Question

Quanto é 4^x -6.2^x +8 =0??? me ajudem urgenteee

Answer 1

Olá Luu,

use 2^x=y, como variável auxiliar:

$4^x-6\cdot2^x+8=0~~.\\ (2^2)^x-6\cdot2^x+8=0\\\\ 2^x=y\\\\ y^2-6y+8=0\\ \Delta=b^2-4ac\\ \Delta=(-6)^2-4\cdot1\cdot8\\ \Delta=36-32\\ \Delta=4\\\\ y= \dfrac{-b\pm \sqrt{\Delta} }{2a}= \dfrac{-(-6)\pm \sqrt{4} }{2\cdot1}= \dfrac{6\pm2}{2}\begin{cases}y'= \dfrac{6-2}{2}= \dfrac{4}{2}=2~~.\\\\ y''= \dfrac{6+2}{2}= \dfrac{8}{2}=4 \end{cases}$

Retomando a variável original, 2^x=y:

$2^x=2~~~~~~~~~~~~~~~~~~2^x=4\\ 2^x=2^1~~~~~~~~~~~~~~~~~2^x=2^2\\ \not2^x=\not2^1~~~~~~~~~~~~~~~~\not2^x=\not2^2\\\\ x=1~~~~~~~~~~~~~~~~~~~~~x=2\\\\\\ \Large\boxed{\boxed{\boxed{S=\{1,2\}}}}\\.$

Tenha ótimos estudos ^^