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I have a problem with understand how function $2^{|\log_{1/2}x|}$ obtains values for the negative $x$ ? I thought that there is the assumption that $x>0$ but wolframalpha shows chart that for negative $x$ also obtains values.

I tried to do it in this way: $2^{|\log_{1/2}x|}$ for $x\in (0,1)$ have formula $y= \frac{1}{x}$ and for $x\in[1,+\infty)$ equals $y=x$

But how it looks for the negative values?

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  • $\begingroup$ If you write$\log_{\frac{1}{2}}(x)$ as $\frac{\ln(x)}{\ln\frac{1}{2}}$, you can extend the argument of the natural log to complex numbers (including negative integers). Wolfram Alpha is then showing the modulus of the principal branch of $\frac{\ln(x)}{\ln\frac{1}{2}}$ for negative x. $\endgroup$ – Bitrex Oct 11 '13 at 22:53
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Disclaimer: I know nothing about this topic.

I think WolframAlpha is interpreting $|\log_{1/2}x|$ as the modulus of a complex logarithm of $x$.

How can we find such for $x$ negative? Let's start with natural logs:

$e^{\ln x} = x$, so $e^{-i\pi-2ki\pi+\ln x}=-x$, so $-\pi i-2k\pi i+\ln x=\ln(-x)$, so $\ln x=\ln(-x) + \pi i + 2k\pi i$. The modulus of that is $\sqrt{(\ln(-x))^2+(\pi + 2k\pi)^2}$, for whatever value of $k$ is conventional. Raising $1/2$ to that unwieldy power does not look likely to give anything pretty. That said, when $|x|$ is very large, the value will be very close to $-x=|x|$.

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  • $\begingroup$ Why the complex logarithm? This is a pretty basic question, chances are the OP hasn't even studied complex analysis yet. $\endgroup$ – DonAntonio Oct 11 '13 at 23:27
  • $\begingroup$ @DonAntonio, because the answer to the question of why WolframAlpha gives something for negative values is precisely that. I figured that to make a proper answer I'd expand it with a bit of calculation to see where the values come from, despite the fact that, as mentioned, I know nothing about complex analysis myself. $\endgroup$ – dfeuer Oct 11 '13 at 23:32
  • $\begingroup$ Oh, I see your point, @dfeuer...nice. +1 $\endgroup$ – DonAntonio Oct 11 '13 at 23:34
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$$\log_{1/2}x>0=\log_{1/2}1\iff x<1$$

But since we're talking of a logarithmic function here it must be that $\;x>0\;$ , so...

Further hint:

$$\log_{1/2}x=\frac{\log x}{\log\frac12}=-\frac{\log x}{\log 2}>0\iff\ldots$$

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