Week 1 Extra Examples Unit 1 Algebraic Contents Subset of Real Numbers Symbol Name N Natural Number Z Integers Q Rational Number

I

Irrational Numbers

R

Real Numbers

Descriptions Counting Numbers Natural numbers, 0 and negative Represented as a/b, where a and b are integers and b ≠ 0 . decimal represented as repeating or terminating. Represented as nonrepeating and nonterminating decimal numbers.

Examples 1, 2, 3, 4,…. …, -1, 0, 1, 2, …. 3, 0, 1, -2/5, 2.4, -2.24242424

2, 4 7 , π , 2.12347

Rational numbers and irrational numbers

Example 1 What kind of number is each of the following? a) 7 b) 2/7 c) 3 π d) -5 e) -3/7 f) π /9 Answer a) 7 is a natural number, an integer, a rational number and a real number. b) 2/7 is rational and real. c) 3 π is irrational and real. π is not a rational number. d) Integer, rational and real number e) Rational and Real number f) Irrational and Real number

Properties of real number Associative (a + b) + c = a + (b + c) (ab)c = a(bc) Commutative a+b=b+a ab = ba Inverse a + (-a) = (-a) + a = 0 a(1/a) = (1/a)a = 1 Distributive a(b + c) = ab + ac (a + b)c = ac + bc

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Example 2 Name the property illustrated in each of the following examples: a) (3 + 4) + 8 = (4 + 3) + 8 b) (3 + 4) + 8 = 3 + (4 + 8) c) (3 + 4) + 8 = 8 + (3 + 4) d) (2 · 5)a = (5 · 2)a e) 2(5a) = (2 · 5)a f) (3a)b = 3(ab) g) x(b + c) = (b + c)x h) (3a + 4b) + 5c = 3a + (4b + 5c) i) (x + y)(a + b) = (x + y)a + (x + y)b j) If a + b = 0, then b = -a k) -1/3 + 1/3 Answer a) Commutative property b) Associative property c) Commutative property d) Commutative property e) Associative property f) Associative property g) Commutative property h) Associative property i) Distributive property j) Inverse property k) Inverse property

Example 3 Use a calculator to evaluate: a) -9(-3) + (-5) -------------3(-4) – 5(2) b) 42 ÷ 8 + 32 ÷ 3 c) [-7 + (-9)] · (-4) – 8(3) d) 36 ÷ 4 · 3 ÷ 9 + 1 -------------------9 ÷ (-6) · 8 – 4

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Answer a) -1 b) 5 c) 40 d) – 0.25

Indices, Surds and Logarithms 1. An index number is written in the form an where a ∈ R and n ∈ Z. a is called the base and n the index. Example, 23, 45, 1.35. Rule of Indices a m ⋅ a n = a m+ n a m ÷ a n = a m− n (a m ) n = a mn

Examples 32 · 34 = 36 33 ÷ 32 = 31 (52)3 = 56

(ab) n = a n ⋅ b n

(2 · 3)2 = 22 · 32 = 36

1

1

an = n a

42 = 2 4 = 2

ab = n a ⋅ n b 1 a−n = n a

9⋅4 = 2 9 ⋅2 4 = 6 1 1 2− 3 = 3 = 2 8

n

2

n

2

an a   = n b b m n

a = a n

n

32 9 3   = 2 = 2 4 2 2 3

5 = 3 52 = 3 25

m

a na = b nb

m n

a =n

3

m

a = mn a

4 34 = 7 37

2 3

x0 = 1

10 = 3 2 10 = 6 10

80 = 1

2. An irrational number in the form

n

p where n ∈ N and p ∈ Q is called a surd. For

example: 2 , 2 12 . Properties of surds p ⋅ q = pq

p q

=

Examples 2 ⋅ 3 = 2⋅3 = 6

2 = 7

p q

p r ± q r = ( p ± q) r

3. The conjugate of

a + b is

2 7

2 3 ± 5 3 = (2 ± 5) 3

a − b , and

3

(

a+ b = a−b

Example

(

)(

) ( a) − ( b) 2

a− b =

)(

2

) ( 3) − ( 2 )

