The problem is:
Given this equation find the possible values of $x$. $\log_{2}\left ( \frac{x+2}{x-1} \right )+\log_{2}\left [ (x+2)(x-1) \right ]=2$
First I defined the domain of the function.
The denominator $x-1$ must not be zero: $x-1\neq0$.
The fraction $\frac{x+2}{x-1}$ must be bigger than zero: $\frac{x+2}{x-1}>0$.
The product $(x+2)(x-1)$ must be bigger than zero: $(x+2)(x-1)>0$.
In the end I concluded that the domain should be $\left ]-\infty,-2\right [\cup \left ]1,+\infty\right[$.
Then I sarted to solve:
$\begin{align*} \log_{2}\left ( \frac{x+2}{x-1} \right )&+\log_{2}\left [ (x+2)(x-1) \right ]=2 \\ &\iff \log_{2}(x+2)-\log_{2}(x-1)+\log_{2}(x+2)+\log_{2}(x-1)=2\\ &\iff 2 \cdot \log_{2}(x+2)=2\\ &\iff \log_{2}(x+2)=1 \\ &\iff x+2=2^1\\ &\iff x=2-2\\ &\iff x=0. \end{align*}$
This result is not in the domain! Where I miss?