Решить неравенство: (5/7)^4х-3>1

$(\frac{5}{7})^{4x-3}\ \textgreater \ 1\ \textless \ =\ \textgreater \ (\frac{5}{7})^{4x-3}\ \textgreater \ (\frac{5}{7})^0\to4x-3\ \textless \ 0\to x\ \textless \ \frac{3}{4}$

или

$(\frac{5}{7})^{4x}-3\ \textgreater \ 1\ \textless \ =\ \textgreater \ (\frac{5}{7})^{4x}\ \textgreater \ 4\ \textless \ =\ \textgreater \ (\frac{5}{7})^{4x}\ \textgreater \ (\frac{5}{7})^{log_{\frac{5}{7}}4}\to4x\ \textless \ log_{\frac{5}{7}}4\to\\x\ \textless \ \frac{log_{\frac{5}{7}}4}{4}\ \textless \ =\ \textgreater \ x\ \textless \ \frac{1}{4}log_{\frac{5}{7}}4\ \textless \ =\ \textgreater \ x\ \textless \ log_{\frac{5}{7}}4^{\frac{1}{4}}\ \textless \ =\ \textgreater \ x\ \textless \ log_{\frac{5}{7}}\sqrt{2}$