3+ 2 3− 2 = = 3− 2 =1

2

2

4. If y = a x , then log a y = x where a > 0 and a ≠ 1. If log a y = x , then y = a x . Laws of Logarithm log a xy = log a x + log a y

log a

Examples log10 4 ⋅ 3 = log10 4 + log10 3 1 log 5 = log 5 1 − log 5 3 3

x = log a x − log a y y

log a x b = b log a x

log 3 8 3 = 3 log 3 8

log a a = 1

log 4 4 = 1

log a 1 = 0

log 4 1 = 0

log a a = x log a x log b x = log a b

log 4 4 3 = 3

x

log 2 6 =

log10 6 log10 2

5. Log M or lg M means log10 M . The natural logarithm, ln M = log e M . Example 4

a) (x 4 y 2 ) = 1 x ≠ 0, y ≠ 0 1 1 b) 10 − 2 = 2 = 100 10 1 c) x − 4 = 4 x −2 a a −2 1 1 b3 b3 d) −3 = ⋅ −3 = 2 ⋅ = 1 b b a 1 a2 e) (3a 5 )(4a −3 ) = (3 ⋅ 4)(a 5 a −3 ) = 12a 2 0

6 y −2 3 y −2−( −4) 3 y 2 = = f) 4 4 8 y −4 g)

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− 2 x 3 − ( −2 x ) 3 = −2 x 3 − (−2) 3 x 3 = −2 x 3 − (−8) x 3 = −2 x 3 + 8 x 3 = 6 x 3 Example 5 −3

2 −2

a) (3 x y )

−2

=3 x y 6

−4

x6 = 4 9y

6 y −2 3 y −2−( −6) 3 y 4 = = b) 4 4 8 y −6 c) 6 5 ⋅ 6 6 = 6 5+ 6 = 611 d) (2k ) 2 ⋅ (2k ) 2 = (2k ) 4  x2 e)  4 y

  

−2

=

x −4 y 8 = y −8 x 4

2

2

2

 (ab) 3   a 3b 3   b3  b6 f)  4  =  4  =   = 2 a  a   a   a 

Example 6 4a −3b −5 2a −3− ( −5) 2a 2 a) = = 8 6a − 5 b 3 3b 3− ( −5) 3b 1 b) (a + b) −3 = ( a + b) 3 3 c)    y

−5

2

5

y5 y5  y =  = 5 = 243 3 3 1

3⋅

1

d) 8 3 = (8 3 ) 2 = (2 3 ) 2 = 2 2 = 4 2 3

1 3

3⋅

1 3

e) 27 = (27 ) = (3 ) 2 = 3 2 = 9 5 3

2

3.

1 3 5

f) (−8) = (−2 ) = (−2) 5 = −32  13  4a g)  1  a2 

1

1 1 2  42 a6 2 2  = 1 = 1 −1 = 1  a4 a 4 6 a 12 

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Example 7 Solve the following equations: a) b)

c) 2 3a −3a = 81 2

−3 a

= 34

4 a−3 = 8

3a

91−a = 9 3

(2 2 ) a−3 = 2 3

a 2 − 3a = 4

1− a = 3

2a − 6 = 3

a 2 − 3a − 4 = 0

a = 1 − 3 = −2

2a = 9

(a − 4)(a + 1) = 0 a = −1

a=

d)

9 2

a=4

e) 4 −2 a

(2 ) 23 2 a = 2 −8 a−3 a = −8a − 3 9 a = −3 2a =

a=−

6 a .2 a

−

1 2

= 12a

2 1 − 3 2

= 12a

1 3

f) 2 a + 21− a = 3 2 2a + a − 3 = 0 2 let ,

x = 2a 2 x+ −3= 0 x 2 x − 3x + 2 = 0 ( x − 1)( x − 2) = 0 x =1 x=2 2 = 1 = 2 ,a = 0 a

2 3

0

2 a = 2 = 21 , a = 1

g)

