Rules for cancelling fractions with exponents

Question

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!

Answer 1

You are correct. The following holds.

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

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

Answer 2

$$X_2^{1-\alpha}=\frac{1}{X_2^{\alpha -1}}$$ also $$X_2^{-\alpha}=\frac{1}{X_2^\alpha}$$ also use the property $$x^ay^a=(xy)^a$$ you get $$\frac{\alpha}{1-\alpha}\frac{(\frac{X_1}{X_2})^{\alpha-1}}{(\frac{X_1}{X_2})^\alpha}$$

Answer 3

I assume that $X_1\neq 0,~X_2\neq 0$ and $\alpha\neq 1$.

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