$$ log_{ \frac{1}{4} } \frac{2x}{3} > log{ \frac{1}{4} } (5x+2) $$
ОДЗ: $$ \left\{{{\frac{2x}{3} > 0} \atop 5x+2 > 0}\right. \\ \left\{{{ x > 0} \atop {x > -0.4}}\right. \\ x > 0 \\ \frac{2x}{3} < 5x+2 \\ 2x < 15x+6 \\ -13x < 6 \\ x < - \frac{6}{13} 0 $$