首先设loga(b)=k,那loga(1/b)=-k>loga(√b)=k/2>logb(a^2)=1/[loga^2(b)]=2/k
即-k>k/2>2/k,然后分步解这这不等式:
-k>k/2得k2/k即k(k+2)(k-2)>0用数轴标根法后得k属于(-2,0)∪(2,+∞)
-k>2/k得k
首先设loga(b)=k,那loga(1/b)=-k>loga(√b)=k/2>logb(a^2)=1/[loga^2(b)]=2/k
即-k>k/2>2/k,然后分步解这这不等式:
-k>k/2得k2/k即k(k+2)(k-2)>0用数轴标根法后得k属于(-2,0)∪(2,+∞)
-k>2/k得k