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Solve : (log)(0. 3)(x^2-x+1)>0...

Solve : `(log)_(0. 3)(x^2-x+1)>0`

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`log_(0.3) (x^2-x+1) gt 0`
As `0.3 lt 1`, so, sign will change.
`=> x^2-x+1 lt (0.3)^0`
`=> x^2-x+1 lt 1`
`=>x^2-x lt 0`
`=>x(x-1) lt 0`
`:. x in (0,1)` is the solution for the given equation.
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