<s>[INST] Pre-Calculus: Fractions with Exponents If $x = \frac{2}{3}$ and $y = \frac{1}{9}$ find the value of $\dfrac{x^3y^2}{xy^5}$
I've tried working it out multiple ways, but my answer is wayyyy off. I'm not sure how to deal with this problem.
Regards,
 [/INST] We could simply, but to avoid this we can also just do the math. Since $x=\frac{2}{3}$ and $y=\frac{1}{9}$, we have
$$
x^3=\left(\frac{2}{3}\right)^3=\frac{8}{27}\text{ and }y^2=\left(\frac{1}{9}\right)^2=\frac{1}{81} \text{ and }y^5=\left(\frac{1}{9}\right)^5=\frac{1}{59049}
$$
then we have
$$
\frac{x^3y^2}{xy^5}=\frac{\frac{8}{27}\frac{1}{81}}{\frac{2}{3}\frac{1}{59049}}=\frac{\frac{8}{2187}}{\frac{2}{177147}}=\frac{8}{2}\cdot \frac{177147}{2187}=4\cdot 81=324
$$
While this works, simplfying works better first. 
$$
\frac{x^3y^2}{xy^5}=x^3y^2(xy^5)^{-1}=x^3y^2x^{-1}y^{-5}=x^{3-1}y^{2-5}=x^2y^{-3}=\frac{x^2}{y^3}
$$
But since we have the values of $x$ and $y$, we have
$$
\frac{(2/3)^2}{(1/9)^3}=\frac{\frac{4}{9}}{\frac{1}{729}}=\frac{4}{9}\cdot \frac{729}{1}=4\cdot 81=324
$$
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