ba101 engineering mathematic chapter 2 standard form, index & logritm
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
1 Prepared By : Azmanira Muhamed
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 107, 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
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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 and23p
can be simplified as
x6y5, 6
2
b
a and
9p
1012
= 10 x 10 10 .
12 times
base
index
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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Activity 1.0
TRY THESE QUESTIONS!
1. Rewrite these numbers in index form:
i. a. 2000000 b. 138000000000
ii. c. 0.00082 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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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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FEEDBACK for Activity 1.0
1. a. 2 x 106 b. 1.38 x 1011
a. c. 8.2 x 10-4 d. 1.5 x 10-8
2. 1.43 x 109
3. 40 000 000 cm, 4 x 107 cm
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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1.1 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
iii. n
nn
b
a
b
a
; b 0
4. Negative Index i. a n =
na
1 ; a 0
ii. m
n
n
m
a
b
b
a
; a 0 and b 0
5. Zero Index
a0 = 1 ; a 0
6. Fraction Index
n mn
m
aa
Table 1.1: Rules of Index Operations
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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Example 1.1
Simplify the following expressions.
a. 34 x 35 b. 85 82
c. (48 )3 d. ( 25 ) -3
e. 70 f. 32
27
SOLUTION:
Expressions Index Rule Used Solutions
a. 34 x 35 am x an = a m + n 34 + 5 = 3 9
b. 85 82 am an = a m - n 85 2 = 8 3
c. ( 48 )3 i. ( a m ) n = a mn 424
d. ( 25 ) -3
i. ( a m ) n = a mn
ii. a n = na
1 ; a 0
( 2 5 ) -3 = 2 -15
2 15 = 152
1
e. 70 a0 = 1 ; a 0 1
f. 3
2
27 n mn
m
aa 927273 23
2
Example 1.2
A sum of RM10,000 is saved in a bank at 8% interest compounded monthly. The total sum J
after t years is given as, J =
t12
12
08.0110000
. What is the total sum after
a. 6 months b. 5 years
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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SOLUTION
a. 6 months = 0.5 year. Therefore t = 0.5.
J =
6
12
08.0110000
= 10406.73
b. t = 5.
J =
)5(12
12
08.0110000
=14898.46
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
8 Prepared By : Azmanira Muhamed
Activity 1.1
TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION..!
1 Simplify
a. a5 x a 6 b. 3 x 3 8 x 3-4
c. z 2 y5 x y 2 d. 4n x 16 2n x 32 2n x 8 -n
2. Simplify
a. m 12 m 3 b. 2 8 2 4
c. z 7 x z 6 z 5 d. 25 n 5 2n x 125 2n
3 Simplify
a. ( x 3 ) 5 b. ( 3x 4) 2
c. ( 2x2 y 3 z )5 d. ( 10 3 ) 4
e. 3 ( ab 2 ) 4 f. ( 2m 2 ) ( 4n )3
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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Feedback for Activity 1.1
4. a. a11 b. 35
c. z2 y 3 d. 2 3n
5. a. m 9 b. 24
c. z 8 d. 5 10n
6. a. x 15 b. 9x 8
c. 32 x 10 y 15 z 5 d. 1012
e. 843
ba f. 3
2
32n
m
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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1.2 LOGARITHMIC EXPRESSIONS
In this section, you will learn about the relationship between indices and logarithms. Lets
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) 0 1 2 3 4 5 6 7
Number of amoeba (y) 1 2 4 8 16 32 64 128
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
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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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.
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 = ?
Index Form Logarithmic Form
index
y = ax x = log a y
base
If y = a x , then x = log a y
If x = log a y , then y = a x
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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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 log 7 49 = x
then 49 = 7x
72 = 7x
therefore x = 2
b. Assuming log 2 0.5 = x
then 0.5 = 2 x
0.5 = = 2 1 = 2 x
therefore x = -1
c. Assuming log 9 3 = x
then 3 = 9 x
3 = ( 3 2 )x
31 = 32x
therefore 1 = 2x
= x or x = 0.5
d. Assuming log 10 0.001 = x
then 0.001 = 10 x
x1010
10
1
1000
1 33
therefore x = -3
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
13 Prepared By : Azmanira Muhamed
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 b. log 7 343
c. log 8 4 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
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
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Feedback for Activity 1.2
1. 5
2. 3
3.
