ba101 engineering mathematic chapter 2 standard form, index & logritm

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.



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



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.



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



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



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



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



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



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



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



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



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



Activity 1.2


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




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



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



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



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



Activity 1.4


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




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



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



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



Activity 1.5


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



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



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 )



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

ba101 engineering mathematic chapter 2 standard form, index & logritm

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