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I have an expression that I need to simplify, I know the answer (wolframalpha) but I'm not sure of the rule that gets me there.

$\dfrac{(\alpha) X_1^{\alpha -1} X_2^{1-\alpha}}{(1-\alpha)X_1^\alpha X_2^{-\alpha}}$

Basically I know that on the $X_1$ side of the fraction it cancels down to $\frac{1}{x_1}$ and the $X_2$ side cancels to $\frac{x_2}{1}$ leaving me with the alphas still outside and $\frac{X_2}{X_1}$.

But I don't know the exponent rule that allows me to do this cancellation. I just want to be sure I understand it fully.

Thanks.

Edit: Thank you very much to both of you! You've really helped with an upcoming exam. This is my first time posting here, I will fill out my profile and hopefully I can help other people from here on in! Thanks again!

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  • $\begingroup$ If I say that (X^a) (X^b) = X^(a+b) or (X^a) / (X^b) = (X^a) (X^(-b)) = X^(a-b), does this remember something to you ? If not, just post and I shall try something else. $\endgroup$ – Claude Leibovici Dec 31 '13 at 11:19
  • $\begingroup$ @ClaudeLeibovici, post an answer and elaborate more. StephenB looks like you received the answer you were looking for, please accept an answer $\endgroup$ – Malachi Jan 8 '14 at 14:48
  • $\begingroup$ @StephenB. What I said is almost identical to what other said better in the answer. I was trying to find a way to refresh your memory with the simplest rules. If there is anything specific you would like me to clarify, just precise it. Cheers. $\endgroup$ – Claude Leibovici Jan 8 '14 at 18:37
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I assume that $X_1\neq 0,~X_2\neq 0$ and $\alpha\neq 1$.

$$\frac{\alpha X_1^{\alpha -1} X_2^{1-\alpha}}{(1-\alpha)X_1^{\color{red}{\alpha}} X_2^{\color{blue}{-\alpha}}}\longrightarrow\frac{\alpha X_1^{(\alpha -1)-\color{red}{\alpha}} X_2^{{1-\alpha}-(\color{blue}{-\alpha})}}{(1-\alpha) }=\frac{\alpha}{(1-\alpha)}X_1^{-1}X_2^{1}=\frac{\alpha X_2}{(1-\alpha) X_1} $$

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    $\begingroup$ I'd set $X_1>0$ and $X_2>0$, if $\alpha$ is not known to be an integer. $\endgroup$ – egreg Dec 31 '13 at 11:33
  • $\begingroup$ @egreg: I think this question is arisen in probability. Isn't it? $\endgroup$ – mrs Dec 31 '13 at 11:39
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You are correct. The following holds.

$$\frac{x^a}{x^b}=x^{a-b}.$$

In your example, $$\frac{x_1^{\alpha-1}}{x_1^\alpha}=x_1^{(\alpha-1)-\alpha}=x_1^{-1}=\frac {1}{x_{1}}.$$ $$\frac{x_2^{1-\alpha}}{x_2^{-\alpha}}=x_2^{(1-\alpha)-(-\alpha)}=x_2^{1-\alpha+\alpha}=x_2^1=x_2.$$

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  • $\begingroup$ I hope you don't mind if I edited it kindly. ;-) $\endgroup$ – mrs Dec 31 '13 at 11:42
  • $\begingroup$ No, I don't. It's better:) $\endgroup$ – mathlove Dec 31 '13 at 11:43
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$$X_2^{1-\alpha}=\frac{1}{X_2^{\alpha -1}}$$ also $$X_2^{-\alpha}=\frac{1}{X_2^\alpha}$$ also use the property $$x^ay^a=(xy)^a$$ you get $$\frac{\alpha}{1-\alpha}\frac{(\frac{X_1}{X_2})^{\alpha-1}}{(\frac{X_1}{X_2})^\alpha}$$

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