Найдите 2x-y/y если 8^y=5 и 4^x=125

У = log8(5) = (1/3)*log2(5)
x = log4(125) = (1/2)*3*log2(5) = (3/2)*log2(5)
Тогда: (2x-y)/y =
= ((2/3)*log2(5) - (3/2)*log2(5))/((3/2)*log2(5)) =
= (2/3)*(2/3) - 1 = 4/9 - 1 = -5/9
ответ: -5/9.

8^y=5
2^(3y)=5
3y=log(2)5
y=1/3*log(2)5
4^x=125
2^(2x)=125
2x=log(2)125
x=1/2*3*log(2)5=3/2*log(2)5
(2x-y)/x=2x/y-1=(2*3/2*log(2)5):(1/3*log(2)5-1=3log(2)5:1/3*log(2)5-1=9-1=8