Solve ${\log _{0.3}}({x^2} - x + 1) > 0$.

Hint: Use the properties of log and solve the given equation.
We are going to start by removing the log from the given equations and by the properties the subscript goes to the other side of the sign and there are changes in the sign due to the log properties and then we solve the equations and get the solutions of the equations which have been asked in the question.

Complete step by step solution:
First, we are given the equation for which we have to solve is
${\log _{0.3}}({x^2} - x + 1) > 0$
To proceed further, we need to eliminate the log from the equation, when we remove the log , the $0.3$ in the subscript goes to the other side and having the power of zero and also the sign between the LHS and RHS gets changed accordingly, which gives us
\[{x^2} - x + 1 < {\left( {0.3} \right)^0}\]
On more simplification, we get
\[{x^2} - x + 1 < 1\]
Now, we are going to equate the solution to zero by sending everything to either LHS or RHS, which gives us and then further simplify it
\[
  {x^2} - x + 1 - 1 < 0 \\
  {x^2} - x < 0 \\
\]
Now, we are going to take common since they are like terms in the equation. Then, we get
$x(x - 1) < 0$
From the above we can tell that the function is always negative, as it is less than zero.
So, the value of x lies between 0 and 1 and not exactly 0 or 1.

Note: When we draw a number line, we understand that the given function lies negative between the interval of 0 and 1 and is not exactly the value 0 and 1, as it is not equal to zero in the given equation.