Number of integers `lt=10` satisfying the inequality `2(log)_(1/2)(x-1)lt=1/3-1/((log)_(x^2-x)8)` is............

Answer

9

Explanation

We must have ` x gt 1`
` 2 log_(1//2) (x-1) le 1/3 - 1/(log_(x^(2)-x)8)`
` or 1/3 - (log_(2)(x^(2)-x))/3 + 2 log_(2) (x-1) ge 0`
` or log_(2)2-log_(2)(x^(2)-x)+6 log_(2) (x-1) ge 0`
` or log_(2). (2(x-1)^(6))/(x(x-1)) ge 0`
` or (2(x-1)^(5))/x ge 1`
Putting x - 1 , we have ` y gt 0`. Thus,
`(2 y^(5))/(y+1) - ge 0`
` or (2 y^(5)-y-1)/(y+1) ge 0`
` or (2y^(5) - 2y+y-1)/(y+1) ge 0`
` or (2y(y^(4)-1)+y-1)/(y+1) ge 0`
` or ((y-1)[2y(y+1)(y^(2)+1)+1))/(y+1) ge 0`
` or (y-1)/(y+1) ge or y ge 1`
` rArr x ge 2`