Question

474 ( ) 548 42. - 8 how do I arrive at this answer.
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Answer 1

9514 1404 393

Answer:

(64x^28)/(5y^22)

Explanation:

The applicable rules of exponents are ...

(a^b)^c = a^(bc)

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

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First, eliminate the outside exponent on the left factor. Then use the above rules to combine factors with the same base.

`\displaystyle\left((4x^4)/(5y^6)\right)^5\cdot\left((5^4y^8)/(4^2x^(-8))\right)=((4^5x^(20))(5^4y^8))/((5^5y^(30))(4^2x^(-8)))=4^(5-2)5^(4-5)x^(20-(-8))y^(8-30)\\\\=4^35^(-1)x^(28)y^(-22)=\boxed{(64x^(28))/(5y^(22))}`