4 2 x − 3(4 x ) + 2 = 0 (4 x ) 2 − 3(4 x ) + 2 = 0 let , u = 4 x u 2 − 3u + 2 = 0 (u − 2)(u − 1) = 0 u=2 u =1 1

1 2 x 0 4 =1= 4 , x = 0 4x = 2 = 42 , x =

6

1 6

Example 8 1

16 = 16 2 = 4

a)

1 4

b) 16 = 16 = 2 4

1

c) 3 1000 = 1000 3 = 10 d)

1 6

1 6

6

e) f)

6

g)

3

64  64  64 2 = =  = 1 729  729  3 729 6 5 10 = 50 = 25 × 2 = 25 2 = 5 2 2

1 2 6

1 1

3

2 6

1 3

16a b = [(4a b) ] = (4a b) = (4a b) = 3 4a 2 b 4

2

2

2

1

27 = [(33 ) 2 ] 3 = 3 6 = 3 2 = 3

Example 9 a) 5 3 + 6 3 = (5 + 6) 3 = 11 3 b) 23 ab 2 − 73 ab 2 = (2 − 7)3 ab 2 = −53 ab 2 2 ( 10 − 3) = 2 10 − 2 × 3 = 20 − 3 2 = 2 5 − 3 2

c)

d) ( 2 − 3)( 2 + 5) = 2 2 − 3 2 + 5 2 − 15 = 2 + 2 2 − 15 = 2 2 − 13 Example 10 Rationalizing Denominators 4 4 2 4 2 a) = ⋅ = =2 2 2 2 2 2 b)

3 5a

=

3 5a

⋅

5a 5a

=

3 5a 5a

Example 11 a) 5a + 1 + 1 = a

b) a + 25 − a = 1

5a + 1 = a − 1

a + 25 = a + 1

( 5a + 1) 2 = (a − 1) 2

(

5a + 1 = a 2 − 2a + 1

a + 25 = a + 2 a + 1

a − 7a = 0 a ( a − 7) = 0 a=0 a=7

2 a = 24

2

a + 25

a = 12 a = 144

7

) =( 2

)

a +1

2

Example 12 Change each logarithm form to an equivalent index form: a) b) c) 1 1 log 2 ( ) = −2 log 25 5 = log 2 8 = 3 4 2 3 1 1 8=2 = 2 −2 5 = 25 2 4

d) 49 = 7

e) 2

log 7 49 = 2

f) 1 = 5−1 5 1 log 5 ( ) = −1 5

3= 9 log 9 3 =

1 2

g) y = log b x

g) log 5 125 = 3

x = by

5 3 = 125

Example 13 Translate each logarithm statement into an equivalent exponential statement. a) log 100,000 = 5 b) log 0.001 = −3 1 c) log 2   = −3 8 Answer a)

b)

log 100,000 = 5

log 0.001 = −3

105 = 100,000

10 −3 = 0.001

(The base of the logarithm is understood to be 10)

Example 14 Use the change of base formula to evaluate: log10 7 0.845098 a) log3 7 = = = 1.7712 log10 3 0.477121 log10 0.03 − 1.522879 b) log 0.1 0.03 = = = 1.5299 log10 0.1 −1

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c) 1 log 2   = −3 8 1 2 −3 = 8

Example 15 Solve for: a) 2 log y x = log y 27 + 2 log y 2 − log y 3 3 2

= log y 27 3 + log y 2 2 − log y 3 = log y 9 + log y 4 − log y 3 9⋅4 = log y 12 3

