已知实数X,Y满足2的X+2次方+4的Y次方=2的X+2Y+1次方,则2的X次方+4的Y次方的最小值是多少?如何解.

2^(x+2)+4^y=2^(x+2y+1)得4*2^x+4^y=2*2^x*4^y,于是有
4^y=4*2^x/(2*2^x-1)=2+1/(2^x-1/2)>0
得2^x-1/2>0或2^x-1/2-1.后者不可能.于是x>-1.
令t=2^x+4^y,则4^y=t-2^x,于是有
4*2^x+t-2^x=2*2^x*(t-2^x)得
t=[3*2^x+2*(2^x)^2]/(2*2^x-1)
令2^x=m>2^(-1)=1/2,则有
t=(3m+2m^2)/(2m-1)
=m+2+1/(m-1/2)=(m-1/2)+1/(m-1/2)+5/2
因m>1/2,故m-1/2>0,故
t=2^x+4^y=m+2+1/(m-1/2)=(m-1/2)+1/(m-1/2)+5/2≥2√[(m-1/2)*1/(m-1/2)+]+5/2=9/2
当且仅当(m-1/2)=1/(m-1/2),即2^x=m=3/2（m=-1/2舍去）时,2^x+4^y取最小值9/2.
此时,x=log(2)(3/2),y=log(4)(3)