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

1 Answers 1

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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, the domain is $% 0, 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


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

alt text

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    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!2010-11-20