Solve the following equation:

$$
\log _{4-x}(3-\sqrt{4-x(4-x)})=\frac{1}{2}
$$
The number whose logarithm is $1 / 2$ is the positive square root of the base, so our equation is equivalent to

$$
\sqrt{4-x}=3-\sqrt{4-x(4-x)}
$$

provided that $4-x \neq 1$, because 1 cannot be the base of a logarithm. On the right side, under the square root, a perfect square stands: $4-x(4-x)=(x-2)^{2}$. Therefore, (1) is equivalent to the following:

$$
\sqrt{4-x}=3-|x-2|
$$

Plotting the functions on both sides, we graphically obtain a root around $x=0.8$, which is the $x0$

Since $x_{1}2$ corresponds to the equation $\sqrt{4-x}=3-(x-2)$, which we can rule out. We could also have shown by direct squaring that this equation has no real roots.