Question

determine o conjunto solução da inequação
${4}^{x + 1} {2}^{x - 1} < 128$

Answer 1

Boa noite!

$4^{x +1}.2^{x -1} \ \textless \ 128$

Devemos deixar tudo na mesma base.

$(2^{2})^{x +1}.2^{x -1} \ \textless \ 2^{7}$

$2^{2x +2}.2^{x -1} \ \textless \ 2^{7}$

Temos uma multiplicação de bases iguais, repetimos a base e somamos os expoentes:

$2^{3x +1}\ \textless \ 2^{7}$

Cortando as bases temos:

$3x +1 \ \textless \ 7 \\\\ 3x \textless \ 7 -1 \\\\ 3x \textless \ 6$

$x \ \textless \ \frac{6}{3} \\\\ x \textless \ 2$

$\boxed{\boxed{S:(x \textless 2)}}$

Bons estudos!!

Answer 2

Olá, boa noite ☺

Resolução:

$\\ \text{4}^{\text{x+1}}\cdot \text{2} ^{\text{x-1}} \ \textless \ \text{128} \\ \\ (\text{2}^{\text{2}})^{\text{x+1}}\cdot \text{2} ^{\text{x-1}} \ \textless \ \text{2}^{\text{7}} \\ \\ \text{2}^{\text{2x+2}} \cdot \text{2} ^{\text{x-1}} \ \textless \ \text{2}^{\text{7}} \\ \\ \text{2}^{3x+1}\ \textless \ \text{2}^{\text{7}} \\ \\ \text{3x + 1} \ \textless \ \text{7} \\ \\ \text{3x} \ \textless \ \text{7 - 1} \\ \\ \text{3x} \ \textless \ \text{6} \\ \\ \text{x} \ \textless \ \dfrac{6}{3} \\ \\ \text{x} \ \textless \ \text{2 }$

Bons estudos :)