Chapter 1_FUNDAMENTAL CONCEPTS OF ALGEBRA

CHAPTER ONE
FUNDAMENTAL CONCEPTS OF ALGEBRA
Introduction
In this chapter, we will work with numbers. Therefore we need to know about this
number and understand its properties. After being introduced to the number systems,
we can look at the operation of the numbers such as exponents, surd and logarithm.
Algebra is about operations and the order of applying these operations in solving
problems in quadratic equations and partial fractions.
Objectives
After completing these tutorials, students should be able to:
 Express each of the given numbers as a quotient
a
.
b
 Write each of the given inequalities in interval notation and show them on the
real number line.
 Simplify the given index expressions.
 Solve the following index equations.
 Express in terms of the simplest possible surds.
 Rationalize the denominator and simplify.
 Express in logarithm form.
 Express in index form.
 Simplify the given logarithm expressions.
 Solve the given logarithm equations.
2
Question 1
Express each of the numbers as a quotient
(a) 1.666666666……
Solution:
Let x = 1.6666…
10 x = 16.66…
Equation (2)-(1)
9x =15
15
x
9
5
x
3
a
b
(1)
(2)
(b) 2.718181818……
Solution:
Let x = 2.71818… (1)
10 x = 27.1818… (2)
1000 x = 2718.18.. (3)
Equation (3)-(2)
990x = 2691
2691
x
990
299
x
110
Question 2
Write each of the following inequalities in interval notation and show them on the real
number line.
(a)
2 x6
Solution:
2
(b)
6
3  x  7
Solution:
-3
7
3
(c)
2  x  0
Solution:
-2
(d)
0
3  x  2
Solution:
-3
(e)
2
3 x
Solution:
3
(f)
x  1
Solution:
-1
Question 3
Show each of the following intervals on the real number line.
(a)
[2,3]
Solution:
-2
(b)
3
(4,4)
Solution:
-4
4
4
(c)
(,5]
Solution:
5
(d)
[1, )
Solution:
-1
(e)
(2, 3)  (0, 6)
Solution:
-2
(f)
6
[6, 2)  (3, 7)
Solution:
-3
2
Question 4
Evaluate
(a)
1
 
2
2
Solution:
2
1
2
2 1
 
 
 22
4
2
5
(b)
27
1
3
Solution:
27
(c)

1
3
1
 1 3
 
 27 

3

1
3
1
27
1
 16  2
 
 49 
Solution:
1
16
 16  2
 49   49
 
4

7
1
(d)
(2.56) 2
Solution:
( 2.56)
1
2
1
 64  2
 
 25 
1
 25  2
 
 64 
25
64


5
8
6
(e)
31  22  40
Solution:
1
31  22  40  .4.1
3
4

3
(f)
1
1
92  82
2
1
2
Solution:
1
2
9 8
2
1
2
1
2
1
2 2

3  .  2 
2

1
3 2
3 .2
1
2
3
2
1
22
 3.2
3 1

2 2
 3( 2)
6
Question 5
Simplify the following expressions:
(a)
3n  2  9n  27n
Solution:
3n  2  9n  27 n
 3n  2  32 n  33n
 3n  2 2 n 3n
 32
9
7
(b)
2
n
3
4  8  16
n
1
n
4
Solution:
2
n
1
4n  8 3 16 4
n
2
n
3
1
n
4 4
   2 
 2 2 n  23
 22 n  2 n  n
 2n
Question 6
Solve the following equations:
(a)
9 x  27
Solution:
9 x  27
32 x  33
2x  3
3
x
2
(b)
2 x 1 
1
64
Solution:
1
2 x 1 
64
1
2 x 1  6
2
x 1
2  2 6
x  1  6
x  7
8
(c)
4 x  2  1281 x
Solution:
4 x  2  1281 x
2 
2
x2
 
 27
1 x
22( x  2 )  27 (1 x )
2x  4  7  7x
9x  3
x
(d)

2
5
1
3
1
16
Solution:
2

1
x 5
16
x

2
 1 5 1
 x   24
 
2 x
4
2
5
2
lg x
5
20 lg 2
lg x 
2
lg x  10 lg 2
4 lg 2 
lg x  lg 210
x  210
x  1024
(e)
7 x  496  2 x  0
2
Solution:
2
7 x  496 2 x  0
7 x  496 2 x
2
 
