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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
CHAPTER 2: STANDARD FORM, INDEX AND LOGARITHMS
1.0
INTRODUCTION
Understanding index and logarithms is important because it is used widely in research,
science, finance and engineering. For example, it can be used to show how substances are
formed, multiply and decay in the natural world.
Index Expressions
Figure 1.1 shows our planet Earth orbiting the Sun. The distance between the Earth and the
Sun is about 93 million kilometers. You can write this number as 93,000,000 km.
Figure 1.1: Distance between Earth and Sun
This number is obviously long to write and hard to read. You can also write this number as
9.3 x 107km. Numbers written this way is called indices. Usually only numbers that are too
small and too big are stated in index form. Referring to the number 9.3 x 10 7, the number 10
is called the base and the number 7 is called the index.
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
1012 = 10 x 10 …………10 .
12 times
base
index
In general, an means a multiplied by itself n times.
An index expression is in simple form if there is:
 No repeating base
 No negative index
For example
x2y5 x 4 , a2 b –6 and
p3
2
can be simplified as
x6y5,
2
a2
p9
6 and
b
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Activity 1.0
TRY THESE QUESTIONS!
1. Rewrite these numbers in index form:
i.
ii.
a. 2000000
c. 0.00082
b. 138000000000
d. 0.000000015
2. The rate of reproduction of a particular insect is about 1430000000 a month. Write this in
index.
3. First, express 400 kilometers in centimeters. Next, state this in index.
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BA101/CHAPTER2
FEEDBACK for Activity 1.0
1.
a. 2 x 106
a. c. 8.2 x 10-4
2.
1.43 x 109
3.
b. 1.38 x 1011
d. 1.5 x 10-8
40 000 000 cm, 4 x 107 cm
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STANDARD FORM, INDEX AND LOGARITHMS
1.1
BA101/CHAPTER2
INDEX RULE
There will be times when you need to add, subtract, multiply or divide two or more index
numbers. The rules provided in Table 1.1 are most useful. Study it carefully.
Given that a, b, m and n are real numbers.
Rule
Statements
1. Multiplication
am x an = a m + n
2. Division
am  an = a m - n
3. Power
i. ( a m ) n = a mn
ii. (ab)n = an b n
n
an
a
iii.    n ; b  0
b
b
4. Negative Index
i. a –n =
a m b n
; a  0 and b  0

b n a m
ii.
5. Zero Index
6. Fraction Index
1
;a0
an
a0 = 1
; a0
m
a n  n am
Table 1.1: Rules of Index Operations
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Example 1.1
Simplify the following expressions.
a.
34 x 35
c.
(48 )3
e.
b.
85  82
d.
( 25 ) -3
70
f.
27
2
3
SOLUTION:
a.
b.
c.
d.
e.
f.
Expressions
34 x 35
85  82
( 48 )3
(
25
)
-3
a0 = 1
70
27
Index Rule Used
am x an = a m + n
am  an = a m - n
i. ( a m ) n = a mn
i.
( a m ) n = a mn
1
ii. a –n = n ; a  0
a
2
3
a
m
n
 a
n
Solutions
34 + 5 = 3 9
85 – 2 = 8 3
424
( 2 5 ) -3 = 2 -15
1
2 –15 = 15
2
; a0
1
2
3
27  3 27 2  9
m
Example 1.2
A sum of RM10,000 is saved in a bank at 8% interest compounded monthly. The total sum J
 0.08 
after t years is given as, J = 100001 

12 

a.
6
6 months
12t
. What is the total sum after
b.
5 years
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
SOLUTION
a.
6 months = 0.5 year. Therefore t = 0.5.
 0.08 
J = 100001 

