Question

关于x的方程log√2(x)+log1/2(x-1)=log2(a)有解,求实数a的取值范围

Answer 1

log√2(x)+log1/2(x-1)=2*log2(x)-log2(x-1)=log2(x²/(x-1))

∴log2(x²/(x-1))=log2(a)

即:x²-ax+a=0

△=a²-4a≥0解得a≥4或者0≥a

提示:log[a^n]_[b^m]=m/nlog(a)