1=> 2(x+1)\ \textgreater \ 1 \\ x+1\ \textgreater \ 0,5 \\ x\ \textgreater \ -0,5 " alt="\displaystyle 16^{x+1}\ \textgreater \ 4 \\ (4^{2})^{x+1}\ \textgreater \ 4^{1} \\ 4^{2(x+1)}\ \textgreater \ 4^{1} \\ 4>1=> 2(x+1)\ \textgreater \ 1 \\ x+1\ \textgreater \ 0,5 \\ x\ \textgreater \ -0,5 " align="absmiddle" class="latex-formula">
x∈(-0,5; ∞)
Ответ: (-0,5; ∞)