Найдите 2x-y/y если 8^y=5 и 4^x=125
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
а как вы получили (2х-у)/х = 2у/х-1?
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.