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A melhor resposta!
2014-08-18T20:51:25-03:00
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
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