Question

Como resolver essas inequeções?
√3^x > 9


$\sqrt[3]{2 ^{x} } \ \textless \ 1/4$

√(0, 1 )^x > 10

Answer 1

$\sqrt{3^x} \ \textgreater \ 9 \\ \\ 3^{ \frac{x}{2} } \ \textgreater \ 9 \\ \\ 3^{ \frac{x}{2} } \ \textgreater \ 3^2 \\ \\ \frac{x}{2} \ \textgreater \ 2 \\ \\ x\ \textgreater \ 4 \\ \\ \\ \sqrt[3]{2^x} \ \textless \ \frac{1}{4} \\ \\ 2^{ \frac{x}{3} } \ \textless \ \frac{1}{2^2} \\ \\ 2^{ \frac{x}{3} } \ \textless \ 2^{-2} \\ \\ \frac{x}{3} \ \textless \ -2 \\ \\ x\ \textless \ -6 \\ \\ \\ \sqrt{(0,1)^x} \ \textgreater \ 10 \\ \\ \sqrt{ (\frac{1}{10} )^{x} } \ \textgreater \ 10 \\ \\ (\frac{1}{10} )^{ \frac{x}{2} } \ \textgreater \ 10 \\ \\ 10^{- \frac{x}{2} } \ \textgreater \ 10 \\ \\ - \frac{x}{2} \ \textgreater \ 1 \\ \\ -x\ \textgreater \ 2 \\ \\ x\ \textless \ -2$