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How to find the domain of the function $\sqrt{ \log_{\frac{1}{2}} x}$ ?

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up vote 9 down vote accepted

Assuming the result is real, we must have $\log _{\frac{1}{2}}x\geq 0$. Since $\log _{\frac{1}{2}}x\geq 0\Leftrightarrow 0<x\leq 1$, the domain is $% 0<x\leq 1$, with $x\in \mathbb{R}$.

Added 2: $\log _{\frac{1}{2}}x=\frac{\log x}{\log \frac{1}{2}}=\frac{\log x}{\log 1-\log 2}=\frac{\log x}{0-\log 2}=-\frac{\log x}{\log 2}$

$\log _{\frac{1}{2}}x\geq 0\Leftrightarrow-\frac{\log x}{\log 2}\geq 0\Leftrightarrow\log x\le 0\Leftrightarrow 0<x\leq 1.$

Added: plot of $\log_{\frac{1}{2}}x$ (green) and $\sqrt{\log_{\frac{1}{2}}x}$ (blue).

alt text

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I am new to logarithms,rather learned it recently,I understood ur explanation.I want you ask you could you please explain this step: "we must have $\log _{\frac{1}{2}}x\geq 0$".What is the rule exactly for logarithms ? – Quixotic Nov 20 '10 at 12:53
Also why I am getting negative values in Table[Log[.5, x], {x, 0, 100}] ... I am confused :( – Quixotic Nov 20 '10 at 13:03
"we must have $\log_\frac{1}{2}x\ge 0$" because the radicand has to be greater or equal to zero, otherwise you would have a complex result. And I assumed that it is real. For instance $\sqrt{\log _{\frac{1}{2}}4}=\sqrt{-2}=i\sqrt{2}$. – Américo Tavares Nov 20 '10 at 13:07
The function $\log_{\frac{1}{2}}x$ is negative for $x\gt 1$, positive for $x\lt 1$ and equal to zero for $x=1$. – Américo Tavares Nov 20 '10 at 13:13
Thanks, now I understand! @J.M: Proof is like this : $\log _{\frac{1}{2}}x=\frac{\log x}{\log \frac{1}{2}}=\frac{\log x}{\log 1-\log 2}=\frac{\log x}{0-\log 2}=-\frac{\log x}{\log 2} = -\log_2 x $ Actually,I was not considering the square root! – Quixotic Nov 20 '10 at 14:29

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