7 в квадрате (2х-16) больше 1/49

[latex]7^2(2x-16)\ \textgreater \ \frac{1}{49}\\\\2x-16\ \textgreater \ \frac{1}{49\cdot 7^2}\\\\2x\ \textgreater \ 16+\frac{1}{2401}\\\\2x\ \textgreater \ \frac{5764802}{2401}\\\\x\ \textgreater \ \frac{2882401}{2401}\\\\ili\\\\7^{2(2x-16)}\ \textgreater \ \frac{1}{49}\\\\7^{4x-32}\ \textgreater \ 7^{-2}[/latex] [latex]4x-32\ \textgreater \ -2\\\\4x\ \textgreater \ 30\\\\x\ \textgreater \ 7,5[/latex] P.S.  Условие надо писать чётко, чтобы не было двоякости.

[latex]7 ^{2(2x-16} )\ \textgreater \ \frac{1}{49} \\ \\ 7 ^{4x-32}\ \textgreater \ 7 ^{-2} \\ \\ 4x-32\ \textgreater \ -2\\4x\ \textgreater \ -2+32\\4x\ \textgreater \ 30\\x\ \textgreater \ 7.5[/latex] x ∈ (7.5;+∞)