x - Kantonsschule Solothurn

3. Potenzen
39. a) a 12 ⋅ a 3 b) x 5 ⋅ x c) 4x 2 ⋅ 5x 3 ⋅ 6x 4 d) x 2n+1 ⋅ x 11 – n
6 3
40. a) (-2) 1 ⋅ (-2) 3 ⋅ (-2) 5 4 7 2 ⎛ 2 ⎞ ⎛ 2 ⎞
b) 5 ⋅ 5 ⋅ 5 c) ⋅ ⎜ ⎟⎠ ⋅ ⎜ ⎟⎠ d) (-1) 11 ⋅ (-1) 13 ⋅ (-1) 15
3 ⎝ 3 ⎝ 3
5 5 5
41. a) 15 3 ⋅ 2 3 b) 0,01 6 ⋅ 1000 6 3 3 ⎛ 1 ⎞ ⎛ 2 ⎞ ⎛ 3 ⎞
c) 12 ⋅ 3 d) ⎜ ⎟⎠ ⋅ ⎜ ⎟⎠ ⋅ ⎜ ⎟⎠
⎝ 2 ⎝ 3 ⎝ 4
3
3
42. a) ((2a) 3 ⋅(3b) 2 ) 4 b) (5a 3 ⋅4b 4 ) 3 c) ( 5 +1) 4 ( 5 -1) 4 ⎛ 2ab ⎞ ⎛ 2 3
3a bc ⎞
d)
⎜
2
⎟ ⋅ ⎜ ⎟
⎝ 3c ⎠
⎜ 4b ⎟
⎝ ⎠
4 4
43. a) 6 9 : 3 9 7 7 ⎛ 3 ⎞ ⎛ 9 ⎞
b) 8 : 2 c) ⎜ ⎟⎠ : ⎜ ⎟⎠ d) (6 3 ⋅ 3 3 ) : 36 3
⎝ 4 ⎝ 8
n
(4abc)
n n
5
2n
(x + ) (x − 1)
(2 + 2)
(x − 1)
44. a)
b)
c)
d)
n
12 n
(2ac)
(x − 1)
5
2n
( 2 + 1)
(1 − x)
45. a) 5 12 : 5 9 b)
10 6
2 : 2 c) a 2n : a 2n+1 d) (a+b) 7 : (a+b) 4
46. a)
( 4
2
)
2 - (2 4 ) 2 b) (-a 3 ) 4 c) ((-a) 4 ) 3 d) (a 1000 + a 1001 ) : a 999
47. Löse in der Grundmenge N:
a) 3 6 ⋅ 3 2 = 3 n b) 9⋅2 7 + 7⋅2 7 = 2 x c) 5⋅5 20 + 20⋅5 20 = 5 x
d) 2 a+4 - 8⋅2 a = 2 x e) (5 3 ) 4 : (5 2 ) 5 = 5 x f) (100 100 ) 100 = 10 x
48. a) 2 -10 b) 2 -4 – 4 -2 c) 1 –1 – (- 1) -1 d) 5⋅2 –3 + 6⋅2 –2 - 4⋅2 -4
3
2 −
4
⎛ ⎞
3 − 2
⎛ ⎞ ⎛ 3 ⎞
−
49. a) ⎜ ⎟
b) ⎜ ⎟ + ⎜ ⎟ c) ( 3) −8
10
⎛ 1 ⎞
−
d) ⎜ ⎟
⎝ 3 ⎠
⎝ 4 ⎠ ⎝ 4 ⎠
⎝ 2 ⎠
50. a) 2⋅3 0 - 3⋅2 0 - 3 0 ⋅2 0 + (2 0 ⋅3 0 ) b) (2 3 ) 0 – (2 0 ) 3 – (3 0 ) -2 + (3 -2 ) 0
51. a) a 5 ⋅ a - 3 ⋅ a 0 b) a - 2 : a - 6 c) (1 + x) 3 : (1 + x) 5 d) (x – y) - 5 : (x – y) 6
52. a) (3 - 2 ) – 3 b) ((- 4) - 4 ) - 4 c) 5 - 3 ⋅ 3,2 – 3 d) ( 2) −5
⋅ ( 8) −5
53. a) 0,6 - 2 : 0,75 - 2 b)
5
⎛ 2 ⎞
−
⎜ ⎟ :
⎝ 3 ⎠
5
⎛ 2 ⎞
−
⎜ ⎟
⎝ 9 ⎠
c)
4
⎛ 9 ⎞
⎜ ⎟⎠
⎝ 8
:
4
⎛ 3 ⎞
−
⎜ ⎟
⎝16
⎠
d) 6 - 4 :
4
⎛ 6 ⎞
−
⎜ ⎟
⎝ 7 ⎠
54. Löse in der Grundmenge Z:
1
a) = 2
x
1
b)
512
6
100
= 10 x c) x 3 = - 8 d) x 6 = 2 – 6
e) x 4 = x 16 f) x 6 = 16 3 g) 5 8 ⋅ x = 5 3 h) 2 - 12 ⋅ 2 - x = 2 6
i) (2 2 ) x x
2
= 1 k)
( 2
2
)
= 2 l)
( x )
2 = 2 m) (2 x ) 2 = 16
1