Exercise 1D

1 a x3 ÷ x−2 = x3 − (−2)
 x
2
3 2 1
= x5 l = x3  x2
x
b x5 ÷ x7 = x5 − 7 2 1

 = x−2 x 3 2

1
3 5 3 5
 x 6

c x x = x
2 2 2 2

= x4 6x

1
3 3

x  2 a 25 2 = 25
2 2 2
d = x 2

=5
= x3

 81
3 3
5 5
b 812 =
x 
3
e 3 3
x 3

 x5  = 93
= 729
f 3x0.5 × 4x−0.5 = 12x0.5 + (−0.5) 1
= 12x0 c 27 3 = 3 27
= 12 =3
2 1 2 1
g 9 x 3  3x 6 = 3x 3
 1
6
d 4−2 =
1 42
= 3x 2 1
=
16
7 2 7 2

h 5x 5  x 5 = 5x 5 5
1
= 5x  1
e 9 2
= 1

4 −5 4 + (−5) 92
i 3x × 2x = 6x
1
= 6x−1 =
9
1 1 1
j x x = x x
3 2 3 =
1 1
3

= x 2 3
1
f (−5)−3 =
5
= x 6
( 5)3
= ( 6 x )5 1
=
125
 x   x 
3 4 3 4
k 3
= x x 2 3
0
3
3 4
 g   =1
x 2 3 4
17
 x6
 
3 3
4
h 1296 4 = 1296
 ( 6 x )17
= 63
= 216

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1
   x  4 x3
3
3
25 x 4 x3
 25  2 3 d = 3 3
2 i   = x3 x x
 16 
3
 16  = x  4  x3  3
13

53 = x2  4 x0
= 3 1
4 = 2 4
125 x
=
64
2x  x 2 2x x 2
e = 4 4
x4
 27 
2 3
2 x x
 27  3 = 2  x  x2  4
1 4
j   =
 8
2
 8  3 = 2x3  x2
2 1
32 = 3 2
= x x
22
9 3 3
=  4  2  4  4 2
3
4 f  x4  =  x
9   9 
 
1 1
6 5 8 6
k     = x
5 6 27
5
 9 x 2  15 x 5 9 x 2 15 x5
6 g = 
3x3 3x3 3x3

  = 3  x2  3  5  x5  3
2 2
 3
 343  3 512
l = = 3x1  5x2
 
 343 
2
 512  3 3
=  5x 2
x
82
= 2
7 5 x  3x 2 5x 3x 2
64 h = 
= 15 x3 15 x3 15 x3
49 1 1
=  x1  3   x2  3
1 1 3 5
 64x 
10 
3 a 10 2
= 64  x 2 1 2 1 1
= x  x
= 8x5 3 5
1 1
= 2
5 x3  2 x 2 5 x3 2 x 2 3x 5 x
b =  5
x5 x5 x 1
= 5  x3  5  2  x 2  5 4 a 81 = 4 81
4

= 5 x 2  2 x 3 = 3
5 2
= 2 3 1 1
x x   4
b x(2 x ) = x  2  x
3 4 4 3

4
1 1  1

125x  = 24 x
12  3
c 12 3
= 125  x 3 3
1

= 5x 4 = 16x 3

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 1
1
 1 3
5 a y =  x3 
3

8 
1 3 1
= 3 x 3
8
x
=
2

2
1 2 1  1 3 
b y =  x 
2 28 
1 1
= 
2  1 3 2
 x 
8 
1 64
=  3 2
2 x
32
= 6
x
= 32x 6

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