In the expression

$\displaystyle \frac{16x^3-24x^2}{8x^4-12x^3}=\frac{8x^2(2x-3)}{4x^3(2x-3)}=\frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}$

I understand everything but the last step, how is the $\displaystyle \frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}$?

If it were me I would think the correct thing to do in this case would be to apply the exponent property$\displaystyle \frac{a^m}{a^n}=a^{m-n}$, thus equaling $\displaystyle 2x^{2-3} =\frac{1}{2x}$

Thanks!