1.0 Standard Form
State positive numbers in standard form
1. Standard form is a way of writing a number in the form A×10n, where 1≤ A<10 and the index n is an integer.
2. A number in standard form consists of two parts
A×10n
3. A positive value of n shows a big number (¿1 ) whereas a negative value of n shows a small number (¿1 ).
4. Another term for standard form is scientific notation.
Example 1
State each of the following numbers in standard form.
1. 74002. 23 810 000
Solution:
1. 7400=7.4×1000
¿7.4×103
2. 23 810 000
¿2.381×10 000 000
¿2.381×107
A number between 1 and 10 inclusive 1
A power of 10
Example:
1. State each of the following numbers in standard form.I. 0.036
II. 0.00000452
Solution:
I. Move the decimal point of a given number to the right in order to produce a number A such that 1≤ A<10
II. The number of places that the decimal point has moved is the negative value of n
a. 0.036=3.6×10−2
b. 0.0 0 00 04 52
Note: move to the right , index n is negative
Perform operations of addition, subtraction, multiplication and division, involving any numbers and state the answers in standard form
Example:
Calculate the value of each of the following by stating its answer in standard form.
a. 3.1×105+4×104
b. 7×105−1.8×10−8
c. (2.4×103 )× (9×107 )
d.4.8×1012
6×10−5
Solution:
a. 3.1×105+4×104
¿3.1×105+4×10−1×105
¿3.1×105+0.4×105
¿ (3.1+0.4 )×105
¿3.5×105
104=10−1×105
4×10−4=0.4
b. 7×105−1.8×10−8
¿7×10−6−1.8×10−2×10−6
¿7×10−6−0.018×10−6
¿ (7−0.018 )×10−6
¿6.982×10−6
c. (2.4×103 )× (9×107 )¿(2.4×9)×(103×107)¿21.6×1010
¿2.16×101×1010
¿2.16×1011
d.4.8×1012
6×10−5
¿ 4.86×
1012
10−5
¿0.8×1017
¿8×10−1×1017
¿8×1016
2. Laws of Indices
a. Below are the basic laws of indices
Laws Statement
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. ( ab )
n
=an
bn ; b 0
4. Negative Index
i. a –n =
1
an ; a 0
ii.
a−m
b−n=bn
am ; a 0 dan b 0
5. Nil Index a0 = 1 ; a 0
6. Fraction Indexamn=
n√am
Example 1
Find the values of the following
823×16
(−1 14 )÷32
25
Solution:
¿ (23 )23× (24 )
−54 ÷ ( 25 )
25
¿22×2−5÷22
¿22+(−5 )−2
¿2−5
Change the numbers to the same base
Use the laws of indices to simplify the multiplication and division
a−n= 1
an
¿ 1
25= 1
32
B. Use laws of indices to simplify algebraic expressionsThe laws of indices can be used to simplify algebraic expressions.
Example 2
Simplify each of the following:
a. (b4 c−6d2 )12
b. 4n+1×2n÷1634n
c.75p+2
33−p×152 p+1×52
Solution:
a. (b4 c−6d2 )12=(b4 )
12× (c−6 )
12× (d2)
12
¿b2×c−3×d
¿ b2dc3
b. 4n+1×2n÷1634n
¿ (22 )n+1×2n÷ (24 )
34n
¿22n+2×2n÷ (24 )34n
¿22n+2+n−3n
¿22
¿4
c.75p+2
33−p×152 p+1×52
¿ 75p+2
33−p× (3×5 )2 p+1×52
¿ 3p+2×52 (p+2)
33−p×32 p+1×52 p+1×52
¿3p+2−(3− p )−( 2p+1)×52 (p+2)−(2 p+1)−2
¿3−2×51
¿ 59
Use 2 as the same base throughout
Add and subtract the indices using the laws of indices
3 LOGARITHMS AND LAWS OF LOGARITHMS
A. Change equations in index forms to logarithm forms and vice versa1. In the equation ¿ax , the number x is called an index.2. This equation can be written as
N=ax
log aN=x
Then, x is the logarithm of N to the base a
3. In general ,N=ax log aN=x ,
Where a>0 , a≠1
A. Find logarithms of numbers1. A logarithm to the base 10 is called a common logarithm and is written as
log 10. In calculator, it is normally marked as log.2. A scientific calculator can be used to find the logarithm or antilogarithm of
a number.
B. Laws of logarithm
Let M and N is the real positive numbers
1. log a MN = log a M + log a N2. log a M/ N = log a M – log a N
3. log a (M) c = c log a M4. log aa = 15. log a a0 = 0 log aa = 0
6. log N M =
log aM
log aN ( Penukaran daripada asas N kepada asas a)
Example 2
Express the following expressions using the laws of logarithms.
a. log a x 2 y 3 b. log a 3 b 3/2
c. lg
1
100b2c.
lg √( ab3
c)
Solution:
Using the laws of logarithms,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 + 3/2 log b
c. log
1
100b2 = log 1 – ( log 100 + log b2 )
= 0 – log 100 – 2 log b = - ( log 100 + 2 log b )
d. log √( ab
3
c)=
log( ab3c )1/2
= ½ ( log ab3 – log c )= ½ log ab3 – ½ log c )= ½ log a + ½ log b3 – ½ log c= ½ log a + 3/2 log b – ½ log c
Example 4:
Given that log 10 2 = 0.3010 dan log 10 3 = 0.4771 , find the value of
a. log 10 8 b. log 10 18 c. log 10 0.6
Solution:
a. log 10 8 = log 10 2 3 = 3 log 10 2 = 3 ( 0.3010 )
= 0. 9010b. log 10 18 = log 10 (2 x 9 ) = log 10 2 + log 10 9
Let M and N is the real positive numbers
1. log a MN = log a M + log a N2. log a M/ N = log a M – log a N
3. log a (M) c = c log a M4. log aa = 15. log a a0 = 0 log aa = 0
6. log N M =
log aM
log aN ( Penukaran daripada asas N kepada asas a)
Laws of addition
= log 10 2 + log103 2
= log 10 2 + 2 log 103 = 0.3010 + 2 ( 0.4771)
= 1.2552
c. log 10 0.6 = log 10 (
610 ) = log 10 6 – log 10 10
= log 10 ( 2 x 3 ) – 1 = log 10 2 + log 10 3 – 1 = 0.3010 + 0.4771 – 1 = - 0 . 2219
C. Equations involving indices and logarithms
Let x and y are the real positive numbers.
If ax = ay then x = y If log a x = log a y , then x = y
Example 5
Solve the following equations: a. 7x = 12 b. 3 5x – 8 = 9 x + 2
Solution:
a. 7x = 12Take the log for the both side of equation:
x log 7 = log 12
x=log12log7
=1. 0790 .845
=1 .277
b. 3 5x – 8 = 9 x + 2
3 5x – 8 = (3 2 ) x + 2
3 5x – 8 = 3 2x + 4
5x – 8 = 2x + 45x – 2x = 4 + 83x = 12 x = 4
Same base
( am )n = amn
If ax = ay then x = y
Example 6:
Solve the following equations for the x: a. log 2 ( 7 + x ) = 3 b. log 50 + log x = 2 + log ( x – 1 ) c. log 2 x = log x 16 = 0
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
Tukar bentuk log kepada bentuk indeks
log = log asas 10
log a + log b = log ab
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