Помогите срочно !!!решить уравнения

1) ОДЗ : - 2x - 7 > 0 ⇒ x < - 3,5
$log _{ \frac{1}{3} }(-2x-7)=3\\\\-2x-7= \frac{1}{27}\\\\-2x=7 \frac{1}{27}\\\\x=- \frac{190}{54} =-3 \frac{14}{27}$
$-3 \frac{14}{27} \ \textless \ -3,5$
Ответ : $-3 \frac{14}{27}$

2) ОДЗ :
$\left \{ {{x+5\ \textgreater \ 0} \atop {4x-7\ \textgreater \ 0}} \right. \\\\ \left \{ {{x\ \textgreater \ -5} \atop {x\ \textgreater \ 1,75}}\\\\ x\ \textgreater \ 1,75 \right. } \\\\\\log _{7} (x+5)=log _{7}(4x-7)\\\\x+5=4x-7 \\\\-3x =-12\\\\x=4$

3) ОДЗ : 8x + 2 > 0 ⇒ x > - 0,25

$log _{5} (8x+2)-log _{5}6=log _{5}10\\\\log _{5} \frac{8x+2}{6}=log _{5}10\\\\ \frac{8x+2}{6}=10\\\\ \frac{4x+1}{3}=10\\\\4x+1=30\\\\4x=29\\\\x=7,25$