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I have a question about absolute values. Does the following hold true? $$\left|\log^2{x}\right|=\log^2x$$

In one problem my textbook removes the absolute from $\left|\log^{2/3}x\right|$.

In theory yes, because the absolute value of $|x|=x$ if x is positive or equal to $0$. Sorry for the dumb question and thanks.

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    $\begingroup$ It's not dumb to ask questions! $\endgroup$ – Jakobian Aug 28 '18 at 9:30
  • $\begingroup$ For $\left|\log^{2/3}x\right|$ we can remove the absolute value assuming $\log x\ge 0$ that is $x\ge 1$. $\endgroup$ – gimusi Aug 28 '18 at 9:33
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Your question has nothing to do with logarithms. For every real function $f$ it is true that $(\forall x\in D_f):\bigl\lvert f^2(x)\bigr\rvert=f^2(x)$ simply because $f^2(x)\geqslant0$.

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Yes of course for $x>0$ since $\log^2 x \ge 0$ we have

$$\left|\log^2{x}\right|=\log^2x$$

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Sometime ago I had the same frustration. Generally, we do not accept no integers powers of negative numbers such as $(-1)^{3/2}$ or $(-1)^{\pi}$ e.t.c, but in case where we have a rational number with odd denominator such as $2/3$, $6/5$ etc, it is possible to write $(-1)^{2/3}$ which of course is equal to $\left((-1)^2\right)^{1/3}=\left((-1)^{1/3}\right)^2$ .

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    $\begingroup$ Sadly there is no of course here: Wolfram Alpha gives $(-1)^{2/3} = -\frac12 + \frac{\sqrt{3}}{2} i$ but $\left((-1)^2\right)^{1/3} =1$ $\endgroup$ – Henry Aug 28 '18 at 9:58
  • $\begingroup$ @Henry So If I undertasnd correctly, $\log^{2/3}x$ is imaginery, in case $x\lt1$ ? $\endgroup$ – dmtri Aug 28 '18 at 10:29
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    $\begingroup$ I would say possibly complex rather than imaginary $\endgroup$ – Henry Aug 28 '18 at 11:25
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    $\begingroup$ For positive $x$ they are usually taken to be the same. For negative $x$ there are likely to be issues of multiple or principal values $\endgroup$ – Henry Aug 28 '18 at 14:56
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    $\begingroup$ If your function $f(x)=x^{1/3}$ is defined solely for $f:\mathbb R \to \mathbb R$ you may be correct when $x$ is negative and real, but if it is defined for $f:\mathbb C \to \mathbb C$ on principal values you may be incorrect when $x$ has a negative real part and a zero imaginary part. Which is why my original comment said possibly complex $\endgroup$ – Henry Aug 28 '18 at 17:32

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