Question

Реши неравенство. решить.log5(2x_1)> 3.log3(4x-3)log0,3(5x-2).

Answer 1

1)
$log_5(2x-1)\ \textgreater \ 3$
ОДЗ: $2x-1\ \textgreater \ 0;x\ \textgreater \ 0.5$
$log_5(2x-1)\ \textgreater \ log_5 125$
$2x-1\ \textgreater \ 125$
$x\ \textgreater \ 63$

2) 
$log_3(4x-3)\ \textless \ log_3(2x+1)$
ОДЗ:
$\left \{ {{4x-3\ \textgreater \ 0} \atop {2x+1\ \textgreater \ 0}} \right. ; \left \{ {{x\ \textgreater \ 0.75} \atop {x\ \textgreater \ -0.5}} \right. ;x\ \textgreater \ 0.75$
$4x-3\ \textless \ 2x+1$
$2x\ \textless \ 4$
$x\ \textless \ 2$
С учетом ОДЗ: $(0,75;2)$

3)
$log_{0.3}(3x+5)\ \textgreater \ log_{0.3}(5x-2}$
ОДЗ:
$\left \{ {{3x+5\ \textgreater \ 0} \atop {5x-2\ \textgreater \ 0}} \right. ; \left \{ {{x\ \textgreater \ -5/3} \atop {x\ \textgreater \ 0.4}} \right. ; x\ \textgreater \ 0.4$
$3x+5\ \textless \ 5x-2$
$2x\ \textgreater \ 7$
$x\ \textgreater \ 3.5$