Question

resolva a equação exponencial 2^(x+1). 4^(x-1) ≤ 1/32

Answer 1

Explicação passo-a-passo:

$\sf 2^{x+1}\cdot4^{x-1} \le \dfrac{1}{32}$

$\sf 2^{x+1}\cdot(2^2)^{x-1} \le \dfrac{1}{32}$

$\sf 2^{x+1}\cdot2^{2x-2} \le \dfrac{1}{32}$

$\sf 2^{x+1+2x-2} \le \dfrac{1}{2^5}$

$\sf 2^{3x-1} \le 2^{-5}$

$\sf 3x-1 \le -5$

$\sf 3x \le -5+1$

$\sf 3x \le -4$

$\sf x \le \dfrac{-4}{3}$