NÅGRA KOMMENTARARER TILL LÖSNINGARNA

2407 f
4 3 5 4+ 3 5
(2 ⋅ 2 ) = (2 ) =
7 5 75 ⋅ 35
(2 ) = 2 = 2
2409a−c
10 ⋅ 10 = 10 = 10
2 5 2+ 5 7
2409 f
2 3 5 2+ 3 5
(10 ⋅ 10 ) = (10 ) =
5 5 55 ⋅ 25
(10 ) = 10 = 10
2410c
(5 ⋅ x) 4
625x
= 5 ⋅ x = 625⋅
x
4 4 4 4
2411 f
6 6
4 4
= = 4 = 4
1
4 4
61 − 5
2412e
t t t t t t
t = t
2 5
⋅ ⋅ =
1
⋅
2
⋅
5
=
1+ 2+ 5 8
2413d − b
t ⋅ t = t ⋅ t = t = t
15 15 1 15+ 1 16
2414b−d
( s ) = s = s
a 4 a⋅4 4a
LÖSNINGAR DEL B
2408e−a−c
55 5 5 5 5
5 = 5
⋅
2
⋅
3
=
1
⋅
2
⋅
3
=
1+ 2+ 3 6
2409b−d
(10 ) = 10 = 10
5 2 52 ⋅ 10
2410a−d
5555 5
55 ⋅ 5
4
⋅ ⋅ ⋅ 4−2 2
= = 5 = 5
2
2410 f
2 3 3 2 3 3
(5 x y) = 5 ⋅( x ) ⋅ y =
6 3 6 3
125⋅x ⋅ y = 125x
y
2412a−c t ⋅ t = t = t
4 6 4+ 6 10
2412 f
( t t ) ( t t )
( t ) = ( t ) = t = t
2
⋅
3
=
1
⋅
2 3
=
1+ 2 3 3 3 33 ⋅ 9
2413 f − a−e 8 8 8
t t t
= = =
2 3 2+ 3 5
t ⋅t
t t
8−5 3
t = t
2414e
2m m 3m
s ⋅s ⋅s ⋅ s =
2m m 3m 1
s ⋅s ⋅s ⋅ s =
s = s
2m+ m+ 3m+ 1 6m+ 1
2408 f − b−d (5 5 5 ) (5 5 5 )
(5 ) = (5 ) = 5 = 5
⋅
2
⋅
3 4
=
1
⋅
2
⋅
3 4
=
1+ 2+ 3 4 6 4 64 ⋅ 24
2409e
10 10 10 10 10 10
10 = 10
35
⋅
6
⋅
14
=
1
⋅
6
⋅
14
=
1+ 6+ 14 21
2410b−e
4
5 4−2 = 5 2
5
2411a−b−c−d
−e
8
2 8−3 5
= 2 = 2
3
2
2412b−d
( t ) = t = t
6 4 64 ⋅ 24
2413c
( x ) = x = x
5 2 5⋅2 10
2414a−c
s ⋅ s = s = s
a 2a a+ 2a 3a
2414 f
( s s ) ( s s )
( s ) = s =
5(1+ 2 m) s
5+ 10m
=
s
⋅
2m 5
=
1
⋅
2m 5
=
1+ 2m 5 (1+ 2 m)5
⋅