Question

(CN) - Se x=7^200, y= 1024^40x3^100 , z= 16^25x625^50, pode-se afirmar que:

a) x < y < z
b) x < z < y
c) y < x < z
d) y < z < x
e) z < x < y

Answer 1

Olá, Wanderson.

$x=7^{200}=7^{100+100}=7^{100}\cdot7^{100}=(7\cdot7)^{100}=\boxed{49^{100}}\\\\ y=1024^{40}\cdot3^{100}=(2^{10})^{40}\cdot3^{100}=2^{400}\cdot3^{100}=(2^{4})^{100}\cdot3^{100}=\\\\=(16\cdot3)^{100}=\boxed{48^{100}}\\\\ z=16^{25}\cdot625^{50}=(2^4)^{25}\cdot(5^4)^{50}=2^{100}\cdot5^{200}=2^{100}\cdot5^{100+100}=\\\\=2^{100}\cdot5^{100}\cdot5^{100}=(2\cdot5\cdot5)^{100}=\boxed{50^{100}}$

$\therefore\boxed{y<x<z}$

Resposta: letra "c"