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,

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
$$
\frac{(2/3)^2}{(1/9)^3}=\frac{\frac{4}{9}}{\frac{1}{729}}=\frac{4}{9}\cdot \frac{729}{1}=4\cdot 81=324
$$
