<s>[INST] Where is the mistake in this logarithmic equation? 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?
 [/INST] If $x>1$, then $\log_2\left(\dfrac{x+2}{x-1}\right) = \log_2(x+2)-\log_2(x-1)$.
If $x<-2$ then $\log_2\left(\dfrac{x+2}{x-1}\right) = \log_2\left(\dfrac{-x-2}{1-x}\right) = \log_2(-x-2)-\log_2(1-x)$.
A similar thing applies to the logarithm of the product.
</s>