Question

(UFRGS)O valor de x que verifica a equação é:
$\sqrt{} {4}^{ \times + 1} = \frac{1}{ {16}^{ \times + 1} }$
a)-1
b)-1/2
c)0
d)1/2
e)1

​

Answer 1

Resposta:

$\boxed{\mathtt{A}}$

Explicação passo-a-passo:

$\\ \displaystyle \mathsf{\sqrt{4^{x + 1}} = \frac{1}{16^{x + 1}}} \\\\\\ \mathsf{\left ( \sqrt{4^{x + 1}} \right )^2 = \left ( \frac{1}{16^{x + 1}} \right )^2} \\\\\\ \mathsf{4^{x + 1} = \left ( 16^{x + 1} \right )^{- 2}} \\\\ \mathsf{4^{x + 1} = 16^{- 2x - 2}} \\\\ \mathsf{4^{x + 1} = (4^2)^{- 2x - 2}} \\\\ \mathsf{4^{x + 1} = 4^{- 4x - 4}}$

Igualando as bases,

$\\ \displaystyle \mathsf{x + 1 = - 4x - 4} \\\\ \mathsf{x + 4x = - 4 - 1} \\\\ \mathsf{5x = - 5} \\\\ \boxed{\boxed{\mathsf{x = - 1}}}$