Question

Reduza a uma só potência:
a) (5^4)^2 =
b) (7^2)^4 =
c) (3^2)^5 =
d) (4^3)^2 =
e) (9^4)^4 =
f) (5^2)^7 =
g) (6^3)^5 =
h) (7^2)^3 =
i) (4^4)^5=
j) (8^3)^5=
k) (2^7)^3 =

Answer 1

Em Potência de Potência , conservamos a base e multiplicamos os expoentes .

generalizando :

$( {a}^{n}) {}^{m} = a {}^{(n \times m)}$

a)

$( {5}^{4} ) {}^{2} = 5 {}^{(4 \times 2)} = {5}^{8}$

b)

$( {7}^{2} ) {}^{4} = 7 {}^{(2 \times 4)} = {7}^{8}$

c)

$( {3}^{2} ) {}^{5} = 3 {}^{(2 \times 5)} = {3}^{10}$

d)

$( {4}^{3} ) {}^{2} = {4}^{(3 \times 2)} = {4}^{6}$

e)

$( {9}^{4} ) {}^{4} = {9}^{(4 \times 4)} = {9}^{16}$

f)

$( {5}^{2}) {}^{7} = {5}^{(2 \times 7)} = {5}^{14}$

g)

$( {6}^{3} ) {}^{5} = {6}^{(3 \times 5)} = {6}^{15}$

h)

$( {6}^{3} ) {}^{5} = {6}^{(3 \times 5)} = {6}^{15}$

i)

$( {4}^{4} ) {}^{ 3} = {4}^{(4 \times 3)} = {4}^{12}$

j)

$( {8}^{3} ) {}^{5} = {</strong><strong>8</strong><strong>}^{(3 \times 5)} = {</strong><strong>8</strong><strong>}^{15}$

k)

$( {2}^{7} ) {}^{3} = {2}^{(7 \times 3)} = {2}^{21}$