12 

b.
6
t = 5.
 0.08 

J = 100001 
12 

7
= 10406.73
12( 5 )
=14898.46
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Activity 1.1
TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION……..!
1
Simplify
a.
c.
2.
a5 x a 6
z 2 y5 x y –2
Simplify
a. m 12  m 3
c. z 7 x z 6  z 5
3
Simplify
a. ( x 3 ) 5
c. ( 2x2 y 3 z )5
e. 3 ( ab 2 ) –4
8
b.
d.
b.
3 x 3 8 x 3-4
4n x 16 2n x 32 –2n x 8 -n
2 8  24
d. 25 n  5 –2n x 125 2n
b. ( 3x 4) 2
d. ( 10 3 ) 4
f. ( 2m 2 )  ( 4n )3
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BA101/CHAPTER2
Feedback for Activity 1.1
4.
5.
6.
a. a11
c. z2 y 3
b. 35
a. m 9
c. z 8
b. 24
a. x 15
c. 32 x 10 y 15 z 5
b. 9x 8
3
e.
4 8
a b
9
d. 2 –3n
d. 5 10n
d. 1012
m2
f.
32n 3
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STANDARD FORM, INDEX AND LOGARITHMS
1.2
BA101/CHAPTER2
LOGARITHMIC EXPRESSIONS
In this section, you will learn about the relationship between indices and logarithms. Let’s
now consider the reproduction rate of amoeba, a single cell living organism that reproduces
by replicating itself. If one new amoeba needs one day to replicate itself into 2 amoebas,
there will be 4 amoebas after 2 days, 8 amoebas after 3 days, and so on.
Time in days(x)
Number of amoeba (y)
0
1
1
2
2
4
3
8
4
16
5
32
6
64
7
128
The index equation y = 2x can be used to represent the rate of reproduction of this amoeba.
Conversely, this equation can also be written as x = log2 y , a logarithmic equation.
log2 y is read as log of
y to the base 2
log is a short form of logarithm
10
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STANDARD FORM, INDEX AND LOGARITHMS
Index Form
BA101/CHAPTER2
Logarithmic Form
index
y = ax

x = log a y
base
Note:
Observe that the base is still the same after changing index form to logarithmic form.
Generally, if a is a positive number and y = a x, then x is equals to the logarithm of y to
the base a.
If y = a x , then x = log a y
If x = log a y , then y = a x
Equation 1.1
Example 1.3
Rewrite the following numbers in logarithmic form, (base given).
a. 100 base10
b. 64 base 4
c. 64 base 2
d. 125 base 5
e. 81 base 3
Sample solutions
a. If 100 = 10 2, then 2 = log 10 100
b. If 64 = 4 3, then 3 = log 4 64
c. If 64 = 2 6, then 6 = ?
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
d. If 125 = 5 3, then ? = log 5 125
e. If 81 = 3 4, then 4 = log ? 81
Example 1.4
Determine these values:
a. log 7 49
b. log 2 0.5
c. log 9 3
d. log 10 0.001
Solution:
a. Assuming
then
therefore
b. Assuming
then
therefore
c. Assuming
then
therefore
d. Assuming
then
therefore
12
log 7 49 = x
49 = 7x
72 = 7 x
x = 2
log 2 0.5 = x
0.5 = 2 x
0.5 = ½ = 2 – 1 = 2 x
x = -1
log 9 3 = x
3 = 9x
3 = ( 3 2 )x
31 = 32x
1 = 2x
½ = x or x = 0.5
log 10 0.001 = x
0.001 = 10 x
1
1
 3  10 3 10 x
1000 10
x = -3
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Activity 1.2
TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION……..!
1. Given that 32 = 25, determine the value of log 2 32
2. Given that 1/8 = 2 –3 , determine the value of log 2 1/8
3. Given that 8 = 64 , determine the value of log 64 8
4. Given that 0.001 = 10 –3 , determine the value of log 10 0.001
5. Calculate the value of
a.
log 3 81
c. log 8 4
b. log 7 343
d. log 27 9
6. Convert the following into logarithmic form
a. 32 = 25
b. 50 = 10 1.699
c. a = x2
d. x-3 = 0.3
7. Find the value of x given that
13
a. log 3 1 = x
b. log 7 49 = x
c. log 10 0.001 = x
d. log 5 25 = x
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Feedback for Activity 1.2
1. 5
2. –3
3. ½
4. –3
5. a. x = 4
c. x = 2/3
6. a. log 2 32 = 5
c. log x a = 2
7. a. x = 0
c. x = -3
14
b. 3
d. 2/3
b. log 10 50 = 1.699
d. log x 0.3 = -3
b. x = 2
d. x = 2
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STANDARD FORM, INDEX AND LOGARITHMS
1.3
BA101/CHAPTER2
LOGARITHM RULE
As in index, you will need to perform basic algebra operations on numbers in
logarithm. Table 1.2 shows the logarithm rule.
Assume M and N are positive real numbers
1.
2.
3.
4.
5.
log a MN = log a M + log a N
log a M/ N = log a M – log a N
log a (M) c = c log a M
log a a = 1
log a a0 = 0 log a a = 0
6. log N M =
log a M
log a N
( to convert base N to base a)
7. a log a N = N
8. log101 = log a (for base 10 only)
Table 1.2: Rules of Logarithm Operations
Example 1.5
Write the following expressions as addition or subtraction of logarithms.
a. log a x 2 y 3
c. log
1
100b 2
b. log a 3 b 3/2
c. log (
ab 3
)
c
Solution:
a. log a x 2 y 3 = log a x2 + log a y 3
= 2 log a x + 3 log a y
RULE 1 and 3
b. log a 3 b 3/2 = log a 3 + log b 3/2
3
= 3 log a + log b
2
15
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STANDARD FORM, INDEX AND LOGARITHMS
c. log
BA101/CHAPTER2
1
= log 1 – ( log 100 + log b2 )
2
100b
= 0 – log 100 – 2 log b
= - ( log 100 + 2 log b )
= -2 (1 + log b )
1/ 2
 ab 3 
ab 3

