Simplifying Logarithms I am Preparing for an Exam, and this is one of my Exercises that I tried to solve, I know that the answer is : $-2<x\le1\frac{1}{2}$. 
but Every time I try to solve, my answer is  Different from the Right answer.
(I need to find the domain of the function)
$$y=\sqrt{\log_2(5-x)+\log_\frac{1}{2}(x+2)}$$
 A: I guess you want to find the domain of the function:
Everything the log is applied on should be  $>0$
$$5-x>0$$
$$x<5$$
And:
$$x+2>0$$
$$x>-2$$
Everything under the square root must be $\ge0$
So:
$$\log_2{(5-x)}+\log_{\frac{1}{2}}{x+2}\ge0$$
$$\log_2{(5-x)}+\frac{\log_2{x+2}}{\log_2{\frac{1}{2}}}\ge0$$
$$\log_2{(5-x)}-\log_2{(x+2)}\ge0$$
$$\log_2{\frac{5-x}{x+2}}\ge\log_2 1$$
$$\frac{5-x}{x+2}\ge1$$
$$\frac{5-x}{x+2}-1\ge0$$
$$\frac{5-x-x-2}{x+2}\ge0$$
$$\frac{-2x+3}{x+2}\ge0$$
$$-2\le x\le1.5$$
Now, If we do an intersection to everything, we get:
$$-2<x \le 1.5$$
A: $y = \sqrt{ \log_2 (5-x) + \log_{1/2}(x+2)}$
The logarithms are defined only for $x < 5$ and $x > -2$ respectively. 
One needs $\log_2(5-x) + \log_{1/2} (x+2) \geq 0$ so that one can take the square root. Using $\log_{1/2}(a) = -\ln(a) / \ln(2)$ this inequality becomes $\ln(5-x) \geq \ln(x+2)$. Take the exponential on both sides: $5-x \geq x+2$ that is, $x \leq 3/2$.
A: Note when dealing with, 
$$
\sqrt{\log_2(5-x)+\log_\frac{1}{2}(x+2)}
$$
Doing a change of base, we obtain:
\begin{align*}
&\sqrt{\log_2(5-x)+\log_\frac{1}{2}(x+2)} = \sqrt{\log_2(5-x)-\log_{2}(x+2)}=\\
&\sqrt{\log_2\left(\frac{(5-x)}{(x+2)}\right)} \Longrightarrow \\
&\frac{(5-x)}{(x+2)} \ge 1 \\
\end{align*}
Can you take it from here?
