log1/2(x−1)≤13−1logx2−x8
x−1>0⇒x>1,x2−x>0 and x2−x≠1⇒x(x−1)>0⇒x<0 or x>1x2−x≠1⇒x2−x−1≠0⇒x≠1±√52
∴x∈(1,∞)−{1+√52}
Now,
log1/2(x−1)≤13−1logx2−x8⇒−log2(x−1)≤13−log2(x2−x)3⇒−3log2(x−1)≤1−log2(x2−x)⇒log2(x2−x)−log2(x−1)3≤1⇒log2(x2−x)(x−1)3≤1
⇒log2x(x−1)2≤1[∵x>1]
As the base of the log is greater than 1, so
⇒x(x−1)2≤2⇒x≤2x2−4x+2⇒2x2−5x+2≥0⇒(2x−1)(x−2)≥0⇒x≤12 or x≥2
Therefore, the required solution is,
x≥2
Hence, the required number of intergral solution is 9