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!
Best Answer
I assume that $X_1\neq 0,~X_2\neq 0$ and $\alpha\neq 1$.
$$\frac{\alpha X_1^{\alpha -1} X_2^{1-\alpha}}{(1-\alpha)X_1^{\color{red}{\alpha}} X_2^{\color{blue}{-\alpha}}}\longrightarrow\frac{\alpha X_1^{(\alpha -1)-\color{red}{\alpha}} X_2^{{1-\alpha}-(\color{blue}{-\alpha})}}{(1-\alpha) }=\frac{\alpha}{(1-\alpha)}X_1^{-1}X_2^{1}=\frac{\alpha X_2}{(1-\alpha) X_1} $$