/x-4y/+（2y—1)^2=0,求log2(y/x)的值2lg[(x-y)/2]=lgx+lgy,求x/y的值lg[3(x^3)]-lg[3(y^3)]=9,求x/ylog2(x)+log2(y)=0,x^4+y^4=194,求log2(x+y)值

Question

/x-4y/+（2y—1)^2=0,求log2(y/x)的值
2lg[(x-y)/2]=lgx+lgy,求x/y的值
lg[3(x^3)]-lg[3(y^3)]=9,求x/y
log2(x)+log2(y)=0,x^4+y^4=194,求log2(x+y)值

Answer 1

1./x-4y/+（2y—1)^2=0,==>x=4y,2y-1=0=>x=2,y=1/2
log2(y/x)= log2(1/4)= -2
2.2lg[(x-y)/2]=lgx+lgy=lgxy,
[(x-y)/2]^2=xy,x^2+y^2-6xy=0
(x/y)^2+1-6x/y=0
x/y =2.（-3舍）
3.lg[3(x^3)]-lg[3(y^3)]=9,
lg[3(x^3)/3(y^3)]=9=lg10^9
(x/y)^3 =10^9
x/y=10^3=1000
4.log2(x)+log2(y)=0==>log2(xy)=0==>xy=1
x^4+y^4=194,==>(x^2+y^2)^2-2x^2y^2=194
==>x^2+y^2=14=(x+y)^2-2xy
==>x+y=4
log2(x+y)=log2(4)=2