√log2x−1−12log2(x3)+2>0
√log2x−1−32log2(x)+2>0[∵logxm=mlogx]
Let log2x be t2+1.
∴ the question looks like t−32(t2+1)+2>0
⇒2t−3t2−3+4>0
⇒3t2−2t−1<0
⇒3t2−3t+t−1<0
⇒(t−1)(3t+1)<0
So, t∈(−13,1)
Hence, log2x∈(109,2)
So, x∈(2109,4)∈(21.11111,4)∈(2.16,4)
In this range, only one integer exists, i.e. 3