ter3k4w8x

2022-10-08

How does ${(\mathrm{log}\sqrt{x})}^{2}=\frac{1}{4}(\mathrm{log}x{)}^{2}\phantom{\rule{thickmathspace}{0ex}}?$

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

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

Maddox Koch

Beginner2022-10-09Added 7 answers

Recall:

$\mathrm{log}\left({a}^{b}\right)=b\mathrm{log}a$

Here, that means that

$\mathrm{log}(\sqrt{x})=\mathrm{log}{x}^{1/2}=\frac{1}{2}\mathrm{log}x$

So

${(\mathrm{log}\sqrt{x})}^{2}=\underset{{\left(\frac{a}{b}\right)}^{c}=\frac{{a}^{c}}{{b}^{c}}}{\underset{\u23df}{{\left(\frac{\mathrm{log}x}{2}\right)}^{2}=\frac{(\mathrm{log}x{)}^{2}}{{2}^{2}}}}=\frac{1}{4}(\mathrm{log}x{)}^{2}$

$\mathrm{log}\left({a}^{b}\right)=b\mathrm{log}a$

Here, that means that

$\mathrm{log}(\sqrt{x})=\mathrm{log}{x}^{1/2}=\frac{1}{2}\mathrm{log}x$

So

${(\mathrm{log}\sqrt{x})}^{2}=\underset{{\left(\frac{a}{b}\right)}^{c}=\frac{{a}^{c}}{{b}^{c}}}{\underset{\u23df}{{\left(\frac{\mathrm{log}x}{2}\right)}^{2}=\frac{(\mathrm{log}x{)}^{2}}{{2}^{2}}}}=\frac{1}{4}(\mathrm{log}x{)}^{2}$

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