1) The document provides examples of solving various types of exponent operations including equal bases, powers of powers, quotient of powers of equal bases, powers of products and powers of quotients. 2) It also notes some exponent operations such as negative exponents, powers of products and powers of quotients that the author does not know how to solve. 3) The examples are provided in Spanish and involve solving for the exponent k in various exponent equations.

1. Productode potenciasde igual base
5𝑘.53 = 57
5𝑘
=
57
53
5𝑘
= 57−3
5𝑘
= 54
k = 4
(−3)2
.(−3) = (3)𝑘
(−3)2+1
= (−3)𝑘
(−3)3
= (−3)𝑘
3 = k
7.7𝑘
= 71
.7
Falto un paso
7.7𝑘
= 71
K=1
Potencia de una potencia
(52
)5
= 5𝑘
510
=5𝑘
10=k
(11𝑘
) = 1124
11𝑘.4
= 1124
k. 4 = 24
k =
24
4
𝑘 = 6
[(
1
3
)
2
]
4
= (
1
3
)
𝑘
(
1
3
)
2.4
= (
1
3
)
𝑘
2.4 = K
8 = k
Cociente de potencia de igual base
57
54
= 5𝑘
57−4
= 5𝑘
53
= 5𝑘
3 = k
(−4)4
(−4)𝑘
= (−4)𝑘
(−4)4
= (−4)2
. (−4)𝑘
4 = 2. k

2. 4
2
= 𝑘 → 2 = k
7𝑘
72
= 7
7𝑘
= 7. 72
7𝑘
= 71+2
7𝑘
= 73
K=3
114
11𝑘
= (11)−2
↑
No se cómo se hace
Potencia de exponente negativo
La de exponente negativo no sé
cómo se resuelve profeso
−(7)−3
=
1
𝑘3
(
7
5
)
−2
= (
7
5
)
𝑘
(k)−3
=
1
(6)3
(
𝑘
3
)
−4
= (
3
−5
)
𝑘

3. Potencia de un producto
La de potencia de un producto no
sé cómo se resuelve profeso
(4 . 7)3
= 4𝑘
.7𝑘
(8 . 𝑘)4
= 84
.74
(6 .9)𝑘
= 85
.75
(
2
7
.
3
5
)
7
= 𝑘7
.
37
5

4. Potencia de un cociente
La de potencia de un cociente no
sé cómo se resuelveprofeso
(
2
3
)
5
=
25
3𝑘
(
−3
4
)
𝑘
=
−27
64
(
4
𝑘
)
3
=
8
125
(
5
2
)
6
=
52+𝑘
64