Question

alguém sabe?? me ajudem por favor :(

Answer 1

Resposta:

Explicação passo-a-passo:

Vamos lá!

a)

$3x^{x-5}>(3^{3})^{1-x} \\3x^{x-5}>3^{3(1-x)} \\3x^{x-5}>3^{3-3x} \\\\x-5>3-3x\\4x>8\\x>2$

b)

$5^{x+12} <5^{2} \\\\x+12<2\\x<-10$

c)

$(\frac{1}{8})^{x+3} \leq (\frac{1}{8})^{8}\\\\x+3 \leq 8\\x\leq 5$

Bons esudos, abraço!

Answer 2

Explicação passo-a-passo:

a)

$3^{x-5} >27^{1-x} \\\\3^{x-5} >(3^{3}) ^{1-x}\\\\3^{x-5} >3^{3.(1-x)}\\\\\text Como\ a\ base\ \'e\ maior\ que\ 1(3>1) se\ mant\'em\ o\ sinal.\\\\x-5>3.(1-x)\\\\x-5>3-3x\\\\x+3x>3+5\\\\4x>8\\\\\boxed{x>2}\\\boxed{\boxed{S=\{x\ \in R/x>2\}}}$

b)

$5^{x+12} <25\\\\5^{x+12} <5^{2} \\\\\text Como\ a\ base\ \'e\ maior\ que\ 1(5>1) se\ mant\'em\ o\ sinal.\\\\x+12<2\\\\x<2-12\\\\\boxed{x<-10}\\\\\boxed{\boxed{S=\{x\ \in R/x<-10\}}}$

c)

$\bigg(\dfrac{1}{8}\bigg) ^{x+3}\leq \bigg(\dfrac{1}{8}\bigg) ^{8}\\\\\text Como\ a\ base\ est\'a\ entre\ 0\ e\ 1(0<\frac{1}{8}<1 ) se\ inverte\ o\ sinal. \\\\x+3\geq 8\\\\x\geq 8-3\\\\\boxed{x\geq 5}\\\\\boxed{\boxed{S=\{x\ \in R/x\geq 5\}}}$