= log y x = 12

b) 3 2 log y 4 − log y 8 + 2 log y 2 = log y x 2 3 3 2

2 3

log y 4 − log y 8 + log y 2 2 = log y x log y 8 − log y 4 + log y 4 = log y x log y 8 = log y x x =8 c) log 2 ( y 2 − 5 y + 2) = 3 y 2 − 5 y + 2 = 23 y2 − 5y − 6 = 0 ( y − 6)( y + 1) = 0 y=6 y = −1

d) 5 2 y = 13 log 5 2 y = log 13 2 y log 5 = log 13 (2 log 5) y = log 13 y=

log 13 = 0.7968 2 log 5

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e) 2 3 y −2 = 5

log 2 3 y − 2 = log 5 log 5 (3 y − 2) = log 2

Remember:

log 5 ≠ log 5 − log 2 log 2

1 log 5   y =  2 + 3 log 2  y = 1.4406 Example 16 Translate each exponential into an equivalent logarithmic statement. a) 101.8751 = 75 b) e3.2189 = 25 1 2

c) 16 = 4 Answer a)

b)

c)

10 = 75 log 75 = 1.8751

e = 25 ln 25 = 3.2189

16 2 = 4

1

1.8751

3.2189

log16 4 =

1 2

Example 17 Evaluate each of the following. a) log 100 b) log 0.001 1 c) log 3 27 1 d) ln  e Answer a)

b)

c)

d)

log 100

log 0.001

log 3

log10 10 2 = 2

log10 10 −3 = −3

1 27 log 3 3−3 = −3

1 ln  e ln e −1 = −1

Example 18 a) log 20 − log 4 b) log 2 + log 8 − log 2 c) 3 ln 2 + 2 ln 3

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d) 2 ln(a + 1) − ln(a + 2) Answer 20 = log 5 4 2(8) b) log 2 + log 8 − log 2 = log = log 8 2 c) 3 ln 2 + 2 ln 3 = ln 2 3 + ln 32 = ln(2 3.32 ) = ln 72

a) log 20 − log 4 = log

 (a + 1) 2 d) 2 ln(a + 1) − ln(a + 2) = ln(a + 1) 2 − ln(a + 2) = ln  a+2 Example 19 Solve these equations. a) log 5 (a + 2) + log 5 (a − 2) = 1 b) log 2 (log 2 a) = 1

  

2

c) 2 x −1 = 10 d) 5(e 2 x − 2) = 15 Answer a) log 5 (a + 2) + log 5 (a − 2) = 1

log 5 [(a + 2)(a − 2)] = 1 (a + 2)(a − 2) = 51 a2 − 4 = 5 a =9 a = ±3 2

c) 2 2 x −1 = 10 ln 2 x

2

−1

b) log 2 (log 2 a) = 1 21 = log 2 a log 2 a = 2 22 = a a=4 d)

= ln 10

( x 2 − 1) ln 2 = ln10 ln 10 x2 −1 = ln 2 2.3026 x2 −1 ≈ 0.6931 2 x − 1 ≈ 3.3219 x 2 = 4.3219 x = ± 4.3219 x ≈ ±2.0789

5(e 2 x − 2) = 15 e2x − 2 = 3 e2x = 5 ln e 2 x = ln 5 2 x = ln 5 ln 5 x= 2 x ≈ 0.805

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Example 20 Write each expression as a sum and/or a difference of logarithms with all variables to the first degree. a) log 5a 2 b 3 b) ln 6a 4 b 2 3p c) ln 5q Answer a) log 5a 2 b 3 = log 5 + log a 2 + log b 3 = log 5 + 2 log a + 3 log b b) 1 1 1 1 4 2 4 2 2 ln 6a b = ln(6a b ) = ln 6a 4 b 2 = (ln 6 + ln a 4 + ln b 2 ) = (ln 6 + 4 ln a + 2 ln b) 2 2 2 1 = ln 6 + 2 ln a + ln b 2

c) ln

3p = ln 3 p − ln 5q = (ln 3 + ln p ) − (ln 5 + ln q ) = ln 3 + ln p − ln 5 − ln q 5q

Example 21 Solve each logarithm equation. a) ln(3a + 1) − ln(5 + a ) = ln 2 b) ln(5a + 2) = ln 4 + ln(a + 3) c) log 3 (6a − 2) = 2 d) ln(a + 9) − ln a = 1 Answer a) ln(3a + 1) − ln(5 + a) = ln 2 3a + 1 ln = ln 2 5+a 3a + 1 =2 5+a 3a + 1 = 10 + 2a a=9

b) ln(5a + 2) = ln 4 + ln(a + 3) ln(5a + 2) = ln[4(a + 3)] 5a + 2 = 4(a + 3) 5a + 2 = 4a + 12 a = 10

c)

