Question

2 log 8 (-x) > log 8 (5-4x)

8 - основание 

Answer 1

2$log_<var>{8}</var>$(-x)>$log_<var>{8}</var>$(5-4x)

$log_<var>{8}</var>$$-<var>x^{2}</var>$>$log_<var>{8}</var>$(5-4x)

$x^{2}$>5-4x

$x^{2}$-5+4x=0

$x^{2}$+4x-5=0

D=36

x1=1

x2=-5

x пренадлежит (-$infty$;0)$cup$ (0;+$infty$)