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