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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)

$$y=\sqrt{\log_2(5-x)+\log_\frac{1}{2}(x+2)}$$

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  • $\begingroup$ So you get that $1. 5-x >0; 2. x+2 >0; 3. \log_2(5-x) + \log_{\frac1{2}}(x+2) \ge 0$. The third condition can be rewritten as $\log_2 {\frac{5-x}{x+2}}\ge 0$ or $\frac{5-x}{x+2}\ge 1$, or $\frac{3-2x}{x+2}\ge 0$ $\endgroup$ – W_D Sep 9 '13 at 6:54
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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: $$-2<x \le 1.5$$

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$y = \sqrt{ \log_2 (5-x) + \log_{1/2}(x+2)}$

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, $x \leq 3/2$.

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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?

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