Решите неравенство log7 (x-3,5) + log7 (x-2) меньше или равно 1.

ОДЗ:
$\left \{ {{x-3,5\ \textgreater \ 0} \atop {x-2\ \textgreater \ 0}} \right. \ \Leftrightarrow \ \left \{ {{x\ \textgreater \ 3,5} \atop {x\ \textgreater \ 2}} \right. \ \Leftrightarrow \ x\ \textgreater \ 3.5$

Решение:
$log_7 (x-3.5) + log_7 (x-2) \leq 1 \\ log_7[(x-3.5)(x-2)] \leq log_77 \\ (x-3.5)(x-2) \leq 7 \\ x^2-5.5x+7 \leq 7 \\ x^2-5.5x \leq 0 \\ x(x-5.5) \leq 0 \\ \\ x=0; \ x=5.5 \\ \\ ++++[0]----[5.5]++++\ \textgreater \ x \\ \\ x \in [0;5.5]$

c учетом ОДЗ:
$x \in (3.5;\ 5.5] \\ \\ OTBET: \ x \in (3.5;\ 5.5]$