12

log 3 (6a − 2) = 2 6a − 2 = 3 2 6a − 2 = 9 6a = 11 a=

11 6

d) ln(a + 9) − ln a = 1 a +9 ln  =1  a  a+9 = e1 a a + 9 = ea a − ea = −9 (1 − e)a = −9 −9 a=− ≈ 5.2378 e −1

Example 22 Solve these exponential equations without using logarithms. a) 3 a −1 = 9 b) 16 −a + 2 = 8 c) 25 −3a = 3125 2 1 d) 2 a − 4 a = 16 Answer a)

b)

16

3 a −1 = 9 a −1

3 =3 a −1 = 2 a=3

2

c) −a+2

=8

25

d) −3 a

= 3125

1 16

2a

2

−4a

=

2

−4a

= 2 −4

(2 4 ) − a + 2 = 2 3

(5 2 ) −3a = 5 5

2a

2 4( − a + 2 ) = 2 3 − 4a + 8 = 3 5 a= 4

5 −6a = 55 − 6a = 5 5 a=− 6

a 2 − 4 a = −4

13

a 2 − 4a + 4 = 0 (a − 2) 2 = 0 a=2

Example 23 Use logarithms to solve these exponential equations. a) 2 a = 5 b) 4 3a −1 = 3 a − 2 c) e 2 a = 7 d) 10e 5 a −7 = 5 Answer a) 2a = 5 ln 2 a = ln 5 a ln 2 = ln 5 ln 5 ln 2 1.6094 a≈ 0.6931 a ≈ 2.3219 a=

c)

e 2a = 7 ln e 2 a = ln 7 2a = ln 7 a=

ln 7 ≈ 0.973 2

b)

4 3a −1 = 3a −2 ln 4 3a −1 = ln 3 a − 2 (3a − 1) ln 4 = (a − 2) ln 3 3a ln 4 − ln 4 = a ln 3 − 2 ln 3 3a ln 4 − a ln 3 = ln 4 − 2 ln 3 (3 ln 4 − ln 3)a = ln 4 − 2 ln 3 ln 4 − 2 ln 3 a= ≈ −0.265 3 ln 4 − ln 3 d) 10e 3a −7 = 5 5 e 3a −7 = 10 3a −7 e = 0.5 ln e 3a −7 = ln 0.5 3a − 7 = ln 0.5 3a = 7 + ln 0.5 7 + ln 0.5 a= ≈ 2.102 3

Example 24 a) log a 3 y = log a 3 + log a y y b) log a = log a y − log a 5 5 c) log a y 7 = 7 log a y ab log a = log a ab − log a pq = log a a + log a b − (log a p + log a q ) pq d) = log a a + log a b − log a p − log a q

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Example 25 Assume that log 6 7 ≈ 1.09 and log 6 5 ≈ 0.90. Use the properties of logarithm to find each of the following: a) log 6 35 = log 6 (7 ⋅ 5) = log 6 7 + log 6 5 ≈ 1.09 + 0.90 = 1.99 5 b) log 6 = log 6 5 − log 6 7 ≈ 0.90 − 1.09 = −0.19 7 c) log 6 53 = 3 log 6 5 ≈ 3(0.90) = 2.70 d) log 6 6 = 1 e) log 6 1 = 0

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References Barnett, R.A., Ziegler, M.R. and Byleen, K.E. (2005). Mathematics for Matriculation. Malaysia:McGraw-Hill. Dugopolski, M. (2006). Intermediate Algebra. McGraw –Hill. How, G.A. (2005). Q & A STPM Mathematics S&T. Longman. Lial, M.L., Hungerford, T.W. and Holcomb, J.P. (2007). Mathematics with Applications. Addison Wesley.

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