 = ½ ( log ab3 – log c )
log
d. log (
)=
c
c


3
= ½ log ab – ½ log c
= ½ log a + ½ log b3 – ½ log c
3
= ½ log a + log b – ½ log c
2
Example 1.6
Rewrite the expressions below as a single logarithm
a. log 2 + log 3
b. log 4 + 2 log 3 – log 6
c. 2 log x + 3 log y – log z
d. ½ log a + log a2 b – 2 log ab
Solution:
a. log 2 + log 3 = log ( 2 x 3 ) = log 6
b. log 4 + 2 log 3 – log 6 = log 4 + log 32 – log 6
= log ( 4 x 9 ) – log 6
36
= log
 log 6
6
c. 2 log x + 3 log y – log z = log x2 + log y 3 – log z
 x2 y3 

= log
 z 
d. ½ log a + log a2 b – 2 log ab = log a 1/2 + log a 2 b – log ( ab )2
log
16
a
a1 / 2 a 2 b
= log
2
b
(ab)
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Example 1.7
Given log 2 = 0.3010 and log 3 = 0.4771 , find the value of:
a. log 8
b. log 18
c. log 0.6
Solution:
a. log 8
= log 2 3
= 3 log 2
= 3 ( 0.3010 )
= 0. 903
b. log 18
= log (2 x 9 )
= log 2 + log 9
= log 2 + log 3 2
= log 2 + 2 log 3
= 0.3010 + 2 ( 0.4771)
= 1.2552
c. log 0.6
= log (
6
)
10
= log 6 – log 10
= log ( 2 x 3 ) – 1
= log 2 + log 3 – 1
= 0.3010 + 0.4771 – 1
= - 0 . 2219
17
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Activity 1.4
TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION……..!
1. Express the following as a combination of log x, log y or log z
a. log x 3 y 2
b. log
c. log xy
x3
y2
d. log
x2 y
3
z
2. Express the following as single logarithm
a. log5 14 – log5 21 + log5 6
1
c. 4 log 3 2  log 3 25
2
3
log 7 9  2 log 7 6
2
2
8
d. 2 log 5  log 5
3
9
b.
3. Determine the values of
a. log4 9 – log436
b. 2 log 2 5 – log2 100 + 3 log2 4
4. Given that log a 3 = 0.477 and log a 5 = 0.699, find the value of
a. log a 15
c.
18
log a 45
log a 9a
b. log a 35
d. loga 0.6
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Feedback for Activity 1.4
1. a. 3 log x + 2 log y
b. 3 log x – 2 log y
c. ½ log x + ½ log y
d. 2 log x + log y –
3
log z
2
2. a. log5 4
b. log7 ¾
c. log3 80
d. log5 ½
3. a. –1
b.
4
4. a. 1.176
b. 0.8265
c. 0.846
d. –0.222
19
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STANDARD FORM, INDEX AND LOGARITHMS
1.4
BA101/CHAPTER2
INDEX AND LOGARITHMIC EQUATIONS
Congratulations! You have already understood and know how to use Index Rule and
Logarithm Rule to solve simple problems involving Index and Logarithm expressions. Now we
move on to solving simple equations involving Index and Logarithm. The following rule is very
important.
Assuming that x and y are real numbers
 If
 If
ax = ay , then x = y
log a x = log a y , then x = y
Example 1.8
Solve the following equations
a. 7x = 12
b. 3 5x – 8 = 9 x + 2
Solution:
a.
7x = 12
Log both sides,
x log 7 = log 12
log 7 0.845
x

