Question

log x na base 2 -1/log x na base 2 =1

Answer 1

E aí Mateus,

$log_2x- \dfrac{1}{log_2x}=1\\\\ (log_2x\cdot log_2x)-1=(1\cdot log_2x)\\ (log_2x)^2-1=log_2x\\ (log_2x)^2-log_2x-1=0\\\\ log_2x=\theta\\\\ \theta^2-\theta-1=0\\\\ \Delta=(-1)^2-4\cdot1\cdot(-1)\\ \Delta=1+4\\ \Delta=5$

$\theta= \dfrac{-(-1)\pm \sqrt{5} }{2\cdot1}= \dfrac{1\pm \sqrt{5} }{2}$

Retomando a variável original:

$log_2x=\theta\\\\ log_2x= \dfrac{1+\sqrt{5} }{2}~~~~~~~~~~~~~~~log_2x= \dfrac{1- \sqrt{5} }{2} \\\\ x=2^{ \tfrac{1+ \sqrt{5} }{2} }~~~~~~~~~~~~~~~~~~~~~x=~2^{ \tfrac{1- \sqrt{5} }{2} }\\\\\\ \large\boxed{\boxed{\boxed{S=\left\{2 ^{ \tfrac{1+ \sqrt{5} }{2} },~2^{ \tfrac{1- \sqrt{5} }{2} }\right\}}}}.\\.$

Ótimos estudos véio ;D