Logarithmic inequality and properties of logarithms So the inequality:
$$2 \cdot \log_{\sqrt3}{(1-x)} - \log_\sqrt3 {(3-x)} \lt 2$$
Can be written as:
$$\log_{\sqrt3}{(1-x)^2} - \log_\sqrt3 {(3-x)} \lt 2$$
?????????????
I have tried both on Wolfram|Alpha and both gave different intervals of x, although the logarithm property of exponent is: $\log_a{b}^n = n \cdot \log_a{b}$
The interval of solutions for the first inequality is:
$$\frac{-1-\sqrt{33}}{2} \lt x \lt 1$$
While for the second one:
$$\frac{-1-\sqrt{33}}{2} \lt x \lt 1 \quad \lor \quad 1 \lt x \lt \frac{\sqrt{33}-1}{2}$$
I guess the reason is the different field of existence "I am not sure what's it called in English, it's basically $\log_a{f(x)} \qquad f(x) \gt 0$ to make the logarithm valid". Or in other words...
First inequality:
$$\begin{cases}  
\frac{-1-\sqrt{33}}{2} \lt x \lt \frac{\sqrt{33}-1}{2} \lor x \gt 3 & \mbox{Solution of inequality between the arguments of the logarithms}\\ x \lt 1 & \mbox{Condition of the first argument} \\ x \lt 3 & \mbox{Condition of the second argument}\end{cases} $$
Second one:
$$\begin{cases} 
\frac{-1-\sqrt{33}}{2} \lt x \lt \frac{\sqrt{33}-1}{2} \lor x \gt 3 & \mbox{Solution of inequality between the arguments of the logarithms} 
\\ (1-x)^2 \gt 0 & \mbox{Condition of the first argument} 
\\ x \lt 3 & \mbox{Condition of the second argument} 
\end{cases}$$
So it will be:
$$
\begin{cases}
\frac{-1-\sqrt{33}}{2} \lt x \lt \frac{\sqrt{33}-1}{2} \lor x \gt 3
\\ x \neq 1
\\ x \lt 3
\end{cases}
$$
So the question is why that difference happens though I only used a logarithmic property, and how to decide the values of {$x$} to be sure that the logarithm is valid, I mean why can't I choose $(1-x)^2 \gt 0$ in the first one too since they're basically the same?

Note that throughout, $x\neq 1$ and $x\neq 3$ applies. 
$$\begin{align}
2\log_\sqrt{3}(1-x)-\log_\sqrt{3}(3-x)&<2\\
\log_\sqrt{3}\frac{(1-x)^2}{3-x}&<\log_\sqrt{3}3\\
\frac {(1-x)^2}{3-x}&<3\\
\frac {(1-x)^2}{3-x}-3&<0\\
\frac {x^2+x-8}{3-x}&<0\\
\frac {x^2+x-8}{x-3}&>0\\
\frac {(x-\alpha)(x-\beta)}{x-3}&>0\\
\text{where $\alpha=\frac {-1-\sqrt{33}}2=-3.372, \beta=\frac{-1+\sqrt{33}}2=2.372$}&\\
\begin{cases}-3.372<x<2.372\;\; (x\neq 1) \\x>3\end{cases}\quad\blacksquare\end{align}$$
