Question

$4^{x+1} -33*2^{x}+8=0$

Answer 1

Olá,

use as propriedades da exponenciação: 

$4^{x+1}-33\cdot2^x+8=0\\ 4^x\cdot4^1-33\cdot2^x+8=0\\ (2^2)^x\cdot4-33\cdot2^x+8=0\\ 4\cdot(2^x)^2-33\cdot2^x+8=0\\\\ 2^x=y\\\\ 4y^2-33y+8=0\\\\ \Delta=(-33)^2-4\cdot4\cdot8\\ \Delta=1.089-128\\ \Delta=961\\\\ y= \dfrac{-(-33)\pm \sqrt{961} }{2\cdot4}= \dfrac{33\pm31}{8}\begin{cases}y_1= \dfrac{2}{8}= \dfrac{1}{4}\\\\ y_2= \dfrac{64}{8}=8 \end{cases}\\\\\\ 2^x=y_1~~\Rightarrow~~2^x= \dfrac{1}{4}~~\Rightarrow~~\not2^x=\not2^{-2}~~\Rightarrow~~x_1=-2$ 

$2^x=y_2~~\Rightarrow2^x=8~~\Rightarrow~~\not2^x=\not2^3~~\Rightarrow~~ x_2=3$

Portanto:

$\huge\boxed{\boxed{\text{S}=\{-2,3\}}}$

Answer 2

Boa noite!


Solução!


$4^{x+1}-33.2^{x}+8=0\\\\\\ 4^{x}.4^{1}-33.2^{x}+8=0\\\\\ 2^{2x}.2^{2}-33. 2^{x}+8=0 \\\\\ (2^{x})^{2}.4-33.2^{x}+8=0$


$Fazendo!\\\\\ 2^{x}=y\\\\\\\ (y)^{2}.4-33.y+8=0\\\\\\ 4y^{2}-33y+8=0$


$y= \dfrac{33\pm \sqrt{(-33)^{2}-4.4.8 } }{2.4}\\\\\\\\ y= \dfrac{33\pm \sqrt{1089-128 } }{8}\\\\\\\\ y= \dfrac{33\pm \sqrt{961} }{8}\\\\\\\\ y= \dfrac{33\pm 31 }{8}\\\\\\\\\\\\\ y_{1}= \dfrac{33+31}{8}= \dfrac{64}{8}=8\\\\\\\\\ y_{2}= \dfrac{33-31}{8}= \dfrac{2}{8}= \dfrac{1}{4}$


$Retomando!\\\\\\ 2^{x}=y_{1} \\\\\\ 2^{x}=8 \\\\\\ 2^{x}=2^{3} \\\\\\ \boxed{x=3}\\\\\\\\ 2^{x}=y_{2}\\\\\\\ 2^{x}= \dfrac{1}{4}\\\\\\\ 2^{x}= \dfrac{1}{2^{2} }\\\\\\\ 2^{x}= 2^{-2}$


$Retomando!\\\\\\ 2^{x}=y_{1} \\\\\\ 2^{x}=8 \\\\\\ 2^{x}=2^{3} \\\\\\ \boxed{x=3}\\\\\\\\ 2^{x}=y_{2}\\\\\\\ 2^{x}= \dfrac{1}{4}\\\\\\\ 2^{x}= \dfrac{1}{2^{2} }\\\\\\\ 2^{x}= 2^{-2} \\\\\\\ \boxed{x=-2}$


$\boxed{Resposta~~S=\{-2,3\}}$

Boa noite!
Bons estudos!