4. 3
5. a. x = 4 b. 3
c. x = 2/3 d. 2/3
6. a. log 2 32 = 5 b. log 10 50 = 1.699
c. log x a = 2 d. log x 0.3 = -3
7. a. x = 0 b. x = 2
c. x = -3 d. x = 2
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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1.3 LOGARITHM RULE
As in index, you will need to perform basic algebra operations on numbers in
logarithm. Table 1.2 shows the logarithm rule.
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 b. log a 3 b 3/2
c. 2100
1log
b c. )(log
3
c
ab
Solution:
a. log a x 2 y 3 = log a x2 + log a y 3
= 2 log a x + 3 log a y
b. log a 3 b 3/2 = log a 3 + log b 3/2
= 3 log a + 2
3log b
Assume M and N are positive real numbers
1. log a MN = log a M + log a N
2. log a M/ N = log a M log a N
3. log a (M) c = c log a M
4. log a a = 1
5. log a a0 = 0 log a a = 0
6. log N M = N
M
a
a
log
log ( to convert base N to base a)
7. a log a N = N
8. log101 = log a (for base 10 only)
RULE 1 and 3
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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c. 2100
1log
b = log 1 ( log 100 + log b2 )
= 0 log 100 2 log b
= - ( log 100 + 2 log b )
= -2 (1 + log b )
d. )(log3
c
ab=
2/13
log
c
ab= ( log ab3 log c )
= log ab3 log c
= log a + log b3 log c
= log a + 2
3log b log c
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. 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
= 6log6
36log
c. 2 log x + 3 log y log z = log x2 + log y 3 log z
=
z
yx 32log
d. log a + log a2 b 2 log ab = log a 1/2 + log a 2 b log ( ab )2
2
22/1
)(log
ab
baa=
b
alog
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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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 (10
6) = log 6 log 10
= log ( 2 x 3 ) 1
= log 2 + log 3 1
= 0.3010 + 0.4771 1
= - 0 . 2219
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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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. 2
3
logy
x
c. xylog d. 3
2
logz
yx
2. Express the following as single logarithm
a. log5 14 log5 21 + log5 6 b. 6log29log2
377
c. 25log2
12log4 33 d.
9
8log
3
2log2 55
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 b. log a 35
c. aa
a
9log
45log d. loga 0.6
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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Feedback for Activity 1.4
1. a. 3 log x + 2 log y
b. 3 log x 2 log y
c. log x + log y
d. 2 log x + log y 2
3 log z
2. a. log5 4
b. log7
c. log3 80
d. log5
3. a. 1
b. 4
4. a. 1.176
b. 0.8265
c. 0.846
d. 0.222
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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1.4 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.
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
783.0
079.1
845.0
12log
7logx
b. 3 5x 8 = 9 x + 2
3 5x 8 = (3 2 ) x + 2
3 5x 8 = 3 2 x + 4
5x 8 = 2x + 4
5x 2x = 4 + 8
3x = 12
x = 4
Assuming that x and y are real numbers
If ax = ay , then x = y If log a x = log a y , then x = y
Similar base
( am
)n = a
mn
If ax = a
y , then x = y
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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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 ) = 2 3
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
x
x2
22
log
16loglog
(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
Convert to index form
log = log base 10
log a + log b = log ab
2 = log 10 100
Convert to similar base
Convert to index form
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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Activity 1.5
TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION..!
1. Solve these equations
a. 32x = 8 b. 2x 3 = 4x + 1
c. 3 4x = 27 x 3 d. xx 26255
2
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
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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FEEDBACK for Activity 1.5
1 a. 32x = 8
(25)x = 23
5x = 3
x =5
3
b. x = -5
c. x = -9
d. x = 6 or 2
2. a. x = 4 or -9
b. x = 3
c. x = 15
1
d. x = 39
1
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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2
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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 b. 3 x 9 x 1 = 27 2x 1
c. log 2 4x = 8 log x 2 d. 5 log x 6 log x 96 = 4
2. Simplify the expressions below:
a. 30
1log
9
10log2
3
5log3 222
b. 3log2log5log 253
3. Given that log 10 5 = p, express the following in term of p.
a. log 10 250 b. log 10 0.5
c. log 5 10 d. log 5 1000
4. 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
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SOLUTIONS TO SELF ASSESSMENT
1. a. x = -5 b. x = 1/3
c. x = 16
1or 4 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 = 5
3
b. x = 5
9
c. x = 5 or -25
d. y = 12 or -4
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