How do you simplify #((3^-2 x^5 y^-5)^10 (3^3 x^-5 y^2)^3)/((3^5 x^-3 y^4)^2)#?

Question

Hazel Anglin · Accepted Answer

#( x^41 )/(3^21 y^52#

Simplify

#((3^-2 x^5 y^-5)^10 (3^3 x^-5 y^2)^3)/((3^5 x^-3 y^4)^2)#

This problem isn't hard conceptually; it's just painstaking.

Clear the parentheses by raising all the powers inside the parentheses by the powers outside. To raise a power to a power, you multiply.

#(3^-20 x^50 y^-50 3^9 x^-15 y^6)/(3^10 x^-6 y^8#

Clear the negative exponents by flipping them to the opposite side of the fraction bar and reversing the negative signs to positive

#( x^50 3^9 y^6 x^6)/(3^10 y^8 3^20 y^50 x^15#

Group like bases to make it easier to multiply them

#(3^9 x^50 x^6 y^6 )/(3^10 3^20 x^15 y^8 y^50 #

Multiply like bases To multiply exponents, you add

#(3^9 x^56 y^6 )/(3^30 x^15 y^58 #

Reduce by cancelling To divide exponents, you subtract

#( x^(56-15) )/(3^(30-9) y^(58-6) #

same as

#( x^41 )/(3^21 y^52# #larr# answer

Ethan Adkins · Answer

undefined To simplify the expression $(3^{-2} x^5 y^{-5})^{10} \times (3^3 x^{-5} y^2)^3 / (3^5 x^{-3} y^4)^2$, we can use the properties of exponents:

$$
\begin{align*}
&= (3^{-20} x^{50} y^{-50}) \times (3^9 x^{-15} y^6) / (3^{10} x^{-6} y^8) \
&= 3^{-20+9+10} x^{50-15-6} y^{-50+6-8} \
&= 3^{-1} x^{29} y^{-52} \
&= \frac{x^{29}}{3y^{52}}
\end{align*}
$$