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I'm blanking on how to do this, I learned it so long ago, any help would be greatly appreciated. More interested on how to do it than the answer.

I'm blanking on how to do this, I learned it so long ago, any help would be greatly-example-1
User Gwiazdorrr
by
8.1k points

1 Answer

12 votes

Answer:


(16 y^(22))/(x^(10)z^(10))

Explanation:

Given expression is ,


\sf\longrightarrow \bigg((2x^3y^(-5)z^8)/(8x^(-2)y^6z^3)\bigg)^(-2)

This would be simplified using the law of exponents , some of which I will use here are ,


  • (an)^m = a^m n^m

  • (a^m)/(a^n)=a^(m-n)


  • a^m a^n = a^(m+n)

  • a^(-n) = (1)/(a^n)

Using the above laws ,


\sf \longrightarrow \bigg[ (2)/(8) \bigg((x^3)/(x^(-2))\bigg)\bigg((y^(-5))/(y^6)\bigg)\bigg((z^8)/(z^3)\bigg) \bigg]^(-2)

Using the second law mentioned above , we have,


\sf \longrightarrow \bigg[ (1)/(4)(x^(3+2))(y^(-5-6))(z^(8-3))\bigg]^(-2)

Simplify ,


\sf \longrightarrow \bigg[(1)/(4) x^5y^(-11)z^5\bigg]^(-2)

Using the first law mentioned above , we have,


\sf \longrightarrow \bigg[ (1)/(4^(-2)) x^(5(-2)) y^(-11(-2)) z^(5(-2))\bigg]

Simplify,


\sf \longrightarrow 4^2x^(-10)y^(22) z^(-10)

Finally using the fourth law mentioned above , we have ,


\sf \longrightarrow \boxed{\bf (16 y^(22))/(x^(10)z^(10))}

Option K is the correct answer.

User Satyam Pathak
by
8.6k points
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