Question

Se $\frac{1}{8} \leq 2^{-x}$ , então vale para x a afirmação:
a) X ∈ R e x $\frac{1}{3}$
b) X ∈ R e x > $\frac{1}{3}$
c) X ∈ R e x ≤ $\frac{1}{3}$
d) X ∈ R e x ≤ 3
e) X ∈ R e x ≥ 3

Answer 1

Explicação passo-a-passo:

$\frac{1}{8} \leq 2^{-x}$

$\frac{1}{2^{3}} \leq 2^{-x}$

$2^{-3} \leq 2^{-x}$

$2^{-x} \geq 2^{-3}$

−x ≥ −3

x ≤ 3

S = {x ∈ ℝ | x ≤ 3}

Answer 2

$\sf{se~a~base~\acute{e}~maior~que~1}\\\sf{a~desigualdade~se~mant\acute{e}m}\\\sf{se~a~base~est\acute{a}~entre~0~e~1}\\\sf{a~desigualdade~se~inverte}$

$\sf{vale~lembrar~que~2^{-x}=(2^{-1})^x=\left(\dfrac{1}{2}\right)^x}\\\sf{portanto}\\\sf{\dfrac{1}{8}\leq2^{-x}\implies\left(\dfrac{1}{2}\right) ^3\leq\left(\dfrac{1}{2}\right)^x}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf{x\leq3}}}}}}$