7 x  72
2
6 2 x
x 2  12  4 x
x 2  4 x  12  0
( x  2)( x  6)  0
x  2,  6
9
(f)
(4 x ) x  4  8 x
Solution:
(4 x )x  4  8x
( 2 2 x ) x  2 2  23 x
22 x  223 x
2
2 x 2  3x  2  0
( 2 x  1)( x  2)  0
1
x   ,2
2
Question 7
Express in terms of the simplest possible surds:
(a)
50
Solution:
50  25  2
 25  2
5 2
(b)
200
Solution:
200  100  2
 100  2
 10 2
Question 8
Simplify
(a)
2 (3  2 2 )
Solution:
2 (3  2 2 )
3 2 4
10
(b)
( 3  2)( 3  1)
Solution:
( 3  2)( 3  1)
 33 3  2
 53 3
(c)
(3 3  2)(3 3  2)
Solution:
(3 3  2)(3 3  2)
 27  6 3  6 3  4
 23
Question 9
Rationalize the denominator and simplify
(a)
2
32
Solution:
2
32
(b)

2

2 32
32

32
16
32
32

32
5
2 5
Solution:
5
2 5


5
2 5

2 5
2 5
10  5 5
1
 10  5 5
11
(c)
2
2 3 3
Solution:
2
2 3 3


2

2 3 3
2 3 3 2 3 3
64 3
12  9
64 3
3
4
 2
3
3
1
1

2 1
2 1

(d)
Solution:
1
1

2 1
2 1


2 1 2 1
2 1
2 2
1
2 2
Question 10
Express in logarithm form
(a)
53  125
Solution:
53  125
lg 5 125  3
12
(b)
1331  (121)
3
2
Solution:
3
1331  (121) 2
3
lg121 1331 
2
(c)
e y  33
Solution:
e y  33
ln 33  y
Question 11
Express in index form
(a)
log 3 27  3
Solution:
log 3 27  3
27  33
(b)
log x y  2
Solution:
log x y  2
y  x2
(c)
ln b  3
Solution:
ln b  3
b  e 3
13
Question 12
Evaluate
(a)
1
log 9 9 2
Solution:
1
log 9 9 2
1
log 9 9
2
1

2

(b)
log 1 4
2
Solution:
log 1 4
2
1
 log 1  
4
2 
1
2
1
 log 1  
2
2 
1
 2 log 1
2
2
 2
Question 13
Express in terms of log a, log b and log c
(a)
log abc
Solution:
log abc
 log a  log b  log c
14
(b)
log
a
b 2c
Solution:
a
log 2
bc
 log a  log b 2  log c
 log a  2 log b  log c
(c)
log
a
b
Solution:
a
log
b
1
 a 2
 log  
b
1
a
 log  
2
b
1
1
 log a  log b
2
2
(d)
 a3 
ln  
b
Solution:
 a3 
ln  
 b 
 ln a 3  ln b
 3 ln a  ln b
15
Question 14
Simplify
(a)
3 ln( x  7)  ln x
Solution:
3 ln( x  7)  ln x
 ln( x  7)3  ln x
 ( x  7)3 
 ln 

 x 
(b)
log 2  log 6  log 4
Solution:
log 2  log 6  log 4
(c)
 2 6 
 log 

 4 
 log 3
1
log 25  2 log 3  2 log 6
2
Solution:
1
log 25  2 log 3  2 log 6
2
1
 log 25 2  log 32  log 62
 log 5  log 9  log 36
 5  36 
 log 

 9 
 log 20
(d)
log 9
log 3
Solution:
log 9
log 3

log 32
log 3

2 log 3
log 3
2
16
Question 15
Solve each equation
(a)
log 4 ( x 2  9)  log 4 ( x  3)  3
Solution:
log 4 ( x 2  9)  log 4 ( x  3)  3
 x2  9 
log 4 
  3 log 4 4
 x3 
 x2  9 
3
log 4 
  log 4 4
 x3 
x2  9
 43
x3
x 2  9  64( x  3)
x 2  64 x  201  0
( x  67 )( x  3)  0
x  67, 3
(b)
log 2 ( x 2  1)  log 4 x 2  1
Solution:
log 2 ( x 2  1)  log 4 x 2  1
log 2 ( x 2  1) 
log 2 x 2
1
log 2 4
log 2 x 2
1
2
2 log 2 ( x 2  1)  log 2 x 2  2
log 2 ( x 2  1) 
 ( x 2  1)2 
log 2 
2
2
 x

2
2
( x  1)
4
x2
( x 2  1)2  4 x 2
x4  2x2  1  4x2
x4  2x2  1  0
( x 2  1)( x 2  1)  0
x  1,  1