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So as the title says it all:

How does $\;\left(\log \sqrt x\right)^2 = \frac 14(\log x)^2 \;?$

To be specific, why the removal of root, and how do we get 4 in denominator?

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    $\begingroup$ In general $\log(a^b)=b\log a$. $\endgroup$ – André Nicolas Aug 27 '14 at 17:45
  • $\begingroup$ I think that the RHS (logx/2)^2 actually mean $((1/2)\log x)^2$. This is equal to the LHS. $\endgroup$ – mike Aug 27 '14 at 17:47
  • $\begingroup$ Well, thank you. Its now cleared! I hope someone edits the title with MathJax. $\endgroup$ – Swetank Aug 27 '14 at 17:51
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Recall: $$\log\left(a^b\right) = b\log a$$

Here, that means that $$\log (\sqrt x) = \log x^{1/2} = \frac 12 \log x$$

So $$\left(\log \sqrt x\right)^2 = \underbrace{\left(\frac {\log x}2\right)^2= \frac {(\log x)^2}{2^2}}_{\large \left(\frac ab\right)^c = \frac{a^c}{b^c}} =\frac 14(\log x)^2$$

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