 0.783
log 12 1.079
b.
3 5x – 8 = 9 x + 2
3 5x – 8 = (3 2 ) x + 2
Similar base
3 5x – 8 = 3 2 x + 4
( am )n = amn
5x – 8 = 2x + 4
If ax = ay , then x = y
5x – 2x = 4 + 8
3x = 12
x=4
20
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Example 1.9
Solve the following logarithmic equations
a. log 2 ( 7 + x ) = 3
b. log 50 + log x = 2 + log ( x – 1 )
c. log 2 x = log x 16
Solution:
a. log 2 (7 + x ) = 3
(7+x )= 23
7+x=8
x=1
b. log 50 + log x = 2 + log ( x – 1 )
log 50x = log 100 + log (x – 1 )
log 50x = log 100(x – 1 )
50x = 100x – 100
100 = 50 x
x=2
c. log 2 x = log x 16
log 2 16
log 2 x 
log 2 x
Convert to index form
log = log base 10
log a + log b = log ab
2 = log 10 100
Convert to similar base
(log 2 x )2 = log 2 16
(log 2 x )2 = 4
(log 2 x ) =  2
log 2 x = 2 or log 2 x = -2
x = 2 2 or x = 2 – 2
x = 4 or ¼
21
Convert to index form
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
Activity 1.5
TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION……..!
1. Solve these equations
2
a. 32x = 8
b. 2x – 3 = 4x + 1
c. 3 4x = 27 x – 3
d. 5 x  2562 x
2
Solve these equations
a. log 6 x + log 6 ( x + 5 ) = 2
b. 5 log x6 - log x 96 = 4
c. 2 log 3 + log 2x = log ( 3x + 1 )
d. log 25 – log x + log ( x + 1 ) = 3
22
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
FEEDBACK for Activity 1.5
1
a. 32x = 8
(25)x = 23
5x = 3
3
x =
5
b. x = -5
c. x = -9
d. x = –6 or 2
2.
23
a. x = 4 or -9
b. x = 3
1
c. x =
15
1
d. x =
39
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
SELF ASSESSMENT 1
Another round of congratulations to you for making it so far. You are very close to mastering this unit.
Attempt all questions in this section and check your solutions with the answers provided in
SOLUTIONS TO SELF ASSESSMENT given after this.
If you face any problems you cannot solve, please discuss with your lecturer.
1.
Solve the following equations:
a. 10 x = 0.00001
c. log 2 4x = 8 log x 2
2.
b. 3 x 9 x – 1 = 27 2x –1
d. 5 log x 6 – log x 96 = 4
Simplify the expressions below:
a. 3 log 2
5
10
1
 2 log 2
 log 2
3
9
30
b. log 3 5  log 5 2  log 2 3
3.
Given that log 10 5 = p, express the following in term of p.
a. log 10 250
c. log 5 10
4.
b. log 10 0.5
d. log 5 1000
Solve the following equations:
a. 3 log 2 + log ( 4x – 1 ) = log ( 7 – 8x )
b. 2 log 15 + log ( 5 – x ) – log 4x = 2
c. log 5 x 2 = 1 + log 5 ( 25 – 4x )
d. log 2 y 2 = 3 + log 2 ( y + 6 )
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STANDARD FORM, INDEX AND LOGARITHMS
BA101/CHAPTER2
SOLUTIONS TO SELF ASSESSMENT
1.
a. x = -5
1
c. x =
or 4
16
b. x = 1/3
d. x = 3]
e. 3 log 2 + log ( 4x – 1 ) – log ( 7 – 8x ) = 0
2.
a. -3
b. –0.954
3.
a. 2p + 1
b. p – 1
c. 1/p
d. 3/p
4.
a. x =
3
5
9
5
c. x = 5 or -25
d. y = 12 or -4
b. x = 
25
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