Решите пожалуйста. Даю 30 баллов

Решите задачу:

$log_{1/5}(3x+4) \geq -2\; ,\; \; \; ODZ:\; \; 3x+4\ \textgreater \ 0\; ,\; \; x\ \textgreater \ -\frac{4}{3}\\\\log_{1/5}(3x+4) \geq log_{1/5}(\frac{1}{5})^{-2}\\\\3x+4 \leq 25\\\\3x \leq 21\\\\x \leq 7\\\\-\frac{4}{3}\ \textless \ x \leq 7\\\\Naimenshee\; celoe:\; \; x=-1\; .$

$2)\quad lg(x^2+x-20)\ \textless \ lg(4x-2)\; ,\\\\ODZ:\; \left \{ {{x^2+x-20\ \textgreater \ 0} \atop {4x-2\ \textgreater \ 0}} \right. \; \left \{ {{(x+5)(x-4)\ \textgreater \ 0} \atop {x\ \textgreater \ 0,5}} \right. \; , \left \{ {{x\in (-\infty ,-5)\cup (4,+\infty )} \atop {x\in (0,5\; ;+\infty )}} \right. \to \\\\ x\in (4,+\infty )\\\\x^2+x-20\ \textless \ 4x-2\\\\x^2-3x-18\ \textless \ 0\\\\x^2-3x-18=0\; \; \; \to \; \; \; x_1=-3\; ,\; \; x_2=6\; (teorema\; Vieta)\\\\+++(-3)---(6)+++$

$x\in (-3,6)\\\\ \left \{ {{x\in (4,+\infty )} \atop {x\in (-3,6)}} \right. \; \; \Rightarrow \; \; \; x\in (6,+\infty )$