express the following in terms of m
1. Logp 4 2. Log8p
3. Log4p64 4.Logp
256
1
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5.5 EQUATION INVOLVING LOGARITHMS
1. Solve the equation :
Log 10( x 5 ) = log10 ( x 1 ) + 2
Solution
Log 10( x 5 ) = log10 ( x 1 ) + 2
Log 10( x 5 ) log10 ( x 1 ) = 2
Log 101
5
x
x= 2
1
5
x
x= 10
2
x 5 = 100( x 1)x 5 = 100x 100
99x = 95
x =99
95
2. Solve the equation:
3x
= 52x + 1
Solutionxlog103 = (2x + 1)log105
xlog103 = 2x log105 + log105xlog103 2x log105 = log105
x(log103 2 log105) = log105
x =5log23log
5log
1010
10
x = 0.7591
EXERCISE 5.5
Solve each of the equation
1. 5x
= 27 2. 5x
= 0.8
3. 3x +
= 18 4. 5x
= 4x +
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5. 2x3
x= 5
x +6. 3
x. 4
x += 6
5.6 SPM QUESTIONS
SPM 2005( Paper 1)
1. Solve the equation 1)12(log4log 33 xx [ 3 marks]
SPM 2005( Paper 1)
2. Given that pm log and rm 3log , express
4
27log
mm in term of pand r
[ 4 marks]
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SPM 2004(Paper 1)
3. Given that 5log 2 = m and 5log = p, express 5log 4.9 in term of m and p
[ 4 marks]
SPM 2003(Paper1)
4. Given that 2log T - 4log V = 3 , express T in term of V. Express T in term of V
[ 4 marks]
SPM 2003(Paper1)5. Solve the equation xx 74 12 [ 4 marks]
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5.7 ASSESSMENT
Answer All the questions Time : 30 minutes
1. Solve the equation 4log3log2)48(log 555 x
2. Given x = mk3log and nk2log . Find the value of 24log k in term ofm and n
3. Given that ,4loglog 93 VT express T in terms of V
4. Solve the equation xx 95 12
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ANSWERS
EXERCISE 5.2
1. 7 4 -1.1938 7.9
1
2.2
35.
2
18. 5
3. -2 6. 6
EXERCISE 5.3
1. 3 5. 2 logab logac
2. 4 6. 4.913. 2 7. a) 2.727 c) 1.834
4. logx2
3b) - 0.203
EXERCISE 5.4.11. 0.7924 3. -3.3222. 2.096 4. -1.887
EXERCISE 5.4.2
1. 1.465 3. 2.19752. 1.302 4. -1.548
EXERCISE 5.4.3
1.m
13.
m1
3
2. m32 4.
m4
EXERCISE 5.5
1. 2.048 4. 20.642. -0.1386 5. 8.827
3. 1.630 6. 0.1632
SPM QUESTIONS
1. x =2
34. T = 8 v
2. 3r 2p + 1 5. 1.677
3. 2p - m -1
ASSESSMENT
1. 5 3. T =v
81
2.nm
31 4. -1.575
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