Solve: `(x-1)/((log)_3(9-3^x))lt=1.`

Explanation

We have `(x-1)/(log_(3)(9-3^(x))-3) le 1`
For this, we must have ` 9-3^(x) gt 0 or 3^(x) lt 9 or x lt 2`
The given expression can be expressed as:
` ((x-1))/(log_(3)(9-3^(x))-log_(3) 27) le 1 `
` rArr ((x-1))/(log_(3)((9-3^(x))/27))le 1`
` rArr (x-1)*log_(((9-3^(x))/27)) 3 le 1`
` rArr log_(((9-3^(x))/27))(3^(x-1)) lt 1`
As ` x lt 2, 0 lt (9-3^(x))/27 lt 1`
We have ` 3^(x-1) ge (9-3^(x))/27`
` rArr 9 xx 3^(x) ge 9 - 3^(x)`
` rArr 10 xx 3^(x) ge 9`
` rArr x ge log_(3) 0.9`
Therefore, ` x in [log_(3) 0.9, 2)`