Решите ! log2 (4x-3)+log1/8 125 = log1/2 x + log4 0,04

Доп. условие x > (3/4)
log2 (4x-3) - (1/3)log2 (125) = -log2 (x) + (1/2)log2 (1/25)
log2 (4x-3) - log2 (125)^(1/3) = -log2 (x) + log2 (1/5)
log2 ((4x-3)/5) = - log2 5x
(4x-3)/5=1/(5x)
4x^2-3x-1=0
x=1 и x= - (1/4) -не удовлетворяет по доп условию
ответ: 1