跪求高手已知a>0,且a≠1,函数f(x)=IogaX,x属于〔2,4〕的值域为〔m,m...答:所以Ioga2=m,Ioga4=m+1,即Ioga2=m,2Ioga2=m+1,两式相减,Ioga2=1,所以a=2,同理当a<1时是减函数,Ioga2=m+1,Ioga4=m,即Ioga2=m+1,2Ioga2=m,Ioga2=-1,a=1/2 综上a=1/2或2
且a≠1,函数f(x)=IogaX,x属于〔2,4〕的值域为〔m,m=1〕求a值答:若a>1,是增函数,所以Ioga2=m,Ioga4=m+1,即Ioga2=m,2Ioga2=m+1,两式相减,Ioga2=1,所以a=2,同理当a