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?

  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.