NÅGRA KOMMENTARARER TILL LÖSNINGARNA

2415a
x
2 = 2⋅2 x
2 = 2 ⋅2
x
2 = 2
x = 1001
1000
1 1000
1001
( identifiering ger)
2422a
1 1
10 100
10 ^ (-2)
− 2
10 = = = 0,01
2
2423b−a
8= 222 ⋅ ⋅ = 2
3 st
2424d
(1010) (1010) (10 ⋅ 10 ) = (10 ) =
4( ⋅−2) −8
1
10 = 10 = 8
10
0,00000001
(10 ∗10
^ 3) ^ (-2)
=
3
⋅
3 −2 =
1
⋅
3 −2
=
1 3 −2 4 −2
2425e
0 2 − 3 0+ 2 + ( −3)
5 ⋅5 ⋅ 5 = 5 =
−1
1 1
5 = = = 1 5
1
5 5
2426a
2 ⋅ 2 = 2 = 2 = 2
5 − 6 5 + ( −6) 5−6 −1
2426d
−5
2
= 2 = 2
7
2
−− 5 7 −12
LÖSNINGAR DEL B
2415b−c
x 300 300 300
3 = 3 + 3 + 3
x
3 = 3⋅3 x
3 = 3 ⋅3
x
3 = 3
x = 301
3 st termer
300
1 300
301
( identifiering ger)
2422c
1 3 1
5
3∗5^(-2) 1 25
−2
35 ⋅ = 3⋅ = ⋅ = 0,12
2
2423d
− c
1 1 1
18= = = = 2 3
8 2 ⋅2⋅2 2
2424 f
3 st
2 0 20 ⋅ 0
(10 ) = 10 = 10 = 1
∗ allting upphöjt till 0 blir 1
∗
−3
2425 f
(5 5 5 ) (5 5 5 )
(5 ) = (5 ) = 5 =
1 1
5 125
⋅
0
⋅
2 −1 =
1
⋅
0
⋅
2 −1
=
1+ 0+ 2 −1 3 −1 3( ⋅ −1)
−3
5 = = = 1125
3
2426b
(2 ) = 2 = 2
−3 −2 ( −3)( ⋅ −2)
6
2426e
2 ⋅ 2 = 2 = 2
−6 6 − 6+ 6 0
2421a−b
1 1
7 7
2121c−d
1 1
= = =
4 16
−1
7 = = = 1 7
1
−2
4 116
2
2422d
1 8 1 8
⋅ = ⋅ = ⋅ = =
2 1 16 16
8∗2^(-4) 36
−4
82 8 0,5
4
2424b
− 1
=
2
1
= =
8
3 1
(2 ) 1 8
3
2425b
(5 ) = 5 = 5
1
=
1
= 1 625
2 −2 2( ⋅−2) −4
4
5 625
2426c
−5
2
= 2 − 7
2
= 2 = 2
−5 −( −7) − 5+ 7 2
2426 f
10
2
− 10
2
= 2 = 2 =
2
10 −− ( 10) 10+ 10 20