Question

Determine o conjunto solução das seguintes inequações:
16ˣ - 4²ˣ⁻¹ - 2²ˣ⁻¹>0

4ˣ - 3x2ˣ≤40

Answer 1

Resposta: a) b)estáaíaaíasaíaaíasaíaaíasaí

Explicação passo a passo:

Answer 2

Resposta:

$x>\frac{ln\frac{2}{3} }{2ln2} \\\\Se ~preferir\\\\x>\frac{ln2-ln3}{2ln2}$

$b)x\geq 3$

Explicação passo a passo:

$16^x-4^2^x^-^1-2^2^x^-^1>0\\\\(2^4)^x-(2^2)^2^x^-^1-2^2^x^-^1>0\\\\(2^x)^4-2^4^x^-^2-2^2^x^-^1>0\\\\(2^x)^4-\frac{2^4^x}{2^2} -\frac{2^2^x}{2} >0\\\\(2^x)^4-\frac{(2^x)^4}{4} -\frac{(2^x)^2}{2} >0\\\\2^x=y\\\\y^4-\frac{y^4}{4} -\frac{y^2}{2}>0\\\\4y^4-y^4-2y^2>0 \\\\3y^4-2y^2>0$

Como 2ˣ = y > 0, podemos dividir ambos os lados da igualdade por y², mantendo o sentido da desigualdade.

$3y^2-2>0\\\\Ra\acute izes\\\\3y^2-2=0\\\\3y^2=2\\\\y^2=\frac{2}{3} \\\\y=-\sqrt{\frac{2}{3} } ~ou~y=\frac{2}{3}\\\\y<-\sqrt{\frac{2}{3} } ~~ou~y>\sqrt{\frac{2}{3} } \\\\2^x<-\sqrt{\frac{2}{3} } ~n\tilde ao~serve\\\\2^x>\sqrt{\frac{2}{3} }$

Quadrando

$(2^x)^2>\frac{2}{3}$

$2^2^x>\frac{2}{3}$

Aplicando logaritmo neperiano aos dois membros:

$ln2^2^x>ln\frac{2}{3} \\\\2xln2>ln2-ln3\\\\x>\frac{ln2-ln3}{2ln2}$

$b)4^x-3.2^x\leq 40\\\\(2^x)^2-3.2^x-40\leq 0\\\\2^x=y\\\\y^2-3y-40\leq 0\\\\Ra\acute izes\\\\y^2-3y-40=0\\\\\Delta=(-3)^2-4.1(-40)\\\\\Delta=9+160\\\\\Delta=169\\\\y=\frac{3\pm13}{2} \\\\y=-5 ~ ou~y=8\\\\y\leq -5~ou ~y\geq 8\\\\2^x\leq -5~ n\tilde a o~serve\\\\2^x\geq 8\\\\2^x\geq 2^3\\\\x\geq 3$