Question

Решите пожалуйста неравенство

Answer 1

$log_{0,7} (5x-3)\ \textgreater \ log_{0,7}(4-x) \\ 5x-3\ \textless \ 4-x \\ 6x\ \textless \ 7 \\ x\ \textless \ 1\frac{1}{6} \\ \\ ODZ: \\ 5x-3\ \textgreater \ 0 \\ 5x\ \textgreater \ 3 \\ x\ \textgreater \ 0,6 \\ \\ 4-x\ \textgreater \ 0 \\ -x\ \textgreater \ -4 \\ x\ \textless \ 4 \\ \\ Otvet:x \in (0,6;1\frac{1}{6})$

Answer 2

$log_{0.7}(5x-3)\ \textgreater \ log_{0.7}(4-x)$
ОДЗ:
$5x-3\ \textgreater \ 0$
$4-x\ \textgreater \ 0$

$x\ \textgreater \ 0.6$
$x\ \textless \ 4$
x∈(0.6;4)

$5x-3\ \textless \ 4-x$
$6x\ \textless \ 7$
$x\ \textless \ 1 \frac{1}{6}$
С учетом ОДЗ
Ответ: (0,6; 1$\frac{1}{6} )$