Lets solve the following example
[tex]\frac{15x^4y^{-9}}{15x^2(y^{-3})^3}[/tex]We can note that y is raised to the negative power of -9. Then, by the inverse law of exponents, we have
[tex]\frac{15x^4y^{-9}}{15x^2(y^{-3})^3}=\frac{15x^4}{15x^2(y^{-3})^3\cdot y^9}[/tex]because
[tex]y^{-9}=\frac{1}{y^9}[/tex]Now, we can see that
[tex]\frac{x^4}{x^2}=x^{4-2}=x^2[/tex]where we used the division rule of exponents, then our last result can be written as
[tex]\frac{15x^4y^{-9}}{15x^2(y^{-3})^3}=\frac{15x^2}{15(y^{-3})^3\cdot y^9}[/tex]Now, by the power to a power rule, we can see that
[tex](y^{-3})^3=y^{(-3)\cdot3}=y^{-9}[/tex]then, our last result can be written as
[tex]\frac{15x^2}{15y^{-9}\cdot y^9}[/tex]But
[tex]y^{-9}\cdot y^9=y^{-9+9}=y^0=1[/tex]Then, the last result can be written as
[tex]\frac{15x^2}{15(1)}=\frac{15x^2}{15}[/tex]Finally, since 15 divided by 15 is one, we get
[tex]\frac{15x^2}{15}=x^2[/tex]Therefore,
[tex]\frac{15x^4y^{-9}}{15x^2(y^{-3})^3}=x^2[/tex]and we used the 3 properties.
