Resolva :.... Pergunta de ideia dePauloVictorMoraus

January 2020 1 38 Report

Resolva :

$( \frac{1}{243} ) ^{2} \ \textgreater \ ( \frac{1}{27} ) ^{ x^{2}-1 }$

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superaks

Verified answer

Olá Paulo,

$\mathsf{\Big(\dfrac{1}{243}\Big)^2\ \textgreater \ \Big(\dfrac{1}{27}\Big)^{x^2-1}\Rightarrow (3^5)^{-2}\ \textgreater \ (3^3)^{-(x^2-1)}\Rightarrow (3)^{-10}\ \textgreater \ (3)^{-3x^2+3}}\\\\=\\\\\mathsf{-10\ \textgreater \ -3x^2+3\cdot (-1)\Rightarrow 10\ \textless \ 3x^2-3\Rightarrow 13\ \textless \ 3x^2\Rightarrow x^2\ \textgreater \ \dfrac{13}{3}}\\\\=\\\\\boxed{\mathsf{x>\sqrt{\dfrac{13}{3}}}}$

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