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I am doing this question, and I need to simplify the following in order for me to complete the question. Have I done anything wrong below?

Given $\left(\frac{1}{7\alpha\sqrt\pi}\right)^{1/2}$, $\alpha=\sqrt{\frac{h}{mw}}$, am I correct to simplify it to:

$$\left(\frac{1}{7\sqrt{\frac{h\pi}{mw}}}\right)^{1/2}$$

and can I write it as:

$$\frac{1}{\sqrt7}\left(\frac{1}{\sqrt{\frac{h\pi}{mw}}}\right)^{1/2}$$ and further simplifying it: $$\frac{1}{\sqrt7}\left(\left({{\frac{h\pi}{mw}}}\right)^{-1/2}\right)^{1/2}$$ $$\frac{1}{\sqrt7}\left({{\frac{h\pi}{mw}}}\right)^{-1/4}$$ $$\frac{1}{\sqrt7}\left(\frac{1}{\left({{\frac{h\pi}{mw}}}\right)^{1/4}}\right)$$ $$\frac{1}{\sqrt7}\left(\frac{1}{\left({{\frac{h\pi}{mw}}}\right)}\right)^{1/4}$$ $$\frac{1}{\sqrt7}\left(\frac{mw}{h\pi}\right)^{1/4}$$

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    $\begingroup$ Why did you remove most of the question? The answer that you accepted no longer makes sense now. In this particular case I don't see any reason for making this change; if you do need to make changes to questions, please mark them clearly as edits if they make one of the answers appear wrong. $\endgroup$ – joriki Jan 18 '12 at 0:48
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Everything you wrote is correct, though you could have proceeded directly from $$ \frac{1}{\sqrt{7}}\left(\frac{h\pi}{mw}\right)^{-1/4} $$ to $$ \frac{1}{\sqrt{7}}\left(\frac{mw}{h\pi}\right)^{1/4}. $$ A negative exponent on a fraction results in "flipping" the fraction. You don't need to form a compound fraction every time you wish to change the sign of an exponent.

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  • $\begingroup$ Thanks for your help, yeah Im doing quite a long question so I just wanted to make sure I did it correctly before proceeding. $\endgroup$ – Ray Jan 18 '12 at 0:25

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