Transcript C1.1 Algebra and functions 1

AS-Level Maths:
Core 1
for Edexcel
C1.1 Algebra and
functions 1
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The index laws
Using and manipulating surds
Contents
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Index notation
Simplify:
a to the power of 5
a × a × a × a ×= aa5
a5 has been written using index notation.
The number a is
called the base.
an
The number n is called
the index, power or
exponent.
In general:
n of these
an = a × a × a × … × a
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Index notation
Evaluate the following:
0.62 = 0.6 × 0.6 =0.36
34 = 3 × 3 × 3 × 381
=
(–5)3 = –5 × –5 × –5 =–125
When we raise a
negative number to
an odd power the
answer is negative.
27 = 2 × 2 × 2 × 2 × 2 × 2 ×
128
2=
(–1)5 = –1 × –1 × –1 × –1 × –1–1
=
(–4)4 = –4 × –4 × –4 × –4256
=
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When we raise a
negative number to
an even power the
answer is positive.
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The multiplication rule
When we multiply two terms with the
same base the indices are added.
For example:
a4 × a2 =(a × a × a × a) × (a × a)
=a×a×a×a×a×a
= a6 = a (4 + 2)
In general:
am × an = a(m + n)
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The division rule
When we divide two terms with the
same base the indices are subtracted.
For example:
a5
÷
a2
a×a×a×a×a 3
= a = a (5 – 2)
=
a×a
2
4p6
÷
2p4
4×p×p×p×p×p×p2
= 2p = 2p(6 – 4)
=
2×p×p×p×p
In general:
am ÷ an = a(m – n)
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The power rule
When a term is raised to a power and the result
raised to another power, the powers are multiplied.
For example:
(y3)2 = y3 × y3
(pq2)4 = pq2 × pq2 × pq2 × pq2
= (y × y × y) × (y × y × y)
= p4 × q (2 + 2 + 2 + 2)
= y6 = y3×2
= p4 × q8
= p4q8 = p1×4q2×4
In general:
(am)n = amn
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Using index laws
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Zero and negative indices
Using and manipulating surds
Contents
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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The zero index
Any number or term divided
by itself is equal to 1.
Look at the following division:
y4 ÷ y4 = 1
But using the rule that xm ÷ xn = x(m – n)
y4 ÷ y4 = y(4 – 4) =y0
That means that
y0 = 1
In general:
a0 = 1
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(for all a ≠ 0)
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Negative indices
Look at the following division:
b2
÷
b4
1
1
b×b
=
= 2
=
b
b×b×b×b b×b
But using the rule that am ÷ an = a(m – n)
b2 ÷ b4 = b(2 – 4) =b–2
That means that
b–2 =
1
b2
In general:
a–n =
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1
an
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Negative indices
Write the following using fraction notation:
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1
u
This is the
reciprocal of u.
1)
u–1 =
2)
2n–4 =
2
n4
3) x2y–3 =
x2
y3
4) 5a(3 –
b)–2 =
5a
(3  b)2
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Negative indices
Write the following using negative indices:
2
1) = 2a–1
a
x3
2) 4 = x3y–4
y
p2
3)
= p2(q + 2)–1
q+2
3m
4) 2
= 3m(n2 – 5)–3
3
(n  5)
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Fractional indices
Contents
Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Fractional indices
Indices can also be fractional. For example:
1
2
What is the meaning of a ?
Using the multiplication rule:
1
2
1
2
a × a =a
1+1
2 2
= a1
=a
a= a × a
But
1
2
So
a = a
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1
2
a is the square
root of a.
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Fractional indices
Similarly:
1
3
1
3
1
3
a ×a ×a = a
1+1 +1
3 3 3
= a1
=a
a = 3 a × 3 a ×3 a
But
1
3
1
3
So
a =3a
a is the cube
root of a.
In general:
1
n
a =na
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Fractional indices
2
3
What is the meaning of a ?
2
3
We can write a as a
2 31
.
Using the rule that (am)n = amn, we can write
2
3
1
3
a  (a )  3 a 2
2
3
We can also write a as a
1 2
3
2
3
2
.
1
3
a  (a )2  ( 3 a )2
In general:
m
n
n
a = a
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m
or
m
n
a =
 a
n
m
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Fractional indices
Evaluate the following:
1) 16
5
4
2) (0.125)
5
4
4
5
16 = ( 16)
 32
3) 36
 32
(0.125) = 8
2
3
36
 32
 32
=
1
( 36 )3
= 25
= ( 3 8)2
= 32
= 22
1
= 3
6
=4
=
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1
216
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Summary of the index laws
Here is a summary of the index laws for all rational exponents:
m
n
a ×a = a
m
n
a ÷a =a
m n
(a ) = a
( m+n )
a
( m n )
mn
1
a =a
a = 1 (for a  0)
0
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n
1
= n
a
1
2
a = a
1
n
n
a = a
m
n
n
a = a
m
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Solving equations involving indices
Contents
Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Solving equations involving indices
We can use the index laws to solve certain types of equation
involving indices. For example:
Solve the equation 25x = 1255 – x.
25x = 1255 – x
(52)x = (53)5 – x
52x = 53(5 – x)
2x = 3(5 – x)
2x = 15 – 3x
5x = 15
x=3
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Examination-style questions
Contents
Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Examination-style question 1
6+ 3
Show that
can be written in the form a + b 2 where
6 3
a and b are integers. Hence find the values of a and b.
Multiplying top and bottom by 6 + 3 gives



3 
6+ 3
6
6+
So
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 = 6+2 6 3 +3
63
3
6+ 3
9 + 2 18
18 can be written as 3 2.
=
3
9+6 2
=
3
=3+2 2
a = 3 and b = 2
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Examination-style question 2
a) Express 32x in the form 2ax where a is an integer to be
determined.
b) Use your answer to part a) to solve the equation
x
32 = 2
x2
32 = 25
So
32x = (25)x
Using the rule that (am)n = amn
32x = 25x
b) Using the answer from part a) this equation can be written as
5x
x2
2 =2
5x = x2
5x – x2 = 0
x (5 – x) = 0
x = 0 or x = 5
a)
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