GCSE MATHEMATICS Higher Tier, topic sheet. INDICES

GCSE MATHEMATICS Higher Tier, topic sheet.
INDICES
Rule i) x m × x n = x m + n . Rule ii) x m ÷ x n = x m n .
Rule iii) ( x m )
= x mn . Rule iv) x 0 = 1.
Rule v) x n = 1 n n
n
. Rule vi) x = x , the n th root of x
x
1
1. Evaluate
a) 4 2 , b)
1
3
8 .
2. Evaluate the following.
a)
1
2
25 , b) 3 2 , c) 2 3 × 2 6 , d)
2
3
8 ,
e) 1000 0 3
4
, f) 16 g) 1 4 .
3. Simplify
a) t 3 × t 5 ,
b) p 6 ÷ p 2 ,
c)
a × a
a
3 2
.
4. a) Work out the value of 2 1 3
x when x = 8.
2
b) Mary wanted to work out ( ) 1 3
64 but she had lost her calculator.
Explain how she could do this calculation mentally.
5. a)
3
2
x can be written as x × 1 2
x . Use this result to factorise
1
2
3
2
b) Simplify x + x .
2 3
x + x
x
1
3
2 2
+ x .
2
x y
6. Express in the form x a y b .
2
xy
State clearly the values of a and b.
7. Simplify
ac
6 4
2 5
ac .
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8. a) Evaluate i) 3 4 , ii) 9 0 , iii)
1
2
49 .
b) Simplify
4ac×
3ab
2
6ab
3 3
.
9. Simplify fully
2ab
× 6ab
3
4ab
3 2 4 2
.
2
10. Simplify ( 3xy ) 4
.
11. Simplify a) x 5 × x 2 , b) y 5 ÷ y 2 .
12. a) Multiply out and simplify ( x + 6 ) 2
.
b) Triangle ABC has a right angle at B.
B
x + 6 x – 6
C
A
2 11
Find the value of x.
You must explain clearly how you obtain your answer.
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SOLUTIONS / ANSWERS.
1. a) 4 2 1
=
4 = 1
2
16 .
1
b) 8 3
= 3 8 = 2.
1
2
2. a) 25 = 25 = 5, b) 3 2 1
=
3 = 1 2
9 , c) 23 + 6 = 2 3 = 8, d)
e) 1000 0 3
4
= 1, f) 16 1 1
=
3
4
1 3
16 4
16 = ( )
2
3
8 = ( ) 2 3
8 = 2 2 = 4,
1
=
2 = 1 3
8 , g) 14 = 1 × 1 × 1 × 1 = 1.
3. a) t 3 × t 5 = t 3 + 5 = t 8 . OR t × t × t × t × t × t × t × t which equals t 8 .
b) p 6 ÷ p 2 = p 6 2 = p 4 .
c)
3 2
a × a
=
a
5 1
a a = a5 1 = a 4 .
4. a) 2 × 1 3
8 = 2 × 3 8 = 2 × 2 = 4.
2
b) ( ) 1 3
64 =
2
1
3
64 = ( )
3
2
64 = ( ) 2
3
64 = 4 2 = 16.
5. a)
b)
x
1
3
2 2
+ x = 1 1
2 2
x + x × x = x
1 2 (1 + x)
.
3
1
2 2
x + x
2 3
x + x
This can only be simplified by cancelling.
Now a fraction represents a division and as such, ‘divisions’ will only cancel
‘multiplications’. This means that we should factorise both the numerator and
denominator of the fraction.
x
x
3
1
2 2
+ x
+ x
2 3
=
1
2
x (1 + x)
2
x (1 + x)
which cancels down to give
1
x 2
2
x .
Using the rules of indices, we now get that
6. This requires the use of rule ii).
2
2
1
x y x y
= × = x 2 1 × y 1 2 = x 3 × y 1 .
2 1 2
xy x y
Thus a = 3 and b = 1.
1
x 2
2
x
= 1 2 2
x =
3
2
x .
7.
ac
ac
6 4
2 5
= a 6 2 × c 4 5 = a 4 × c 1 . OR
ac
ac
6 4
2 5
= a × a × a × a × a × a × c × c × c × c
a× a× c× c× c× c×
c
4
a
= (after cancelling!)
c
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8. a) i) 3 4 1
=
3 4
= 1 81 .
ii) 9 0 = 1.
b)
4ac×
3ab
2
6ab
3 3
=
12
4 3
abc
2
6ab
= 2a 2 b 2 c.
9.
2ab
× 6ab
3
4ab
3 2 4 2
=
12ab
3
4ab
7 4
= 3a 6 b.
2
10. ( 3xy ) 4
= (3xy 2 ) × (3xy 2 ) ×(3xy 2 ) ×(3xy 2 ) = 81x 4 y 8 .
11. a) x 5 × x 2 = x 5 + 2 = x 3 , b) y 5 ÷ y 2 = y 5 (2) = y 7 .
12. a) ( x + 6 ) 2
= ( x 6 )( x 6 )
+ + = x 2 + 6 x + 6 x + 6
= x 2 + 2 6 x + 6.
b) Use Pythagoras: ( 2 11 ) 2
= ( x + 6 ) 2
+ ( x 6 ) 2
4 × 11 = {x 2 + 2 6 x + 6} + {x 2 2 6 x + 6}
44 = 2x 2 + 12
hence 32 = 2x 2
16 = x 2 .
This means that x = 4. {x cannot equal 4 since the lengths of the sides
must be positive.}
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