[tex]1)\quad 2^{x+1} \leq 4 \Rightarrow \left\| \begin{array}{c}2^{x+1} \leq 2^2\\$-baza supraunitara \end{array} \right| \Rightarrow x+1 \leq 2 \Rightarrow x \leq 1 \Rightarrow \\ \\ \Rightarrow \boxed{x\in (-\infty,1]} \\ \\ \\ 2)\quad 3^\big{log_\big3 x} \ \textless \ 1 \\ \\ $Conditie de existenta: $x \ \textgreater \ 0 \Rightarrow D = (0,+\infty) \\ \\ 3^\big{log_\big3 x} \ \textless \ 1 \Rightarrow x \ \textless \ 1 \Rightarrow {x\in (-\infty, 1) $\\ \\ Din $(-\infty,1)\cap D \Rightarrow \boxed{x \in (0,1)}[/tex]
[tex]\boxed{\left\| \begin{array}{c}b = a^\big{log_\big a b} \\ \\ $Atunci cand: \left| \begin{array}{c} b \ \textgreater \ 0 \\ a \in (0,1)\cup ( 1, \infty) \end{array} \\ \end{array} \right }$ \rightarrow $ proprietate. $}[/tex]
